PLANE AND SOLID GEOMETRY PLANE AND SOLID GEOME TRY BY ISAAC NEWTON FAILOR PRINCIPAL OF THE RICHMOND HILL HIGH SCHOOL NEW YORK CITY NEW YORK THE CENTURY CO. 1906 COPYRIGHT, 1906, BY THE CENTURY CO. CONTENTS INTRODUCTION PAGX GENERAL TERMS.... 1 GEOMETRIC CONCEPTS........ 3 ANGLES........... 7 AXIOMS........... 11 SYMBOLS AND ABBREVIATIONS....... 13 PLANE GEOMETRY BOOK I. RECTILINEAR FIGURES STRAIGHT LINES.......... 15 PARALLEL LINES..... 20 TRIANGLES........... 30 QUADRILATERALS..... 53 POLYGONS IN GENERAL......... 62 REVIEW QUESTIONS..... 67 ANALYSIS OF THEOREMS..... 69 BOOK II. THE CIRCLE THE CIRCLE.... 78 MEASUREMENT..... 91 THEORY OF LIMITS......... 91 MEASURE OF ANGLES... 94 CONSTRUCTIONS......110 ANALYSIS OF PROBLEMS........ 125 LocI.......126 REVIEW QUESTIONS..... 132 V vi CONTENTS BOOK III. PROPORTION AND SIMILAR POLYGONS THEORY OF PROPORTION. SIMILAR POLYGONS.. NUMERICAL PROPERTIES OF FIGURES CONSTRUCTIONS. BOOK IV. AREAS OF POLYGONS AREAS OF POLYGONS COMPARISON OF POLYGONS CONSTRUCTIONS. PAGE. 141. 154. 164. 184. 192.199.216 BOOK V. REGULAR POLYGONS REGULAR POLYGONS. CONSTRUCTIONS. MAXIMA AND MINIMA SYMMETRY FORMULAS OF MENSURATION SOLID GEOMETRY BOOK VI. STRAIGHT LINES AND PLANES STRAIGHT LINES AND PLANES. DIEDRAL ANGLES.. POLYEDRAL ANGLES. BOOK VII. POLYEDRONS, CYLINDERS, AND CONES 225 239 254 262 266 268 284 296 POLYEDRONS.... PRISMS AND PARALLELOPIPEDS. PYRAMIDS. SIMILAR POLYEDRONS REGULAR POLYEDRONS CYLINDERS.. CONES...303.304.321.338.342.344. 353 CONTENTS vii V11 BOOK VIII. THE SPHERE PAGE THE SPHERE.....365 SPHERICAL ANGLES..... 374 SPHERICAL POLYGONS.. 375 POLAR TRIANGLES.... 379 SYMMETRICAL SPHERICAL POLYGONS.. 382 SPHERICAL SURFACES.... 392 SPHERICAL VOLUMES.... 400 FORMULAS OF MENSURATION...... 411 INDEX. 413 PREFACE IT has been the aim of the author to make this a teachable book. Critical attention has been given to the minutest details. In the earlier pages most corollaries are proved, and most references, including the postulates, are quoted in full. The subject-matter is so arranged that no page must be turned in reading a demonstration. Brevity. -While brevity is not made an end in itself, yet brevity is an element of clearness. The object at all times has been to present the subject in as clear, simple, and direct a manner as possible. The author's long experience in the class-room has taught him that the fewer the words in which a thing can be said the better; elaborate explanations beget confusion. Propositions Omitted. -In the Plane Geometry only those theorems forming links in the chain of logic leading up to the mensuration of the circle, and a few required in the treatment of Solid Geometry, have been admitted. Familiar propositions omitted will be found among the exercises. No theorem, however, has been omitted which is necessary to meet the entrance requirements of colleges and technical schools. Definitions. -Definitions may be defective in many ways, but an essential feature of any definition is that its converse must be true. The reason for this is that we frequently assume the truth of such converses in our reasoning. ix X PREFACE Great pains have been taken to make the definitions clear, concise, and correct. No person can reason correctly on any subject without a clear conception of the meaning of the terms used. If the student is to be broadly grounded in an exact science, he must cognize the exact meaning of the terms peculiar to that science. The point, line, surface, and solid are the fundamental concepts in Geometry. A clear mental grasp of these concepts is all-important to the pupil's progress. Hence, in ~~ 27 to 44 these concepts are developed, illustrated, and explained. Then follow their mathematical definitions. Exercises. -The book contains a large number of exercises for original work on the part of the student. They are well proportioned between theorems, problems, problems of computation, and problems in loci. These exercises have been collected with great care and labor, extending over many years of patient research. Many are original with the author. The others have been gleaned from a common field of wide range. The aim has been to give only exercises which have a direct educational value, or which give a distinct property of some geometric figure. Curios and puzzles have been scrupulously avoided. All these exercises have been carefully graded. Those within the body of each book are, for obvious reasons, based upon the proposition, or propositions, which they immediately follow. Those at the end of each book are, in general, more difficult. It is not expected that any one class will find time to work all these exercises. The judgment of the teacher, guided by local conditions, must be depended upon to make suitable selections. The great value of original work on the part of the student is now universally admitted. PREFACE xi It is the aim of this book to lead the student into this original work so gradually, so systematically, that in a short time he will grow enthusiastic over "originals." Later on full directions are given for attacking original theorems and for analyzing problems. Special attention is called to the large number of numerical exercises. The author believes that numerical exercises are the best means at our command for impressing geometric truths upon the mind of the student. These exercises are not given for the purpose of teaching arithmetic. Hence they are exceedingly simple, and can be solved rapidly. A simple numerical problem illustrates a principle better than a complex one. Moreover, the student in Geometry has no time for complicated mental gymnastics. Limits. -Every principle in limits is given which is required to meet the demands of logic at every point in the Plane and Solid Geometry. The author believes that the subject as here presented is clearly within the mental grasp of the average student. He has found it so in his own classes. Loci.-The subject of loci is a very simple one when properly presented. It is of much greater importance than usually supposed. In the analysis of many problems, its use is invaluable. The author has aimed to give the subject a treatment commensurate with its importance. For obvious reasons, the subject is introduced under circles. Maxima and Minima, also Symmetry, are added as supplementary. Students who take Plane Geometry only may omit these subjects; those who take Solid Geometry may omit Maxima and Minima, but should take Symmetry. xii PREFACE The author takes great pleasure in acknowledging his indebtedness to Dr. Charles Robert Gaston, and to Professor Leland L. Landers, of the Richmond Hill High School, for valuable suggestions, and for their kindness in reading the proof-sheets. GEOMETRY INTRODUCTION GENERAL TERMS 1 A proposition is the expression of a judgment. 2 Propositions are distinguished as definitions, theorems, problems, corollaries, lemmas, axioms, postulates, absurdities, and scholiums. 3 A definition is such a description of an object as serves to distinguish it from all other objects. The converse of a correct definition is always true. 4 A theorem is a truth which may be logically demonstrated. 5 A problem is a question proposed for solution. 6 A corollary is a truth easily deduced from one or more propositions. 7 A lemma is an auxiliary theorem or problem. 8 An axiom is a self-evident truth. 9 A postulate is a self-evident possibility. 10 An absurdity is a self-evident falsity. 11 A scholium is a remark made upon one or more propositions. 12 A demonstration is the proof of a theorem. 13 A solution is the process of solving a problem. 14 A demonstration is direct or indirect. 1 2 GEOMETRY 15 A direct demonstration proves a theorem in either of two ways: 1 By superposing one figure upon another. 2 By a logical combination of definitions, axioms, postulates, and theorems previously demonstrated. 16 An indirect demonstration proves a theorem by proving that the supposition that it is not true involves an absurdity. 17 A theorem consists of two parts, the hypothesis and the conclusion. 18 The hypothesis sets forth the things given or granted, either in the statement of a theorem or in the course of the demonstration. The conclusion sets forth the things to be proved. Thus: Hypothesis: If A is B, Conclusion: C is D. 19 The converse of a theorem is the proposition formed by reversing the hypothesis and conclusion. Thus: Theorem: If A is B, C is D. Converse: If C is D, A is B. 20 The opposite of a theorem is the proposition formed by making the hypothesis and the conclusion negative. Thus: Theorem: If A is B, C is D. Opposite: If A is not B, C is not D. 21 The converse, or the opposite, of a theorem is not necessarily true. Thus: Theorem: All birds are bipeds. Converse: All bipeds are birds. Opposite: All animals not birds are not bipeds. While the theorem, "All birds are bipeds," is true, both its converse and opposite are false. INTRODUCTION 8 22 The logical relation existing between a theorem, its converse, and its opposite, is expressed in the following principles: 1 If the converse of a theorem is true, the opposite is true. 2 If the opposite of a theorem is true, the converse is true. Or, in general, thus: The converse and the opposite of a theorem are either both true or both false. 23 Magnitude is the amount or extent of anything that may be increased, diminished, or measured. 24 Extension is a reaching or stretching out. 25 Space is indefinite extension in every direction. 26 Geometry is the science of space. GEOMETRIC CONCEPTS 27 A block of wood, as represented in the above figure, is a material solid. The space it occupies is a geometric solid, or simply a solid. 28 A solid, in geometry, is a limited portion of space. 29 A careful inspection of the above solid will make the following concepts clear: The boundaries of a solid are surfaces. The boundaries of a surface are lines. The limits of a line are points. The intersection of two surfaces is a line. The intersection of two lines is a point. 4 GEOMETRY 30 A solid has three dimensions: length, breadth, and thickness. A surface has two dimensions only, length and breadth. A line has one dimension only, length. A point has no dimension, and hence no magnitude. It has position alone. 31 The above solid is bounded by six surfaces whose intersections form twelve lines, and the intersections of these lines form eight points. 32 A solid is bounded by surfaces. A surface is bounded by lines. A line is bounded by points. 33 A point is no part of a line. A line is no part of a surface. A surface is no part of a solid. GEOMETRIC CONCEPTS REVERSELY CONSIDERED 34 A line may be traced by a moving point. A surface may be traced by a moving line. A solid may be traced by a moving surface. Whence A line is the path of a moving point. A surface is the path of a moving line. A solid is the path of a moving surface. SCHOLIUM. The line must move out of its own direction, and the surface must move out of its own plane. GEOMETRIC CONCEPTS ABSTRACTLY CONSIDERED 35 A point may be regarded as the limit of a line, as the intersection of two lines; or, apart from all lines, as position alone without magnitude. INTRODUCTION 5 36 A line may be regarded as the limit of a surface, as the intersection of two surfaces, as the path of a moving point; or, apart from all points and surfaces, as extension in one dimension - length, without breadth or thickness. 37 A surface may be regarded as the limit of a solid, as the intersection of two solids, as the path of a moving line; or, apart from all lines and solids, as extension in two dimensions -length and breadth, without thickness. 38 A solid may be regarded as the path of a moving surface; or, apart from all surfaces, as extension in three dimensions - length, breadth, and thickness. 39 The point, line, surface, and solid are the four fundamental concepts in Geometry. GEOMETRIC CONCEPTS DEFINED 40 A point is position without magnitude. 41 A line is extension in one dimension, length. 42 A surface is extension in two dimensions, length and breadth. 43 A solid is extension in three dimensions, length, breadth, and thickness. 44 The magnitude of a point is zero. The magnitude of a line is its length. The magnitude of a surface is its area. The magnitude of a solid is its volume. GEOMETRIC CONCEPTS CLASSIFIED POINTS 45 A point is represented by a dot, designated by a letter, and read by reading the letter. Thus, " Point A." A. 6 GEOMETRY LINES 46 A line is represented by a mark, designated by letters placed at different points on it (usually at its end points), and read by reading the letters. Thus, " The line AB." A B For simplicity, a line may be designated by a single letter. Thus, "The line m." m 47 Lines are distinguished as straight, curved, broken, and mixed. 48 A straight line is a line of unchanging direction, as AB. A - B The word line unqualified means a straight line. 49 A curved line is a line no part of which is straight, as CD. A curved line is usually called a curve. Scholium. A straight line changes its direction at no point; a curved line changes its direction at every point. 50 A broken line is a line composed of c different straight lines, as ABCD. A B D 51 A mixed line is a line composed of G straight and curved lines, as EFGH. F 52 Classification of lines. f Straight. Curved. ( Lines. Broken. Mixed. INTRODUCTION 7 SURFACES 53 Surfaces are distinguished as plane and curved. 54 A plane surface, or simply a plane, is a surface such that the straight line which joins any two of its points lies wholly in the surface. 55 A curved surface is a surface no part of which is plane. 56 Geometry is the science of position, magnitude, and form. FIGURES 57 A geometric figure is any combination of points, lines, surfaces, or solids. 58 A plane figure is a figure having all its parts in the same plane. 59 A rectilinear figure is a figure bounded by straight lines. 60 A curvilinear figure is a figure bounded by curved lines. ANGLES 61 An angle is the figure formed by two straight lines proceeding from a point. 62 The sides of an angle are the two diverging lines. 63 The vertex of an angle is the point in which the sides meet. Thus, the lines AB and AC meeting at B A form the angle BAC. AB and AC are the sides of the angle, and the point A is its vertex. A 64 When there is but one angle at a point, it may be read by reading the letter at the vertex; as, " The angle A." 8 GEOMETRY 65 When two or more angles have the same vertex, each is read by reading the three letters designating it, remembering to read the letter at the vertex between the other two. Thus, at the vertex A are three angles, BAC, CAD, BAD. For simplicity an angle may be designated by a small italic letter placed between its sides -, and near its vertex; as, angles m and n. 66 Adjacent angles are two angles having the same vertex and a common side between them; \ as, angles s and t. 67 Angles are distinguished as right, straight, round, and oblique. Oblique angles are classified as acute, obtuse, and reflex. 68 A right angle is an angle formed by one B straight line meeting another so as to make the adjacent angles equal; as, angles BAC and BAD. - A D 69 A straight angle is an angle whose sides extend in opposite directions from the vertex, forming -.. one straight line; as, angle DEF. E 70 COROLLARY 1. A straight angle is equal to two right angles. 71 COROLLARY 2. A right angle is half a straight angle. 72 A round angle, or perigon, is the entire angular magnitude around a point in a plane; ( ---- S as, BOB. 73 COROLLARY. A round angle is equal to two straight angles or four right angles. 74 An acute angle is an angle less than a right angle; as, the angle C. INTRODUCTION 9 75 An obtuse angle is an angle greater than a right angle and less than a straight angle; as, the angle K. K 76 A reflex angle is an angle greater than a straight angle and less than a perigon; as, angle a.. 77 Classification of angles. Right. Straight. Angles. Round.rAcute. Oblique. i Obtuse. Reflex. ( 78 The complement of an angle is the difference between that angle and a right angle. 79 Complementary angles are two angles whose sum is a right angle. Thus, if GHK is a right angle, the angles a and b are complementary. a a is the complement of b, and b is the complement,. K of a. 80 The supplement of an angle is the difference between that angle and a straight angle. 10 GEOMETRY 81 Supplementary angles are two angles whose sum is a straight angle. Thus, if DEF is a straight angle, the angles s and t are supplementary. s is the supplement of t, and t is the supplement of s. D82 Perpendicular lines are two lines which A form a right angle with each other. Thus, if the angle B is a right angle, AB and BC are perpendicular lines. AB is perpendicular, or a perpendicular, to BC, and BC is perpendicular to AB. B The point B is called the foot of the perpendicular. 83 Oblique lines are two lines which form an oblique angle with each other. Thus, if the D D angle DEF is an oblique angle, / DE and EF are oblique lines. E F Er p MEASUREMENT OF ANGLES 84 The unit of angular measurement is the three hundred sixtieth part of a perigon, and is called a degree. The degree is divided into sixty equal parts, called minutes. The minute is divided into sixty equal parts, called seconds. Degrees, minutes, and seconds are designated respectively by the symbols ~,,. Thus, 18 degrees 25 minutes 37 seconds is written 18~ 25' 37 " The right angle contains 90~. MAGNITUDES 85 The geometric magnitudes are lines, surfaces, angles, and solids. 86 Similar magnitudes are identical in form. 87 Equivalent magnitudes are identical in extent. 88 Equal magnitudes are identical in form and extent. 89 The test of equality of any two magnitudes is that they can be made to coincide. INTRODUCTION 11 90 The object of Geometry is to measure and compare the geometric magnitudes. 91 Geometry is divided into two parts, Plane Geometry and Solid Geometry. 92 Plane Geometry treats of plane figures. AXIOMS 93 There are many axioms. Most of them are not peculiar to Geometry, but permeate all our thinking about magnitudes. They are truths so simple and fundamental that the rational mind grants them upon the mere statement. Euclid calls them " common notions. " The following are the more important axioms used in Geometry: 1 If equals are added to equals, the sums are equal. 2 If equals are added to unequals, the sums are unequal in the same order. 3 If unequals are added to unequals in the same order, the sums are unequal in the same order. 4 If equals are taken from equals, the remainders are equal. 5 If equals are taken from unequals, the remainders are unequal in the same order. 6 If unequals are taken from equals, the remainders are unequal in the reverse order. 7 The doubles of equals are equal. 8 The doubles of unequals are unequal in the same order. 9 The halves of equals are equal. 10 The halves of unequals are unequal in the same order. 11 Things which are equal to the same thing are equal to each other. 12 The whole is greater than any of its parts. 13 The whole is equal to the sum of all its parts. 12 GEOMETRY 14 Only one straight line can be drawn between two points. 15 A straight line is the shortest line between two points. 16 Through the same point only one straight line can be drawn parallel to the same straight line. COROLLARIES TO AXIOM 14 94 COROLLARY 1. Two straight lines which have two points in common coincide and form but one line. 95 COROLLARY 2. Two straight lines can intersect in one point only. 96 COROLLARY 3. Two points determine a straight line. POSTULATES 97 Postulates, like axioms, are numerous. Euclid calls a postulate " a request," and adds, " To my postulates I request, to my axioms I claim, your assent." The following are the postulates most frequently used in Geometry: 1 A straight line can be produced indefinitely in either direction. 2 A straight line can be drawn from any point in any direction to any extent. 3 A geometric figure can be moved from one position to another without changing its form, size, or the relation of its parts. 4 Any magnitude can be bisected. 5 Geometric magnitudes can be added, subtracted, multiplied, or divided. 6 From the greater of two magnitudes a part can be cut off equal to the less. 7 A circle can be drawn with any radius, and from any point as center. PLANE GEOMETRY 14 GEOMETRY 98 + plus, increased by. - minus, diminished by. x multiplied by. -divided by. = equals. is equivalent to. > is greater than. < is less than..~. therefore. Z angle. As angles. 1L perpendicular. SYMBOLS Is perpendiculars. II parallel. Its parallels. A triangle. A triangles. D7 parallelogram. /7 parallelograms. O circle. G circles. Oce circumference. Oces circumferences. 99 ABBREVIATIONS Adj.. Alt.. Ax. Comp. Const. Cor. Def. Ex. Exs. Ext.. Fig. Figs.. Hyp.. Q.E.D. Q.E.F.. adjacent.. alternate.. axiom.. complement.. construction.. corollary.. definition.. exercise.. exercises.. exterior.. figure.. figures.. hypothesis. Int. Opp. Post. Prob. Prop. Pt. Pts.. Quadl. Rt. Scho. Sq. St. Sup...interior. opposite. postulate. problem. proposition. point. points. quadrilateral. right. scholium. square. straight. supplement. quod erat demonstrandum, which was to be proved... quod erat faciendum, which was to be done. PLANE GEOMETRY BOOK I RECTILINEAR FIGURES PROPOSITION I. THEOREM 100 All straight angles are equal. A B c X C At; C' B B HYPOTHESIS. The A ABC and A'B'C' are any two straight A. CONCLUSION. Z ABC = Z A'B'C'. PROOF Place the / ABC on the / A'B'C' so that the vertex B falls on the vertex B', and the line BA along the line B'A'. Post. 3 Then BC will fall along B'C'. ~ 94.'. the s ABC and A'B'C' coincide and are equal. ~ 89 Q. E. D. 101 COROLLARY 1. All right angles are equal. PROOF All st. As are equal. ~ 100 Art. Z is half a st. Z. ~ 71 Whence all rt. A are halves of equal st. As, and.'. equal. " The halves of equals are equal." Ax. 9 102 COROLLARY 2. Only one perpendicular can be drawn to a given line at a given point in the line. PROOF If two such perpendiculars could be drawn, we should have unequal right angles, which is impossible (~ 101). 15 16 PLANE GEOMETRY -BOOK I PROPOSITION II. THEOREM 103 When one straight line meets another, the sum of the two adjacent angles thus formed is equal to two right angles. A B D HYPOTHESIS. The line AD meets the line BC at D. CONCLUSION. Z ADB + L ADC = 2 rt. A. PROOF Z ADB + Z ADC = the st. Z BDC Ax. 13 = 2 rt. /. "A st. Z is equal to two rt. A." ~ 70 Q. E. D. 104 COROLLARY 1. The complements of equal angles are equal. 105 COROLLARY 2. The supplements of equal angles are equal. 106 COROLLARY 3. The sum of all the angles that can be formed at a point in a straight line, and on the same side of the line, is equal to two right angles. For their sum is equal to a straight angle, Ax. 13 which is equal to two right angles. " A straight angle is equal to two right angles." ~ 70 STRAIGHT LINES 17 107 COROLLARY 4. The sum of all the angles that can be formed at a point is equal to four \ right angles. For their sum is equal to a round angle, Ax. 13 which is equal to four right angles. "' A round angle is equal to four right angles." ~ 73 108 COROLLARY 5. If one of the angles formed by the intersection of two straight lines is a right angle, the others are right angles also. d~7 PROOF Z a + L b = 2 rt. s. ~ 103 Let Z a = a rt. L. Hyp. Subtracting, Z b = a rt. L. Ax. 4 Again, Z b + Z c = 2 rt. Zs. ~ 103 Z b = a rt. /. Proved Subtracting, Z c = a rt. L. Ax. 4 Again, Z c + Z d = 2 rt. As. ~ 103 / c = a rt. Z. Proved Subtracting, L d = a rt. Z. Ax. 4 109 DEFINITION. Supplementary adjacent angles are two adjacent angles whose sum is equal to a straight angle; as, As a and b. M 110 COROLLARY. The exterior sides of supplementary adjacent angles are in the same straight line. For a + b = ast. Z. ~109... MN is a st. line. "A straight angle is an angle whose sides extend in opposite directions from the vertex, forming one straight line." ~ 69 18 PLANE GEOMETRY-BOOK I 111 DEFINITION. Vertical angles are the two angles on opposite sides of two intersecting lines; as, Az a \ and b; also, A c and d. EXERCISES 1 Find the complement of an angle of 27~. 2 Find the supplement of an angle of 27~. 3 Of what angle is 58~ the complement? 4 Of what angle is 126~ the supplement? 5 How many degrees in an angle if it is twice its complement? if one-third its complement? 6 How many degrees in an angle if it is five times its supplement? if one-fourth its supplement? 7 How many degrees in each of two complementary angles if the greater exceeds the less by 24~? 8 How many degrees in each of two supplementary angles if the greater exceeds the less by 36~? 9 Find an angle the sum of whose complement and supplement is 102~. 10 An angle is - of a perigon. How many degrees in its complement? in its supplement? 11 The sum of three angles is a perigon. The first is three times the second, and the second is twice the third; how many degrees in each? 12 An angle is 40~ less than its complement. What fraction is it of a right angle? of a straight angle? 13 Three angles, whose sum is a right angle, are in the ratio 2:3:5. How many degrees in each angle? 14 What angle is included between the hands of a watch at 3 o'clock? at 5 o'clock? at 8 o'clock? 15 What angle does the minute hand of a watch describe in 12 min.? in 28 min.? in 1 hr.? 16 What angle does the hour hand of a watch describe in 30 min.? in 1 hr.? in 31 hr.? STRAIGHT LINES 19 PROPOSITION III. THEOREM 112 Vertical angles are equal. C B HYPOTHESIS. AB and CD are two intersecting lines. CONCLUSION. Z m = Z n, and s = t. PROOF Z m is the sup. of Z s, and L n is the sup. of Z s. ~ 103.. / m = Z n. ' The supplements of equal s are equal." ~ 105 Likewise / s = Z t. Q. E. D. EXERCISES 17 If 1 = 38~, find Z3, Z4, Z2. 18 If Z 3 is four times Z 1, find each angle. 19 The straight line which bisects one of two vertical angles bisects the other also. [Let HK bisect A Z AOD; then HK bisects Z COB. PROOF. Z 3 = Z 2, and Z4 = Zlby~ 112. ButZ1 = 2 byhyp.... 3 3 K =Z 4 by Ax. 11.] 20 The bisectors of vertical angles form one straight D line. (Converse of Ex. 19.) 21 The bisectors of supplementary adjacent angles are perpendicular to each other. [Z a = Z a', and Z b = / b' by hyp. To prove Z a + Z b = a rt. Z. What is the a sum of the four A? A a and b are half this sum by hyp.] 22 The bisectors of complementary adjacent angles include an angle of 45~. [Follow the line of proof given in Ex. 21.] 20 PLANE GEOMETRY-BOOK I PROPOSITION IV. THEOREM 113 From a point without a line, only one perpendicular can be drawn to the line. P ' ~. K K HYPOTHESIS. P is a point without the line ED, PK is perpendicular to ED, and PK' is any other line from P to ED. CONCLUSION. PK/ is not perpendicular to ED. PROOF Let PK, remaining constantly perpendicular to ED and in the same plane, move on from P until the point K falls on the point K', PK taking the position of P'K'. "A geometric figure can be moved from one position to another without changing its form, size, or the relation of its parts." Post. 3 Then P'K' is 1 to ED by construction..'. PK' is not L to ED. "Only one perpendicular can be drawn to a given line at a given point in the line." ~ 102 Q. E. D. 114 DEFINITION. Parallel lines are lines in the same plane which cannot meet however far produced. Exercise 23. In the figure above, show that the Z PK K is an acute angle (~ 74), and that the Z PK'D is an obtuse angle (~ 76). PARALLEL LINES 21 PROPOSITION V. THEOREM 115 Two straight lines in the same plane, perpendicular to the same straight line, are parallel. d...,..... - ad b ae to ls eh p r to t l HYPOTHESIS. a and b are two lines each perpendicular to the line m. CONCLUSION. a and b are parallel. PROOF Could a and b, upon being produced, meet at some point, as at X, there would be two perpendiculars drawn from the point X to the line m, which is impossible. "From a point without a line only one perpendicular can be drawn to the line." ~ 113.. a and b cannot meet..a and b are parallel. " Parallel lines are lines in the same plane which cannot meet however far produced." ~ 114 Q. E. D. 116 SCHOLIUM. Proposition V is proved by the indirect method, called "reductio ad absurdum." The truth of the theorem is established by proving that the supposition that the theorem is not true leads to an absurdity. Thus, we suppose that the lines a and b are not parallel, i.e. that they will, if sufficiently produced, meet. The supposition leads to the conclusion that two perpendiculars can be drawn from the same point to the same straight line. The conclusion is false; therefore the supposition is false and must be abandoned. The logic is this: If it is not true that the lines are not parallel, then it is true that the lines are parallel. 22 PLANE GEOMETRY-BOOK I PROPOSITION VI. THEOREM 117 A straight line perpendicular to one of two parallels is perpendicular to the other. G A B E ----— P - C H D HYPOTHESIS. AB and CD are parallel lines, and GH is perpendicular to AB. CONCLUSION. GH is perpendicular to CD. PROOF Through H draw EF J to GH. "A straight line can be drawn from any point in any direction to any extent." Post. 2 Then EF is II to AB. "Two straight lines in the same plane perpendicular to the same straight line are parallel." ~ 115 But CD is II to AB by hyp..'. EF and CD coincide, for each passes through the same point H. "Through the same point only one straight line can be drawn parallel to the same straight line." Ax. 16 But EF is 1 to GH by const..'. CD is L to GH. That is, GH is L to CD. Q. E. D. PARALLEL LINES 23 PROPOSITION VII. THEOREM 118 Two straight lines parallel to a third straight line are parallel to each other. A K L B HYPOTHESIS. The lines H and K are each parallel to the line L. CONCLUSION. H and K are parallel lines. PROOF Draw AB L- to L. Post. 2 Then H and K are I to AB. ~ 117.'. H and K are parallel to each other. ~ 115 Q. E. D. DEFINITIONS 119 A transversal is a straight line T which intersects two or more straight L lines. Thus, the line T is a transversal a to the lines L and L', and the eight angles formed at the points of intersec- v ef tion are named as follows: 120 Interior angles are those within the two lines; as, angles c, d, e, and f. Exterior angles are those without the two lines; as, angles a, b, g, and h. Alternate interior angles are those within the two lines and on opposite sides of the transversal, but not adjacent; as, angles c and f, d and e. Alternate exterior angles are those without the two lines and on opposite sides of the transversal, but not adjacent; as, angles a and h, b and g. Corresponding angles are those similarly situated with respect to the two lines and on the same side of the transversal; as, angles a and e, c and g, b and f, d and h. 24 PLANE GEOMETRY-BOOK I PROPOSITION VIII. THEOREM 121 If two parallels are cut by a transversal, the alternate interior angles are equal. H E A H X B 0/ F K HYPOTHESIS. AB and CD are parallel lines, cut by the transversal EF. CONCLUSION. L x = Z y. PROOF Through 0, the middle point of EF, draw HK I to AB. HK is also ~ to CD. "A straight line. to one of two I)s is I to the other." ~ 117 Apply the figure OKF to the figure OHE so that the equal vertical angles at 0 (~ 112) shall coincide, OK falling along OH, and OF along OE. Post. 3 Then F will fall on E, for OF = OE by construction; and FK will fall along EH, for FK and EH are both I to HK. "From a point without a line, only one I can be drawn to the line." ~ 113.'. the A x and y coincide and are equal. "The test of equality of any two magnitudes is that they can be made to coincide." ~ 89 Q. E. D. PARALLEL LINES 25 122 COROLLARY 1. If two parallels are cut by a transversal, the corresponding angles are equal. E A AB C..\. D F PROOF Let AB and CD be parallel lines cut by the transversal EF. Then L m = L p, being vert. As. ~ 112 But Z r -= p by the theorem. ~ 121.'.Z m = r. Ax. 11 Likewise Z o= t, n =Z s, Z = Z v.. 123 COROLLARY 2. If two parallels are cut by a transversal, the alternate exterior angles are equal. PROOF m = r by Cor. 1. ~ 122 But Z v = Z r, being vert. As. ~ 112.'.Zm= Zv. Ax. 11 Likewise Z n = Z t. 124 COROLLARY 3. If two parallels are cut by a transversal, the interior angles on the same side of the transversal are supplementary. PROOF Z o = Z s by the theorem. ~ 121 Adding Z r = Z r, Iden. Zo+Z r=Zs+ r Ax. 1 = a straight Z. ~ 103.A. s o and r are supplementary. ~ 81 Likewise As p and s are supplementary. 26 PLANE GEOMETRY —BOOK I PROPOSITION IX. THEOREM 125 If a transversal cuts two straight lines making the alternate interior angles equal, the two lines are parallel. (Converse of ~ 121.) v \K E C e to s l c b t t HYPOTHESIS. AB and CD are two straight lines cut by the transversal EF at H and K, making the angles AHK and HKD equal. CONCLUSION. AB and CD are parallel. PROOF Through H draw MN II to CD. Post. 2 Then Z MHK = L HKD. (1) "If two parallels are cut by a transversal, the alternate interior angles are equal." ~ 121 But Z AHK = / HKD. (2) Hyp. From (1) and (2), Z MHK = Z AHK. "Things which are equal to the same thing are equal to each other." Ax. 11.'. AH and MH coincide. That is, AB and MN coincide. "Two straight lines which have two points in common coincide and form but one line." ~ 94 But MN and CD are II by const..'. AB and CD are parallel. Q. E. D. PARALLEL LINES 27 126 COROLLARY 1. (Converse of ~ 122.) If a transversal cuts two straight lines making the corresponding angles equal, the two lines are parallel. E Aim\ B C --- —- D F PROOF Let EF be a transversal to AB and CD, and let Zr =Z m. Now Z p = Z m, being vert. /. ~ 112.-. Zr= Zp. Ax. 11.'. AB and CD are parallel by the theorem. ~ 125 127 COROLLARY 2. (Converse of ~ 123.) If a transversal cuts two straight lines making the alternate exterior angles equal, the two lines are parallel. PROOF Z m = Lv by hyp. But Z r = Z v, being vert. Zs. ~ 112.'. Z m = Z r. Ax. 11.'. AB and CD are parallel by Cor. 1. ~ 126 128 COROLLARY 3. (Converse of ~ 124.) If a transversal cuts two straight lines making the interior angles on the same side of the transversal supplementary, the two lines are parallel. PROOF / o is the sup. of Z r by hyp. Let the student complete the proof. 28 PLANE GEOMETRY-BOOK I PROPOSITION X. THEOREM 129 Two angles whose corresponding sides are parallel are either equal or supplementary. \. A ---\ —. A-K '*-G D An H HYPOTHESIS. AB is parallel to EF, and CD is parallel to GH. CONCLUSION. Angles m, n, and o are equal; angles m and p are supplementary. PROOF Produce, if necessary, the two non-parallel sides AB and HG to intersect at K. Then Z m = Z s, ~ 122 and Z n = Z s. ~ 122.'. m = Z n. Ax. 11 But / n = Z o. ~112./. s m, n, and o are equal. Again, Z n + Zp = 2 rt. A. ~103 Substituting Z m for its equal Z n, we have / m + Zp = 2 rt. s. Q. E. D. 130 SCHOLIUM. The angles are equal when both pairs of corresponding sides extend in the same direction, or in opposite directions, from the vertices. The angles are supplementary when one pair of corresponding sides extends in the same direction, and the other pair in opposite directions, from the vertices. PARALLEL LINES 29 EXERCISES 24 If AB and CD are parallel and Z o = 110~, how many degrees in Z s? A n B 25 If AB and CD are parallel and Zp = 70~, \ how many degrees in Z v? Xc. D 26 If AB and CD are parallel and Z o = 110~, F how many degrees in Zr? 27 If AB and CD are parallel, prove that A n and v are supplementary. 28 If Zp = 75~ and Zr = 75~, are AB and CD parallel? 29 If Z o = 120~ and Z r = 60~, are AB and CD parallel? 30 If AB and CD are parallel and Z in = 82~, how many degrees in each of the other seven angles? 31 X and Y are two parallel lines cut by two s T parallel transversals S and T. Prove Z a = Z r; Z l - nab k/1 -Zg. cd g= ' r eef of P 32 In Ex. 31 prove that Z c + Z o = 2 rt. A. /k r 33 If a straight line intersects two parallels: 1 The bisectors of the alternate interior angles are parallel. [~ 125.] 2 The bisectors of the corresponding angles are parallel. [~ 126.] 3 The bisectors of the interior angles on the same side of the transversal are perpendicular to each other. [Ex. 21, ~ 124.] 34 Perpendiculars to two parallels are parallel. [~ 117, ~ 115.] 35 If AB is 1I to CD, and A = ZC, prove AE Ii to CF. [Draw AE' 11 to CF and prove that P- - AE and AE' coincide.] A< c< 36 If two right angles have two sides parallel, B the other two sides are parallel. 37 X and Y are parallel lines; prove that ZC = Za __ + Zb. [Through the vertex C draw a line parallel to X.] y 38 In a line AB, points C and D are so taken A -B that AD = CB; prove AC = DB. [Ax. 4.] c D 39 Two lines are not parallel if they are respectively perpendicular to two non-parallel lines. 30 PLANE GEOMETRY-BOOK I POLYGONS DEFINITIONS 131 A polygon is a plane surface bounded A by straight lines; as, ABCD. D The sides of a polygon are the bounding lines, whose sum is the perimeter of the B c polygon. The angles of a polygon are the angles formed by its sides, and the vertices of the angles are the vertices of the polygon. Adjacent angles of a polygon are two angles having a common side; as, angles B and C, C and D, etc. 132 A diagonal of a polygon is the straight line joining two vertices not adjacent. 133 An exterior angle of a polygon is the angle included between one side and an ad- jacent side produced; as, angle EDF. I TRIANGLES 134 A triangle is a polygon of three sides. 135 A right triangle is a triangle having a right angle; as, A and B. 136 An oblique triangle is a triangle having all its angles oblique; as, C, D, E, F, and G. TRIANGLES 31 137 An acute triangle is a triangle having all its angles acute; as, C, D, and E. 138 An obtuse triangle is a triangle having an obtuse angle; as, F and G. 139 An equiangular triangle is a triangle having all of its angles equal; as, E. 140 An equilateral triangle is a triangle having all of its sides equal; as, E. 141 An isosceles triangle is a triangle having two of its sides equal; as, A, D, E, and G. 142 A scalene triangle is a triangle having no two of its sides equal; as, B, C, and F. 143 The hypotenuse of a right triangle is the A side opposite the right angle; as, AC. 144 The legs of a right triangle are the sides B_ including the right angle; as, AB and BC. 145 The legs of an isosceles triangle are the two D equal sides; as, DE and DF. 146 The base of a triangle is the side upon which the triangle is assumed to stand. E F Any side of a triangle may be assumed as the base. In the isosceles triangle the unequal side is always the base, unless otherwise stated. 147 The vertical angle of a triangle is the angle opposite the assumed base, and the other two angles of the triangle are called the base angles. 148 The vertex of a triangle is the vertex of G the vertical angle. Thus, in the triangle GHK, if HK is taken as the base, then angle G is the vertical angle, the point G is the vertex of the H X triangle, and angles H and K are the base angles. 32 PLANE GEOMETRY-BOOK I 149 A median of a triangle is a straight line drawn from the vertex of an angle to the middle point of the opposite side. Every triangle has three medians. 150 An altitude of a triangle is the perpendicular drawn from the vertex of an angle to the opposite side. Every triangle has three altitudes. When we speak of the altitude of a triangle, we mean the altitude upon the assumed base. 151 A bisector of an angle of a triangle is a straight line bisecting an angle and terminated by the opposite side. Every triangle has three angle-bisectors. 152 Classification of triangles. F Isosceles. Right. Scalene. T Isosceles. /\ Triangle.! Acute. | Scalene. Oblique. <. Isosceles. { Obtus alene. Scalene. TRIANGLES 33 PROPOSITION XI. THEOREM 153 The sum of the three angles of a triangle is equal to two right angles.* X z --- -^,,, HYPOTHESIS. ABC is any triangle. CONCLUSION. ZA+ B+ C = 2 rt.. PROOF Draw MN through B II to AC, and produce AB and CB, forming the angles a, b, and c. Then Z A = Z a, "corresponding A of II lines," Z B = L b, " being vertical A," and / C = Z c, "( corresponding A of II lines." Adding, LA+ZB+ZC=Za+Z b+ c =2 rt. ZA. ~ 122 ~ 112 ~ 122 Ax. 1 ~ 106 Q. E. D. 154 COROLLARY 1. A triangle can have but one right angle or one obtuse angle. 155 COROLLARY 2. The acute angles of a right triangle are complementary. 156 COROLLARY 3. If two triangles have two angles of the one equal to'two angles of the other, the third angles are equal. 157 COROLLARY 4. Each angle of an equilateral triangle is 60~. * This theorem, one of the most important in geometry, was discovered by Pythagoras. 34 PLANE GEOMETRY-BOOK I PROPOSITION XII. THEOREM 158 An exterior angle of a triangle is equal to the sum of the opposite interior angles. A B.. —D HYPOTHESIS. ACD is an exterior angle of the triangle ABC. CONCLUSION. Z ACD = Z A + Z B. PROOF Z ACD is the sup. of Z ACB. ~ 103 But Z A + Z B is the sup. of Z ACB. ~ 153. ZACD = A+ZB. ~105 Q. E. D. 159 COROLLARY. An exterior angle of a triangle is greater than either opposite interior angle. EXERCISES 40 If two right triangles have an acute angle of the one equal to an acute angle of the other, the other acute angles are equal. 41 If one angle of a triangle is right or obtuse, each of the other two is acute. 42 In an obtuse triangle the sum of the two acute angles is less than a right angle. 43 In an acute triangle the sum of any two angles is greater than a right angle. 44 If one angle of a triangle is 78~, what is the sum of the other two angles? 45 If two angles of a triangle are 65~ and 84~ respectively, find the third angle. 46 If the acute angles of a right triangle are equal, how many degrees in each? TRIANGLES 35 PROPOSITION XIII. THEOREM 160 Any side of a triangle is less than the sum of the other two sides. A HYPOTHESIS. Let AC be the longest side of the triangle ABC. CONCLUSION. AC < AB + BC. PROOF The straight line AC is less than the broken line ABC. "A straight line is the shortest line between two points." Ax. 15.. AC < AB + BC. Q. E. D. 161 COROLLARY. Any side of a triangle is greater than the difference of the other two sides. For AB + BC > AC. ~ 160 Subtract BC = BC, Iden. and AB > AC - BC. Ax. 5 EXERCISES 47 In order to determine whether three given numbers may represent the sides of a triangle, is it necessary to test them by both ~ 160 and ~ 161? 48 Which of the following sets of numbers may represent the sides of a triangle: 7, 8, 13; 6, 7, 15; 9, 16, 27; 12, 17, 26; 3, 5, 8? 49 If one of the exterior angles of a triangle is a right angle, the triangle is a right triangle. 50 If the sum of two exterior angles of a triangle is equal to three right angles, the triangle is a right triangle. 51 Any side of a triangle is less than half the perimeter. [By ~ 160 AC < AB + BC. Add AC to each member and reduce. ] 36 PLANE GEOMETRY-BOOK I PROPOSITION XIV. THEOREM 162 Two triangles are equal if two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. A D B C BE F HYPOTHESIS. ABC and DEF are any two triangles having AB = DE, AC = DF, and Z A =Z D. CONCLUSION. A ABC = A DEF. PROOF Place the A ABC on the A DEF so that the equal A A and D shall coincide, AB falling along DE, and AC along DF. Then B will fall on E, for AB = DE by hypothesis; and C will fall on F, for AC = DF by hypothesis..'. BC coincides with EF. Ax. 14.'. the two A coincide, and are equal. Q. E. D. 163 COROLLARY 1. Two right triangles are equal if their legs are equal, each to each. 164 DEFINITIONS. Mutually equilateral polygons are polygons which have their corresponding sides equal. Mutually equiangular polygons are polygons which have their corresponding angles equal. 165 Homologous lines or angles in mutually equiangular polygons are lines or angles similarly situated. 166 COROLLARY 2. In equal triangles, the homologous sides are equal and the homologous angles are equal. TRIANGLES 37 PROPOSITION XV. THEOREM 167 Two triangles are equal if two angles and the included side of the one are equal respectively to two angles and the included side of the other.* A D B EC E - HYPOTHESIS. ABC and DEF are two triangles having Z B = Z E, Z C = Z F, and BC = EF. CONCLUSION. / ABC = A DEF. PROOF Place the A ABC on the A DEF so that the equal sides BC and EF shall coincide. Then BA will fall along ED, for Z B = ZE by hyp.; and CA will fall along FD, for Z C = Z F by hyp..'. A will fall on D, for "two straight lines can intersect in one point only." ~ 95.. the two A coincide, and are equal. Q. E. D. 168 COROLLARY 1. Two right triangles are equal if the hypotenuse and an acute angle of the one are equal respectively to the hypotenuse and an acute angle of the other. 169 COROLLARY 2. Two right triangles are equal if a leg and an acute angle of the one are equal respectively to a leg and the homologous acute angle of the other. * This proposition is attributed to Thales. 38 PLANE GEOMETRY -BOOK I PROPOSITION XVI. THEOREM 170 If two sides of one triangle are equal respectively to two sides of another, but the included angle of the first is greater than the included angle of the second, the third side of the first is greater than the third side of the second. A A' B /C C' HYPOTHESIS. In the A ABC and A'B'C', AB = A'B', AC = A'C', and L BAC> Z B'A'C'. CONCLUSION. BC > B'C'. PROOF Place the A A'BC' in the position of AB"C, A'C' coinciding with its equal AC. Draw AD bisecting the Z BAB", and join DB". The AADB and ADB" are equal; for AB = AB" by hyp., AD = AD by identity, and Z DAB = Z DAB" by const. "Two triangles are equal if two sides and the included Z of the one are equal respectively to two sides and the included Z of the other." ~ 162.'. DB = DB", "being homologous sides of equal.." ~ 166 Now DC + DB" > B"C. "Any side of a A < the sum of the other two sides." ~ 160.. DC+DB > B"C. Or BC > B"C or B'C'. Q. E. D. TRIANGLES 39 PROPOSITION XVII. THEOREM 171 If two sides of one triangle are equal respectively to two sides of another, but the third side of the first is greater than the third side of the second, the angle opposite the third side of the first is greater than the angle opposite the third side of the second. A Af B C f HYPOTHESIS. In the A ABC and A'B'C', AB = A'B', AC = A'C', and BC > B'C'. CONCLUSION. Z A > Z A'. PROOF The Z A is either equal to, less than, or greater than the ZA'. If / A = Z A', BC = B'C'. ~ 162 If Z A < Z A', BC < B'C'. ~ 170 Both these conclusions contradict the hypothesis that BC > B'C'..LA > ZA'. Q. E. D. EXERCISES 52 If AD be drawn to the middle point of BC in the triangle ABC above, prove that Z ADC > Z ADB. [~ 171.] 53 In the triangle ABC, AD is drawn to the middle point of BC. Then 1 If Z ADB is acute, AC > AB. [~ 170.] BL 2 If Z ADB is obtuse, AC < AB. [~ 170.] 3 If Z ADB is art. 4, AC = AB. [~ 162.] 40 PLANE GEOMETRY-BOOK I PROPOSITION XVIII. THEOREM 172 Two triangles are equal if the three sides of the one are equal respectively to the three sides of the other. A D B C E P HYPOTHESIS. In the A ABC and DEF, AB = DE, AC = DF, and BC = EF. CONCLUSION. The & ABC and DEF are equal. PROOF The L A is either greater than, less than, or equal to the D. If L A > L D, BC > EF. If Z A < Z D, BC < EF. ~ 170 Both these conclusions contradict the hypothesis that BC = EF....A=Z D..-. A ABC =A DEF, "having two sides and the included angle of the one equal respectively to two sides and the included angle of the other." ~ 162 Q. E. D. EXERCISES 54 ABC and DBC are two isosceles triangles on the same base BC. Prove that the triangles ADB and ADC are equal. 55 Prove Ex. 54 by drawing the triangles on opposite B c sides of the base. 56 HK and MN bisect each other at P. Prove A HPM = A KPN. [~ 162.] M TRIANGLES 41 PROPOSITION XIX. THEOREM 173 In an isosceles triangle, the angles opposite the equal sides are equal. s a ss s HYPOTHESIS. ABC is an isosceles triangle, having AB = AC. CONCLUSION. Z B = Z C. PROOF Draw AD bisecting the base BC. The A ADB and ADC are equal; for AD = AD by identity, DB = DC by construction, and AB = AC by hyp. " Two ^ are equal if the three sides of the one are equal respectively to the three sides of the other." ~ 172.. LB=Z C, "being homologous A of equal." Q. E. D. 174 COROLLARY 1. The straight line drawn from the vertex of an isosceles triangle to the middle point of the base bisects the vertical angle and is perpendicular to the base. 175 COROLLARY 2. An equilateral triangle is equiangular. EXERCISES 57 The exterior angles at the base of an isosceles triangle are equal. 58 The triangle formed by joining the middle points of the sides of an isosceles triangle is isosceles. [& 2 and 3 are 7 equal by ~ 162; whence A 1 is isosceles by ~ 166.] 42 PLANE GEOMETRY-BOOK I PROPOSITION XX. THEOREM 176 If two angles of a triangle are equal, the sides opposite are equal, and the triangle is isosceles. A B C HYPOTHESIS. In the triangle ABC, Z B = Z C. CONCLUSION. AB = AC. PROOF Draw AD L to BC. The rt. A ADB and ADC are equal; for AD = AD by identity, and Z B = L C by hyp. "Two rt. A are equal if a leg and an acute Z of the one are equal respectively to a leg and the homologous acute Z of the other." ~ 169.'. AB = AC, "being homologous sides of equal A." ~ 166 Q. E. D. 177 COROLLARY. An equiangular triangle is equilateral. EXERCISES 59 HK is drawn parallel to BC the base of the isosceles A triangle ABC. Prove that the triangle AHK is isosceles. [~~ 122, 176.] J 60 If two exterior angles of a triangle are equal, the triangle L c is isosceles. TRIANGLES 43 61 If the vertical angle of an isosceles triangle is 60~, the triangle is equilateral. 62 The sum of any two exterior angles of a triangle is greater than two right angles. 63 The exterior angle formed by producing the hypotenuse of a right triangle is obtuse. 64 If one angle of a triangle is equal to the sum of the other two, the triangle is a right triangle. 65 If a line drawn from the vertex of a triangle to the mid-point of the base is perpendicular to the base, the triangle is isosceles. [~ 163.] 66 The bisector of the vertical angle of an isosceles triangle bisects the base at right angles. [~ 162.] 67 If the bisector of an angle of a triangle is perpendicular to the opposite side, the triangle is isosceles. [~ 169.] 68 If two isosceles triangles have their vertical angles equal, they are mutually equiangular. 69 The exterior angle at the vertex of an isosceles triangle is twice an angle at the base. [~ 158.] 70 The bisector of the exterior vertical angle of an isosce- les triangle is parallel to the base. [Ex. 69, ~ 126.] 71 Two equal straight lines drawn from the same point to the same straight line make equal angles with the line. [~ 173.] 72 The perpendiculars drawn from the mid-point of the base of an isosceles triangle to the legs cut off equal parts from the legs. [~ 168.] 73 If two points in the base of an isosceles triangle are equidistant from the extremities of the base, they are also equidistant from the vertex. [~~ 162, 166.] 74 The sum of the three exterior angles of a triangle / (one at each vertex) is equal to four right angles. [~ 158.] 75 The triangle formed by the base of an isosceles triangle and the bisectors of the base angles is isosceles. [Ax. 9, ~ 176.] l 76 In the triangle ABC, Z B = 2 Z C. BD bisects Z B and meets AC at D. Prove A DBC isosceles. 44 PLANE GEOMETRY -BOOK I PROPOSITION XXI. THEOREM 178 Two right triangles are equal if the hypotenuse and a leg of the one are equal respectively to the hypotenuse and a leg of the other. A CZ dBB C} HYPOTHESIS. In the rt. A ABC and A'B'C', AC = A'C', and AB - A'B'. CONCLUSION. A ABC = A AB'C', PROOF Place the A so that the equal legs AB and A'B' shall coincide, C and C' falling on opposite sides of AB, and the A A'B'C' taking the position of ABC'. Z CBC' = 2 rt. Zs. Hyp..'. CBC' is a straight line. ~ 110 The A ACC' is an isosceles triangle, for AC = AC'. Hyp....C=Z C'. "In an isosceles A the A opposite the equal sides are equal." ~ 173.. A ABC = A ABC' or A A'B'Ct. "Two right A are equal if the hypotenuse and an acute Z of the one are equal respectively to the hypotenuse and an acute Z of the other." ~ 168 Q. E. D. EXERCISE 77. The altitude of an isosceles triangle bisects the base, the vertical angle, and the triangle. [~ 178.] TRIANGLES 45 PROPOSITION XXII. THEOREM 179 If two sides of a triangle are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side. A B C HYPOTHESIS. In the A ABC, AC is greater than AB. CONCLUSION. Z ABC is greater than Z C. PROOF On AC take AD = AB, and draw BD. Z ABC > Z s. "The whole is greater than any of its parts." Ax. 12 Zs = s'. "In an isosceles A the A opp. the equal sides are equal." ~ 173 Z s'> Z C. "An ext. Z of a A > either opp. int. Z." ~ 159 We now have / ABC > Z s = s' > / C... /ABC > / C. Q E.D. EXERCISES 78 How many degrees in each angle of an isosceles right triangle? 79 The vertical angle of an isosceles triangle contains 74~. How many degrees in each angle at the base? 80 One angle of an isosceles triangle contains 120~. How many degrees in each of the other two angles? 81 The vertical angle of an isosceles triangle contains 65~. How many degrees in an exterior angle at the base? 46 PLANE GEOMETRY-BOOK I PROPOSITION XXIII. THEOREM 180 If two angles of a triangle are unequal, the sides opposite are unequal, and the greater side is opposite the greater angle. A D HYPOTHESIS. In the A ABC, Z ABC > Z C. CONCLUSION. AC is greater than AB. PROOF Suppose BD drawn making Z DBC = Z C. "From the greater of two magnitudes a part can be cut off equal to the less." Post. 6 Then DC = DB (1). "If two A of a A are equal, the sides opp. are equal, and the A is isosceles." ~ 176 Adding AD to each member of (1), we have AC = AD + DB. Ax. 1 But AD + DB > AB. "Any side of a A < the sum of the other two sides." ~ 160.'. AC > AB. Q.E.D. 181 COROLLARY. In a right triangle the hypotenuse is greater than either leg. EXERCISES 82 Prove Prop. XXIII by the indirect method. 83 In an isosceles triangle a line drawn from the vertex to any point in the base is less than either leg. [~~ 154, 180.] 84 In an isosceles triangle, a line drawn from the vertex to any point in the base produced is greater than either leg. TRIANGLES 47 PROPOSITION XXIV. THEOREM 182 The perpendicular is the shortest line that can be drawn from a point to a straight line. P A D H B HYPOTHESIS. P is any point without the line AB, PD is perpendicular to AB, and PH any other line drawn from P to AB. CONCLUSION. PD is less than PH. PROOF In the rt. A PDH, PD < PH. " In a rt. A the hypotenuse is greater than either leg." ~ 181 Q. E; D. 183 DEFINITION. The distance from a point to a line, or from one point to another, is the shortest distance. Therefore 184 COROLLARY 1. The distance of a point from a line is the length of the perpendicular drawn from the point to the line. 185 COROLLARY 2. The shortest line that can be drawn from a point to a given line is perpendicular to the given line. EXERCISES 85 An altitude of a triangle is less than either of the adjacent sides. 86 The three altitudes of any triangle are together less than the perimeter. [Ex. 85.] 87 The altitude of an equilateral triangle is greater than the half of one side. 88 How many degrees are included between the bisectors of two angles of an equilateral triangle? 48 PLANE GEOMETRY -BOOK I PROPOSITION XXV. THEOREM 186 Every point in the perpendicular bisector of a line is equidistant from the extremities of the line. C P A D HYPOTHESIS. CD is the perpendicular bisector of AB, and P is any point in CD. CONCLUSION. PA = PB. PROOF The rt. A PDA and PDB are equal; for PD is common, and DA = DB by hyp. "Two rt. & are equal if their legs are equal, each to each." ~ 163.'. PA = PB, "being homologous sides of equal Q." ~ 166 Q. E. D. EXERCISES 89 Each angle at the base of an isosceles triangle contains 70~. How many degrees in the exterior angle at the vertex? 90 The bisectors of the two angles at the base of a triangle contain an angle of 111~. How many degrees in the vertical angle? 91 One angle of a triangle contains 52~. How many degrees in the angle included by the bisectors of the other two angles? 92 One angle of a triangle contains 64~. How many degrees in each of the other two angles if one is 23~ larger than the other? 93 The angles of a triangle are in the ratio of 2:3:5. How many degrees in each angle? TRIANGLES 49 PROPOSITION XXVI. THEOREM 187 Every point equidistant from the extremities of a line is in the perpendicular bisector of the line. C D HYPOTHESIS. CD is I to AB, P is any point in CD, and PA = PB. CONCLUSION. CD bisects AB. PROOF The rt. A PDA and PDB are equal; for PD is common, and PA = PB by hyp. "Two rt. & are equal if the hypotenuse and a leg of the one are equal respectively to the hypotenuse and a leg of the other." ~ 178.'. AD = DB, "being homologous sides of equal A." ~ 166 Q. E. D. 188 COROLLARY. Two points each equidistant from the extremities of a line determine the perpendicular bisector of the line. EXERCISES 94 If two isosceles triangles have the same base, the straight line joining their vertices bisects their common base at right angles. Two cases. A 95 The triangles ABC and DBC are on opposite sides B of the same base BC, having AB = BD, AC = DC. Prove AD I to BC. D 96 In a triangle GHK, Z G = 80~, Z H = 70~. What angle does the bisector of the Z K make with HG? 50 PLANE GEOMETRY-BOOK I PROPOSITION XXVII. THEOREM 189 Of two oblique lines drawn from the same point in a perpendicular and cutting off unequal distances from the foot, the more remote is the greater. A / K' D HYPOTHESIS. PD is L to AB, and DH > DK. CONCLUSION. PH > PK. PROOF Take DK' = DK, and draw PK'. Then PK' = PK. "Every point in the perpendicular bisector of a line is equidistant from the extremities of the line." ~ 186 The Z PDH is a rt. L by hyp..-. the Z PK'D is acute. ~ 154.'. the L PK'H is obtuse. ~ 103.-. the L PHK' is acute. ~ 154 That is, the L PK'H > the L PHK... PH > PK' or its equal PK. "If two z of a A are unequal, the sides opp. are unequal, and the greater side is opp. the greater Z." ~ 180 Q. E. D. 190 COROLLARY 1. Only two equal straight lines can be drawn from a point to a straight line. 191 COROLLARY 2. Of two unequal oblique lines drawn from a point in a perpendicular, the greater cuts off the greater distance from the foot of the perpendicular. TRIANGLES 51 PROPOSITION XXVIII. THEOREM 192 Every point in the bisector of an angle is equidistant from the sides of the angle. A I DP B K C HYPOTHESIS. BD bisects the angle ABC, P is any point in BD, PH and PK are perpendiculars respectively to AB and BC. CONCLUSION. PH = PK. PROOF The rt. A PBH and PBK are equal; for PB is common, and Z PBH = Z PBK by hyp. " Two rt. A are equal if the hypotenuse and an acute Z of the one are equal respectively to the hypotenuse and an acute Z of the other." ~ 168.. PH = PK, "being homologous sides of equal &." ~ 166 Q. E. D. 193 COROLLARY. Every point equidistant from the sides of an angle is in the bisector of the angle. PROOF Let PH = PK, and join PB. The rt. A PBH = the rt. A PBK. ~ 178.-. Z PBH = Z PBK. ~ 166.-. PB bisects the angle ABC. 52 PLANE GEOMETRY-BOOK I EXERCISES D 97 A line parallel to the base of an isosceles triangle 4 makes equal angles with the legs. A 98 In the triangle ABC, BA is produced to D making AD = AC. Prove Z D < Z BCD. B c 99 DE is parallel to BC, the base of an isosceles tri- A angle. Prove A 1 and 2 supplementary. D 100 The perpendiculars drawn to each of two paral- B c lels are parallel. A B 101 AB and CD are equal and parallel lines. Prove that AD and BC bisect each other. c — D 102 On the same base and on the same side of it, there can be but one equilateral triangle. 103 If from any point in the bisector of an angle, a line is drawn parallel to one side of the angle and intersecting the other side, the triangle thus formed is isosceles. [Z 1 = Z2= Z 3.] 104 Show by a diagram that Cor. 2, Prop. XV, is not true if the word " homologous " be omitted. 105 Why, in Cor. 1, Prop. XV, is the word "homologous" not inserted before " acute angle'" in the last line? 106 If the vertex A of the triangle ABC is joined to A any two points in the base, as D and E, prove that Z ADB > ZAEB > Z ACB. 107 Two isosceles triangles are equal if the legs and B D E C altitude of the one are equal respectively to the legs and altitude of the other. [Prove by superposition, making the equal altitudes coincide.] A A 108 In an equilateral triangle ABC, M and N are any two points in BC. OM is parallel B/ \ C B Nc to AB, and ON is parallel to AC. Prove the V triangle OMN equilateral. Two cases. D 109 DEF is an equilateral triangle. GE and GF bisect the angles E and F respectively. GH and GK are parallel respectively to DE and DF. Prove that EH = HK =KF. E I K QUADRILATERALS 53 QUADRILATERALS DEFINITIONS * 194 A quadrilateral is a polygon of four sides. A trapezium is a quadrilateral which has no two sides parallel. A trapezoid is a quadrilateral which has two sides parallel, and two non-parallel. An isosceles trapezoid is a trapezoid whose non-parallel sides are equal. Trapezium Trapezoid Isosceles Trapezoid 195 A parallelogram is a quadrilateral whose opposite sides are parallel. A right parallelogram is one whose angles are right angles. An oblique parallelogram is one whose angles are oblique. PARALLELOGRAMS Rectangle Square Rhomboid Rhombus 196 A rectangle is a right parallelogram. A square is an equilateral rectangle. A rhomboid is an oblique parallelogram. A rhombus is an equilateral rhomboid. 197 The bases of a trapezoid are the parallel sides, called its lower and upper bases. The legs of a trapezoid are the nonparallel sides. The median of a trapezoid is the straight line joining the middle points of the legs. * The student should now review ~~ 131-133. 54 PLANE GEOMETRY-BOOK I 198 The bases of a parallelogram are the side on which it stands and the opposite side, called its lower and upper bases. 199 The altitude of a parallelogram or trapezoid is the perpendicular distance between the bases. PROPOSITION XXIX. THEOREM 200 The opposite sides of a parallelogram are equal. A\... X\D B X\ HYPOTHESIS. ABCD is a parallelogram. CONCLUSION. AD = BC, and AB = DC. PROOF Draw the diagonal BD. The A BDA and BDC are equal; ~ 167 for BD is common, x = Z x', and Z y = / y', "being alt. int. A of parallel lines." ~ 121.-. AD = BC, and AB = DC, "being homologous sides of equal A." ~ 166 Q. E. D. 201 COROLLARY 1. A diagonal divides a parallelogram into two equal triangles. 202 COROLLARY 2. Parallels comprehended between parallels are equal. 203 COROLLARY 3. The opposite angles of a parallelogram are equal. QUADRILATERALS 55 PROPOSITION XXX. THEOREM 204 If the opposite sides of a quadrilateral are equal, the figure is a parallelogram.. ABD is a qdrira, AD HYPOTHESIS. ABCD is a quadrilateral, having AD = BC and AB = DC. CONCLUSION. ABCD is a parallelogram. PROOF Draw the diagonal BD. In the A BDA and BDC, BD = BD, Iden. AD = BC, Hyp. and AB = DC. Hyp..'. A BDA = A BDC, "having three sides of each equal respectively." ~ 172..Z x= x', and Z y = Z y', "being homologous A of equal." ~ 166.'. AD is II to BC, and AB is II to DC, "having the alt. int. A equal." ~ 125.'. ABCD is a parallelogram, s' having its opposite sides parallel." ~ 195 Q. E. D. 56 PLANE GEOMETRY BOOK I PROPOSITION XXXI. THEOREM 205 If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram. At xHYPOTHESIS. ABCD is a quadrilateral, having AD equal and parallel to BC. CONCLUSION. ABCD is a parallelogram. PROOF Draw the diagonal BD. In the A BDA and BDC, BD = BD, Iden. AD = BC, Hyp. and L x = Z x', ' being alt. int. A of parallel lines." ~ 121.~. A BDA = A BDC, "having two sides and the included Z of each equal respectively." ~ 162.'. AB = DC, "being homologous sides of equal ~." ~ 166 But AD = BC by hyp..'. ABCD is a parallelogram, "having its opposite sides equal." ~ 204 Q. E. D. EXERCISES 110 Two adjacent angles of a parallelogram are supplementary. 111 The bisectors of two adjacent angles of a parallelogram are perpendicular to each other. [Z a + Z b / = art. Z. ~~ 124, 153.] QUADRILATERALS 57 PROPOSITION XXXII. THEOREM 206 The diagonals of a parallelogram bisect each other. HYPOTHESIS. ABCD is a parallelogram whose diagonals intersect at 0. CONCLUSION. OA = OC, and OD = OB. PROOF In the A AOD and BOC, AD = BC. " The opposite sides of a a7 are equal." ~ 200 L x = Z x', and Z y = Z y', "being alt. int. A of parallel lines." ~ 121.. A AOD = A BOC, "having a side and two adj. A of each equal respectively." ~ 167.-. OA = OC, and OD = OB, "being homologous sides of equal Q." ~ 166 Q. E. D. EXERCISES 112 The bisectors of two opposite angles of a parallelogram are parallel. [By Ex. 111, each is I to the same line.] 113 The bisectors of the four angles of a parallelogram form a rectangle. [Ex. 111.] 114 If two parallelograms have an angle in each equal, they are mutually equiangular. 115 The triangle formed in a parallelogram by the /\ / bisector of an angle is isosceles. [Z 1 = Z 2 = Z 3.] 58 PLANE GEOMETRY-BOOK I PROPOSITION XXXIII. THEOREM 207 Two parallelograms are equal, if two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. * - rsD Al D B C B' HYPOTHESIS. ABCD and A'B'C'D' are two parallelograms, having AB = A'B', AD = A'D', and Z A = Z A'. CONCLUSION. a7 ABCD = 17 A'B'C'D'. PROOF Apply the ABCD to the l A'B'C'D', so that ZA will coincide with its equal Z A', AB falling along A'B', and AD along A'D'. B will fall on B', and D on D'; for AB = A'B', and AD = A'D', by hyp. BC will fall along B'C'. "Through the same point only one st. line can be drawn parallel to the same st. line." Ax. 16 Likewise DC will fall along D'C'..'. C will fall on C'. "Two st. lines can intersect in one point only." ~ 95 The two /7 therefore coincide, and are equal. Q. E; D. EXERCISES 116 Two rectangles are equal if two adjacent sides of the one are equal to two adjacent sides of the other. 117 Two squares are equal if a side of the one is equal to a side of the other. QUADRILATERALS 59 118 If one angle of a parallelogram is 82~, how many degrees in each of the other angles? 119 Two adjacent sides of a parallelogram are 14 feet and 23 feet respectively. Find the perimeter. 120 If one side of a rhombus is 6 inches in length, what is the length of the perimeter? 121 If two adjacent angles of a parallelogram are in the ratio 3:5, how many degrees in each angle? 122 If one angle of a parallelogram is a right angle, the figure is a right parallelogram. 123 If the diagonals of a quadrilateral bisect each other, A D the figure is a parallelogram. [A AOD = A BOC (~ 162)... AD =BC, and Z1 = Z2 (~166)..'. AD is II to BC (~ 125)..-. ABCD is a /7 (~ 205).] A~~~oD 124 The diagonals of a square are perpendicular to each other, K\A and bisect the angles. [In A ABD, AB = AD, AO bisects BD 7 (~ 206)..-. AO is l. to BD and bisects Z A (~ 174).] B 125 The diagonals of a rhombus are perpendicular to each other, and bisect the angles. [Proof identical with that A in Ex. 124.] 126 The diagonals of a rectangle are equal. [Rt. A ABD B = rt. A ABC (~ 163)..-. BD = AC.] 127 The diagonals of a rhomboid are unequal. [Z A is obtuse, and ZB is acute... ZA > Z B. Whence in A ABD and ABC, BD > AC (~ 170).] 128 The diagonals of a rectangle, not a square, are oblique to each other. [In A AOD and AOB, OA is common, OD = OB A D (~ 206), and AD>AB.... ZAOD>ZAOB (~ 171). o That is, AC and BD are oblique.] 129 The diagonals of a rhomboid are oblique to each other. [The proof is identical with that of Ex. 128.] 130 In the parallelogram ABCD, AF bisects ZA, BE D bisects ZB. Prove that AF and BE bisect each other. / [AE = AB = BF (Ex. 115)..-. ABFE is a 0 (~ 205).] 60 PLANE GEOMETRY-BOOK I PROPOSITION XXXIV. THEOREM 208 If three parallel lines intercept equal segments on one transversal, they intercept equal segments on every transversal. A - E \F HYPOTHESIS. AE and BF are transversals to the parallels AB, CD, EF, and AC = CE. CONCLUSION. BD = DF. PROOF Draw HDK 11 to AE. Then HD = AC = CE = DK. ~ 200 A HDB = A KDF; ~167 for HD = DK (proved), Z t = Z t' (~ 112), and Z s = L s' (~ 121)..-. BD = DF. ~ 166 Q. E. D. 209 COROLLARY 1. If a line is parallel to the base of a triangle and bisects one side, it bisects the other side also. Let DE be I to BC, and AD = DB. Draw A HAK II to BC. HK is also II to DE (~ 118). Now HK, DE, and BC intersect equal segments D --- —-F on AB by hypothesis;.~. they intersect equal segments on AC. That is, DE bisects AC. B G C 210 COROLLARY 2. If a line bisects two sides of a triangle, it is parallel to the third side, and is equal to half the third side. Let DE bisect AB and AC. Draw DF 1I to BC, and EG QUADRILATERALS 61 11 to AB. DF passes through E (~ 209), and coincides with DE (~ 94)..-. DE is II to BC. Again: EG bisects BC by ~209, and DBGE is a 7 by const.. DE = BG = BC. 211 COROLLARY 3. The median of a trapezoid is parallel to the bases, and is equal to half the sum of A D the bases. E/ -— "- G -— H Let EF be the median. Draw EH II to B -.C AD. EH is also 11 to BC (~ 118), and bisects DC (~ 208)..-. EF coincides with EH and is II to AD and BC. EF bisects AC at G (~ 208)... EG = 1 BC, and GF = A AD (~ 210)..EF = 1 (AD + BC). EXERCISES 131 Every straight line drawn through the middle A X D point of a diagonal, and terminated by the sides, of a \ parallelogram is bisected at that point. [Draw FOE {I to AD, and apply ~ 208.] B y c 132 In the figure of ~ 208, prove that EF - CD = CD - AB. 133 Prove that the median of a trapezoid is equal to half the sum of the bases by Ex. 132. 134 If a line is parallel to the base of a trapezoid and bisects one leg, it bisects the other leg also. 135 If a line bisects two sides of a triangle, it bisects every line drawn from the third side to the opposite vertex. 136 The line bisecting two adjacent sides of a quadrilateral is parallel and equal to the line bisecting the other two sides. 137 The median of a trapezoid bisects both diagonals. 138 In a right triangle, the mid-point of the hypote- c nuse is equidistant from the three vertices. [Let b = c, z andx=y. Then p islltoz by~ 210... p is to x by ~ 117... a = b = c (~ 186).] x Y 139 In the parallelogram ABCD, E and F are D the mid-points of AD and BC. Prove that AF f \ and EC trisect the diagonal BD. [AFCE is a 0 (~ 205). In the A DAG, EH bisects DG (~ 209). Likewise FG bisects BH in the A BCH.] B -- 62 PLANE GEOMETRY-BOOK I POLYGONS IN GENERAL DEFINITIONS * 212 A triangle (trigon) is a polygon of three sides. A quadrilateral (tetragon) is a polygon of four sides. A pentagon is a polygon of five sides. A hexagon is a polygon of six sides. A heptagon is a polygon of seven sides. An octagon is a polygon of eight sides. A decagon is a polygon of ten sides. A dodecagon is a polygon of twelve sides. A pentadecagon is a polygon of fifteen sides. 213 An equilateral polygon is a polygon which has all its sides equal. An equiangular polygon is a polygon which has all its angles equal. A convex polygon is a polygon in which each angle is less than a straight angle. a F A concave polygon is a polygon which has one or more reflex angles. ( Thus, ABCDEFG is a concave polygon and has one reflex (reentrant) angle, D. B A polygon is considered convex unless otherwise stated. 214 A polygon may be divided into triangles by drawing diagonals from any vertex to all the vertices not adjacent. The number of triangles into which any polygon may be thus divided is evidently equal to the number of sides of the polygon less two; and the number of diagonals thus drawn is three less than the number of sides of the polygon. * The student should now review ~~ 131-133. POLYGONS IN GENERAL 63 PROPOSITION XXXV. THEOREM 215 The sum of the angles of a polygon is equal to two right angles taken as many times, less two, as the polygon has sides. A c --- D HYPOTHESIS. ABCD... is a polygon of n sides. CONCLUSION. ZA+Z B + Z C + D +.= (n- 2) 2rt. F. PROOF The polygon may be divided into (n - 2) A. ~ 214 The sum of the As of the A = the sum of the As of the polygon. The sum of the As of each A = 2 rt. A. ~ 153.-. the sum of the A of the A = (n - 2) 2 rt. As..-. the sum of the As of the polygon = (n - 2) 2 rt. As. Q. E. D. 216 COROLLARY. Each angle of an equiangular polygon of n sides is equal to 2 (n- 2) right angles. n EXERCISES 140 How many diagonals can be drawn from one vertex in a polygon of n sides? 141 What is the sum of the angles of a quadrilateral? of a hexagon? 142 How many degrees in each angle of an equiangular octagon? 143 How many sides has a polygon, the sum of whose angles is 14 right angles? 8 straight angles? 7 perigons? n right angles? 144 How many sides has an equiangular polygon, if the.sum of five of its angles is 8 right angles? 64 PLANE GEOMETRY -BOOK I PROPOSITION XXXVI. THEOREM 217 The sum of the exterior angles of a polygon, formed by producing one side at each vertex, is equal to four right angles. / \ \B C/ec HYPOTHESIS. ABCD... is a polygon of n sides. CONCLUSION. The sum of the exterior A = 4 rt. A. PROOF At each vertex of the polygon there are one interior and one exterior angle whose sum is 2 rt. zs. ~ 103.'. the sum of all the interior and exterior angles of the polygon is 2 n rt.. Subtracting the sum of the interior s, 2 n rt. s - 4 rt. A, ~ 215 4 rt. /s remain as the sum of the exterior As of the polygon. Q. E. D. 218 COROLLARY. Each exterior angle of an equiangular polygon of n sides is equal to 4 right angles. n EXERCISES 145 Prove Prop. XXXVI by drawing from a given point lines parallel to the sides of the polygon. 146 How many sides has a polygon if the sum of the interior angles is eight times the sum of the exterior angles? POLYGONS IN GENERAL 65 147 How many sides has an equiangular polygon if three of its exterior angles are together equal to two right angles? 148 The bisectors of the angles of a triangle meet in a point which is equidistant from the sides of the triangle. A PROOF. Let the angle-bisectors BE and CF intersect in O. Then O is equidistant from AB, AC, and BC F (~ 192); and being equidistant from AB and AC, O is in the bisector of the Z A (~ 193). That is, the three angle- B D c bisectors meet in O. DEFINITION. 0 is called the incenter of the triangle ABC. 149 The perpendicular bisectors of the sides of a triangle meet in a point which is equidistant from the vertices of the triangle. PROOF. Let the _ bisectors FF' and EE' intersect in 0. Then 0 is equidistant from A, B, and C (~ 186); and being equidistant from B and C, O is in the _L B~ bisector of BC (~ 187). That is, the three I bisectors meet in O. DEFINITION. 0 is called the circumcenter of the triangle ABC. 150 The altitudes of a triangle meet in a point. PROOF. Let AD, BE, and CF be the altitudes of - the A ABC. Through the vertices of the A draw Is to the opposite sides, intersecting in G, H, and K. Then BCAG and BCKA are L/ by const.; whence D GA = BC = AK (~ 200).... AD is the ~ bisector I, of GK (~ 117). Likewise, BE and CF are the l X bisectors of GH and HK respectively. That is, the three altitudes meet in O (Ex. 149). DEFINITION. O is called the orthocenter of the triangle ABC. 151 The medians of a triangle meet in a point of trisection. PROOF. Let the medians BE and CF intersect in O.!A Draw BH II to FC meeting AOD produced in H, and join HC. Then in the A ABH, AO = OH (~ 209); whence OE is IItoHC (~ 210)... BOCH is a L7, and BD = DC ( ~ 206)..'. AOD is the third median. That is, the three medians meet in O. Again, OD -= OH (~ 206) = a AO (Ax. 9) = AD. Also, OF = HB (~ 210) = CO (Ax. 9) = I CF. Likewise OE = ~ BE. DEFINITION. 0 is called the centroid of the triangle ABC. 66 PLANE GEOMETRY-BOOK I 152 In what kind of a triangle is the circumcenter within the triangle? without the triangle? in one side of the triangle? 153 In what kind of a triangle does the orthocenter fall within the triangle? without the triangle? coincide with one vertex of the triangle? 154 Is the centroid always within the triangle? 155 In what kind of a triangle do one altitude and one median only coincide? 156 In an equilateral triangle, the incenter, the circumcenter, the orthocenter, and the centroid coincide. 157 In the figure of Ex. 151, prove that the triangles ABC and DEF are mutually equiangular. 158 In the figure of Ex. 151, prove that O is the centroid of the triangle DEF. [Join FE, FD, ED, and let AD intersect FE at G. AEDF is a 7 (~ 210)... AD bisects FE at G (~ 206). That is, in the A DEF, DG is the median to FE. In like manner show that the medians to DF and DE fall on EB and FC respectively.] 159 In the figure of Ex. 150, prove that the triangle GHK is four times the triangle ABC. 160 Why do the perpendicular bisectors of two sides of a triangle intersect? [Show that they are not parallel.] 161 The six angles formed about the orthocenter of an equilateral triangle are equal. 162 In the triangle ABC, prove that the bisectors of the angle A and the exterior angles B and C meet in a point. 163 If a diagonal of a parallelogram bisects the angles, the parallelogram is equilateral. 164 If the diagonals of a parallelogram are perpendicular to each other, the parallelogram is equilateral. 165 If the diagonals of a parallelogram are equal, the figure is a rectangle. 166 If the diagonals of a parallelogram are equal and bisect the angles, the figure is a square. 167 If the diagonals of a parallelogram are equal and perpendicular to each other, the figure is a square. REVIEW QUESTIONS 67 REVIEW QUESTIONS 1 What is a proposition? 2 How are propositions distinguished? Define each. 3 What is a demonstration? a solution? 4 Distinguish between a direct demonstration and an indirect demonstration. 5 Distinguish between the two parts of a theorem. 6 What is the converse of a theorem? the opposite? 7 State the logical relation existing between a theorem, its converse, and its opposite. 8 Define magnitude; extension; space. 9 Distinguish between a material solid and a geometric solid. 10 How many dimensions has a solid? a surface? a line? a point? 11 What are the boundaries of a solid? of a surface? of a line? 12 How may a line be traced? a surface? a solid? 13 Give the abstract conception of a point; of a line; of a surface; of a solid. 14 What are the fundamental concepts in Geometry? Define each. 15 Give the classification of lines and define each. 16 Define a plane surface; a curved surface. 17 Define Geometry (~ 56). 18 Give the classification of angles, and define each. 19 What is the complement of an angle? the supplement? 20 What are perpendicular lines? oblique lines? 21 Define geometric figure; plane figure; rectilinear figure; curvilinear figure. 22 What is the unit of angular measurement? How obtained? 23 What are the geometric magnitudes? 24 Distinguish between similar, equivalent, and equal magnitudes. 25 What is the test of equality of geometric magnitudes? 26 What is the object of Geometry? 27 How is Geometry divided? 28 Of what does Plane Geometry treat? 68 PLANE GEOMETRY- BOOK I 29 Can you prove an axiom? 30 Distinguish between an axiom and a postulate. 31 Can you give an axiom or a postulate not found in the text? 32 How many points determine a straight line? 33 What are supplementary adjacent angles? vertical angles? 34 What are parallel lines? 35 Define a polygon; the sides, angles, and vertices of a polygon. 36 What are adjacent angles of a polygon? 37 What is an exterior angle of a polygon? 38 Give the classification of triangles, and define each. 39 Define an equiangular triangle; an equilateral triangle. 40 Is an equilateral triangle equiangular? why? 41 Is an equiangular triangle equilateral? why? 42 Can a leg of a right triangle be equal to the hypotenuse? 43 What is the base of a triangle? 44 What is the base of an isosceles triangle? 45 What is the vertex of a triangle? 46 Define a median, an altitude, an angle-bisector of a triangle. 47 Define mutually equilateral polygons; mutually equiangular polygons. 48 What are homologous lines in mutually equiangular polygons? 49 Define quadrilateral; trapezium; trapezoid. 50 What does the word " parallelogram " mean? 51 Define rectangle; square; rhomboid; rhombus. 52 Define the bases of a trapezoid; the median; the legs. 53 What is the altitude of a parallelogram or trapezoid? 54 What does the word "polygon" mean? 55 Define pentagon; hexagon; heptagon; octagon; decagon; dodecagon; pentadecagon. 56 Define equilateral polygon; equiangular polygon; convex polygon; concave polygon. 57 How may any polygon be divided into triangles? 58 Write a short essay on the great geometers Thales and Pythagoras, ANALYSIS OF THEOREMS 69 SUGGESTIONS ON THE TREATMENT OF EXERCISES 219 In Arithmetic and Algebra definite rules are given for the solution of problems. No such specific directions can be given for the treatment of original exercises in Geometry. The following general directions, however, will greatly assist the beginner: 1 Draw generalfigures. Thus, if you are dealing with a triangle in general, draw a scalene triangle; if with a parallelogram, draw a rhomboid. Observe that this is always done in the text. 2 Draw figures accurately. An accurate figure frequently suggests a clue to the proof; an inaccurate figure leads into error. 3 Fix clearly in mind the things given about the figure, and the precise thing in the figure to be proved. This direction is fundamental; its disregard is the prime cause of the beginner's frequent failure. Observe under "Hypothesis" and "Conclusion" how carefully and clearly this injunction is observed in every proposition of the text. 4 In the proof make use of every condition given in the hypothesis. Work along lines involving only a part of the things given will end in certain failure. No relation would be given were it not a necessary part upon which to build the proof. Examine the proof of any proposition in the text, and observe that in such proof every condition given in the hypothesis is always used. 5 Begin by supposing the theorem true; then, step by step, trace out the relations which follow this supposition, until some known theorem is reached. Now begin with this known theorem and arrange the steps in reverse order, thus leading up to the required proof. In illustration consider the following 70 PLANE GEOMETRY - BOOK I THEOREM. The bisectors of two adjacent angles of a parallelogram are perpendicular to each other. In the 17 DEFG let DH and EK bisect the D Q As D and E respectively. We are to prove that Ad 77 Z c is a rt. angle. Suppose that Z c is a rt. Z. Then a + Z b s =a rt. L (~ 155). Then Z D+ E = 2 rt. s (Ax. 7). But this last conclusion is a known theorem (~ 124) with which we begin the direct PROOF Z D+-ZE=2rt. A. ~124..Z a4- Z b = a rt. Z. Ax. 9..Z c =art. L. ~ 153.. DH and EK are I lines. Q. E. D. This is the usual method of attacking most exercises in Geometry, and is called Analysis. The reverse process, the orderly building up from known facts until a new truth is obtained, is called Synthesis. It should be observed that the synthetic method of proof is used in most text propositions. 6 When other means fail, especially in converse theorems, try the indirect method. See ~~ 115, 116. 7 Draw such auxiliary lines as may be necessary to give a clue to the line of argument. See the dotted lines in the figures of many of the text propositions; also, Exs. 138, 150, 151. The ability to attack exercises successfully and rapidly implies a comprehensive knowledge of the chief propositions of the text, and a keen recognition of their application at sight. To this end we here remind the student of some of the cardinal truths which should ever be vividly in his mind. Two lines are equal: 1 If they are homologous sides of equal triangles. ~ 166 2 If they are opposite equal angles in a triangle. ~ 176 3 If they are the sides of a square or a rhombus. ~ 196 4 If they are the opposite sides of a parallelogram. ~ 200 ANALYSIS OF THEOREMS 71 Two angles are equal: 1 If they are complements or supplements of equal angles. ~~ 104, 105 2 If they are vertical angles. ~ 112 3 If they are alternate-interior or corresponding angles of parallel lines. ~~ 121, 122 4 If they are homologous angles of equal triangles. ~ 166 5 If they are opposite equal sides in a triangle. ~ 173 6 If they are the opposite angles of a parallelogram. ~ 203 Two triangles are equal: 1 If two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. ~ 162 2 If two angles and the included side of the one are equal respectively to two angles and the included side of the other. ~ 167 3 If the three sides of the one are equal respectively to the three sides of the other. ~ 172 Two right triangles are equal: 1 If their legs are equal, each to each. ~ 163 2 If the hypotenuse and an acute angle of the one are equal respectively to the hypotenuse and an acute angle of the other. ~ 168 3 If a leg and an acute angle of the one are equal respectively to a leg and the homologous acute angle of the other. ~ 169 4 If the hypotenuse and a leg of the one are equal respectively to the hypotenuse and a leg of the other. ~ 178 Two lines are parallel: 1 If they are perpendicular to the same straight line. ~ 115 2 If they are parallel to a third straight line. ~ 118 72 PLANE GEOMETRY- BOOK I 3 If they are cut by a transversal making (a) the alternate interior angles equal; ~ 125 (b) the corresponding angles equal; ~ 126 (c) the alternate exterior angles equal; ~ 127 (d) the interior angles on the same side of the transversal supplementary. ~ 128 4 If they are the opposite sides of a parallelogram. ~ 200 5 If they are the bases of a trapezoid. ~ 197 A quadrilateral is a parallelogram: 1 If the opposite sides are parallel. ~ 195 2 If the opposite sides are equal. ~ 204 3 If two sides are equal and parallel. ~ 205 MISCELLANEOUS EXERCISES 168 If from any point within a triangle lines are drawn to the extremities of any side, the angle included by them is greater than the angle included by the other two sides. / [Zx>Zy>Zz. ~159.] 169 If from any point within a triangle lines are drawn to the extremities of any side, their sum is less than the sum of the other two sides. [PC < PD + DC... PB + PC < BD + DC. For BD substitute the greater value AB + AD, and PB + PC < AB + AC.] s c 170 The sum of the lines drawn from any point within a triangle to the three vertices is less than the perimeter. [By Ex. 169, x + y<b + c (1),y + z < a + c (2), x + z < a + b (3). a \ Add (1), (2), (3), and reduce.] 171 The sum of the lines drawn from any point within a: triangle to the three vertices is greater than half the perimeter. [x + y > a (1), y + z > b (2), x + z > c (3). Add (1), (2), (3), and reduce.] 172 Perpendiculars drawn from any point in the base of an isosceles triangle to the legs make equal angles with the base. [Ex. 40.] MISCELLANEOUS EXERCISES 73 173 Perpendiculars dropped from any two points in the, legs of an isosceles triangle to the base make equal angles with the legs. [Ex. 40.] A 174 BC is the base of the isosceles triangle ABC. BA is produced its own length to D. Prove that DC is per- pendicular to BC. [Z B = Z x. Z D = Z y..-. Z B - B ZD = BCD = art. Z.] 175 A triangle is divided into four equal triangles by the lines joining the mid-points of its sides. [By ~ 201, show that A 1 is equal to each of the & 2, 3, and 4.] /2 176 The triangle formed by joining the mid-points of the sides of an equilateral triangle is equilateral. 177 In the triangle ABC, BE and CD are altitudes A intersecting at H. Prove that the Z DHE is the supplement of Z A. [In the quadl. ADHE, D = Z E = a rt. Z..'. Z H is sup. of Z A (~ 216).] o 178 In the equilateral triangle ABC, AX = BY = CZ. A Prove XYZ an equilateral triangle. [~ 162.] 179 If the base angles of two isosceles triangles are complementary, the vertical angles are supplementary. B y o 180 If the vertical angles of two isosceles triangles are supplementary, the base angles are complementary. 181 The median to the hypotenuse of a right triangle is equal to half the hypotenuse. [CD = EF = DA = 2AB.] 182 If one acute angle of a right triangle is double the E other acute angle, the hypotenuse is double the shorter leg. [In the Fig. of Ex. 181, let Z A = 2 Z B. Then Z A = 60~, and A ADC is equilateral.] 183 The feet of two altitudes of a triangle are equidistant A from the mid-point of the third side. [In rt. A ADC, DM is the median to the hypotenuse AC... DM = 1 AC by t Ex. 181. Likewise in rt. A AEC, EM = X AC.... DM L =EM.] B D 184 The altitudes of an equilateral triangle are equal. 185 In equal triangles the homologous altitudes are equal. 186 In equal triangles the homologous medians are equal. 74 PLANE GEOMETRY-BOOK I 187 In equal triangles the homologous angle-bisectors are equal. 188 Each segment of a side of a triangle made by the altitude is less than the adjacent side of the triangle. [~ 180.] 189 In any triangle a median makes with the side to which it is drawn an acute angle toward the less of the other two sides. [~ 171.] 190 If a median of a triangle is perpendicular to the side to which it is drawn, the triangle is isosceles. 191 In any triangle the bisector of an angle makes with A the opposite side an acute angle toward the less of the othertwosides. [AB<AC. ThenB + Lx = ZC + y. But Z B > Z C..-. Z x < Z y.] o \D 192 In any triangle an angle is acute, obtuse, or right, whnl,:i tlhe median drawn from its vertex is greater than, less than, or equal to half the opposite side. [~ 179.] 193 The six angles into which the altitudes divide the angles of a triangle are equal, two and two. [Ex. 40.] A 194 In any triangle the sum of any two sides is greater than twice the median to the third side. [Complete the /7. The diagonals mutually bisect. AB + BA' >AA'..~. AB B D +AC >AA'..-.AB+AC>2AD.] 195 In any triangle the perimeter is greater than the: sum of the three medians. [Apply Ex. 194.] ' 196 If in a right triangle the altitude and median are drawn to the hypotenuse, the angle which the altitude makes with one l leg is equal to the angle which the median makes with the other leg. [L ABC and ACP are mutually equiangular by Ex. 40. A MBC is isosceles by Ex. 181.] B 197 The angle contained by the altitude and median drawn to the hypotenuse of a right triangle is equal to the difference of the two acute angles of the triangle. [In the Fig. of Ex. 196, Z PCB = ZA by Ex. 40, and Z x' = / B by ~ 173. Subtracting equalities, Z PCM = Z A - Z B.] 198 The angle contained by the bisector of the right angle,nl1 the median drawn to the hypotenuse of a right triangle A is equal to half the difference of the two acute angles of the triangle. [Z ACM = / A by Ex. 181. Z ACH -= ZA+l Z 1/B, for each equals 1 a rt. Z. Subtracting equalities, Z 11CM = Z / A - Z B.] MISCELLANEOUS EXERCISES 75 199 The bisector of the right angle of a triangle bisects the angle contained by the altitudeand median drawn to the hypotenuse. A 200 The sum of the perpendiculars drawn from alny point in the base of an isosceles triangle to the legs is D constant. [To prove PH + PK = CD, a constant. Complete the rectangle HC. A PCK = A PCE by ~ 168..'. PH + PK = PH + PE = EH = CD, an altitude and. a constant.] 201 The difference of the perpendiculars drawn from any point in the base of an isosceles triangle produced to the legs is constant. [To prove PH - PK = CD, a constant. Proof same as in Ex. 200.] E B 202 The sum of the perpendiculars drawn from any A point within an equilateral triangle to the three sides is constant. [Draw MPN II to BC. By Ex. 200, PH + PK = AE... PH + PK + PL = AD, an altitude and E-.'. a constant.] ~D 203 If from any point in the base of an isosceles triangle A parallels are drawn to the legs, the perimeter of the parallelogram thus formed is constant. [DP = DB, and EP = EC. D.. the perimeter of L- =AB + AC, a constant.] 204 If from any point in the bisector of a right angle B p A parallels are drawn to the sides of the angle, the parallelogram thus formed is a square. 205 If from any point in the bisector of an oblique angle parallels are drawn to the sides of the angle, the parallelogram thus formed is a rhombus. 206 The bisectors of the angles of a square meet in a point. [The angle-bisectors coincide with the diagonals.] 207 The bisectors of the angles of a rectangle inclose a square. [A EAB and FDC are isosceles and equal rt. Q. Also A HAD and GBC are isosceles and equal rt. A.] f\AA B 208 The bisectors of the angles of a rhombus meet in a point; the bisectors of the exterior angles of a rhombus inclose a rectangle. 209 The bisectors of the angles of a rhomboid inclose a rectangle; the bisectors of the exterior angles of a rhomboid inclose a rectangle. 76 PLANE GEOMETRY - BOOK I 210 The lines joining the mid-points of the adjacent sides of a rectangle, not a square, inclose a rhombus. [The diagonals ---- of the rectangle are equal..'. by ~ 210 the figure is an equilateral L7, and by ~ 129 its A are equal to the A included by the diagonals of the rectangle which are oblique. ] 211 The lines joining the mid-points of the adjacent sides of a square inclose a square. 212 The lines joining the mid-points of the adjacent sides of a rhombus inclose a rectangle. 213 The lines joining the mid-points of the adjacent sides of a rhomboid inclose a rhomboid. 214 The diagonals of an oblique parallelogram are unequal, and the greater diagonal joins the vertices of the two smaller angles. 215 Each base of an isosceles trapezoid makes equal At angles with the legs. [Draw AE II to DC. Then Z B =ZAEB =Z C.] 216 The opposite angles of an isosceles trapezoid are supplementary. [Z B = Z C (Ex. 215) = sup. of Z D A (~ 124).] 217 The diagonals of an isosceles trapezoid are equal. [Z BAD = Z CDA by Ex. 215; whence & ABD B C and ACD are equal by ~ 162.] 218 The bases of an isosceles trapezoid form, with the segments of the diagonals, two isosceles triangles. [Follow proof of Ex. 217.] 219 The quadrilateral formed by joining the mid-points of the adjacent sides of an isosceles trapezoid is an equilateral parallelogram. [The diagonals are equal by Ex. 217. Now apply ~ 210.] 220 The line joining the mid-points of the diagonals of a trapezoid is equal to half the difference of the bases. 221 If one base of a trapezoid is double the other, the diagonals meet at a point of trisection. [Ex. 151.] 222 A quadrilateral is a parallelogram if its opposite angles are equal. 223 A quadrilateral is a square if its diagonals are equal and bisect each other at right angles. 224 A quadrilateral is a rectangle, not a square, if its diagonals are equal and bisect each other obliquely. MISCELLANEOUS EXERCISES 77 225 A quadrilateral is a rhombus if its diagonals are unequal and bisect each other at right angles. 226 If the mid-points of the adjacent sides of a quadrilateral are joined, the figure thus formed is a parallelogram whose perimeter is equal to the sum of the two diagonals of the quadrilateral. 227 The lines joining the mid-points of the opposite sides of a quadrilateral bisect each other. 228 The bisectors of the angles of any quadrilateral inclose a quadrilateral whose opposite angles are sup- -plementary. [In the A FAD and EBC, the four A at A, D, B, and C are together equal to 2 rt. A, for they are half the four angles of the quadl.. E + Z F = 2 rt. A. ~ 153.] 229 The perimeter of a quadrilateral is greater than the sum of its diagonals. 230 The lines joining the mid-points of the opposite sides, and the line joining the mid-points of the diagonals of a quadrilateral, meet in a point. 231 The maximum number of diagonals which may be drawn in a polygon of n sides is n (n - 3). 2 232 No polygon can have more than three exterior obtuse angles, or more than three interior acute angles. 233 Two lines drawn from two vertices of a triangle to the opposite sides cannot bisect each other. 234 No scalene triangle can be divided into two equal triangles. BOOK II THE CIRCLE DEFINITIONS 220 A circle is a plane surface bounded by a curved line all points of which are equidistant from a point within called the center. The circumference is the curved line which bounds the circle. An arc is any part of the circumference. 221 A radius of a circle is a straight line E\ drawn from the center to any point in the A C B circumference; as, CD. All radii of a circle or of equal circles are equal. A circle may be designated and read, 0( C. 222 A chord of a circle is a straight line joining any two points in the circumference; as, EF. A chord divides the circumference into two arcs. If the arcs are unequal, the less is called the minor arc; and the greater, the major arc. A minor are is called simply an arc. A chord subtends the two arcs into which it divides the circumference. 223 A diameter of a circle is a chord which passes through the center; as, AB. All diameters of a circle or of equal circles are equal. 224 A semicircumference is half a circumference. 225 A quadrant is one-fourth of a circumference. 226 A segment of a circle is a part of a circle bounded by a chord and its arc. A chord divides mll a circle into two segments. If the segments Major are unequal, the less is called the minor seg- Segment ment; and the greater, the major segment. 78 THE CIRCLE 79 227 A semicircle is a segment of a circle whose chord is a diameter. 228 A sector of a circle is a part of the circle bounded by two radii and their intercepted arc. Two radii divide a circle into two sectors, ec called the minor sector and the major sector. l Mar A minor sector is called simply a sector. The Sector angle formed by the radii is called the angle of the sector. 229 A secant of a circle is a straight line of unlimited length which intersects the circumference S in two points; as, SS'. 230 A tangent to a circle is a straight line of unlimited length which touches the circumference at one point only; as, TT'. The point of tangency, or point of contact, T p is the point common to the tangent and the circumference; as, P. 231 The line of centers of two circles is the straight line which joins their centers. 232 Two circles are tangent to each other when their circumferences have one point, and only one, in common. When one circle is within the other, they are tangent internally; when each is without the other, they are tangent externally. D C C' 233 A common tangent to two circles is a tangent to each of them. AB, which does not cut the line of centers CC', is called a common external tangent; DE, which cuts the line of centers CC', is called a common internal tangent. 80 PLANE GEOMETRY-BOOK II 234 A central angle is an angle at the center of a circle; as, Z C. / 235 An inscribed angle is an angle whose vertex is in the circumference and whose sides are chords; as, / H. 236 An inscribed polygon is a polygon whose sides are chords of a circle. In this case, the circle is circumscribed about the polygon. A circumscribed polygon is a polygon whose sides are tangents to a circle. In this case, the circle is inscribed in the polygon. InscribedI Inscribed Polygon Circle Circumscribed Circle Circumscribed Polygon 237 Concyclic points are points through which a circumference may be described. 238 Concentric circles are circles which have the same center. 239 COROLLARY. Circles having equal radii are equal. 240 SCHOLIUM. Two circles having equal radii being equal, that is, identical in form and extent, will, when superposed, coincide; and if one of them is now made to turn about its center as a pivot, the other remaining stationary, the first will continue to coincide with the second at every point in the revolution. Also, if a part of the circumference of a circle be made to slide around on the remaining part of the circumference, it will always fit evenly upon the part to which it is applied. This important property of the circle is fundamental and is called the homeoidality of the curve. Hence the following accurate and concise definition of this most renowned of all geometric figures: A circle is a plane surface bounded by a homeoidal curve. THE CIRCLE 81 PROPOSITION I. THEOREM 241 In the same circle or in equal circles, equal central angles intercept equal arcs; conversely, equal arcs subtend equal central angles. A A' B 0 B 0' HYPOTHESIS. In the equal circles 0 and 0', Z 0 = Z O'. CONCLUSION. Arc AB = arc A'B'. PROOF Superpose the circles so that the equal angles 0 and 0' coincide. Then A coincides with A', and B with B'. "All radii of a circle or of equal circles are equal." ~ 221 That is, the arcs AB and A'B' coincide, ~ 240 and are therefore equal. CONVERSELY HYPOTHESIS. In the equal circles 0 and 0', arc AB = arc A'B '. CONCLUSION. Z 0 = Z O0. PROOF Superpose the circles so that the equal arcs AB and A'B' coincide. Then OA coincides with O'A', and OB with O'B'. Ax. 14 That is, the angles 0 and 0' coincide, and are therefore equal. Q. E. D. 242 COROLLARY. In the same circle or in equal circles, the greater of two central angles intercepts the greater arc, and conversely. 82 PLANE GEOMETRY-BOOK II PROPOSITION II. THEOREM 243 In the same circle or in equal circles, equal chords subtend equal arcs; conversely, equal arcs are subtended by equal chords. HYPOTHESIS. In the equal circles 0 and 0', chord AB = chord A'B'. CONCLUSION. Arc AB = arc A'B'. PROOF Draw the radii OA, OB, O'A', O'B'. The A OAB and O'A'B' are equal; ~ 172 for OA = O'A', OB = O'B', ~ 221 and AB = A'B'. Hyp.... 0 = / O'. ~ 166.'. arc AB = arc A'B'. ~ 241 CONVERSELY HYPOTHESIS. In the equal circles 0 and 0', arc AB=arc A'B'. CONCLUSION. Chord AB= chord A'B'. PROOF Superpose the circles so that the equal arcs AB and A'B' coincide. Then the chords AB and A'B' coincide, and are, therefore, equal. "Only one st. line can be drawn between two points." Ax. 14 Q. E. D. 244 COROLLARY. In the same circle or in equal circles, the greater chord subtends the greater arc, and conversely. THE CIRCLE 83 PROPOSITION III. THEOREM 245 A diameter perpendicular to a chord bisects the chord and its arcs. HYPOTHESIS. In the circle 0, DD' is a diameter perpendicular to the chord AB at E. CONCLUSION. EA = EB, arc D'A = arc D'B, arc DA = arc DB. PROOF Draw the radii OA and OB. The rt. A OEA and OEB are equal; for OE is common, and OA = OB..'. EA = EB, and Z s = Z t... arc DA arc D'B. Also, Z AOD = Z BOD..-. arc DA = arc DB. " In the same ~ equal central A intercept equal arcs." ~ 178 ~ 221 ~ 166 ~ 241 Q. E. D. 246 COROLLARY 1. A diameter bisects a circle and the circumference.* 247 COROLLARY 2. The perpendicular bisector of a chord passes through the center of the circle and bisects the arcs of the chord. EXERCISES 235 A diameter which bisects a chord is perpendicular to the chord. 236 If a point is within, on, or without the circumference of a circle, its distance from the center is less than, equal to, or greater than the radius. * This theorem is attributed to Thales. 84 PLANE GEOMETRY-BOOK II EXERCISES 237 The triangle formed by drawing radii to the end-points of a chord is isosceles. 238 The opposite arcs intercepted by two diameters of a circle are equal. 239 AB is a diameter of a circle, OC a radius, and BD a chord parallel to OC. Show that the arcs AC A ~ B and CD are equal. [ 1 = Z 2 = 3 = Z4.] 240 A circle cannot have more than one center. 241 If the chords AB, BC, CD, DE, are equal, show that the chords AC and CE are equal. [~ 243.] Bs 242 Can two equal circles be concentric and not coincide? 243 Can two intersecting circles have the same center? 244 The diameter CD is perpendicular to the chord AB. / Prove that chords CA and CB are equal. [~~ 245, 243.] 245 If two chords subtend equal angles at the center of a circle, they are equal. [~ 241.] D 246 The parts of a secant intercepted between the circumferences of two concentric circles are equal. [~ 245.] 247 From the same point on the circumference of a circle two, and only two, equal chords can be drawn. 248 Two chords of a circle, one of which is not a diameter, cannot bisect each other. [~ 245.] 249 The line which joins the middle points of two parallel chords passes through the center of the circle. 250 The perpendicular bisectors of the sides of an inscribed polygon meet in a point. 251 If perpendiculars are drawn from the extremities of a diameter to a secant, the segments of the secant between the feet of the perpendiculars and the circumference are equal. THE CIRCLE 85 PROPOSITION IV. THEOREM 248 In the same circle or in equal circles, equal chords are equidistant from the center; conversely, chords equidistant from the center are equal. B A 0 HYPOTHESIS. AB and CD are equal chords of the circle 0, OK is I to AB and OH is L to CD. CONCLUSION. OK = OH. PROOF Draw OA and OC. OK bisects AB, and OH bisects CD. ~ 245 The rt. A OAK and OCH are equal; ~ 178 for OA = OC, being radii, ~ 221 and AK = CH. Ax. 9.'. OK = OH. ~ 166 CONVERSELY HYPOTHESIS. OK = OH. CONCLUSION. AB = CD. PROOF The rt. A OAK and OCH are equal; ~ 178 for OA = OC, being radii, ~ 221 and OK = OH by hyp... AK = CH. ~ 166.. AB = CD. Ax. 7 Q. E. D. 86 PLANE GEOMETRY-BOOK II PROPOSITION V. THEOREM 249 In the same circle or in equal circles, the greater of two chords is nearer the center; conversely, the chord nearer the center is the greater. E t K HYPOTHESIS. In the circle 0, EG > EF, OK is I to EG, OH is I to EF. CONCLUSION. OK is less than OH. PROOF Draw HK. OK bisects EG, and OH bisects EF. ~ 245.-. EK is greater than EH. Ax. 10.'. Z a is greater than Z b. ~ 179.'. Z c is less than Z d; Ax. 6 for Z a + Z c = Z b + d = a rt. L by hyp..-. OK is less than OH. ~ 180 CONVERSELY HYPOTHESIS. OK is less than OH. CONCLUSION. EG is greater than EF. PROOF Z c is less than Z d. ~ 179.. Z a is greater than Z b. Ax. 6.'. EK is greater than EH. ~ 180.'. EG is greater than EF. Ax. 8 Q. E. D. 250 COROLLARY. The diameter of a circle is greater than any other chord. For its distance from the center is zero. THE CIRCLE 87 PROPOSITION VI. THEOREM 251 A tangent to a circle is perpendicular to the radius drawn to the point of contact. H P. P HYPOTHESIS. TT' is a tangent to the circle 0 at P. CONCLUSION. TTt is I to the radius OP. PROOF Every point in TT', except P, is without the circumference by the definition of a tangent. ~ 230.'. OP is the shortest line that can be drawn from O to TT' (Ax. 12), and is therefore L to TT'. ~ 185 That is, TT' is _ to OP. Q. E. D. 252 COROLLARY 1. A straight line perpendicular to a radius at its extremity is tangent to the circle. Otherwise we should have two Is to OP at P. ~ 251 253 COROLLARY 2. A perpendicular to a tangent at the point of contact passes through the center of the circle. Otherwise we should have two Is to TT' at P. ~ 251 EXERCISE 252. Two parallels intercept equal arcs on a circumferThree cases. The parallels may be ( a chord and a tangent (Fig. 1), two chords (Fig. 2), or two tangents (Fig. 3). Prove case 1 by ~~ 251, Fig1 Fig. 2 Fig. 3 17, 1245. Prove cases 2 and 3 by case 1 and Axs. 4 and 1. 88 PLANE GEOMETRY-BOOK II PROPOSITION VII. THEOREM 254 The tangents drawn from a point to a circle are equal, and make equal angles with the line joining the point to the centre of the circle. HYPOTHESIS. PA and PB are tangents to the circle HYPOTHESIS. PA and PB are tangents to the circle O. CONCLUSION. PA = PB, and Z OPA = Z OPB. PROOF Draw the radii OA and OB. Then As OAP and OBP are rt. /s. "A tangent to a 0 is I to the radius drawn to the point of contact." ~ 251 In the rt. A POA and POB, PO is common, and OA = OB, "being radii of the same circle." ~ 221.. A POA = A POB. ~ 178.*. PA = PB, and Z OPA = Z OPB. ~ 166 Q. E. D. EXERCISES 253 In the figure of ~ 254, prove that PO bisects the chord AB at rt. angles. 254 The common internal tangents of two circles are equal. 255 The common external tangents of two circles are equal. [Produce the tangents until they meet.] 256 The bisectors of the angles of a circumscribed quadrilateral meet in a point. THE CIRCLE 89 PROPOSITION VIII. THEOREM 255 If two circles intersect, their line of centers bisects their common chord at right angles. ( -o (X oT — Fig. 1 Fig. 2 HYPOTHESIS. O and 0' (Fig. i) are two circles whose circumferences intersect at A and B. CONCLUSION. 00' bisects AB at right angles. PROOF 0 and 0' are two points each equidistant from A and B. ~ 221.'. 00' is the IL bisector of AB. ~ 188 Q. E. D. 256 COROLLARY 1. If two circles are tangent, the point of contact is in the line of centers. For since A and B are always equidistant from 00', if the ~ O and 0' move apart, A and B will come together and coincide on 00' at T (Fig. 2). 257 COROLLARY 2. If two circles are tangent, they have a common tangent at the point of contact. For a perpendicular drawn to 00' at T is tangent to both circles. ~ 252 EXERCISES 257 What is the position of two circles whose line of centers is (1) greater than the sum of the radii? (2) equal to the sum of the radii? (3) less than the sum but greater than the difference of the radii? (4) equal to the difference of the radii? (5) less than the difference of the radii? (6) zero? 90 PLANE GEOMETRY-BOOK II EXERCISES 258 What is the position of two circles which can have (1) two common external and two common internal tangents? (2) two common external tangents and one common internal tangent? (3) two common external tangents and no common internal tangent? (4) one common external and no common internal tangent? (5) no common tangent? 259 The least chord which can be drawn through a fixed point within a circle is perpendicular to the diameter drawn through that point. [~ 249.] 260 What is the longest chord which can be drawn through a point within a circle? 261 If in a circle one chord bisects a second, both not being diameters, the first chord is greater than the second. 262 In two concentric circles, a chord of the outer circle which touches the inner is bisected at the point of contact. [~~ 251, 245.] 263 In two concentric circles all chords of the outer circle which touch the inner are equal. [~~ 251, 248.] 264 Radii of two circles drawn to the points of contact of a common tangent are parallel. 265 If two non-intersecting lines intercept equal arcs on a circumference, they are parallel. [Converse of Ex. 252.] ^b' a 266 Two parallel chords drawn from the end-points of a diameter are equal. [By Ex. 252, arc b = arc b5..'. arc a = arc a'..-. the chords are equal by ~ 243.] a' b 267 If a quadrilateral is circumscribed about a circle, the sum of one pair of opposite sides is equal to the sum of the other pair. [~ 254.] 268 The median of a circumscribed trapezoid is equal to one-fourth of the perimeter. [Ex. 267, ~ 211.] 269 An inscribed trapezoid is isosceles. [Ex. 252, ~ 243.] 270 If two parallel tangents are drawn to a circle, the straight line which joins the points of contact is a B diameter. A 271 In the annexed figure, AP = AO. Prove that to c Z BOC = 3 Z P. THEORY OF LIMITS 91 MEASUREMENT DEFINITIONS 258 Ratio is the relation of two magnitudes of the same kind. This relation (ratio) is expressed by the number of times the first contains the second. Thus, the ratio of 6 ft. to 3 ft. is 2; the ratio of a to b is a, or a: b, and is read "the b ratio of a to b." The number expressing the ratio of two magnitudes is always abstract. 259 To measure a magnitude is to find its ratio to another magnitude of the same kind, called the unit of measure. 260 Commensurable magnitudes are magnitudes which have a common measure. Thus, a foot and a yard are commensurable, for they have a common measure, the inch, which the first contains 12 times and the second 36 times. Incommensurable magnitudes are magnitudes which have no common measure. THE THEORY OF LIMITS 261 A constant is a quantity whose magnitude remains fixed. 262 A variable is a quantity whose magnitude may take an indefinite number of different values. 263 The limit of a variable is a constant, such that the difference between the variable and the constant can be made less than any assigned quantity, but not equal to zero. The variable is said to approach the constant as a limit. Pt P" P p A I I I B Thus, suppose a point P starting from A moves along the line AB, under the condition that it shall move half the distance 92 PLANE GEOMETRY-BOOK II from A to B the first second, half the remaining distance the second second, and so on indefinitely. Then 1. P can never reach B, for there is always half of some distance between them. 2. P can be made to approach as near to B as we choose, for by continuing the bisection, the distance between P and B can be made less than any assigned quantity. 3. The changing distance AP', AP", AP"', etc., is an increasing variable which approaches the constant AB as a limit; the changing distance P'B, P"B, P"'B, etc., is a decreasing variable which approaches zero as a limit. Again. Consider the isosceles triangle ABC. " Let the legs AB and AC constantly increase but,, always remain equal, and let the base BC remain. '. constant. Then the angles B and C are two in- "A' creasing variables, always equal and each approach- ing a right angle as a limit, which it can never become, for no triangle can have two right angles. B C Let it be clearly observed that the two variables, the angles B and C, always remain equal, and that the limits which they approach, two right angles, are equal. This fact is of great importance, and is formally set forth in the following 264 PRINCIPLE 1. If two variables are always equal, and each approaches a limit, their limits are equal. For since the variables are always equal, they are the same variable approaching a common limit. Their limits are, therefore, identical and equal. 265 PRINCIPLE 2. If a variable can be made less than any assigned quantity, the product of the variable by a constant or a decreasing quantity can be made less than any assigned quantity. Let x be the variable and c a constant or a decreasing quantity. Then cx is their product. Since x decreases indefinitely, cx decreases indefinitely, and therefore can be made less than any assigned quantity. 266 COROLLARY. If a variable can be made less than any assigned quantity, the quotient of the variable by a constant can be made less than any assigned quantity. THEORY OF LIMITS 93 Let x be the variable and c a constant. Then x is their quotient. Now C x = (-)x, which is the product of a variable (x) by a constant (). x Therefore - can be made less than any assigned quantity. ~ 265 267 PRINCIPLE 3. If a variable x approaches a constant c as a limit, mx approaches me as a limit, m being a constant. For c-x can be made less than any assigned quantity. ~ 263.'. m(c - x) = (m - mx) can be made less than any assigned quantity. ~ 265.'. x approaches me as a limit. ~ 263 268 COROLLARY. If a variable x approaches a constant c as a limit, x approaches c as a limit, m being a constant. m m For c- x can be made less than any assigned quantity. ~ 263 m. cx = (-c -_ can be made less than any assigned quantity. ~ 266 m m mit * x approaches c as a limit. ~ 263 m m 269 PRINCIPLE 4. If each of two variables can be made less than any assigned quantity, their product can be made less than any assigned quantity. Let x and y be the variables and c a constant. Now cy can be made less than any assigned quantity. [~ 265.] But x can be made less than c. Hyp.. xy can be made less than any assigned quantity. 270 COROLLARY. If a variable can be made less than any assigned quantity, the square of the variable can be made less than any assigned quantity. Let x = y. Then x2 = xy, which by ~ 269 can be made less than any assigned quantity. 271 PRINCIPLE 5. If a variable x approaches a constant c as a limit, x2 approaches c2 as a limit, and x/x approaches x/ as a limit. For cn - xn = d; and since d can be made less than any assigned quantity by taking x large enough, Xn approaches cn as a limit. ~ 263 If n = 2, x2 approaches c2 as a limit. If n = -, V/x approaches /c as a limit. 94 PLANE GEOMETRY-BOOK II 272 PRINCIPLE 6. If two variables x and y approach two constants b and c as their respective limits, x + y approaches b + c as a limit, and xy approaches be as a limit. For (b + c) - (x + y) = d, and bc - xy = d'; and since d and d' can each be made less than any assigned quantity by taking x and y large enough, x + y approaches b + c as a limit, and xy approaches be as a limit. PROPOSITION IX. THEOREM 273 In the same circle or in equal circles, two central angles have the same ratio as their intercepted arcs. 0 0 B C m A Fig. 1 Fig. 2 HYPOTHESIS. In the circle 0, the central angles AOB and AOC intercept the arcs AB and AC. / AOB arc AB CONCLUSION. / AOB arc AB Z AOC arc AC CASE 1. When the arcs are commensurable (Fig. 1). PROOF Let the arc m, a common measure of the arcs AB and AC, be contained 3 times in AB and 4 times in AC. Then arc AB 3 Then __ — arc AC 4 By drawing radii to the several points of division in the two arcs, the central angles will be divided into 3 and 4 parts, all equal. ~ 241 ZAOB 3. AOB arc AB A Z.AOC 4 o "ZA Ax. rcAC AOC 4 Z AOC are AC MEASURE OF ANGLES 95 CASE 2. When the arcs are incommensurable (Fig. 2). PROOF Since the arcs AB and AC are not commensurable, let the arcs AD and AC be commensurable, arc DB being less than the common measure m. Then Z AOD arc AD Case 1 Z AOC arc AC If m, the unit of measure, be indefinitely diminished, the arc DB, always less than m, will become less than any assigned quantity... the arc AD approaches the arc AB as a limit, l ~ 263 and the L AOD approaches the Z AOB as a limit. j arc AD approaches arc AB a limit, arc AC arc AC ~268 and AOD pproaches AO as a limit. L AOC Z AOC ratios Z AOD -and arc AD But the variable ratios A and ar A are always equal. Z AOC arc AC Case 1 Therefore their limits are equal. That is AOB are AB i AOC are AC "If two variables are always equal, and each approaches a limit, their limits are equal." ~ 264 Q. E. D. The circumference of a circle is divided into 360 equal parts called degrees. Also, a perigon is divided into 360 equal parts called degrees. Therefore a central angle of one degree intercepts on the circumference an arc of one degree; a central angle of 12~ intercepts an arc of 12~. In other words, the numerical measure of a central angle is equal to the numerical measure of its intercepted arc. This important fact is concisely stated as follows: 274 A central angle is measured by its intercepted arc. 96 PLANE GEOMETRY -BOOK II PROPOSITION X. THEOREM 275 An inscribed angle is measured by half its intercepted arc. A A A B D D Fig. 1 Fig. 2 Fig. 3 HYPOTHESIS. BAC is an inscribed angle. CONCLUSION. The Z A is measured by ~ arc BC. PROOF CASE 1. When the center O is in one side of the angle (Fig. 1). Draw OC. Then OC = OA. "Radii of the same circle are equal." ~ 221.. A = ZC.," In an isosceles A the A opposite the equal sides are equal." ~ 173 Zs=LA+ C. "An exterior Z of a A is equal to the sum of the opposite int. A." ~ 158. Z s =2 ZA, or Z A =1 s. But / s is measured by the arc BC. " A central Z is measured by its intercepted arc." ~ 274.*. Z A is measured by ~ arc BC. CASE 2. When the center 0 is within the angle (Fig. 2). Draw the diameter AD. Then Z s is measured by ~ arc BD, MEASURE OF ANGLES 97 and L t is measured by 2 arc DC. Case 1 Adding, Z s + / t is measured by 1 (arc BD + arc DC). Ax. 1 That is, Z BAC is measured by 1 arc BC. CASE 3. When the center 0 is without the angle (Fig. 3). Draw the diameter AD. Then Z DAC is measured by 1 arc DC, and Z DAB is measured by - arc DB. Case 1 Subtracting, Z BAC is measured by I arc BC. Ax. 4 Q. E. D. B A C D Jr D B Fig. 1 Fig. 2 Fig. 3 276 COROLLARY 1. Angles inscribed in the same segment or in equal segments are equal (As A, B, C, Fig. 1). For each is measured by half the same arc. 277 COROLLARY 2. An angle inscribed in a semicircle is a right angle (As A, B, Fig. 2). For it is measured by half a semicircumference. 278 COROLLARY 3. An angle inscribed in a segment greater than a semicircle is an acute angle (Z A, Fig. 3); and an angle inscribed in a segment less than a semicircle is an obtuse angle (Z B, Fig. 3). For the first Z is measured by an arc less than a quadrant, and the second is measured by an arc greater than a quadrant. 279 COROLLARY 4. The opposite angles of an inscribed quadrilateral are supplementary. * The discovery of this theorem is attributed to Thales, who. it is said, out of gratitude sacrificed an ox to the gods. Prop. X, however, is first found in Euclid (about 300 B.C.). 98 PLANE GEOMETRY- BOOK II EXERCISES 272 If in Fig. 1, ~ 275, Z A = 27~, how many degrees in the arc BC? in arc AC? 273 If in Fig. 2, ~ 275, arc AB contains 136~ and arc AC 148~, find L s and Z t. 274 If in Fig. 3, ~ 275, Z DAB = 20~ and arc AC contains 73~, find the Z BAC. A 275 ABC is a triangle inscribed in circle 0. OE is perpendicular to BC. Prove Z BOE = Z A. [OE produced bisects arc BC at D by ~ 245. ~~ 274, C 275.] 276 How many degrees are there in the arcs subtended D by the legs of an inscribed isosceles triangle whose vertical angle contains 50~? 277 If two equal chords are drawn from the same point in the circumference of a circle, the line which bisects their included angle is a diameter. [~~ 243, 246.] 278 P is any point in the arc of a circle of which AB is the chord. Prove that the sum of the angles A and B is A — \ constant for all positions of P. 279 The chord of an arc of 60~ is equal to the radius of the circle. 280 If an inscribed angle is 30~, the chord on which it stands is equal to a radius of the circle. 281 A circle described upon one of the legs of an isosceles triangle as a diameter bisects the base. [~ 277. ] 282 A perpendicular from the center of a circle to a tangent passes through the point of contact. 283 An isosceles triangle has its vertex without a circle and a chord as base; show that the segments of the legs without the circle are equal. 284 AB is a fixed chord of a circle, and P is any point in L either arc; show that the bisector of the angle APB intersects the opposite arc in the same point for all positions of P. 285 If an isosceles triangle has its vertex within a circle, and a chord as base, the legs produced are equal chords. MEASURE OF ANGLES 99 PROPOSITION XI. THEOREM 280 An angle formed by two intersecting chords is measured by half the sum of the intercepted arcs. C —B HYPOTHESIS. The chords AB and CD intersect at E. CONCLUSION. L AEC is measured by f (arc AC + arc DB). PROOF Draw CB. Z AEC = Z C + Z B. " An exterior Z of a A is equal to the sum of the opposite int. A." ~ 158 Z C is measured by ~ arc DB, and Z B is measured by 1 arc AC. " An inscribed Z is measured by half its intercepted arc." ~ 275.'. L AEC is measured by ~ (arc AC + arc DB). Ax. 1 Q. E. D. EXERCISES 286 In Fig. of Prop. XI, if arc AC = 75~, and arc DB = 53~, how many degrees in Z AEC? 287 If two chords of a circle are perpendicular to each other, the sum of two opposite arcs is equal to a semicircumference. 288 BAC is an angle inscribed in a circle, E and F are E the mid-points of the arcs AB and AC, the chord EF intersects AB at G and AC at II. Prove AG = AH. [Prove Z AGF = Z AHE.] B 289 The exterior angle formed by producing a side of an inscribed quadrilateral is equal to the opposite interior angle. [~ 279.] 100 PLANE GEOMETRY-BOOK II PROPOSITION XII. THEOREM 281 An angle formed by a tangent and a chord is measured by half the intercepted arc. C AK T B ' HYPOTHESIS. BA is a chord and BT a tangent. CONCLUSION. The Z ABT is measured by ~ arc AB. PROOF Draw the diameter BC. The rt. / CBT is measured by I arc CAB. The Z CBA is measured by 1 are CA. ~ 275 Subtracting, Z ABT is measured by ~ arc AB. Ax. 4 Q. E. D. EXERCISES 290 Prove that the angle ABT' is measured by half the arc ACEB. 291 If arc AC = 88~, how many degrees in the angle ABT? 292 Prove Prop. XII by this figure, AC being A parallel to TT'. 293 The bisector of an angle formed by a tangent and a chord bisects the intercepted arc. -- ' T 294 Two equal chords drawn from the point of contact make equal angles with the tangent. 295 State and prove the converse of Ex. 294. 296 An angle formed by a tangent and a chord is equal to the angle inscribed in the opposite segment. 297 A tangent drawn through the vertex of an inscribed isosceles triangle makes equal angles with the legs. 298 If a tangent bisects an arc, it is parallel to the chord of the arc. MEASURE OF ANGLES 101 PROPOSITION XIII. THEOREM 282 An angle formed by two secants, two tangents, or a tangent and a secant, intersecting without the circle, is measured by half the difference of the intercepted arcs: _ A C A C A S Fig. 1 Fig. 2 Fig. 3 HYPOTHESIS. CASE 1. ADB and AEC are two secants (Fig. i). CONCLUSION. Z A is measured by ~ (arc BC - arc DE). PROOF Draw EB. Then Z A + Z t = Z s. ~ 158.'. Z A=Z s-Z t. Ax. 4 But Z s is measured by I arc BC, and / t is measured by - arc DE. ~ 275.-. A is measured by I (arc BC - arc DE). Ax. 4 HYPOTHESIS. CASE 2. AB and AC are two tangents (Fig. 2). CONCLUSION. Z A is measured by i (arc BEC - arc BC). PROOF DrawBC. Then A = s - t. ~ 158 But Z s is measured by 1 arc BEC, and / t is measured by I arc BC. ~ 281.'. L A is measured by ~ (arc BEC -arc BC). Ax. 4 Q. E. D. HYPOTHESIS. CASE 3. AC is a tangent and ADB is a secant (Fig. 3). CONCLUSION. Z A is measured by i (arc BC - arc DC). The proof is left to the student. 102 PLANE GEOMETRY-BOOK II EXERCISES 299 Prove Prop. XIII by the following figures: 300 In Prop. XIII, Fig. 1, if arc BC =78~, arc DE = 34~, find angle A. 301 In Prop. XIII, Fig. 2, if angle A = 65~, find the arcs BC and BEC. 302 In Prop. XIII, Fig. 3, if arc DB = 152~, and angle A = 44~, find the arcs BC and DC. 303 In Prop. XIII, Fig. 2, prove the two tangents equal by ~ 281. 304 An angle formed by two tangents is equal to v 180~ less the number of degrees in their intercepted A arc. [Z A is measured by 1 (arc BEC - arc BC) = 1 (360~ - 2 arc BC) = 180~ - arc BC.] 305 If the quadrilateral ABCD is circumscribed about the circle, Z A + Z C is measured by arc HE + arc GF, and Z B + A Z D is measured by arc HG + arc EF. - [Z A is measured by 180 ~- arc HG, I F and Z C is measured by 180~ - arc EF (Ex. 304). /'\ Adding, Z A +C is measured by 360~ - (arc HG+ arc B h EF) = arc HE + arc GF.] 306 In the Fig. of Ex. 305, L HGF = ~ (Z A + Z D). [Z A is measured by 180~ - arc HG, and Z D is measured by 180~ - arc GF (Ex. 304). Adding, Z A + Z D is measured by (3600 - arc HGF) = arc HEF... (L A + Z D) is measured by 1 are HEF; but Z HGF is measured by 2 arc HEF... ZHGF = i (Z A +Z D).] 307 In the Fig. of Ex. 305, prove that Z GHE = ~ (Z A + Z B). 308 AS and AT are tangents to a circle, and P is the i middle point of the arc ST. Prove that PS bisects the angle AST. [~~ 275, 281.] EXERCISES 103 MISCELLANEOUS EXERCISES PROBLEMS OF COMPUTATION 309 A chord is drawn perpendicular to a diameter; one of the arcs thus intercepted is 28~; how many degrees in each of the other arcs? 310 Two chords in a circle are 5 in. and 7 in. respectively; which subtends the larger arc? which is nearer the center? 311 A circle is divided by a chord into two segments. An angle inscribed in one is 82~; what angle is inscribed in the other? 312 What chord divides a circle into segments which contain equal inscribed angles? 313 A chord so divides a circle that an angle inscribed in one segment equals twice an angle inscribed in the other. Find the number of degrees in each arc. 314 Two intersecting chords include an angle of 26~; one of the arcs contains 28~; how many degrees in the arc intercepted by the vertical angle? 315 Two intersecting chords intercept opposite arcs of 17~ and 127~; what angle between the chords? 316 Two secants intercept arcs of 12~ and 74~; what angle is included by the secants? 317 The angle formed by two secants, intersecting without the circle, is 18~; one of the intercepted arcs is 132~; what is the other arc? 318 What angle is formed by two tangents which intercept an arc of 158~? 319 The angle formed by two tangents is 58~; into what two arcs do they divide the circumference? 320 What angle is formed by a tangent and a secant drawn through the center of the circle, if one of the arcs intercepted is 29~? 321 If in Ex. 320 the angle formed by the tangent and the secant is 37~, into what two arcs does the point of tangency divide the semicircumference? 322 The angle formed by a tangent and a secant is 27~; the smaller intercepted arc is 68~; find the other arc. 323 What angle is formed by two secants which intercept arcs of ay0 and i of a circumference? 104 PLANE GEOMETRY —BOOK II 324 What angle is formed by two tangents which intercept an arc 37~ less than a semicircumference? 325 How many degrees in an inscribed angle which stands on 3 of a quadrant? 326 Subtract 26~ 34' 17" from 84~ 16' 38". 327 What is the angle between a tangent and a secant which intercept arcs of a quadrant and a semicircle? 328 Two sides of an inscribed triangle subtend arcs of 128~ and 136~ respectively; how many degrees in each angle of the triangle? 329 The base of an inscribed isosceles triangle subtends an arc of 78~; find the angles of the triangle. 330 Find the angles of an inscribed trapezoid if the bases subtend arcs of 98~ and 162~ respectively. 331 In the last exercise, find the angle formed by the non-parallel sides produced. 332 Find the angles of an inscribed quadrilateral, if three consecutive sides subtend arcs of 76~, 98~, 112~, respectively. 333 In the last exercise find the angle formed by the diagonals. 334 A leg of an inscribed trapezoid subtends an arc of 74~; find the angle between the diagonals. 335 One angle of a circumscribed rhombus is 85~; find each of the intercepted arcs. 336 The line of centers of two tangent circles is 21 inches; the diameter of the larger circle is 22 inches; find the diameter of the smaller circle. 337 The line of centers of two circles tangent internally is 5 inches; the radius of the smaller circle is 14 inches; find the radius of the larger circle. 338 Find the radius of a circle tangent to two concentric circles whose diameters are 12 meters and 20 meters respectively. 339 Find the radius of a circle tangent to two intersecting circles whose line of centers is 14 inches and whose diameters are 16 inches and 22 inches respectively. Four determinate solutions. 340 Find the number of inches in the circumference of a circle in which a central angle of 72~ intercepts an arc of 7 inches. EXERCISES 105 THEOREMS 341 If two circles are tangent externally, the common internal tangent bisects the common external tangent. [~ 254.] 342 Two equal circles intersect at A. CAB and E DAE are straight lines terminated by the circumferences. Prove that the chords CD and EB are D equal. [~~ 112, 275, 243.] 343 A circle can be circumscribed about a right parallelogram. 344 The rectangle is the only parallelogram that can be inscribed in a circle. [~ 203.] 345 A circle can be inscribed in an equilateral parallelogram. Two cases. [Exs. 124, 125, ~ 192.] 346 A parallelogram circumscribed about a circle is equilateral. 347 ABCD is a square inscribed in a circle, and P is any p point in the arc AD. Prove that PB and PC trisect the angle A D APD.(/\ -348 The diagonals of an inscribed trapezoid are equal. 349 In a circle, the chords joining the end-points of two diameters are parallel and equal, two by two. 350 Two circles intersect at A and B. MAN is any m straight line terminated by the circunferences. Prove that the angle MBN is constant. [In the A MBN, A M and N are constant (~ 275);.'. Z B is constant.] 351 The middle points of all parallel chords of a circle are in a straight line. 352 If two equal circles intersect, the intercepted arcs are equal, and conversely. ( c [To prove arc ACB = arc AEB; conversely, if arc ACB= arc AEB, the circles are equal. See ~ 243.] 353 If an equilateral triangle is inscribed in a circle, the triangle formed by drawing tangents through its vertices is equilateral. 354 The circumscribed equilateral triangle is four times the inscribed equilateral triangle. 355 Two equal tangent circles intercept equal chords on any secant drawn through the point of contact. [In -- ) the figure prove the triangles equal.] 106 PLANE GEOMETRY-BOOK II 356 Two circles intersect at S and T; SH and SK s are diameters, one in each circle. Prove that the points H, T, K, are in a straight line. [Z STH = a rt. Z = Z STK.] 357 In Ex. 356, prove that HK is parallel to the line of centers. [~ 255.] 358 Each side of a circumscribed equilateral triangle is bisected at the point of contact. 359 The common external tangents of two unequal circles intersect on the line of centers produced. 360 The common internal tangents of two circles intersect on the line of centers. 361 The line of centers of equal circles is equal to their common external tangent. T T' 362 The line of centers of two unequal circles / \ l, is greater than their common external tangent. C ) [TH is a rectangle... TT' = CH < CC' (~.181).] 363 The common external tangent of two circles is greater than their common inter-. nal tangent. [)raw CD II to TT'. Then ) TT' = CD > CE > SS'.] 364 If a circle be described upon the hy- r potenuse of a right triangle as a diameter, the circumference will pass through the vertex of the right angle. [Prove by the indirect method. See ~ 277.] 365 Two circles are tangent at E, and SET is a s straight line terminated by the circumferences. Prove that the tangents at S and T are parallel. Two cases. t [I)raw the common tangent HEK. Then Z H = Z K, for the isosceles A HES and KET are mutually equi- F T angular..-. SH and TK are 1I by ~ 125. The proof for internal tangency is identical.] 366 Two circles C and C' are tangent,and a line drawn through their point of contact cuts the circumferences at P and P'. Prove that the radii CP and C'P' are parallel. Two cases. [Draw the line of centers, and prove the two triangles thus formed equiangular.] 367 If an isosceles trapezoid is circumscribed about a A HB circle, the chord joining the points of contact of the legs is parallel to the bases. [Z A = Z B (Ex. 215)..-. arc HE = arc HK... BHK Z HKE... EK is lI to AB.] D EXERCISES 107 368 If two equal chords of a circle intersect, the segments of one are equal respectively to the segments of the other. 369 AS and AT are fixed tangents to a circle, B and BC is a third tangent meeting AS at B and AT at C. Prove that the perimeter of the triangle ABC P is constant for all positions of BC. [Show that c the perimeter = AS + AT.] 370 The shortest and the longest line that can be drawn from a given point to a given circumference lie in the diameter that passes through the point. 371 Two circles intersect at A and B. X and Y are any two points on one circumference. XA and I XB produced meet the other circumference at P and y T; YA and YB produced meet it at S and Q. Prove ST and PQ parallel. [Z XAY = Z XBY..'. Z SAP x = Z TBQ..'. arc SP = arc TQ... ST is II to PQ.] 372 If two circles are tangent internally, and the radius of the first equals the diameter of the second, any chord of the first drawn from the point of contact is bisected by the circumference of the second. 373 If two secants are drawn through the point of - S- D contact of two tangent circles, the chords of the in- \ tercepted arcs are parallel. Two cases. [Z C = Z APS = Z BPT = Z D.] 374 If a side of an inscribed equilateral triangle is perpendicular to a radius it bisects the radius. 375 The altitude of an inscribed equilateral triangle is three fourths of the diameter. [Ex. 151.] 376 Two circles intersect at A and B; XAY and A Y PBQ are any two straight lines terminated by the P circumferences. Prove PX and QY parallel. [Z X = sup. of Z ABP, and Z Y = sup. of Z ABQ (~ 279).] 377 Circles inscribed in and circumscribed about a square are concentric. 378 If two circles intersect, the straight line drawn through a point of intersection parallel to the line of centers and terminated by the circumferences is twice the line of centers. 379 If two circles intersect, the longest line which can be drawn through a point of intersection and terminated by the circumferences is the one parallel to the line of centers. 108 PLANE GEOMETRY-BOOK II 380 Two circumferences cannot bisect each other. 381 If the vertical angle of an isosceles triangle inscribed in a circle is twice an angle at the base, the base is a diameter of the circle. 382 If two triangles inscribed in a circle are mutually equiangular, they are also mutually equilateral. 383 Is Ex. 382 true if both triangles are circumscribed about the circle? 384 Two circles intersect at A and B. PAH and As: PBK are straight lines terminated by the circumfer- ences. Prove that the tangent at P is parallel to HK. [Z H = Z ABP (Ex. 289) = Z APT (~~ 275, 281).] 385 The altitude of an equilateral triangle is three times the radius of the inscribed circle. 386 The radius of the circle circumscribed about an equilateral triangle is twice the radius of the inscribed circle. 387 The diameter of a circle inscribed in a right triangle is equal to the difference between the sum of the legs and the hypotenuse. [~ 254.] 388 If a circle is inscribed in a right triangle, and another is circumscribed about it, the sum of the legs of the triangle is equal to the sum of the diameters of the two circles. [~ 277, Ex. 387.] A 389 A circle inscribed in the triangle ABC touches AB at E and AC at D. Prove that the circle inscribed in the triangle AED has the midpoint of the arc ED as its center. 390 If a triangle is circumscribed about a circle, the triangle formed by joining the points of contact is acute. [Z ADF = Z AFD..-. Z ADF is acute (~ 154)..'. Z DEF is acute.] 391 Two circles are tangent externally at P; S and T are the points of contact of their common s external tangent. Prove that the angle SPT is a right angle. [Draw the common tangent HP. The & HSP and HTP are isosceles (~ 254).] 392 If two circles are tangent externally, a circumference described upon their common external tangent as a diameter passes through the point of contact. [Ex. 391.] EXERCISES 109 393 The line of centers of two externally tangent circles is tangent to the circle described upon their common external tangent as a diameter. [The tangent at P passes through the center of 0 C" (~ 254).] 394 If two circles are tangent externally, the circles described upon their common external tangents tangent. [Ex. 393.] 395 H is the mid-point of the common external tangent of two tangent circles whose centers are C and C'. Prove that the angle CHC' is a right angle. 396 If two circles intersect, their common tangent subtends supplementary angles at the points of intersection. [To prove that Z SAT + Z SBT = 2 rt. A. Prove that Z SBT = Z AST + Z ATS.] as diameters are H T 397 Circles described on two sides of a triangle as diameters intersect on the third side or the third side produced. 398 If circles are described on the sides of a quadrilateral as diameters, the common chord of two consecutive circles is parallel to the common chord of the other two circles. [Draw the diagonals of the quad'l and apply Ex. 397.] 399 If two circles intersect, any two parallel -A D straight lines drawn through the points of intersec- c tion and terminated by the circumferences are equal. [A E and F are supplementary..-. CEFD is a 0.] 400 If two circles are internally tangent, the longest chord of the one tangent to the other is perpendicular to the line of centers. 401 If the diagonals of an inscribed quadrilateral inter- A O D sect at the center of the circle, the figure is a right parallelogram. [Z A=ZB = C = D = a rt. Z (~ 277).] 402 The chords of a circle bisecting the arcs subtended by the sides of an inscribed equilateral triangle meet those sides in points of trisection. 403 If three circles are externally tangent, two by two, their common internal tangents meet in a point. 404 If three circles are externally tangent, two by two, the inscribed circle of the triangle formed by joining their centers is the circumscribed circle of the triangle formed by joining their points of contact. 405 Radii which trisect a chord do not trisect the subtended arc. 110 PLANE GEOMETRY -BOOK II PROBLEMS OF CONSTRUCTION Thus far in the Geometry we have supposed the necessary figures constructed. We shall now show how to construct geometric figures with the straight edge and compasses, the only instruments allowed in Euclidean Geometry. PROPOSITION XIV. PROBLEM 283 To draw a perpendicular to a given line from a given point without the line. P A B DATA. P is a given point without the line AB. DATA. P is a given point without the line AB. REQUIRED. To draw a perpendicular to AB from P. SOLUTION From P as a center, and with a radius sufficiently great, describe an arc cutting AB at H and K. From H and K as centers, and with a radius greater than half of HK, describe arcs intersecting at C. Draw PC and produce it to meet AB at D. CONCLUSION. PD is the 1 required. Q. E. F. PROOF. P and C are by construction each equidistant from H and K..-. PD is I to AB. ~ 188 Q. E. D. CONSTRUCTIONS 111 PROPOSITION XV. PROBLEM 284 To draw a perpendicular to a given line from a given point in the line. f,C., "-E AJ-B S --- - Fig. 1 Fig. 2 FIRST METHOD. DATA. P is a point in the line AB (Fig. 1). REQUIRED. To draw a perpendicular to AB at'P. SOLUTION Take PH = PK. From H and K as centers, and with a radius greater than PH, describe arcs intersecting at C. Draw CP. CONCLUSION. CP is the perpendicular required. Q. E. F. PROOF. P and C are by construction each equidistant from H and K..'. CP is ~ to AB. ~ 188 Q. E. D. SECOND METHOD. DATA. P is an end-point of the line SP (Fig. 2). REQUIRED. To draw a perpendicular to SP at P. SOLUTION From C, any point without SP, as a center, and with CP as a radius, describe an arc intersecting SP at D. Draw DC and produce it to meet the arc at E. Join EP. CONCLUSION. EP is the _L required. Q. E. F. PROOF. The angle P is a right angle. ~ 277.-. EP is I to SP. Q. E. D. DISCUSSION. In the second method, the point C must be without the required perpendicular. 112 PLANE GEOMETRY-BOOK II PROPOSITION XVI. PROBLEM 285 To bisect a given straight line. J x DATA. AB is a straight line. REQUIRED. To bisect AB. SOLUTION From A and B as centers, and with a radius greater than the half of AB, describe arcs intersecting at C and D. Join CD. CONCLUSION. CD bisects AB at X. Q. E. F. PROOF. C and D are by construction each equidistant from A and B... CD bisects AB. ~ 188 Q. E. D. PROPOSITION XVII. PROBLEM 286 To bisect a given arc. SOLUTION Bisect the chord of the arc as in ~ 285. ~ 247 CONSTRUCTIONS 113 PROPOSITION XVIII. PROBLEM 287 To bisect a given angle. / Ai DATA. ABC is a given angle. REQUIRED. To bisect the angle ABC. SOLUTION From B as a center, and with any convenient radius, describe an arc cutting the sides of the angle at D and E. From D and E as centers, and with a radius sufficiently great, describe arcs intersecting at O. Join OB. CONCLUSION. OB bisects the angle ABC. Q. E. F. PROOF. OB bisects the chord DE at rt. A. ~ 188.G. OB bisects the arc DE. ~ 247.-. OB bisects the angle ABC. ~ 241 Q. E. D. 288 SCHOLIUM. By the same construction, each half of the angle may be bisected. Hence an angle may be divided into 2, 4, 8, 16, 2", equal angles. EXERCISE 406. To trisect a right angle. [Let ABC be a rt. Z. From B as a center describe an A e arc cutting the sides of the angle at A and C. From A and /\ C as centers, and with AB as a radius, describe arcs cutting the arc AC at D and E. Then EB and DB trisect the B ': Z ABC. For the A ADB and CEB are equilateral by const..'. Z ABD = Z CBE = 60~..'. Z DBC = Z ABE = 30~.] NOTE. There is no method known in Geometry for trisecting any plane angle, and mathematicians believe that the problem is impossible. 114 PLANE GEOMETRY -BOOK II PROPOSITION XIX. PROBLEM 289 At a given point in a straight line, to construct an angle equal to a given angle. F A P G B CDATA. P is a given point in the line AB, and C is a given angle. REQUIRED. To construct an angle at P equal to the angle C. SOLUTION From C as a center, and with any convenient radius, describe an arc cutting the sides of the Z C at D and E. From P as a center, and with CD as a radius, describe an arc GF cutting AB at G. From G as a center, and with the chord DE as a radius, describe an arc cutting the arc GF at H. Join PH. CONCLUSION. The Z HPB = the Z C. Q. E. F. PROOF. The A PGH and CDE are equal by const. ~ 172.'. L HPB = Z C. ~ 166 Q. E. D. EXERCISES 407 To construct an angle of 45~. 408 To construct an angle of 135~. 409 To construct an equilateral triangle, one side being given. 410 To construct an angle of 30~. 411 To trisect an angle of 45~. 412 To construct an angle of 120~; of 150~. 413 To construct an angle of 105~; of 165~. 414 To construct an isosceles right triangle, having a leg given. 415 To construct an isosceles right triangle, having the hypotenuse given. CONSTRUCTIONS 115 PROPOSITION XX. PROBLEM 290 Through a given point to draw a line parallel to a given line. F - --— 1 G'" A B DATA. AB is a given line, and P is a given point. REQUIRED. To draw a line through P parallel to AB. SOLUTION Through P draw any line ED meeting AB at D. Construct the Z EPG equal to the Z PDB. ~ 289 CONCLUSION. FPG is II to AB. ~ 126 Q. E. F. PROPOSITION XXI. PROBLEM 291 Given two angles of a triangle to find the third. cia i:' A B c DATA. A and B are two angles of a triangle. REQUIRED. To find the third angle of the triangle. SOLUTION At any point C in the line MN, construct Z a = Z A, and Z b = B. ~ 289 CONCLUSION. The Z c is the angle required. ~~ 153, 106 Q. E. F. 116 PLANE GEOMETRY-BOOK II PROPOSITION XXII. PROBLEM 292 To construct a triangle when two sides and the included angle are given. K B4 DATA. a and c are two sides of a triangle, and K is their included angle. REQUIRED. To construct the triangle. SOLUTION Take BC equal to a. Construct an / ABC equal to the L K. ~ 289 Take BA = c, and draw AC. CONCLUSION. ABC is the triangle required. Q. E. F. PROOF. The A ABC contains all the given data by construction. Any other A constructed with the same data will be equal to the A ABC, and therefore a repetition of it. ~ 162.'. ABC is the triangle required. Q. E. D. PROPOSITION XXIII. PROBLEM 293 To construct a triangle when a side and two angles are given. H / i —aK B a --- — KC a The solution is left to the student. If the given angles are not both adjacent to the given side, find the third angle by ~ 291. CONSTRUCTIONS 117 PROPOSITION XXIV. PROBLEM 294 To construct a triangle when the three sides are given. A,/ a b c B a C DATA. a, b, c, are the three sides of a triangle. REQUIRED. To construct the triangle. SOLUTION Take BC equal to a. From B as a center, and with c as a radius, describe an arc. From C as a center, and with b as a radius, describe an arc intersecting the first arc at A. Join AB and AC. CONCLUSION. ABC is the triangle required. Q. E. F. EXERCISES To construct a right triangle, having given: 416 The two legs. 417 A leg and the adjacent acute angle. 418 A leg and the opposite angle. 419 The hypotenuse and an acute angle. 420 The hypotenuse and a leg. To construct an isosceles triangle, having given: 421 The base and a leg. 422 The base and an angle at the base. 423 The base and the altitude. 424 The base and the vertical angle. 118 PLANE GEOMETRY- BOOK II PROPOSITION XXV. PROBLEM 295 To construct a triangle when two sides and an angle opposite one of them are given. a a a b A B" B B'- - -.......' B DATA. a and b are two sides of a triangle, and A is the angle opposite a. REQUIRED. To construct the triangle. SOLUTION Construct the Z EAF equal to the Z A. ~ 289 On AE take AC = b, and draw p a I from C to AF. From C as a center, and with a as a radius, describe an arc BB'. CASE 1. When a is less than b. 1 If a is greater than p, the arc BB' will cut AF on the same side of A, at B and B'. Join CB, CB'. CONCLUSION. CAB, or CAB', is the triangle required, for each contains all the given data, and there are two solutions. This is called the ambiguous case. 2 If a is equal to p, the arc BB' will touch AF at H, and there is only one solution, the right triangle b a P CHA. 3 If a is less than p, the arc BB' - "t will not touch AF, and there is no solution. In case 1, the ZA is acute. For a by the data a < b;.. LA > ZB B B (~ 179);.. L A is acute (~ 154). A CONSTRUCTIONS 119 CASE 2. When a is equal to b. E Since a= b, the arc BB' will cut AF at B and A, and there is only one solution, the isosceles triangle CAB. b a In case 2, the Z A is acute, and p is / less than a or b. A B - CASE 3. When a is greater than b. 1 If the Z A is acute, the arc BB' f will cut AF on opposite sides of A, at C B and B'. CAB is the triangle re- a / quired, and there is only one solution. - '/ * The triangle CAB' contains a and b -. — but not the acute angle A. 2 If the Z A is a right angle, the arc BB' will cut AF on opposite sides c of A, at B and B', and there is only one a, "a solution, the right triangle CAB. The right triangle CAB' is equal to B", - B — CAB, and therefore a repetition of it. Hence the right triangle CAB' is not a second solution. 3 If the Z A is obtuse, the arc BB' will cut AF on opposite sides of A, at B and B', and there is only one solu- a tion, the triangle CAB. \ The triangle CAB' contains a and b,,- A -- F but not the obtuse angle A. Q. E. F. EXERCISES To construct an isosceles triangle, having given: 425 A leg and an angle at the base. 426 A leg and the vertical angle. 427 A leg and the altitude. 428 The altitude and the vertical angle. 429 The perimeter and the base. 120 PLANE GEOMETRY-BOOK II PROPOSITION XXVI. PROBLEM 296 To divide a given straight line into any number of equal parts. A- -X DATA. AB is a straight line. REQUIRED. To divide AB into n equal parts. SOLUTION Draw AK making any convenient angle with AB. On AK lay off any convenient length n times, and let E be the nth point of division. Join EB, and through the other points of division in AK draw parallels to EB. CONCLUSION. These parallels will divide AB into n equal parts. ~ 208 Q. E. F. EXERCISE 430. To construct a parallelogram when two sides and the included angle are given. 'C m aaa B s t in nC DATA. m and n are two sides of a 7, and B is their included angle. REQUIRED. To construct the 7. The solution is left to the student. CONSTRUCTIONS 121 PROPOSITION XXVII. PROBLEM 297 To circumscribe a circle about a given triangle. B i. DATA. ABC is a given triangle. REQUIRED. To circumscribe a circle about ABC. SOLUTION Draw the perpendicular bisectors of AB and BC. ~~ 285, 284. Since ABC is not a straight line, these perpendicular bisectors will intersect at some point O. From 0 as a center, and with OB as a radius, describe a circle. CONCLUSION. This is the circle required. Q. E. F. PROOF. 0 is equidistant from A and B, and likewise from B and C. ~ 186.'. O is equidistant from A, B, and C... a circle described from center O and with radius OB passes through the vertices of the triangle. Q. E. D. NOTE. O is called the circumcenter of the triangle ABC. 298 SCHOLIUM. The solution to the above problem enables us to draw a circumference through three given points not in a straight line, and to find the center of a given circle or arc. 299 COROLLARY 1. Only one circumference can be drawn through three points not in a straight line. 300 COROLLARY 2. Two circumferences, or a straight line and a circumference, cannot intersect in more than two points. 122 PLANE GEOMETRY-BOOK II PROPOSITION XXVIII. PROBLEM 301 To inscribe a circle in a given triangle. A D DATA. ABC is a given triangle. REQUIRED. To inscribe a circle in the triangle ABC. SOLUTION Bisect the angles A and B. These bisectors will intersect at some point O. Draw OD I to AB. ~ 283 From 0 as a center, and with OD as a radius, describe a circle. CONCLUSION. This is the circle required. Q. E. F. PROOF. O is equidistant from the three sides of the triangle. ~ 192 Hence the circle 0 is tangent to the three sides of the triangle ABC, ~ 230 and is therefore inscribed in it. ~ 236 NOTE. O is called the incenter of the triangle ABC. Q. E. D. 302 DEFINITION. An escribed circle is a circle which touches one side of a triangle and the other two sides produced. A triangle has three escribed circles, whose centers are called the excenters of the triangle. CONSTRUCTIONS 123 PROPOSITION XXIX. PROBLEM 303 From a given point, to draw a tangent to a given circle. T -------- - - -- -- - Fig. 1 Fig. 2 DATA. O is the center of a circle, and P is a given point. REQUIRED. To draw a tangent from P to the circle O. CASE 1. When P is on the circumference (Fig. 1). SOLUTION Draw the radius OP. Draw TPT' I to OP. ~ 284 CONCLUSION. TT' is the tangent required. ~ 252 Q. E. F. CASE 2. When P is without the circumference (Fig. 2). SOLUTION Draw OP. On OP as a diameter, describe a circumference intersecting the given circumference at H and K. Draw PH, PK, OH, and OK. CONCLUSION. PH and PK are the tangents required, and there are two solutions. Q. E. F. PROOF. As PHO and PKO are right angles. "An angle inscribed in a semicircle is a rt. Z." ~ 277. PH and PK are tangents. ~ 252 Q. E. D. 124 PLANE GEOMETRY —BOOK II PROPOSITION XXX. PROBLEM 304 Upon a given straight line, to describe a segment of a circle which shall contain a given angle. DA Ag t ln a Kis aB DATA. AB is a straight line, and K is a given angle. REQUIRED. To describe a segment of a circle upon AB which shall contain the angle K. SOLUTION Construct the / BAT equal to the Z K. ~ 289 Bisect AB by the 1L DH. ~ 285 From A draw a I to AT intersecting DH at O. ~ 284 From O as a center, and with OA (= OB) as a radius, describe a circle AEB. CONCLUSION. The segment AEB is the segment required. Q. E. F. PROOF. AT is 1 to AO by const..'. AT is a tangent to the circle, "being I to a radius at its extremity." ~ 252.'. the Z BAT is measured by i arc AB, "being an Z formed by a tangent and a chord." ~ 281 But the Z AEB is measured by 1 arc AB. "An inscribed Z is measured by 4 the intercepted arc." ~ 275.. L AEB = Z BAT = /K. QE.D. ANALYSIS OF PROBLEMS 125 THE ANALYSIS OF PROBLEMS* 305 The analysis of a problem is a course of reasoning by which a solution is discovered. In attacking a problem, if no obvious solution presents itself, proceed by analysis as follows: 1. Suppose the problem solved and all necessary constructions made. 2. Study this figure carefully to discover some elementary principle upon which the required construction depends. 3. Make this elementary principle the basis of a synthetic solution. 4. If you fail, draw such auxiliary lines and figures as the nature of the problem may suggest, and proceed as before. Consider the following illustrations: 306 PROBLEM. To construct a triangle, having given the mid-points of its sides. ANALYSIS. Suppose the problem solved, and that ABC is the A required, D, E, and F being the given A mid-points of the sides. Now DF is 11 to BC, DE is I| to AC, and EF is II to AB (~ 210). D.This is the elementary principle which we make the basis of the - E C SOLUTION. Construct the A DEF, and through each vertex draw a line II to the opposite side, intersecting in A, B, and C. ABC is the A required. It remains to prove that E, F, and D are the mid-points of the triangle ABC. PROOF. BE is II to DF, and BD is II to EF by const.. BEFD is a E7. Likewise ECFD is a C7... BE= DF, and EC = DF..'. BE = EC..'. E is the mid-point of BC. Likewise D and F are the mid-points of AB and AC respectively. Q. E. D. * Suggestions on the treatment of exercises, ~ 219, apply to the solution of problems as well as to the proof of theorems. These suggestions should be carefully observed. 126 PLANE GEOMETRY-BOOK II 307 PROBLEM. From a given point in the circumference of a circle to draw a chord which will be bisected by a given chord. ANALYSIS. Let P be the given point in the Oce, AB the given chord. We are required to draw a s chord through P which will be bisected by A the fixed chord AB. Suppose the problem p.T \ solved and that PS is the required chord bisected by AB at H. Now OH is I to PS \ (~ 245). Hence the problem is reduced to this: Find a point H in AB such that the Z PHO is a rt. /. Now an angle inscribed in a semicircle is a right angle (~ 277), and this is the key to the SOLUTION. On the radius PO as a diameter describe a Oce intersecting AB in H and K. Draw the chords PHS and PKT, and they are the chords required. There are, in general, two solutions. PROOF. A PHO and PKO are rt. As (~ 277)..~. PS and PT are bisected at H and K respectively (~ 245). When will there be one solution only? When no solution? 308 Another important aid in the solution of many problems is the intersection of loci. THE LOCUS OF A POINT 309 DEFINITION. The locus of a point is the place where a point may be under a given condition. This place may be a point, a line (straight or curved), or a group of lines. The given condition is called the law of the locus. Locus is a Latin word and means place. The plural is loci. The following illustrations will make the subject clear. LOCI 310 PROBLEM. What is the locus of a point in a plane equidistant from the extremities of a given straight line? Let AB be a given straight line. We are required to find where a point may be so that it is always A B equidistant from A and B. " Where a point may be" is the locus of the point. "So that it is always equidistant from A and B " is the law of the locus. Let LL' be the perpendicular bisector of AB. Then, from ~ 186, we know that the point may be anywhere in LL', and from ~ 187 we know that the point cannot be without LL'. Therefore the answer to the question asked in the problem is the following 311 THEOREM. The locus of a point equidistant from the extremities of a straight line is the perpendicular bisector of the line. Let it be clearly observed that ~ 186 and ~ 187 are here concisely stated in this one theorem in loci. 312 The solution of problems in loci involves the finding and the proving of a theorem in loci, and this theorem always implies two propositions, namely: a theorem and its converse, or a theorem and its opposite. Thus, in ~ 311, we must prove 1 THEOREM. All points in LL' are equidistant from A and B. 2 CONVERSE. All points equidistant from A and B are in LL'. Or, we must prove 1 THEOREM. All points in LL' are equidistant from A and B. 2 OPPOSITE. All points not in LL' are not equidistant from A and B. 128 PLANE GEOMETRY-BOOK II NOTE. Since the converse and the opposite of a theorem are either both true or both false (~ 22), the demands of Logic are fully met by proving the theorem and either the converse or the opposite. The choice we make is largely a matter of convenience, depending upon the nature of the problem under consideration. 313 Propositions in loci may be given in the form of problems or theorems. In ~ 310 the proposition is given as a problem. It might have been given in the form of the following THEOREM. The locus of a point in a plane equidistant from the extremities of a straight line is the perpendicular bisector of the line. The theorem tells what the locus is and requires the student to prove why it is so. The problem requires the student to find out for himself what the locus is, and then prove why it should be so. NOTE. In this work the author gives most propositions in loci as problems for the reason that the discovery of the locus is the most interesting and profitable part of the exercise. A few propositions in loci are now given in form. 314 PROBLEM. What is the locus of a point equidistant from the sides of an angle? B D A X DATA. BAC is a given angle. REQUIRED. To find the locus of a point equidistant from AB and AC. SOLUTION Draw AD bisecting the Z BAG. ~ 287 CONCLUSION. AD is the locus required. Q. B. F. LOCI 129 1 To prove that every point in AD is equidistant from AB and AC. PROOF. Every point in the bisector of an angle is equidistant from the sides of the angle. ~ 192 Q. E. D. 2 To prove that every point equidistant from AB and AC is in AD. PROOF. Every point equidistant from the sides of an angle is in the bisector of the angle. ~ 193.'. AD is the locus of a point equidistant from AB and AC. Q. E.D. 315 PROBLEM. To find the locus of a point which bisects any line drawn from a given point to a given line. P L G LK A D E DATA. P is a given point and AB a given line. REQUIRED. To find the locus of a point which bisects any line drawn from P to AB. SOLUTION From P draw any line to AB, as PC, and bisect it at G. Through G draw LL' II to AB. ~ 290 CONCLUSION. LL' is the locus required. Q. E.F. 1 To prove that every point in LL' satisfies the law of the locus. PROOF. Let H be any point in LL' and draw PHD to AB. Then in the A PCD, GH is II to CD, Const. and PG = GC. Const... PH= HD. ~ 209.. H satisfies the law of the locus. Q. E. D. 130 PLANE GEOMETRY-BOOK II 2 To prove that every point not in LL' does not satisfy the law of the locus. PROOF. Let S be any point not in LL', and draw PSE to AB, intersecting LL' at K. Then in the A PDE, HK is II to DE, Const. and H is the mid-point of PD. Proved..-. K is the mid-point of PE. ~ 209.'. S is not the mid-point of PE..'. S does not satisfy the law of the locus. Q. E. D. NOTE. In this problem, the law of the locus —that the point shall bisect any line drawn from P to AB -suggests the drawing of some line from P to AB, as PC, and the bisection of it at G, thus locating one point in the required locus. In like manner other points, H, K, etc., may be located through which the required locus must pass. The student must soon discover by inspection that the points G, H, K, etc., lie in a straight line parallel to AB. Having thus discovered the locus, it remains for himl to prove why it should be so. This method of analysis should be followed in the solution of all problems in loci. 316 PROBLEM. To find the locus of the vertex of the right angle of a right triangle whose hypotenuse is constant. DATA. AC is the hypotenuse of the right triangle ABC. REQUIRED. To find the locus of the vertex B. SOLUTION A C On AC as a diameter describe a circumference ABCD. CONCLUSION. Oce ABCD is the locus required. Q. E. F. PROOF. 1 AC subtends a rt. / at any point in the Oce ABCD. ~ 277 2. AC subtends an acute / at any point without, and an obtuse L at any point within, the Oce ABCD. ~ 278 That is, every point in the Oce ABCD satisfies the law of the locus, and no other point does..-. The Oce ABCD is the locus required. E. D. LOCI 131 EXERCISES Find the locus of a point: 431 Which is at a given distance from a given line. 432 Which is equidistant from two parallel lines. 433 Which is equidistant from two intersecting lines. 434 Which is at a given distance from a given point. 435 Which bisects the radius of a given circle. 436 Equidistant from the vertices of a triangle. 437 Equidistant from the sides of a triangle. 438 Equidistant from two concentric circumferences. 439 Which is at a given distance from a given circumference. Find the locus of the center of a circle: 440 Which passes through two given points. 441 Which touches a given line at a given point. 442 Which touches a given circle at a given point. 443 Which touches each of two given parallel lines. 444 Which touches each of two given intersecting lines. 445 Which has a given radius and touches a given line. 446 Which has a given radius and passes through a given point. 447 Which has a given radius and touches a given circle. Find the locus of the mid-point of a chord of a circle: 448 Which remains parallel to a fixed line. 449 Which has a constant length. 450 Which passes through a fixed point in the circumference. 451 Which passes through a fixed point within the circle. 452 Which produced passes through a fixed point without the circle. Find the locus of the vertex of a triangle: 453 Whose base is fixed, and whose altitude is constant. 454 Whose base is fixed, and whose vertical angle is constant, 132 PLANE GEOMETRY -BOOK II REVIEW QUESTIONS 1 Define circle; circumference; arc; radius; chord; diameter; semicircumference; quadrant; segment; semicircle; sector; secant; tangent. 2 Why are all radii of a circle or of equal circles equal? 3 Why are all diameters of a circle or of equal circles equal? 4 Why are circles equal which have equal radii? 5 Is a semicircle a segment or a sector? 6 What is the line of centers of two circles? 7 When are two circles tangent to each other? 8 Define a common tangent to two circles. 9 Define central angle; inscribed angle; inscribed polygon; circumscribed polygon. 10 What are concentric circles? 11 What is the greatest chord that can be drawn in a circle? 12 How does a chord differ from a secant? 13 What is ratio? unit of measure? 14 When are two magnitudes commensurable? when incommensurable? 15 What is a constant? a variable? the limit of a variable? 16 Explain how a central angle is measured by its intercepted arc. 17 What instruments are allowed in geometric constructions? 18 How many points determine a circle? 19 Define escribed circles. 20 In general, how many circles can be drawn tangent to three intersecting lines? 21 What is the incenter of a triangle? the circumcenter? an excenter? 22 What is the orthocenter of a triangle? the centroid? 23 What is the analysis of a problem? 24 What is the locus of a point? 25 What is the law of a locus? 26 What is the locus of a point equidistant from the extremities of a straight line? EXERCISES 133 MISCELLANEOUS EXERCISES PROBLEMS OF CONSTRUCTION 455 In a given line, to find a point equidistant from two given points. [~ 311.] 456 To find in one side of a triangle a point equidistant from the other two sides. [~ 314.] 457 To find a point equidistant from the extremities of a given line and at a given distance from a given point. [Let AB be the given line, P the given point, d the xi given distance. ~ 311 gives one locus of the required p point, and Ex. 434 gives another. x and xI, the inter- sections of these loci, are the points required. ] A --- -— ' s To construct an equilateral triangle, having given: 458 The altitude. 459 The perimeter. 460 Three points through which the sides must pass. 461 The radius of the circumscribed circle. 462 The radius of the inscribed circle. To construct an isosceles triangle, having given: 463 The altitude and an angle at the base. 464 The base and the radius of the circumscribed circle. 465 The base and the radius of the inscribed circle. 466 The perimeter and the base. 467 The perimeter and the base angles. 468 The perimeter and the altitude. [The annexed figure shows the solution of the A last two exercises. In each MN is the perimeter.. — / — K In 467, construct ZM = ZN = a base angle. \M Draw BH and CK bisecting at right A AM and AN respectively. Then ABC is the triangle required. In 468, AD, the altitude, is the - bisector of MN. Join AM and AN, and draw BH and CK as before. Then ABC is the A required.] 134 PLANE GEOMETRY-BOOK II To construct a right triangle, having given: 469 The hypotenuse and the altitude upon the hypotenuse. - 470 A leg and the altitude upon the hypotenuse. 471 The median and the altitude drawn to the hypotenuse. 472 One leg and the radius of the inscribed circle. To construct a triangle, having given: 473 The base, the vertical angle, and the altitude. [In problems giving the base and the vertical angle, invariably construct upon the base a segment of a circle containing the vertical angle (~ 304).] 474 The base, the altitude, and an angle at the base. 475 A side, an adjacent angle, and the radius of the circumscribed circle. 476 Two sides and the altitude upon the third side. 477 Two sides and the median to one of them. [Construct a triangle with the median, the half of its side, and the other given side as sides.] 478 The angles and the radius of the circumscribed circle... [~~ 275, 281.] k l 479 The angles and the radius of the inscribed circle. / 480 One side and the medians to the other two sides. [Always keep in mind Ex. 151 when dealing with medians. For a method of trisecting a line, see Ex. 109.] 481 The base, an angle at the base, and the sum of the two remaining sides. [The isosceles A, as an auxiliary figure in the solution of problems, is frequently used.] 482 The base, an angle at the base, and the difference of the other two sides. 483 The base, the vertical angle, and the median to the base. [See note to Ex. 473.] 484 Two sides and the median to the third side. [With the two sides as sides and twice the median as one diagonal, construct a 0.] 485 The angles and the perimeter. [Solved as in Ex. 467, but make A M and N equal to half the given angles respectively.] EXERCISES 135 486 The base, the altitude, and the radius of the circumscribed circle. 487 The base, an angle at the base, and the sum of the other two sides. 488 The base, the vertical angle, and the altitude upon one of the remaining sides. 489 The base and the altitudes upon the remaining sides. 490 The angles and the sum of two sides. 491 The altitude, the angle-bisector, and the median drawn from the same vertex. 492 The three medians. To construct a square, having given: 493 The side. 494 The diagonal. 495 To inscribe a square in a given circle. 496 To circumscribe a square about a given circle. 497 To inscribe a circle in a given square. 498 To circumscribe a circle about a given square. 499 To inscribe a square in a given sector which is one fourth of a circle. 500 To inscribe a square in a right triangle. To construct a rectangle, having given: 501 Two adjacent sides. 502 A side and a diagonal. 503 The diagonals and their included angle. To construct a rhombus, having given: 504 A side and an angle. 505 The diagonals. 506 One angle and one diagonal. 507 The base and the altitude. To construct a rhomboid, having given: 508 One side and the diagonals. 509 The diagonals and their included angle. 510 An angle, one side, and one diagonal. 511 An angle, a side, and the perimeter. 136 PLANE GEOMETRY -BOOK II To construct an isosceles trapezoid, having given: 512 Two adjacent sides and their included angle. 513 The altitude and the bases. 514 One angle and the bases. 515 A diagonal and the bases. To construct a trapezoid, having given: 516 The four sides. 517 The bases and the base angles. 518 The bases and the diagonals. To construct a circle with a given radius: 519 Which shall pass through two given points. 520 Tangent to a given circle at a given point. 521 Tangent to two intersecting lines. Four solutions. 522 With its center in a given line, and tangent to a second given line. To construct a circle which shall: P 523 Pass through two given points and have its ' center in a given line. B [~ 186 is of great importance in the solution of many problems.] 524 Have its center in a given line, and touch another given line at a given point. [~ 251.] 525 Touch a given line at a given point, and pass through a second given point. 526 Touch two given lines, one of them at a given point. 527 Have its center in a given line, pass through a given point in that line, and touch a given circle. / 528 Touch a given line AB at a given point P, so that tangents drawn to the circle from two. points K and H in AB may be parallel. 529 In the annexed figure, point out six pairs of equal angles. 530 To find a point in one side of a triangle from which parallels drawn to the other two sides and limited by them are equal. EXERCISES 137 531 From a point without a circle, to draw a line to the concave circumference which will be bisected at the convex circumference. 532 To draw a common tangent to two given circles. In general, two cases, each having two solutions. -..-.. D I:A _ ',, - \l '. Fig. 1 Fig. 2 Let C and C' be the centers of the two ~ whose radii are r and r' respectively. CASE 1. (Fig. 1.) From center C and with r r' as a radius, describe a circle to which draw the tangent C'D (~ 303). Join CD and produce it to meet the Oce at A. Draw the radius C'B 11 to CA. Join AB and it is a common external tangent. Let the student give the proof and draw the second tangent. CASE 2. (Fig. 2.) From center C and with r + r' as a radius, describe a circle to which draw the tangent C'D. Join CD cutting the Oce of circle C at A. Draw the radius C'B II to CD but in the opposite direction. Join AB, and it is a common internal tangent. Let the student give the proof and draw the second tangent. 533 To inscribe a circle in a given sector. 534 From three given points as centers, to \ describe three circles, each tangent to the other ' two. 535 With three given radii, to describe three circles, each tangent to the other two. 536 To find the shortest broken line between two D given points which shall touch a given line. c [AB the given line, C and D the given points. / CPD is the broken line required.] -.'f B 537 To construct a square whose sides, produced:if necessary, shall pass through the vertices of a given quadrilateral. Six solutions. 138 PLANE GEOMETRY- BOOK II PROBLEMS IN LOCI If the base of a triangle is fixed, and its vertical angle is constant, find the locus of: 538 The foot of the altitude upon the base. 539 The foot of each altitude upon the two changing sides. 540 The center of the circumscribed circle. 541 From a fixed point in the circumference of a given circle a chord is drawn and produced its own length to a point P. Find the locus of P. 542 A straight rod slides down between a vertical wall and a horizontal floor. Find the locus of its midpoint. 543 A straight line equal to the side of a square so moves around in the square that its end-points are always in the sides of the square. Find the locus of its mid-point. 544 Find the loci of the mid-points of the sides of an inscribed triangle of constant form. 545 0 is the center of a given circle, OP a line equal to a diameter of the circle, and PT a tangent. Find the locus of the center of the circle circumscribed about the triangle OPT. 546 Find the locus of the extremity of a given tangent drawn to a circle. 547 A given straight line moves parallel to a fixed line with one end in a given circumference. Find the locus of the other end. 548 Find the locus of a point such that tangents drawn from it to a given circle shall contain a constant angle. 549 From a fixed point tangents are drawn to a system of concentric circles. Find the locus of the points of contact. 550 On a given line AB, a segment of a circle is described in which ACB is any inscribed angle. On CA, CP is taken \ equal to CB. Find the locus of P. 551 Two equal tangent circles so move that each constantly touches a side of a right angle. Find the locus of the point of contact between the two circles. 552 A rhombus is constructed on a given straight line as base. Find the locus of the intersection of its diagonals. EXERCISES 139 553 Find the locus of the incenter of an inscribed triangle whose base is fixed. [Let I be the incenter. Z A is constant (~ 275); A -. Z B + Z C is constant (~ 153);.-. ZB + ZC is constant;.. / BIC is constant; that is, as A traces the arc BAC, I traces the arc BIC. Hence the locus of I is B -7 C the arc BIC.] 554 Find the locus of the orthocenter of a triangle whose base and vertical angle are given. 555 Find the locus of the centroid of a triangle whose base and vertical angle are given. 556 Two circles intersect at A and B. Through A any straight line CD is drawn terminated by the circumferences. Find the locus of the mid-point of CD. A B c 557 Find the locus of the mid-point of a secant drawn from a given point to a given circumference. [K is Al the given point, C the given 0, KA any P secant, P its mid-point. Bisect KC at K c O. Then 0 is a fixed point, and OP = i CA (~ 210) = a constant..'. the locus of P is the Oce OP.] 558 ABC is a triangle having the base BC and the vertical angle A constant. BA is produced to P so that BP = BA + AC. Find the locus of P. 559 AB is a fixed chord in a circle, and AC is any other chord. If the parallelogram ABC is completed, find the locus of the intersection of its diagonals. 560 Find the locus of the intersection of two chords of a given circle drawn from two fixed points in the circumference and intercepting an arc of constant magnitude. 561 AB is a fixed diameter of a given circle. Any chord AC is produced to M so that CM = BC. Find the locus of M. 562 AB is a fixed diameter of a given circle, AC any chord, CD a tangent, BD a perpendicular to CD meeting AC produced at P. Find the locus of P. BOOK III PROPORTION AND SIMILAR POLYGONS DEFINITIONS 317 A proportion is an equality of ratios. 318 A proportion may be written in three different forms; a c as, =; a: b=c:d; a:b:: c: d. b d Each form is read, "the ratio of a to b is equal to the ratio of c to d;" or, more briefly, " a is to b as c is to d." 319 The terms of a proportion are the four quantities compared. 320 The extremes of a proportion are the first and fourth terms; as, a and d. 321 The means of a proportion are the second and third terms; as, b and c. 322 The antecedents of a proportion are the first and third terms; as, a and c. 323 The consequents of a proportion are the second and fourth terms; as, b and d. 324 In the proportion a: b = c: d, d is called the fourth proportional to a, b, and c. 325 In the proportion a: b = b: c, c is called the third proportional to a and b; and b is called the mean proportional between a and c. 326 A continued proportion is a series of equal ratios; as, a: b =: d = e:f=g: h. 327 SCHOLIUM. In Geometry it is assumed that the four quantities of a proportion are represented by their numerical measures. 140 THEORY OF PROPORTION 141 PROPOSITION I. THEOREM 328 In any proportion the product of the extremes is equal to the product of the means. HYPOTHESIS. a: b = c: d. CONCLUSION. ad = be. PROOF a c b ~d ~ 318 Clearing of fractions, ad = bc. Q. E. D. EXERCISES 563 Find the value of x if x:6 = 8:3. 564 Find the value of x if 5: 4 = x: 8. 565 Find the value of x if a: b = c: x. 566 Find the value of x if a: x = b: c. 567 Find the value of x if x: 2x = 3 x: 12. 568 Find the value of x if x: x2 = xs: 81. PROPOSITION II. THEOREM 329 The mean proportional between two quantities is equal to the square root of their product. HYPOTHESIS. a: b = b: c. CONCLUSION. b = /ac. PROOF b2= ac. ~ 328 Extracting the square root of each number, b = Vc. Q. E.D. EXERCISES 569 Find the value of x if 4: x = x:9. 570 Find the value of x if a: x = x: c. 571 Find the mean proportional between 3 and 27. 142 PLANE GEOMETRY BOOK III 572 Find the mean proportional between m and n. 573 Find the third proportional to 2 and 6. 574 Find the third proportional to m and n. PROPOSITION II. THEOREM 330 If the product of two quantities is equal to the product of two others, either two may be made the extremes and the other two the means of a proportion. HYPOTHESIS. ad = bc. CONCLUSION, a: b = c: d. PROOF ad = be. Hyp. Dividing each member by bd, a c b d or a: b =c: d. ~ 318 Q. E. D. EXERCISE 575. Separate 12 into two pairs of factors which will form a proportion. PROPOSITION IV.. THEOREM 331 If four quantities are in proportion, they are in proportion by alternation; that is, the first term is to the third as the second is to the fourth. HYPOTHESIS. a: b = c: d. CONCLUSION. a: c = b: d. PROOF ad = be. ~ 328.'. a:c=b:d. ~ 330 Q. E. D. NOTE. The student should observe that when we make b and c the means by ~ 330, we are at liberty to take either for the second term and the other for the third. Obviously be = cb. THEORY OF PROPORTION 143 PROPOSITION V. THEOREM 332 If four quantities are in proportion, they are in proportion by inversion; that is, the second term is to the first as the fourth is to the third. HYPOTHESIS. a: b = c: d. CONCLUSION. b: a =d: c. PROOF ad = bc. ~ 328 Making b and c the extremes and a and d the means, ~ 330 b: a=d: c. Q. E.D. EXERCISE 576. Transform the proportion a: b = c: d so that each term becomes the first. PROPOSITION VI. THEOREM 333 If four quantities are in proportion, they are in proportion by composition; that is, the sum of the first two terms is to the second term as the sum of the last two terms is to the fourth term. HYPOTHESIS. a: b = c: d. CONCLUSION. a + b: b = c +d:d. PROOF a=c. ~318 b d Adding 1 to each member, a C a+l= -+1. Ax. 1 b d Whence, a+ b c+d b d or a+b: b=c+d:d. ~ 318 Q. E. D. 144 PLANE GEOMETRY- BOOK III PROPOSITION VII. THEOREM 334 If four quantities are in proportion, they are in proportion by division; that is, the difference of the first two terms is to the second term as the difference of the last two terms is to the fourth term. HYPOTHESIS. a: b = c: d. CONCLUSION. a-b:b= c -d: d. PROOF a=. ~318 b d Subtracting 1 from each member, a C a —1=C-1. Ax. 4 b d a-b c-d Whence a- - b d or a-b:b=c —d:d. ~318 Q. E. D. EXERCISES 577 If a:b=c:d, provethata+b:a = c + d:c. [b: a=d:c (~332); bd or b (~318). a c.b. 1+=1+-d (Ax. 1). Whence a+b = c+d a c 578 Verify ~ 333 and Ex. 577 in the proportion 3: 7 = 6:14. 579 If m - n: n = 3: 5, find the ratio of m to n. 580 If a: b = c: d, prove that a - b: a = c - d:c. 581 Verify ~ 334 and Ex. 580 in the proportion 7: 2 = 14: 4. 582 Ifm + n: n = 7:4, find the ratio of m to n. THEORY OF PROPORTION 145 PROPOSITION VIII. THEOREM 335 If four quantities are in proportion, they are in proportion by composition and division; that is, the sum of the first two terms is to their difference as the sum of the last two terms is to their difference. HYPOTHESIS. a: b = c: d. CONCLUSION. a+b:a-b = c+d:c-d. PROOF a+b c+d 333 b d a-b c-d and b- - d ~ 334 b d Dividing member by member, a + b _ c+ d a-b c-d' or a+b:a-b=c+d:c-d. ~318 Q. E. D. EXERCISES 583 Verify ~ 335 in the proportion 8: 5 = 16: 10. 584 If x + y:x- y = 8: 2, find the ratio of x to y. 585 If x- y:13=x-y: 3, find the ratio of x to y. PROPOSITION IX. THEOREM 336 In a continued proportion, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. HYPOTHESIS. a:b=c:d=e:f. CONCLUSION. a: b = a + c +e: b d +f. 146 PLANE GEOMETRY- BOOK III PROOF ab = ba, Iden. ad = be, ~ 328 and af= be. ~ 328 Adding, ab + ad + af= ba + bc + be; Ax. 1 or a (b + d +f) = b (a + c + e)...a: b = a +c+ e: b + d +f ~ 330 Q. E. D. EXERCISES 586 Verify ~ 336 in the continued proportion 6: 3 = 10: 5 = 14: 7. 587 If a:b =c: d = e:f= 21: 7, find the value of a + c + e b+d+ f PROPOSITION X. THEOREM 337 The products of the corresponding terms of two or more proportions are in proportion. HYPOTHESIS. a:b= c:dand e:f=g:h. CONCLUSION. ae: bf = cg: dh. PROOF a c b d' and e=g. and -e g ~318 f h Multiplying member by member, ae cg. bf dh' or ae: bf cg: dh. ~ 318 Q. E. D. EXERCISE 588. Verify ~ 337 in the proportions 2: 3 = 10:15, and 7:4=14:8. THEORY OF PROPORTION 147 PROPOSITION XI. THEOREM 338 Like powers or like roots of the terms of a proportion are in proportion. HYPOTHESIS. a: b = c: d. CONCLUSION. a: b" = c: dn, and /a: /b = -/: {/d. PROOF a c 8 ac. ~318 b d Raising each member to the nth power, an cn bn d~ Taking the nth root of each member, or an: bn = c": d, and /a: V/b = t/c: /d. ~ 318 Q. E. D. EXERCISES 589 Verify ~ 338 in the proportion 4: 9 = 16: 36. 590 If 3: y3 = 27: 125, find the ratio of x to y. 591 Ifv/x: 2 = V/y: 3, find the ratio of x to y. PROPOSITION XII. THEOREM 339 Equimultiples of two quantities have the same ratio as the quantities themselves. HYPOTHESIS. a and b are any two quantities. CONCLUSION, ma: mb = a: b. 148 PLANE GEOMETRY —BOOK III PROOF a a _b=_~-~. Iden. b b Multiplying both terms of the first member by m, ma a mb b' or ma: mb =a: b ~ 318 Q. E. D. PROPOSITION XIII. THEOREM 340 A line parallel to one side of a triangle divides the other two sides proportionally. A A D e D E E E:' B' C B C Fig. 1 Fig. 2 HYPOTHESIS. In the triangle ABC, DE is parallel to BC. CONCLUSION. BD: DA = CE: EA. CASE 1. When BD and DA are commensurable (Fig. 1). PROOF Let m, a common measure of BD and DA, be contained 2 times in BD and 3 times in DA. Then BD: DA = 2: 3 (1). Through the points of division in BA, draw parallels to BC. These parallels divide CA into 5 equal parts, of which CE contains 2 and EA 3. ~ 208.'. CE: EA = 2: 3 (2). From (1) and (2), BD: DA = CE: EA. Ax. 11 CASE 2. When BD and DA are incommensurable (Fig. 2). THEORY OF PROPORTION 149 PROOF Let B'D and DA be commensurable, BB' being less than m, the unit of measure. Draw B'C' II to BC. Then B'D: DA = C'E: EA. Case 1 If m, the unit of measure, be indefinitely diminished, BB', always less than m, will become less than any assigned quantity..-. B'D approaches BD as a limit, and. C'E approaches CE as a limit. ~ 263.. B'D: DA approaches BD: DA as a limit, and C'E: EA approaches CE: EA as a limit. ~ 268 But the variable ratios B'D: DA and C'E: EA are always equal. Case 1 Therefore their limits are equal. That is, BD: DA = CE: EA. ~ 264 Q. E. D. 341 SCHOLIUM. By the theorem, AD: DB = AE: EC. ~ 340 By composition, AB: AD = AC: AE. ~ 333 In like manner, AB: DB = AC: EC. By alternation, AB: AC = DB: EC, and AB: AC = AD: AE. ~ 331 From the last two, AD: AE = DB: EC. In the following pages, ~ 340 is given as authority for any of these proportions. 342 COROLLARY. If two lines are cut by a series of parallels, the corresponding segments are proportional. B' \' For AD: AE = BD: CE, and AD: AE = DF: EG. Fr Also, AF: AG = DF: EG, and AF: AG = FH: GK..'. BD: CE = DF: EG = FH: GK. H -- 150 PLANE GEOMETRY -BOOK III PROPOSITION XIV. THEOREM 343 If a straight line divides two sides of a triangle proportionally, it is parallel to the third side. A HYPOTHESIS. In the A ABC, DE is so drawn that AB: AD AC: AE. HYPOTHESIS. In the A ABC, DE is so drawn that AB: AD = AC: AE. CONCLUSION. DE is I| to BC. PROOF Draw DG II to BC intersecting AC at E'. Then AB:AD = AC: AE'. ~ 340 But AB:AD = AC:AE. Hyp.. '. AE' = AE. Ax. 11.*. DE and DE' coincide. ~ 94 But DE' is II to BC by construction..'. DE is II to BC. Q E. D. EXERCISES In the triangle ABC, DE is parallel to BC: 592 If AD = 5, DB = 10, and AE = 4, find EC. 593 If AB = 12, AD = 8, and AC = 9, find AE. 594 If AD = a, AB = b, and AE = c, find AC. 595 If AD = a, AE = b, and EC = c, find DB. 596 If AD = 4, DB = AE, and EC = 9, find AE. 597 If AD = 2 EC, DB = 2, and AE = 9, find AD. 598 If AD: EC = 2: 3, DB = 6, and AE = 16, find AD and EC 599 If AD = ~ AE, and DB = 7, find EC. 600 If AD: AE = 3: 5, and DB = 6, find EC. THEORY OF PROPORTION 151 DEFINITIONS 344 A line is divided internally when the point of division is between the extremities of the line. A line is divided externally when the point of division is in the line produced. Thus, AB is divided internally at P, A B and A'B' is divided externally at P'. In either case, the segments of the iP A' B' line are the distances from the point of division to the extremities of the line. Thus, the segments of AB are PA and PB, and the segments of A'B' are P'A' and PB'. PA and PB are called the internal segments of AB. P'A' and P'B' are called the external segments of A'B'. A line is equal to the sum of its internal segments or the difference of its external segments. 345 A line is divided harmonically when it is divided internally and externally in the same ratio. 346 PROBLEM. To divide a given line harmonically. II I l l I I, P' A P B DATA. AB is a given line. REQUIRED. To divide AB harmonically in the ratio of 2 to 5. SOLUTION 1 Divide AB into 7, (5 + 2), equal parts, and take two parts from A to the right, which gives P the internal point of division. 2 Divide AB into 3, (5 - 2), equal parts, and take two of these parts from A to the left, which gives P' the external point of division. CONCLUSION. AB is divided harmonically at P and P'. PROOF. PA: PB = 2: 5, Const. and P'A: P'B = 2: 5. Const..-. PA: PB = P'A:P'B. Q. E.n. 152 PLANE GEOMETRY -BOOK III PROPOSITION XV. THEOREM 347 The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the adjacent sides. B D C HYPOTHESIS. In the A ABC, AD bisects the Z A. CONCLUSION. AB: AC = DB: DC. PROOF Draw BE II to AD and meeting CA produced at E. Then AE: AC = DB: DC (1). "A line parallel to one side of a A divides the other two sides proportionally." ~ 340 Also, Z z = Z x, "being alt.-int. A of 11 lines," ~ 121 = Z y by hypothesis, = Z E, "being corresponding A of 1I lines." ~ 122.. z = ZE. Ax. 11.'. AE = AB. ~ 176 Substituting AB for its equal AE in (1), AB: AC = DB: DC. Q. E. D. EXERCISES AD bisects the angle A in a triangle ABC: 601 If AB = 6, AC = 8, BC = 7, find BD and DC. 602 If AB = 15, AC 20, BC = 7, find BD and DC. 603 If BC = a, AC = b, AB = c, find BD and DC. 604 If AB = 8, BD = 4, )C = 5, find AC. 605 If AB = 6, AC = 12, BD = 5, find BC. 606 If AB = DC, AC = 8, BD = 2, find AB. THEORY OF PROPORTION PROPOSITION XVI. THEOREM 348 The bisector of an exterior angle of a triangle divides the opposite side externally into segments which are proportional to the adjacent sides. D 'BA HYPOTHESIS. In the A ABC, AD' bisects the exterior Z BAE. CONCLUSION. AB: AC = D'B: D'C. PROOF Draw BF II to D'A and meeting AC at F. Then AF: AC = D'B: D'C (1). ~ 340 Also, Z o = Z n, " being alt.-int. A of 11 lines," ~ 121 = Z m by hypothesis, = L p, " being corresponding A of 11 lines.'' ~ 122..Zo = / p. Ax. 11.'. AF = AB. ~ 176 Substituting AB for its equal AF in (1), AB: AC = D'B: D'C. Q. E. D. EXERCISES AD' bisects the exterior angle at A in the triangle ABC: 607 If AB = 7, AC = 13, BC = 12, find DI'B. 608 If AB = 5, AC = 8, D'B = 10, find BC. 609 If AB = 9, AC = D'B, DIC = 16, find BC. 610 The bisectors of an interior angle and the adjacent exterior angle of a triangle divide the opposite side harmonically. 611 If AD and AD' bisect the angles at A in the triangle ABC, DID is divided harmonically at B and C. [Apply ~ 331 to Ex. 610.] 4 PLANE GEOMETRY-BOOK III SIMILAR POLYGONS DEFINITIONS 349 Similar polygons are polygons which have their homologous angles equal and their homologous sides proportional. 350 The ratio of similitude of two similar polygons is the ratio of any two homologous lines. PROPOSITION XVII. THEOREM 351 Mutually equiangular triangles are similar. A A' D - ~~E B C B C' HYPOTHESIS. In the triangles ABC and A'B'C', ZA = ZA', Z B = Z B', and LC = C'. CONCLUSION. The A ABC and A'B'C' are similar. PROOF Apply the A A'B'C' to the A ABC so that the Z A' shall coincide with the Z A, B'C' taking the position of DE. Then ZADE = Z B. Hyp..'. DE is II to BC. ~ 126.-. AB: AD = AC: AE. ~ 340 Or AB: A'B' = AC: A'C'. By applying the Z B' to the Z B, we may prove that AB: A' = BC: B'C'... AB: A'B' = AC: A'C'= BC: B'C'..'. the A are similar. ~ 349 Q. E. D. 352 COROLLARY 1. Two triangles are similar if two angles of the one are equal respectively to two angles of the other. SIMILAR POLYGONS 155 353 COROLLARY 2. Two right triangles are similar if an acute angle of the one is equal to an acute angle of the other. PROPOSITION XVIII. THEOREM 354 If two triangles have an angle of the one equal to an angle of the other, and the including sides proportional, they are similar. A A' B C B C' HYPOTHESIS. In the triangles ABC and A'B'C', Z A = A', and AB: A'B' = AC: A'C'. CONCLUSION. The A ABC and A'B'C' are similar. PROOF Apply the A A'B'C' to the A ABC so that the Z A' shall coincide with the / A, B'C' taking the position of DE. Then AB: AD = AC: AE. Hyp..'. DE is II to BC. ~ 343.-. LADE = Z B, and AED = Z C. ~122.-. the A ABC and ADE are similar. "Mutually equiangular A are similar." ~ 351.'. the A ABC and A'B'C' are similar. Q. E. D. EXERCISES 612 The sides of two similar triangles are 6, 10, 14, and 3, 5, 7, respectively. What is the ratio of similitude? 613 The ratio of similitude of two similar triangles is 3. If the sides of the second triangle are 7, 11, 15, find the sides of the first. PLANE GEOMETRY-BOOK III 156 PROPOSITION XIX. THEOREM 355 Triangles are similar if their sides are respectively proportional. A A' B C BC HYPOTHESIS. In the triangles ABC and A'B'C', AB: A'B' = AC: A'C' = BC B'C'. CONCLUSION. The A ABC and A'B'C' are similar. PROOF On AB take AD = A'B', and on AC take AE = A'C', and draw DE. Then AB: AD = AC: AE. Hyp..'. the A ABC and ADE are similar. ~ 354.-. AB: AD = BC: DE. ~ 349 Or AB: A'' = BC: DE (1). But AB: A'B' = BC: BC' (2). Hyp. In (1) and (2) the first three terms are identical..'. the fourth terms DE and B'C' are equal..-. the A ADE and A'B'C' are equal. ~ 172 But the A ABC and ADE are similar. Proved.'. the A ABC and A'B'C' are similar. Q. E. D. 356 SCHOLIUM. From ~~ 351 and 355, we see that in triangles either condition of similarity (~ 349) involves the other. This is true in no other polygons. SIMILAR POLYGONS 157 PROPOSITION XX. THEOREM 357 Triangles whose sides are respectively parallel, or respectively perpendicular, are similar. E cC' ' A A BX c F C Fig. 1 Fig. 2 HYPOTHESIS. In the triangles ABC and A'B'C', AB, AC, and BC are respectively parallel (Fig. 1), or perpendicular (Fig. 2), to A'B', A'C', and B'C'. CONCLUSION. Triangles ABC and A'B'C' are similar. PROOF In Fig. 1, produce B'C' to meet BA produced at E. Also produce B'A' to meet AC at G and BC at F. Then Z BAC = Z B'GA (~ 121) = L B'A'C' (~ 122)..'. / BAC = B'A'C'. Ax. 11 Also, B, = Z B'. ~ 203.-. A ABC and A'B'C' are similar. ~ 352 In Fig. 2, produce A'B' to meet AB at E. In the right A DAE and DA'F, Z ADE = L A'DF. ~ 112.'. Z A=L A'. ~ 156 Likewise in the right A CFG and C'HG, / C = L C'. ~ 156.-. the A ABC and A'B'C' are similar. Q. E. D. 358 SCHOLIUM. In the similar triangles, the parallel sides, or the perpendicular sides, are homologous. EXERCISE 614. If in Fig. 1, AA' and CC' intersect in P, the triangles PAC and PA'C' are similar. 158 PLANE GEOMETRY -BOOK III PROPOSITION XXI. THEOREM 359 In two similar triangles, homologous altitudes have the same ratio as any two homologous sides. A A' B C C' HYPOTHESIS. The triangles ABC and A'B'C' are similar, and the altitudes AD and A'D' are homologous. CONCLUSION. AD: A'D' = AB: A'B'. PROOF B Z B', "being homologous A of similar Q." ~ 349.'. the rt. A ADB and A'D'B' are similar, "having an acute Z in each equal." ~ 353.'. AD: A'D' = AB: A'B'. ~ 349 Q. E. D. EXERCISES 615 The base of a triangle is 34 in. and its altitude 28 in. If the homologous base of a similar triangle is 51 in., find the homologous altitude. 616 The homologous altitudes of two similar triangles are 5 and 7. If the base of the first is 9, find the base of the second. 617 The homologous angle-bisectors of two similar triangles have the same ratio as any two homologous sides. 618 The homologous medians of two similar triangles have the same ratio as any two homologous sides. 619 The radii of the inscribed circles of two similar triangles have the same ratio as any two homologous sides. 620 The radii of the circumscribed circles of two similar triangles have the same ratio as any two homologous sides. SIMILAR POLYGONS 159 PROPOSITION XXII. THEOREM 360 If two parallels are cut by three or more transversals passing through the same point, the corresponding segments are proportional. JD C' B' At A/ B' C' \D' A B C D HYPOTHESIS. The parallels AD and A'D' are cut by the transversals OA, OB, OC, OD, at A and A', B and B', C and C', D and D'. CONCLUSION. AB = B = CD A'B' B'C' C'D' PROOF Since A'D' is II to AD, the A OAB and OA'B', OBC and OB'C', OCD and OC'D' are similar. " Mutually equiangular A are similar." ~ 351 AB = OB BC = OC _ CD A'B' OB' BC' kOC'1 C'D" "being homologous sides of similar Q." ~ 349 AB BC CD.-~......L L ~ Ax. 11 A'B' B'C' C'D' Q. E. D. EXERCISES 621. If in the above figure AB = 15, BC = 9, CD = 12, and AB' = 10, find B'C' and C'D'. 622 The sides of a triangle are 9, 15, 18. In a similar triangle the side homologous to 9 is 6. Find the other two sides. 160 PLANE GEOMETRY-BOOK III 623 Isosceles triangles are similar if their vertical angles are equal. 624 Two isosceles triangles are similar if a base angle in one is equal to a base angle in the other. 625 Find the fourth proportional to 24, 20, 34. 626 Find the third proportional to 15 and 45. 627 Find the mean proportional between 4 and 121. 628 In a triangle ABC, DE is parallel to BC. If AD: DB = 3: 2, and BC = 15, find DE. 629 In a triangle ABC, DE is parallel to BC. If AD: DB = 3: 5, and AC = 176, find AE and EC. 630 The sides of a triangle are 3, 5, 7. In a similar triangle the side homologous to 3 is 8; find the remaining sides. 631 The homologous bases of two similar triangles are 21 ft. and 19 ft. The altitude of the first is 14 ft.; find the altitude of the second. 632 How high is a monument which casts a shadow 192 feet long, when a vertical staff 5 feet high casts a shadow 3 feet long? 633 The homologous bases of two similar triangles are 6 and 7. The median to the base of the first is 5; find the median to the base of the second. [Ex. 618.] 634 The homologous bases of two similar triangles are 10 and 14. The angle-bisector to the base of the first is 8; find the homologous anglebisector in the second. [Ex. 617.] 635 If two triangles are similar, prove that the radii of the inscribed circles have the same ratio as the radii of the circumscribed circles. [Exs. 619, 620.] 636 The sides of two similar triangles are 14, 30, 40, and 28, 60, 80. The radii of the inscribed and circumscribed circles of the first are 4 and 25. Find the radii of the inscribed and circumscribed circles of the second. 637 The sides of a triangle are 8, 12, 15. Compute the segments of the sides made by the bisectors of the angles. 638 The sides of a triangle are a, b, c. Compute the segments of the sides made by the bisectors of the angles. 639 Compute the segments of a line 16 in. long, divided harmonically in the ratio of 3 to 5. SIMILAR POLYGONS 161 PROPOSITION XXIII. THEOREM 361 The perimeters of two similar polygons have the same ratio as any two homologous sides. E A D A' HYPOTHESIS. P and P' are the perimeters of the similar polygons ABCDE and A'B'C'D'E'. CONCLUSION. P: P' = AB: A'B'. PROOF AB BC CD DE EA A'B' B'C' -C'D D'E' E'A' ~ 349 AB + BC + CD- +DE + EA AB A'B' + B'C' + C'D' + D'E' + E'A A'B' ~ 336 Or P: P -AB: A'B'. Q.E.D. EXERCISES 640 The perimeters of two similar polygons are 32 and 40. If a side of the first is 8, compute the homologous side of the second. 641 Homologous sides of two similar polygons are 5 and 8. If the perimeter of the first is 35, compute the perimeter of the second. 642 The sides of a polygon are 6, 8, 10, 12, 14. Compute the perimeter of a similar polygon whose side homologous to 6 is 9. 643 Compute the sides of a rectangle if the perimeter is 80 and the ratio of the sides 3: 5. [Let x = one side; then 40 - x = the other side..:. 40-x = 3: 5.] 644 Compute the sides of a rectangle if the perimeter is p and the ratio of the sides a b. 162 PLANE GEOMETRY- BOOK III PROPOSITION XXIV. THEOREM 362 If two polygons are composed of the same number of triangles, similar each to each, and similarly placed, the polygons are similar. E D A' HYPOTHESIS. The A ABC, ACD, and ADE are similar respectively to the A A'B'C', A'C'D', and A'D'E', and similarly placed. CONCLUSION. The polygons ABCDE and A'B'C'D'E' are similar. PROOF Z B= B'. ~349 Also, / s = s', and / t = t'. ~ 349 Adding, Z C= C'. Ax. 1 In like manner it may be shown that the other corresponding angles of the polygons are equal. That is, the polygons are mutually equiangular. AB BC (AC CDetc. Again, B = B' C= etc A'B'1 B C1 J C'IY ~ 349 That is, the homologous sides of the polygons are proportional..'. the polygons are similar. "Similar polygons are polygons which have their homologous angles equal and their homologous sides proportional." ~ 349 Q. E. D. SIMILAR POLYGONS 163 PROPOSITION XXV. THEOREM 363 If two polygons are similar, they are composed of the same number of triangles, similar each to each, and similarly placed. E A ----------------— ~ --- -— 1 B C BA C HYPOTHESIS. The polygons ABCDE and A'B'C'D'E' are similar. CONCLUSION. The A ABC, ACD, and ADE are similar respectively to the A A'B'C', A'C'D', and A'D'E'. PROOF ZB=ZB'. ~349 Also, BA: B'A' = BC: B'C'. ~ 349.'. the A ABC and A'B'C' are similar. ~ 354 Again, Z BCD = Z B'C'D', and / BCA = Z B'C'A'. ~ 349 Subtracting, Z ACD = Z A'C'D'. Ax. 4 Again, BC: B'C' = AC: A'C', and BC: B'C'= CD: C'D'. ~ 349.'. AC: A'C' = CD: C'D'. Ax. 11.'. the A ACD and A'C'D' are similar. ~ 354 Likewise the A ADE and A'D'E' are similar. Q. E. D. 364 COROLLARY. Two similar polygons are equal, if any two homologous lines are equal. 164 PLANE GEOMETRY -BOOK III PROPOSITION XXVI. THEOREM 365 If in a right triangle an altitude is drawn to the hypotenuse: 1 The two triangles thus formed are similar to the whole triangle and to each other. 2 The altitude is the mean proportional between the segments of the hypotenuse. 3 Each leg of the right triangle is the mean proportional between the hypotenuse and the adjacent segment. A B C HYPOTHESIS. In the rt. A BAC, AD is I to BC, the hypotenuse. CONCLUSION. 1 The A BDA, ADC, and BAC are similar. PROOF The rt. A BDA and BAC have the L B common;.'. they are similar. ~ 353 The rt. A ADC and BAC have the Z C common;.'. they are similar. ~ 353 The A BDA and ADC are similar, for each is similar to the A BAC. Proved 2 AD is the mean proportional between BD and DC. PROOF In the similar A BDA and ADC, BD: AD = AD:DC. ~ 349 NUMERICAL PROPERTIES 165 3 BA is the mean proportional between BC and BD, and CA is the mean proportional between CB and CD. PROOF In the similar A BAC and BDA, BC: BA B: BD. ~ 349 In the similar A BAC and ADC, CB: CA= CA: CD. ~ 349 Q. E. D. 366 COROLLARY 1. The squares of the legs of a right triangle are proportional to the adjacent segments of the hypotenuse. PROOF From the proportions in the third part of the theorem, BA2 = BC x BD, and CA2 = CB x CD. ~ 328 BA2 BCxBI) BD CA2 CB xCD CD Q. E. D. 367 COROLLARY 2. The altitude upon the hypotenuse of a right triangle is equal to the product of the legs divided by the hypotenuse. PROOF In the similar A BAC and BDA, BC: BA = CA: AD. ~ 349 Whence BC x AD= BA x CA. ~ 328..AD BA x CA BC Q. E. D. 368 COROLLARY 3. The perpendicular drawn from any point in the circumference of a circle to the diameter is the mean proportional between the seg- / ments of the diameter. 166 PLANE GEOMETRY-BOOK III PROPOSITION XXVII. THEOREM 369 The square of the hypotenuse of a right triangle is equal to the sum of the squares of the two legs. HYPOTHESIS. BC is the hypotenuse of the rt. A BAC. CONCLUSION. BC2= AB2+ AC2. PROOF Draw AD I to BC. Then AB = BC X BD, and AC2 = BC x DC. 2 2 Adding, AB2 + AC2 = BC (BD + DC) = BC2. 370 COROLLARY 1. The square of either leg of angle is equal to the difference of the squares of nuse and the other leg. ~ 365 Ax. 1 Q. E. D. a right trithe hypote1 I 371 COROLLARY 2. The side and the diagonal of ds a square are incommensurable. d2 = s2 + 2 =.'. d = s2. 372 DEFINITION. The projection of MN is that part of MN included between the feet of the perpendiculars drawn from A and B to MN. Thus, CD is the projection of AB upon MN. a line AB upon a line B I C - -D — el NUMERICAL PROPERTIES 167 EXERCISES 645 Find the hypotenuse of a right triangle whose legs are 3 and 4. 646 The hypotenuse of a right triangle is 15 and one leg is 9. Find the other leg. 647 Find the diagonal of a rectangle whose sides are 5 and 12. 648 Find the diagonal of a square whose side is 10. 649 3, 4, 5, are the sides of a right triangle. Why are any equimultiples of 3, 4, 5 (such as 6, 8, 10; 9, 12, 15; 12, 16, 20, etc.), the sides of a right triangle? 650 The sides of a right triangle are 8, 15, 17. Find the altitude upon the hypotenuse. [~ 367.] 651 The legs of a right triangle are 9 and 12. Find their projections upon the hypotenuse. 652 How long is a ladder that reaches a window 24 feet high when one end of the ladder rests on the ground 18 feet from the house? 653 Find the width of a street, if a ladder 29 feet long reaches from the same point in the street a window on one side 20 feet from the ground and a window on the other side 21 feet from the ground. 654 Two flag poles whose heights are 40 feet and 70 feet stand 100 feet apart. Compute the distance from the top of one to the top of the other. 655 Compute the altitude of an isosceles triangle in which the base is 16 and a leg 17. 656 Compute the leg of an isosceles triangle in which the base is 24 and the altitude 35. 657 Find the altitude of an equilateral triangle whose side is 14 in. 658 Find the altitude of an equilateral triangle whose side is a. 659 If a is a side and h the altitude of an equilateral triangle, show that a 2h. 3 660 Compute the side of an equilateral triangle whose altitude is 100 feet. [Use the formula of Ex. 659.] 661 The squares of the hypotenuse and either leg of a right triangle are proportional to the hypotenuse and the adjacent segment (~ 365). 662 The chord drawn from any point in the circumference of a circle to the extremity of a diameter is the mean proportional between the diameter and the adjacent segment (~ 365). 168 PLANE GEOMETRY-BOOK III PROPOSITION XXVIII. THEOREM 373 In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides, diminished by twice the product of one of those sides and the projection of the other side upon it. A A C Ai -B ~a C B x -D Fig. 1 Fig. 2 HYPOTHESIS. In the A ABC, the Z B is acute, and BD is the projection of AB upon BC. Let AB = c, AC = b, BC = a, AD = h, BD =, DC = y. CONCLUSION. b2 = a2 + c2 - 2 ax. PROOF In Fig. 1, y = a- x; in Fig. 2, y = x - a. In both cases, y2 = a + x2 - 2 ax. Adding h2 to both members, y2 + h2 =a2+ x2 + h2_ -2 ax (1). Ax. 1 But y2 + h2 = b2, and X2 + h2 = c. ~ 369 Substituting b2 and c2 for their equals in (1), b2 = a2 + c2 - 2 ax. Q. E. D. EXERCISES 663 If in a triangle a = 14, b = 15, and c = 13, find the projections of b and c upon a. 664 If a = 21, b = 10, and c = 17, find the projections of b and c upon a. NUMERICAL PROPERTIES 169 PROPOSITION XXIX. THEOREM 374 In an obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of those sides and the projection of the other side upon it. A B — Y —'D HYPOTHESIS. In the A ABC, the Z C is obtuse, and CD is the projection of AC upon BC produced. Let AB = c, AC = b, BC = a, AD = h, BD = z, CD =y. CONCLUSION. c2 = a2 + b2 + 2 ay. PROOF x = a + y. Squaring, x2 = a2 + y2 + 2 ay. Adding h2 to both members, x2 + h2 = a2 + y2 + h2 + 2 ay (1). Ax. 1 But X2 + h2 = c2, and y2 + h2 = b2. ~ 369 Substituting c2 and b2 for their equals in (1), c2= a2 + b2 + 2 ay. Q. E. D. 375 COROLLARY. From the last three theorems, it follows that an angle of a triangle is acute, right, or obtuse, according as the square of the opposite side is less than, equal to, or greater than, the sum of the squares of the other two sides. 170 PLANE GEOMETRY -BOOK III PROPOSITION XXX. THEOREM 376 In any triangle, if a median is drawn to the base: 1 The sum of the squares of the other two sides is equal to twice the square of half the base, increased by twice the square of the median. 2 The difference of the squares of the other two sides is equal to twice the product of the base and the projection of the median upon the base. A B C HYPOTHESIS. In the A BC, BC is the base, h the altitude, m the median to BC, and n the projection of m upon BC; a, b, and c are the sides, and b is greater than c. CONCLUSION. - b2 + c2 = 2 ( + 2 m2. 2 b2- c2= 2 an. PROOF Since b > c, the Z t is obtuse, and the Z s is acute. ~ 171.. b2 = 2+ m2 + an (1), ~ 374 and c2 = -+ m2 - an (2). ~ 373 (1) + (2) gives b2 + c2 = 2 ()+ 2 m2. Ax. 1 (1) - (2) gives b2 - c2 = 2 an. Ax. 4 Q. E. D. NUMERICAL PROPERTIES 1T71 PROPOSITION XXXI. THEOREM 377 If two chords intersect within a circle, the product of the segments of one is equal to the product of the segments of the other. A C HYPOTHESIS. The chords AB and CD intersect at P. CONCLUSION. PA x PB = PC x PD. PROOF Draw AC and DB. Then Z A = Z D, "each being measured by half the arc CB,"' ~ 275 and / C = L B, "each being measured by half the arc AD." ~ 275.'. the A PAC and PDB are similar. "Two & are similar if two angles of the one are equal respectively to two angles of the other." ~ 352.-. PA:PD = PC:PB. ~ 349 Whence PA x PB = PC x PD. Q E. D. 378 COROLLARY. If two chords intersect within a circle, their segments are reciprocally proportional. EXERCISES 665 If PA = 3, PB = 6, and PC = 2, find PD. 666 If PA = 2, PB = 8, and PC = PD, find PC. 667 If AB = 15, CD - 13, and PC = 4, find PA. 172 PLANE GEOMETRY -BOOK III PROPOSITION XXXII. THEOREM 379 If from a point without a circle a tangent and a secant are drawn, the tangent is the mean proportional between the whole secant and its external segment. HYPOTHESIS. PT is a tangent and PAB a secant of the circle ATB. CONCLUSION. PB: PT = PT: PA. PROOF Draw TA and TB. The A PBT and PTA are similar; ~ 352 for Z B = Z PTA, "each being measured by half the arc AT," ~~ 275, 281 and Z P is common... PB: PT = PT: PA. Q. E. D. 380 COROLLARY. If from a point without a circle a secant is drawn, the product of the whole secant and its external segment is constant. For PB x PA = PT2 = a constant. EXERCISES 668 If PA = 8, AB = 9, find PT. 669 If PT = 8, PB = 16, find PA. 670 PT = 10, and PA = AB. Find PA. NUMERICAL PROPERTIES 173 PROPOSITION XXXIII. THEOREM 381 In any triangle, the product of two sides is equal to the square of the bisector of their included angle, increased by the product of the segments of the third side. A E HYPOTHESIS. In the A ABC, AD bisects the Z A and divides BC into the segments x and y. Let AD = t and DE = s. CONCLUSION. bc = t2 + y. PROOF Circumscribe a circle about the A ABC. ~ 297 Produce AD to meet the Oce at E, and draw BE. In the A BAE and DAC, E = C, "each being measured by half the arc AB," ~ 275 and Z g = Z k. Hyp..'. the A BAE and DAC are similar. ~ 352.. c:t=t+s:b. ~ 349.. bc = t (t + s) ~ 328 = t2 + ts. But ts = xy. ~ 377... bc = t2 + y. Q. E.D. 174 PLANE GEOMETRY-BOOK III PROPOSITION XXXIV. THEOREM 382 In any triangle, the product of two sides is equal to the product of the diameter of the circumscribed circle and the altitude upon the third side. A HYPOTHESIS. AE is a diameter of the circle circumscribed about the triHYPOTHESIS. AE is a diameter of the circle circumscribed about the triangle ABC, and AD is the altitude upon BC. Let AD = h and AE = d. CONCLUSION. bc = dh. PROOF Join EC. Then Z B= Z E, "each being measured by half the arc AC." ~ 275.. the rt. A ABD and AEC are similar, "having an acute Z in each equal." ~ 353.'.c:d=h:b. ~349.-. be = dh. Q.E.D. 383 SCHOLIUM. If the three sides of a triangle are given, we can compute the altitudes by ~~ 369, 373, and 374; the medians by ~ 376; the bisectors of the angles by ~ 381; and the diameter of the circumscribed circle by ~ 382. EXERCISES 671 In the triangle ABC, AB = 13, AC = 20, and the projection of AB upon BC = 5. Find the diameter of the circumscribed circle. 672 The sides of 9 triangle are 5, 5, and 6. Find the radius of the circumscribed circle. EXERCISES 175 MISCELLANEOUS EXERCISES PROBLEMS OF COMPUTATION 673 To compute the altitudes of a triangle in terms of the sides. A B D C SOLUTION. In the A ABD, h2 = c2 - 32 (1). ~ 369 In the AABC, b2 = a2 + c2- 2a x BD. ~ 373 Whence BD a2 + c2 - b2 2a Substituting the value of BD in (1), h2 = 2 _(a2 +c2- b2)2 4 a2 4 a2c2 - (a2 + 2-b2)2 4 a2 (2 ac + a2 + 2 - b2) (2 ac - a2 - c2+ b2) 4 a2 = (a + c + b)(a + c - )(b+ a - c)(b - a + c) (2) 4 a2 Let s denote half the perimeter. Then a + b + c =2 s, a + c - b = 2 (s - b), b + a - c = 2 (s - c), b - a + c = 2 (s - a). Substituting these values in (2), 2 2s x 2 (s - a) x 2 (s -b) x2 (s -c) 4 a2 Simplifying, h2 = 4 (s - a)( - b) (s - c) a2 Extracting the square root, h = 2-s (s -a) (s -b)(s -c). a 176 PLANE GEOMETRY- BOOK III 674 To compute the medians of a triangle in terms of the sides. SOLUTION. 2 rn2 + 2() b2 + c2 ~ 376 Whence m = V- /2(b2 + c2)a2. / /m 675 To compute the angle-bisectors of a triangle in terms of the sides. B a SOLUTION. Let t, t', t", be the bisectors of the angles A, B, C, respectively. Then t2 = bc - xy (1). ~ 381 Nowx:y=c:b. ~ 347..x + y:x= c+b:c. ~ 333 B Or a: x = c + b:c. Whence x = -ac and likewise, y b ~328 b + c b - c Substituting these values in (1), bc a2bc (b + c)2 = bc[1 - (b + c)2] be [(b + c)2- a2] (b + c)2 _ be (b + c + a)(b + c-a) (b + c)2 _ x 2 s x 2 (s - a) (b + c)2 4 bcs (s- a). (b + c)2 Extracting the square root, 2 t = bcs (s- a) b +Likewise t Likewise t' = + cacs (s - b), aand /aCS as - c). 2 and t"- = -Vabs (s - c). EXERCISES 177 676 To compute the radius of the circle circumscribed about a triangle in terms of the sides of the triangle. SOLUTION. bc = dh ~ 382 =2R h. But h = - (s (- a)(s - b)(s - c). aba ~63 6 -DWhence R = abc 4 ' s (s - a)(s -b)(s - c) 677 The following numbers represent the sides of triangles: 2,:;, 4; 3, 4, 5; 5, 6, 7; 5, 7, 9; 3, 6, 7; 5, 5, 8; 20, 21, 29; 48, 55, 73. Is the largest angle in each acute, right, or obtuse? [~ 375.] 678 Compute the altitude of an isosceles triangle in which the base is 90 ft. and a leg 53 ft. 679 The projections of the legs of a right triangle upon the hypotenuse are 3 in. and 6 in. Compute the altitude and the legs. 680 If the diameter of a circle is 20 in., compute the chord which bisects a radius at right angles. 681 Compute the altitude upon the hypotenuse of a right triangle in terms of the legs. [h= c (~ 367). But a = vb2 + c2.. h= be a -/b + c2 " 682 Compute the altitude upon the hypotenuse of a right triangle whose legs are 5 and 12. 683 Compute the legs of an isosceles right triangle in terms of the hypotenuse. 684 Compute the legs of an isosceles right triangle whose hypotenuse is 16 feet. 685 Compute the sides of a right triangle which are three consecutive even numbers. 686 Compute the legs of a right triangle which are as 7: 24, if the hypotenuse is 100. 687 The homologous bases of two similar triangles are 21 feet and 19 feet. The altitude of the first is 14 feet. Find the altitude of the second. 688 The sides of a rectangle are 65 and 72. Find the diagonal. 178 PLANE GEOMETRY-BOOK III 689 Compute the distance between the opposite bases of a baseball diamond, the bases being 90 feet apart. 690 The radius of a circle is 13 in. Compute the least and the greatest chord that can be drawn through a point 5 in. from the center. 691 If a chord 80 in. long is 9 in. from the center of a circle, compute the radius. 692 If a chord 30 in. long is 20 in. from the center of a circle, how long is a chord which is 7 in. from the center? 693 Two chords of a circle are 32 in. and 66 in. long. The first is 63 in. from the center; how far from the center is the second? 694 Find the product of the segments of a chord drawn through a point 4 in. from the center of a circle whose radius is 9 in. 695 Of two intersecting chords of a circle, the segments of one are 15 and 28; find the second chord if one segment is 21. 696 Of two intersecting chords of a circle, the segments of one are a and b; find the second chord if one segment is c. 697 The radius of a circle is 8 in., to which two tangents, each 15 in. long, are drawn from the same point. Compute the length of the chord joining the points of tangency. 698 The radii of two concentric circles are 24 and 26. Compute the chord of one circle tangent to the other. 699 In a circle whose radius is 85, two parallel chords are 72 and 102. Compute the distance between the chords. Two solutions. 700 From the same point a tangent and a secant are drawn to a circle. The tangent is 15 in., and the secant 25 in. If the secant passes through the center of the circle, find the radius. 701 A diameter is perpendicular to a chord. The chord is 18 in. and one segment of the diameter is 3 in. Find the diameter. 702 From the same point a tangent and a secant are drawn to a circle, the secant passing through the center. If the radius is 39 and the tangent 80, find the secant and its external segment. 703 From the same point without a circle two secants are drawn whose external segments are 5 and 6. If the internal segment of the first is 13, find the internal segment of the second. 704 The radius of a circle is 28cm. Compute the tangent drawn from a point 53cm from the center. EXERCISES 179 705 The radius of a circle is 12cm. Compute the tangent drawn from a point 25cm from the circumference. 706 A secant and its external segment are 81cm and 25cm. Find the tangent drawn from the same point. 707 The radii of two intersecting circles are 26m and 30m, and their common chord is 48m. Compute their line of centers. 708 Compute the tangent-chord of two concentric circles whose radii are r and r'. [Let c be the tangent-chord. Then c = Vr2- r... c = 2 r2-r'2.] 2 709 Find the tangent-chord of two concentric circles whose radii are 73m and 55m. 710 Compute the altitudes of a triangle whose sides are 6, 8, 12. 711 Compute the medians of a triangle whose sides are 3, 5, 7. 712 Compute the angle-bisectors of a triangle whose sides are 8, 10, 12. 713 Compute the radius of the circle circumscribed about a triangle whose sides are 14, 30, 40. 714 Two sides and one diagonal of a parallelogram are Q D respectively 7, 9, 12. Compute the other diagonal. [AO 6 = 12 - 2 = 6, the median to BD in the A ABD. Substitut- 7 ing these values in the formula of Ex. 674, BD is found.] B 715 The bases of a trapezoid are 5 and 8, and the altitude is 4. Compute the altitudes of the two triangles formed by producing the legs of the trapezoid until they meet. 716 The bases of an isosceles trapezoid are 8 and 14, and a leg is 5. Compute the altitude and the diagonal. 717 Compute the sides of a rectangle whose perimeter is 28 and diagonal 10. 718 The sides of a triangle are 65, 72, 97. Is the largest angle acute, right, or obtuse? 719 A chord of a circle is equal to the radius r. Compute its distance from the center. 720 Compute a chord of a circle perpendicular to the radius r at its mid-point. 180 PLANE GEOMETRY- BOOK III 721 The chord of an arc is 25 ft., and the chord of twice the arc is 48 ft. Compute the diameter of the circle. 722 A tangent to a circle is 5 ft., and the distance from the extremity of the tangent to the circumference is 3 ft. Compute the radius. 723 Find the radius of a circle in which two parallel chords on opposite sides of the center are 40m and 48m, and the distance between them is 22m. 724 Find the radius of a circle in which two parallel chords on the same side of the center are 72m and 150m, and the distance between them is 37m. 725 Two secants to a circle, drawn from the same point without, are 15 and 21. If the external segment of the first is 7, find the external segment of the second. 726 The radius of a circle is 33cm. Find the distance of a point from the circumference, if a tangent drawn from the same point is 56cm. 727 Compute the common exterior tangent of two circles whose radii are 13cm and 22cm, and whose line of centers is 41cm. 728 Compute the common interior tangent of two circles whose radii are 9cm and 11cm, and whose line of centers is 29cm. 729 Compute the tangents drawn from a point to a circle in which the radius is 13m, and the chord joining the points of tangency is 24cm. 730 Compute the tangents drawn from a point to a circle in which the radius is r, and the chord joining the points of tangency bisects the radius at right angles. 731 The radii of two intersecting circles are 9 and 11, and their line of centers is 14. Find their common chord. 732 Two tangents are drawn from the same point to two concentric circles whose radii are 36 and 40. If the tangent to the inner circle is 77, find the tangent to the outer circle. 733 The radii of two circles tangent internally are 9 and 14. Find their tangent-chord: 1 Parallel to the line of centers. 2 Perpendicular to the line of centers. 3 Drawn from the extremity of that diameter of the larger circle which passes through the centers. EXERCISES 181 THEOREMS 734 Tangents drawn to two intersecting circles from any point in their common chord produced are equal. [~ 379.] 735 The common chord of two intersecting circles produced bisects their common tangents. 736 AB and CD are two chords of a circle which intersect at E. Prove the triangles AEC and DEB similar. 737 A square DEFG is inscribed in a right triangle ABC, E falling in AB, F in BC, D and G in the hypotenuse AC. D Prove DG the mean proportional between AD and CG. [A ADE and FGC are similar.] E 738 If in a right triangle one leg is twice the other, the B F square of the hypotenuse is five times the square of the shorter leg. 739 Two altitudes of a triangle are inversely proportional to the corresponding bases. 740 In any triangle draw the three altitudes, and point out from the figure six pairs of similar triangles. 741 AB is a diameter of a circle, BT a tangent, AC a chord produced to meet the tangent at H. Prove AC the third proportional to All and AB. 742 The square of a line is equal to four times the square of half the line. 743 In a parallelogram, the sum of the squares of the four sides is equal to the sum of the squares of the diagonals. [~~ 206, 376, Ex. 742.] 744 On OA, the radius of a circle, a semicircle is p described. AT is a tangent, OBCF is a secant intersecting the semicircumference at B, the circumference at C, and the tangent at F. BH is perpendicular to OA. Prove OF: OC = OB: OH. o A 745 In the triangle ABC, AD bisects BC, ED bisects the angle ADB, and FD bisects the angle ADC. Prove EF parallel to BC. [~ 347.] 746 The median to the base of a triangle bisects every line parallel to the base and terminated by the sides of the triangle. 182 PLANE GEOMETRY - BOOK III 747 The tangent and the perpendicular drawn from any point in the circumference of a circle to a diameter divide the diameter harmonically. 748 If two circles are tangent internally, all chords of the larger circle drawn from the point of contact are divided proportionally by the circumference of the smaller circle. 749 If two circles are externally tangent, their common external tangent is the mean proportional between their diameters. 750 If from a point without a circle two secants are drawn, they are reciprocally proportional to their external segments. [~ 380.] 751 If in a right triangle one leg is double the other, the altitude upon the hypotenuse divides the hypotenuse into segments which are in the ratio 1:4. 752 AC is the hypotenuse of the right triangle ABC, and D is the mid-point of BC. Prove AC2 = AD2 + 3 DC2. [~ 374.] 753 If in a right triangle one acute angle is double the other acute angle, the square of one leg is three times the square of the other leg. 754 Two circles intersect at A and B. Tangents at A intersect the circumferences at C and D. Prove that the ratio of AC to AD is equal to the ratio of the radii. 755 The line of centers of two circles, produced if necessary, meets their common tangent at P. The line of centers drawn from P intersects the circumference at A, B, C, and D in order. Prove that PA x PD = PB x PC. Two cases. Fig. 1 Fig'. 2 [Case 1. When the common tangent is external (Fig. 1). Case 2. When the common tangent is internal (Fig. 2). In either case the A POT and PO'T' are similar. EXERCISES 183 Whence PO: OT = PO':O'T'. By division and composition, PA: PB = PC:PD. ~ 335. PA x PD = PB x PC.] 756 AB is a chord of a circle fixed in position, and P is the mid-point of its arc. PH is any chord intersecting AB in K. Prove that PH x PK is constant. [Draw the diameter PD intersecting AB in C. The rt. A PCK and PHD are similar.] 757 The sum of the squares of the segments of two perpendicular chords of a circle is equal to the square of the diameter. 758 The diagonals of a trapezoid divide each other in the same ratio. 759 If a line is drawn through the centroid of a triangle D parallel to one side, it intersects the other two sides in points of trisection. A', 760 If in the triangle ABC the median AM is bisected 0 in 0, BO produced intersects AC in a point of trisection. B - 761 In any triangle the distance of the centroid from a given line is one third the sum of the distances of the three vertices from the line. [Bisect AO at S. Then AS = SO = OM. Draw p A AD, SE, OF, BG, MH, CK, each ~ to PQ the given line. Then OF = - (SE + MH) = - (AD + OF + BG + CK). FB G o...OF = (AD + BG + CK)... OF= (AD + BG + CK).] Q 762 A line drawn through the intersection of the diagonals of a trapezoid, parallel to the bases, and terminated by the non-parallel sides, is bisected by the diagonals. 763 The perpendicular from a point P in the circumference of a circle to a chord is the mean proportional between the perpendiculars from P to the tangents drawn from the extremities of the chord. 764 In an inscribed quadrilateral, the product of the diagonals is equal to the sum of the products of the opposite sides. 765 The sum of the squares of the four sides of a quadrilateral is equal to the sum of the squares of the diagonals, increased by four times the square of the line joining the middle points of the diagonals. 184 PLANE GEOMETRY-BOOK III PROPOSITION XXXV. PROBLEM 384 To divide a given straight line into segments proportional to any number of given lines..X /I, m 4-> C- G a > AV C H DATA. m, n, p, and AB are given lines. REQUIRED. To divide AB into segments proportional to m, n, and p. SOLUTION Draw the indefinite line AX, making any convenient angle with AB. On AX take AC = m, CD = n, DE =p, and draw EB. Through C and D, draw parallels to EB, intersecting AB in G and H. CONCLUSION. G and H are the points of division required. Q. E. F. PROOF. Through A draw a parallel to CG. Then AG: m = GH: n = HB:p. "If two lines are cut by a series of Ils, the corresponding segments are proportional." ~ 342 Q. E. D. EXERCISES 766 Bisect a given line by the method of ~ 384. 767 Divide a line 15 inches long into segments proportional to 5, 6, 7. 768 Divide a line 23 inches long into segments proportional to 5, 8, 11. CONSTRUCTIONS 185 PROPOSITION XXXVI. PROBLEM 385 To find the fourth proportional to three given straight lines. mI AA. m n a p are tr i i DATA. m, n, and p are three given lines. REQUIRED. To find the fourth proportional to m, n, and p. SOLUTION Draw AB and AC making any convenient angle. On AB take AD = mn, and DE = n. On AC take AF =p. Join FD. Through E draw EG II to DF. CONCLUSION. FG is the fourth proportional required. Q. E. F. PROOF. AD: DE = AF: FG. "A line parallel to one side of a triangle divides the other two sides proportionally." ~ 340 Substituting m, n, and p for AD, DE, and AF respectively, m: n =p: FG. Q. E. D. EXERCISES 769 Construct x, if x = b. a 770 If a: b = c:x, find x in terms of a when b = 2 a, and c = 2 b. 771 If a: b = c: x, find x in terms of a when b = c = 2 a. 772 Find the fourth proportional to 3 in., 5 in., 7 in. 186 PLANE GEOMETRY-BOOK III PROPOSITION XXXVII. PROBLEM 386 To find the third proportional to two given lines. F GC A D B DATA. m and n are two given lines. REQUIRED. To find the third proportional to mn and n. SOLUTION Draw AB and AC making any convenient angle. On AB take AD = m, and DE = n. On AC take AF = n. Draw DF. Through E draw EG II to DF. CONCLUSION. FG is the third proportional required. Q. E. F. PROOF. AD: DE = AF: FG. "A line parallel to one side of a triangle divides the other two sides proportionally." ~ 340 Substituting m and n for their equals, m: n = n: FG. Q. E. D. EXERCISES 773 Construct x, if x ba 774 If a:b = b:x, find x in terms of a when = 2 a. 775 Find the third proportional to 7 in. and 9 in. 776 If a:b = b x, compute x when b = Va. ___ CONSTRUCTIONS 187 PROPOSITION XXXVIII. PROBLEM 387 To find the mean proportional between two given straight lines. c, ~ _/' \; \ A D B DATA. m and n are two given lines. REQUIRED. To find the mean proportional between m and n. SOLUTION On any line AX take AD = m, and DB = n. On AB as a diameter describe a semicircumference. Draw a perpendicular to AB at D, intersecting the circumference at C. CONCLUSION. CD is the mean proportional required. QO.O. F. PROOF. AD: CD = CD: DB. "The perpendicular drawn from any point in the circumference of a circle to the diameter is the mean proportional between the segments of the diameter." ~ 368 Substituting m and n for AD and DB, m:CD =CD:n. Q. E. D. 388 DEFINITION. A straight line is divided in extreme and mean ratio, when one of its segments is the mean proportional between the whole line and the other segment. C' A C 1 Thus, the line AB is divided internally at C in extreme and mean ratio, if AB: AC = AC: CB; and externally at C' in extreme and mean ratio, if AB: AC' = AC': C'B. 188 PLANE GEOMETRY -BOOK III PROPOSITION XXXIX. PROBLEM 389 To divide a given line in extreme and mean ratio. / is f I -------- --------— A C B DATA. -AB is a given line. REQUIRED. To divide AB in extreme and mean ratio. SOLUTION Draw OB I to AB, and equal to the half of AB. With center 0, and radius BB, describe a circle. Draw the secant ADOE. On AB, and BA produced, take AC = AD, and AC' = AE. CONCLUSION. AB is divided internally at C, and externally at C', in extreme and mean ratio. Q. E. F. PROOF. AE: AB = AB: AC (= AD) (1). ~ 379 By division, AE - AB: AB = AB - AC: AC. ~ 334 ButAE - AB = AD = AC; and AB - AC = CB..-. AC: AB = CB: AC. By inversion, AB: AC = AC: CB. ~ 332 Again. From (1), AE + AB: AE = AB + AC: AB (2). ~ 333 But AE = AC',.. AE + AB == C'B; and AB + AC = ED + DA = AE = AC'. Substituting these values in (2), C'B: AC' = AC': AB. Or AB: AC' = AC: C'B. ~ 332 Q. E. D. CONSTRUCTIONS 189 PROPOSITION XL. PROBLEM 390 On a given line to construct a polygon similar to a given polygon. D - ' C A --- —------- -B A'l B'H DATA. ABCDE is a given polygon, and A'B' a given line. REQUIRED. To construct on A'B' a polygon similar to ABCDE. SOLUTION Draw the diagonals BE and BD. Construct the / A' = to the / A, and the Z A'B'E' = to the / ABE. ~ 289 Then the A ABE and A'B'E' are similar, "having two A of the one equal respectively to two A of the other." ~ 352 In like manner construct the A E'B'D' and D'B'C' similar to the A EBD and DBC respectively. CONCLUSION. A'B'C'D'E' is the polygon required. Q..F PROOF. The two polygons are composed of the same number of similar triangles, similarly placed. Const. Therefore the polygons are similar. "If two polygons are composed of the same number of triangles, similar each to each, and similarly placed, the polygons are similar." ~ 362 Q. E. D. 190 PLANE GEOMETRY- BOOK III MISCELLANEOUS EXERCISES PROBLEMS OF CONSTRUCTION 777 To divide one side of a triangle into segments proportional to the other two sides. 778 To construct two straight lines, having given their sum and ratio. 779 To trisect a given straight line. 780 To find the mean proportional between the diagonal and the side of a square. 781 ABC is a right triangle, right-angled at C. To find a point P in BC produced so that PC: AC = AC: BC. 782 To inscribe in a circle a triangle similar to a given triangle. [~~ 281, 275.] 783 To circumscribe about a circle a triangle similar to a given triangle. 784 To inscribe a square in a semicircle. [Draw the tangent AC = to AB, and join CO. Corn- plete the rt. 7I DEFG and it is the required square. PROOF. The rt. A ODG and OEF are equal (~ 178).. GO = OF. The A CAO and DGO are similar (~ 351). But CA = 2AO by const..'. DG = 2GO = GF... DGFE \ is a square.] A. o o B 785 To inscribe a square in a given triangle. [Draw AD I to BC, and on AD construct the A_. square ADEF. Join BF intersecting AC in G, / and complete the rt. 7 GHKL. Then in the A BFE and BFA, BF: BG = FE: GL = FA: GH / \ (~ 340). But FE = FA by const... GL = GH. B r L D.-..E.'. GHKL is a square.] 786 In a given triangle, to inscribe a- rectangle similar to a given rectangle. [Follow the construction of the last Ex., except on the altitude AD construct a rectangle similar to the given rectangle.] 787 In a given semicircle to construct a rectangle similar to a given rectangle. [Follow the construction of Ex. 784, except for the tangent AC take the fourth proportional to ~ b, a, and r; b being the base and a the altitude of the given rectangle, and r the radius of the given semicircle.] EXERCISES 191 788 To describe a circle through two given points and tangent to a given line. 789 Having given one of the segments of a line divided in extreme and mean ratio, to construct the line. 790 To construct a triangle similar to a given triangle and having a given perimeter. [Divide the given perimeter into three parts proportional to the three sides of the given triangle (~ 384).] 791 To inscribe a square in a given segment of a circle. [Follow the construction of Ex. 784, except draw AC equal and l to the chord of the segment.] 792 A, D, and B are three points in a straight line. To find a point P in the line, so that A, D, B, and P are harmonic points. [Ex. 747.] 793 To draw a chord in a given circle parallel to a given chord and having a given ratio to it. PROBLEMS IN LOCI 794 Find the locus of the intersection of the diagonals of the parallelogram formed by drawing lines from any point in the base of a triangle parallel to the other two sides. 795 Find the locus of a point from which a given straight line is seen at a given angle. 796 Find the locus of a point such that the sum of the squares of its distances from two fixed points shall be constant. 797 Find the locus of a point which divides all chords of a given circle into segments whose product is equal to a given square. 798 From a given point A, a secant is drawn intersecting the circumference of a circle in B. In AB, or AB produced, a point P is so taken that the ratio of AP to AB is constant. Find the locus of P. 799 From a fixed point E, any secant is drawn to a fixed circle intersecting the circumference in M and N. Tangents drawn at M and N intersect at P. Find the locus of P. 800 Through a fixed point E within a circle, any chord MN is drawn. Tangents drawn at M and N intersect at P. Find the locus of P. BOOK IV AREAS OF POLYGONS DEFINITIONS 391 The unit of surface is a square whose side is a linear unit. 392 The area of a surface is the number of units of surface it contains. 393 Equivalent surfaces are surfaces which have the same area. 394 SCHOLIUM. The terms triangle, rectangle, etc., are frequently used for area of triangle, etc. PROPOSITION I. THEOREM 395 Two rectangles having equal altitudes are to each other as their bases. A. D F, H e m I __ _ __ __ _ __ _ _i! B c F G HYPOTHESIS. The rectangles ABCD and EFGH have equal altitudes, AB and EF. CONCLUSION. ABCD: EFGH = BC: FG. CASE 1. When the bases are commensurable. PROOF Let m, a common measure of the bases, be contained 5 times in BC and 3 times in FG. Then BC: FG = 5:3 (1). 192 AREAS OF POLYGONS 193 At the points of division in the bases draw Is. These Is divide ABCD into 5 and EFGH into 3 rectangles, all equal. ~ 207.'. ABCD:EGFH = 5:3 (2). From (1) and (2), ABCD: EFGH = BC: FG. CASE 2. When the bases are incommensurable. A D E K It B C F G PROOF Let BC and FL be commensurable, LG being less than m, the common measure of BC and FL. Draw LK L to FG. Then EFLK: ABCD = FL: BC. Case 1 If m, the unit of measure, be indefinitely diminished, LG, always less than m, will become less than any assigned quantity..-. FL approaches FG as a limit, and EFLK approaches EFGH as a limit. ~ 263 EFLK EFGH.ABD approaches AFBGHas a limit, X~ A~E~CD ABCD FL FG and - approaches -F as a limit. ~ 268 BC BC But the variable ratios EFLK and are always equal. Case 1 ABCD BC Therefore their limits are equal. ~ 264 That is, EFGH FG ABCD BC Q. E. D. 396 COROLLARY. Two rectangles having equal bases are to each other as their altitudes. 194 PLANE GEOMETRY-BOOK IV PROPOSITION II. THEOREM 397 Two rectangles are to each other as the products of their bases and altitudes. ---------------------—, a R a' Rf aA Q L-.: b b' b HYPOTHESIS. R and R' are two rectangles, b and b' their bases, a and a' their altitudes. CONCLUSION. R: R'= ax b: a'x b'. PROOF Construct the rectangle Q with base b and altitude a'. Then R: Q = a: a' (1). "Rectangles having equal bases are to each other as their altitudes." ~ 396 Also, Q: R' = b: b' (2). "Rectangles having equal altitudes are to each other as their bases." ~ 395 Multiplying (1) and (2), term by term, R:R'= a x b: a' x b'. ~337 Q. E. D. 398 SCHOLIUM. It should be understood that by the product of two lines we mean the product of their numerical measures. EXERCISES 801 What is the ratio of a square yard to a square foot? 802 Find the ratio of a rectangle 10 in. by 24 in. to a rectangle 6 in. by 8 in. 803 Find the ratio of two rectangles which have equal bases, and whose altitudes are 42 in. and 6 in. 804 The ratio of two rectangles having equal altitudes is 8: 5. If the base of the second is 15 feet, find the base of the first. AREAS OF POLYGONS 195 PROPOSITION III. THEOREM 399 The area of a rectangle is equal to the product of its base and altitude. a R t b HYPOTHESIS. R is a rectangle, b its base, and a its altitude. CONCLUSION. The area of R = a x b. PROOF Let U be the unit of surface. Then X =-axb. U 1 x 1 "Two rectangles are to each other as the products of their bases and altitudes." ~ 397 But R = the area of R. ~ 392 U... the area of R = a x b. Q. E. D. 400 SCHOLIUM. If the linear unit is an exact divisor of the base and the altitude of a rectangle, the truth of Prop. III is made evident by the annexed diagram. Thus, if the base contains 5 linear units and the altitude 3, the rectangle can be divided into 3 x 5, or 15, squares, each equal to the unit of surface. EXERCISES 805 The base of a rectangle is 17 in. and its altitude is 27 in. Compute its area. 806 The area of a rectangle is 792 sq. in., and its altitude is 24 in. Compute the base. 196 PLANE GEOMETRY-BOOK IV PROPOSITION IV. THEOREM 401 The area of a parallelogram is equal to the product of its base and altitude. E..A F D ~I ~ ~~~ a B C HYPOTHESIS. ABCD is a L7, b its base, and a its altitude. CONCLUSION. The area of the L7 ABCD = a X b. PROOF Draw BE 11 to CF, meeting DA produced at E. Then EBCF is a rectangle, having the same base and altitude as the L7 ABCD. Polygon EBCD = polygon EBCD (1). Iden. Rt. A EAB = rt. A FDC (2), "having the hypotenuse and a leg in each equal." ~ 178 Subtracting (2) from (1), 7 ABCD c rect. EBCF. Ax. 4 But the area of the rect. = a x b. ~ 399.'. the area of the 7 = a x b. Q. E. D. 402 COROLLARY 1. Parallelograms having equal bases and equal altitudes are equivalent. 403 COROLLARY 2. Parallelograms having equal bases are to each other as their altitudes. 404 COROLLARY 3. Parallelograms having equal altitudes are to each other as their bases. 405 COROLLARY 4. Two parallelograms are to each other as the products of their bases and altitudes. AREA OF POLYGONS 197 PROPOSITION V. THEOREM 406 The area of a triangle is equal to half the product of its base and altitude. AHYPOT. / HYPOTHESIS. ABC is a triangle, b its base, and a its altitude. CONCLUSION. The area of the A ABC = ~ a x b. PROOF Complete the 0 ABCD. The A BC = 1 the D ABCD. ~ 201 But the area of the 0 ABCD = a x b. ~ 401.*. the area of the A ABC = a x b. Q. E. D. 407 COROLLARY 1. Triangles having equal bases and equal altitudes are equivalent. 408 COROLLARY 2. Triangles having equal bases are to each other as their altitudes. 409 COROLLARY 3. Triangles having equal altitudes are to each other as their bases. 410 COROLLARY 4. Two triangles are to each other as the products of their bases and altitudes. EXERCISES 807 The area of a rhombus is equal to half the product of its diagonals. 808 If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angles supplementary, they are equivalent. [~ 407.] 198 PLANE GEOMETRY -BOOK IV PROPOSITION VI. THEOREM 411 The area of a trapezoid is equal to the product of its altitude and half the sum of the bases. HYPOTHESIS. ABCD. CONCLUSION. b and b' are the bases, and a is the altitude of the trapezoid The area of ABCD == a (b + b'). PROOF Draw the diagonal BD. Then the area of the A BDC = I a x b, and the area of the A BAD = ~ a X b'. Adding, the area of ABCD = I a (b + b'). ~ 406 Ax. 1 Q. E. D. 412 COROLLARY. The area of a trapezoid is equal to the product of its altitude and median. ~ 211 413 SCHOLIUM. The area of any polygon may be found by dividing the polygon into triangles. In following method is preferred. Draw the longest diagonal. To this diagonal draw perpendiculars from the remaining vertices. The polygon is thus decomposed into right triangles and trapezoids, whose bases and altitudes are measured. The areas of the practice, however, the I T-1 ------- --- TI I~~~~~~~~~~~~~~~~ I~~~~~~~~~ I right triangles and trapezoids are then computed by the previous theorems. COMPARISON OF POLYGONS 199 PROPOSITION VII. THEOREM 414 Two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. HYPOTHESIS. CONCLUSION. The A ABC and ADE have the Z A common. AABC AB x AC AADE AD x AE PROOF Draw DC. The A ABC and ADC, having the same vertex C and their bases in the same line AB, have the same altitude. AABC AB A ADC AD ( ) "k having equal altitudes are to each other as their bases." ~ 409 Likewise, A ADC AC (2). A ADE AE Multiplying (1) and (2), member by member, ABC AB x AC 3 ADE AD x AE Q. E. D. EXERCISE 809. If, in the above figure, AB = 10, AC = 12, AD = 6, and AE = 5, find the ratio of the triangle ABC to the triangle ADE. 200 PLANE GEOMETRY-BOOK IV PROPOSITION VIII. THEOREM 415 Two similar triangles are to each other as the squares of any two homologous sides. A' A B C B' - ~xC HYPOTHESIS. A ABC and A'B'C' are similar. A ABC AB2 CONCLUSION. ABC -AB A A'B'C' A-B72 PROOF A ABC AB x BC AB BC. 414 A A'B'C' A'B' x B'C' A'B' B'C BC AB But ~349 B'C' A'B ' Substituting this value of BC in (1) B'C ' A ABC AB AB AB2 A 'BC AB'C A'B' A'B~',2 Q. E. D. EXERCISES 810 Two similar triangles are to each other as the squares of any two homologous altitudes. 811 The homologous bases of two similar triangles are as 2: 1. Find the ratio of their areas. 812 Two homologous altitudes of similar triangles are 3 in. and 5 in. Find the ratio of their areas. 813 The areas of two similar triangles are 12 sq. ft. and 27 sq. ft.; the base of the first is 6 ft. Find the homologous base of the second. 814 The sides of a triangle are 3, 5, 6. Find the sides of a similar triangle whose area is 9 times as large. COMPARISON OF POLYGONS 201 PROPOSITION IX. THEOREM 416 Two similar polygons are to each other as the squares of any two homologous sides. E -D A' B C B' C' HYPOTHESIS. ABCDE and A'B'C'D'E' are similar polygons whose areas are S and S'. CONCLUSION. S: S' = AB2: AB2. PROOF By drawing the diagonals AC, AD, A'C', and A'D', the polygons are divided into the same number of similar triangles, similarly placed. ~ 363 Whence Ap f AC Aq /AD> Ar Ap' \A'C'2/V Aq' k\A'D'2V Ar' "Two similar & are to each other as the squares of any two homologous sides." ~ 415 Ap _Aq Ar. Ax. 11 Ap' A q' A r' Whence Ap+-A +Ar A_ p = AB2 ~ WhenceA ' + A q' + A r Ap' AB ~ 336 AP'+ Aq'+A rl -'AB'2 That is, S: S' = AB2: A'B'2. QE. D. 417 COROLLARY. Two similar polygons are to each other as the squares of any two homologous lines. 202 PLANE GEOMETRY- BOOK IV EXERCISES THEOREMS 815 The square on the sum of two lines is equivalent to the sum of the squares on the lines increased by twice the rectangle of the lines. Let AB and BC be two lines, and AC their sum. On AC construct the square ACDE. On AE take AG J D ab b2 = AB. Draw GH II to AC, and BK II to AE. Then ao - AF is a square on AB, FD is a square on BC, BH and a ab GK are rectangles whose dimensions are AB and BC. Therefore the square on AC is equivalent to the sum of the squares on AB and BC increased by twice the rectangle of AB and BC. This theorem is expressed algebraically, thus: (a + b)2 = a2 + 2 ab + b2. 816 The square on the difference of two lines is equivalent to the sum of the squares on the lines diminished by twice the rectangle of the lines. Let AB and BC be two lines, and AC their difference. Construct the squares ABDE, BCFG, ACKH. Produce HK to meet BD in L. Then HD and FL are rectangles whose dimensions are AB and BC. Therefore the square on AC is equivalent to the sum of the squares on AB and BC diminished by twice the rectangle of AB and BC. This theorem is expressed algebraically, thus: (a - b)2 = a2 - 2 ab + b2. D Hr K L B A C F -- 817 The rectangle of the sum and difference of two lines is equivalent to the difference of the squares on the lines. Let AB and AC be two lines. Construct the X D L squares ABDE and ACFG. Prolong GF to K, making FK = AB, and complete the rect. KD. Then rect. CH = rect. HL, and the rect. GL has for its dimen- P H sions AB + AC and AB - AC. Therefore the rect. GL is equivalent to rect. GD + rect. CH, which is the A c B difference of the squares on AB and AC. This theorem is expressed algebraically, thus: (a + b) (a - b) = a2 - b2. PYTHAGOREAN THEOREM 203 PROPOSITION X. THEOREM 418 The square on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two sides. D G E K L H HYPOTHESIS. AH is the square on the hypotenuse, AD and PF are the squares on the legs of the right triangle ABC. CONCLUSION. AH = AD + BF. PROOF Draw BL II to CH, and join BK and CE. ABG and CBD are straight lines. ~ 110 The A ABK and AEC are equal; for AB = AE, being sides of the sq. AD, AK = AC, being sides of the sq. AH, and Z BAK = Z EAC, each being equal to the Z BAC plus a rt. Z. ~ 162 But the A ABK - half the rect. AL, for each has the same base AK, and the same altitude AJ; and the A AEC - half the sq. AD, for each has the same base AE, and the same altitude AB..'. the rect. AL = the sq. AD (1). Ax. 11 In like manner, by joining AF and BH, we may prove that the rect. CL - the sq. BF (2). Adding (1) and (2), AH = AD + BF. Ax. 1 Q. E.D. 204 PLANE GEOMETRY-BOO0K IV 419 COROLLARY. The square on either leg of a right triangle is equivalent to the difference of the squares on the hypotenuse and on the other leg. 420 SCHOLIUM. Prop. X is the most celebrated theorem in Geometry. It is also one of the most important. It was discovered by Pythagoras (about 530 B.C.), and in honor of his illustrious name is called the Pythagorean Theorem. It is also known as the 47th proposition of Euclid, that being its number in the first book of Euclid's Elements. Tradition has it that Pythagoras was so delighted over his great discovery that he sacrificed one hundred oxen to the gods. The proof of Pythagoras has been lost. That in the text is due to Euclid (about 300 B.C.), and is the most elegant demonstration extant. For centuries the theorem has been a favorite with mathematicians; hence many different proofs have appeared, the figures for the best of which are here given as exercises for the curious; they should not be required of the student. The student, however, should seek out an original proof of his own. EXERCISES Prove the theorem of Pythagoras by the following figures. In each, ABC is the right triangle. A few suggestions are added on the construction and the order of proof. [See ~ 420.] 818 The rt. A 1, 2, 3, 4, also the rectangles BD G _ f l F and BG, are equal by const. 3 \ 2 Prove ACHK a square. Then from the whole K figure take away the four rt. A, and the square on \ B AC remains; take away their equivalent, the two rec- N tangles, and the squares on AB and BC remain. 4 NOTE. This proof is very old. A 819 The four rt. A are equal by const. Prove ACHK c N a a a square. \ Then b2 (a c)2 - 4 A ABC =o(a + )2- 2 ac. \ c Now apply Ex. 815. a NOTE. Another proof of great antiquity. a A c B PYTHAGOREAN THEOREM 205 820 Within the square on AC, the A ABC is repeated three times. Then BD is (AB - BC)2. Now apply Ex. -816. NOTE. This elegant proof is given by Bhaskara, a Hindu mathematician, born A.D. 1114. Bhaskara simply draws the figure and says " Look! " taking it for granted that the reader will readily see that b2 = 4 c + (c - a)2. 2 821 A classic demonstration and one of the simplest. Let ACHK be a square. Draw DHF II to BC, and draw the Is AG, CF, KD, KE. Prove that BF and ED are squares on BC and AB, and that the four rt. A are equal. Observe that K AE = BC = BG, whence AB=EG. 822 Complete the rect. BGMD. Produce KA to O, HC to N, join MB, and draw BL I to HK. Prove MBL a straight line. Then sq. BE a / ABMO ~ rect. AL, and sq. BF E- D CBMN - rect. CL. NOTE. A celebrated proof of classic flavor. 823 Let 0 be the intersection of the diagonals of the larger square BE. Through 0 draw XY I to AC, and UV II to AC. Let L, M, N, and P be the mid-points of the sides of the square AH. Draw LS and NQ 11 to BC, MT and PR II to AB. Prove that SQ is a square equal to the square BF, and that the four quadrilaterals into which the square BE is divided are equal to the four quadrilaterals around the square SQ. NOTE. This is one of the most ingenious proofs extant. 824 Construct the A HLK = to the A ABC. Draw DG, EB, BF, and BL. Prove EBF a straight line, then prove that the quadrilaterals EFGD, EFCA, BLKA, and LBCH are equal. A 0 I\ 0 H M 'NP' I, I l T; K X D.Xk K D,'I E, I Xf \,',J I, K —, — II~ A^ / K\ I,~ r NOTE. The construction of this figure is finely conceived. 206 PLANE GEOMETRY-BOOK IV 825 Draw HN II to BC, KO II to AB. Join BO D and produce it to meet HK in M. Prove A KOH = A ABC; and that ABOK and CBOH are 7, whence BM E is II to CH. \ \ F Next prove A ONB = A ABC, whence ON = AB, A c and BN = BC. Then rect. AM C7 ABOK ~ sq. BE,- and rect. CM - 7 CBOH sq. BF. K M 826 Use ~ 379, observing that CD = b + c, and CE = b - c. NOTE. ~~ 421 and 422 are inserted for the curious; the student should not be required to read them. ~ 421 is the rule of Maseres. ~ 422 is original with the author.\ _ 421 PROBLEM. To find three whole numbers which represent the sides of a right triangle. ANALYSIS. Let a, b, and c be the sides of a right triangle, a being the hypotenuse. Then any three numbers which satisfy the equation a2 = b2 + c2 (1) will represent the sides of a right triangle. Let x and y be any two numbers. Then (x2 + y2)2 = x4 + 2 x2y2 + y4, and (X2 - y2)2 = 4_ 2 2y2 + y4. Subtracting, (x2 + y2)2 - (x2 - y2)2 = 4 x2y2 = (2 xy)2. Transposing, (x2 + y2)2 = (X2 _ y2)2 + (2 xy)2 (2). Comparing (1) and (2), we observe that a, b, and c correspond respectively to x2 + y2, x2 - y2, and 2 xy.. when x and y are any two numbers, x2 + y2, x2 - y2, and 2 xy are three numbers which represent the sides of a right triangle. Hence the following SOLUTION. Take any two whole numbers. Then the sum of their squares, the difference of their squares, and twice their product give three numbers which represent the sides of a right triangle. Thus, let x = 3 and y = 2. Then x2 + y2 = 13, x2 - y2 = 5, and 2 xy = 12..~. 13, 5, and 12 represent the sides of a right triangle. Most of the sets of numbers derived by this method are equimultiples of other sets, and hence denote the sides of similar right triangles. This gives rise to the following DISSIMILAR RIGHT TRIANGLES 207 422 PROBLEM. To find sets of whole numbers which represent the sides of dissimilar right triangles. ANALYSIS. Let x and y be any two numbers. Then x2 + y2, x2 - y2, and 2 xy are three numbers which represent the sides of a right triangle. ~ 421 1 x2 + y2, x2 y2, and 2 xy must be relatively prime; for if they contain a common factor, they are equimultiples of another set, and the two triangles are similar. 2 The number 2 xy is always even, for it contains the factor 2. Therefore x2 + y2 and x2 - y2 cannot both be even. 3 If x and y are both odd or both even, x2 + y2 and x2 - y2 are both even. For the square of an odd number is odd, and the sum or difference of two odd numbers is even. Also, the square of an even number is even, and the sum or difference of two even numbers is even. Therefore x and y cannot both be odd or both be even; that is, one must be odd and the other even. 4 If x and y contain a common factor, x2 + y2, x2 - y2, and 2 xy will each contain that factor, which violates (1). Therefore x and y must be relatively prime. We therefore have the following SOLUTION. Take any two numbers relatively prime, the one odd, the other even. Then the sum of their squares, the difference of their squares, and twice their product give three numbers which denote the sides of a right triangle dissimilar to any other right triangle whose sides are represented by numbers relatively prime. We thus find 16, and only 16, sets of whole numbers denoting the sides of 16 dissimilar right triangles in which no number exceeds two digits, as follows: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 x 2 3 4 4 5 5 6 6 7 7 7 8 8 8 9 9 y= 1 2 1 3 2 4 1 5 2 4 6 1 3 5 2 4 These 16 sets with their equimultiples give 50 sets of numbers denoting the sides of 50 right triangles in which no number exceeds two digits. 208 PLANE GEOMETRY —BOOK IV MISCELLANEOUS EXERCISES PROBLEMS OF COMPUTATION 827 To compute the area of an equilateral triangle in terms of its sides. SOLUTION. Let S be the area, a the side, and h the altitude. Then h2 = a2 a2 3 ~ 370 a 4 4 a h.... h= Now S =a x h. ~ 406 2.*. S= x a / a2 2 2 4 828 To compute the area of a triangle in terms of its sides. SOLUTION. S =2 x h. ~ 406 2 / But h = - Vs (s a)(s - b) (s-c). Ex. 673 a a a2.'. S=2 x a s s- a)(s -b )(s-c) = V/s (s - a)(s - b)(s - c). 829 To compute the area of a triangle in terms of its sides and the radius of the circumscribed circle. SOLUTION. Let R denote the radius of the circumscribed circle. Now2 R x h= b x c. ~ 382.'.2Rxaxh=axbxc. But a x h = 2 S. ~ 406..4 R x S = abc.. abc 4R COROLLARY. R = abc. 4S EXERCISES 209 830 To compute the area of a triangle in terms of its sides and the radius of the inscribed circle. SOLUTION. Let r denote the radius of the in- a scribed circle. Then the area of the A BOC = 1 a x r, the area of the A AOC = a b x r, and the area of the A AOB = c x r. ~406 B a 0 Adding, S = (a + b + c) x r = sr. 831 To compute the radius of the circle inscribed in a triangle in terms of the sides. SOLUTION. S = sr. Ex. 830 Whence r = S /s(s- a)(s - b)(s - c) (s -a)(s - b)(s- c). 832 Compute the area of a right triangle whose legs are 16 ft. 833 Compute the area of an isosceles right triangle whose hypotenuse is 22 in. 834 Compute the area of an isosceles right triangle in terms of the legs. 835 Compute the area of an isosceles right triangle in terms of the hypotenuse. 836 Compute the area of a right triangle whose legs are 9 in. and 16 in. 837 Compute the area of an isosceles triangle whose base is 90 in. and whose leg is 53 in. 838 Compute the area of an equilateral triangle whose side is 16 in. [Ex. 827.] 839 Compute the area of a triangle whose sides are 13, 14, 15. 840 The altitude of a triangle is 8 in., and the base is 14 in. Compute the area. 210 PLANE GEOMETRY-BOOK IV 841 The base of a triangle is 16 in., and the area is 144 sq. in. Compute the altitude. 842 The bases of two equivalent triangles are 24 in. and 38 in. Find the ratio of their altitudes. 843 Compute the radii of the inscribed and the circumscribed circles of the triangle whose sides are 13, 14, 15. [First compute the area (Ex. 828), then find r from the formula r = S - s (Ex. 831), and R from the formula R = abc - 4 S (Ex. 829).] 844 The sides of a triangle are 14, 30, 40. Compute S, R, and r. 845 The sides of a triangle are 72, 80, 136. Compute S, R, and r. 846 The sides of a triangle are 66, 78, 120. Compute S, R, and r. 847 The sides of a triangle are 30, 30, 48. Compute S, R, and r. 848 The sides of a triangle are 40, 40, 48. Compute S, R, and r. (In the last five exercises, all answers are whole numbers.) 849 The homologous altitudes of two similar triangles are 7 in. and 14 in. The first triangle is what part of the second? 850 The altitude of a triangle is 7 yd. What is the homologous altitude of a similar triangle 25 times as large? 851 The bases of a trapezoid are 17 ft. and 25 ft., and the altitude is 11 ft. Compute the area. 852 The area of a trapezoid is 315 sq. yd., and the bases are 13 yd. and 17 yd. Compute the altitude. 853 The area of a trapezoid is 700 sq. ft., one base is 40 ft., and the altitude is 20 ft. Find the other base. 854 Find the side of a square equivalent to a trapezoid whose altitude is 7 in. and whose median is 28 in. 855 Find the side of a square equivalent to a rectangle whose dimensions are 12 and 27. 856 Find the side of a square which contains 40 sq. yd. and 1 sq. ft. 857 Find the area of a square whose perimeter is 28 ft. 858 Find the side of a square equivalent to a parallelogram whose base is 52, and whose altitude is 13. 859 Find the side of a square equivalent to a triangle whose base and altitude are 45 and 10 respectively. EXERCISES 211 860 Find the side of a square equivalent to an equilateral triangle whose side is 10 yd. 861 The dimensions of a rectangle are 65 and 72. Find (1) the area; (2) the perimeter; (3) the diagonal. 862 Find the area of a rectangle in which the base is 45 yd. and the diagonal 53 yd. 863 The area of a parallelogram is 180 sq. in. and the base is 15 in. Find the altitude. 864 The base and the altitude of a triangle are 80 and 60. Find the base of an equivalent parallelogram whose altitude is 50. 865 Find the area of a rhombus in which the perimeter is 60 ft. and the altitude 12 ft. 866 Find the area of a rhombus in which the diagonals are 28 and 18. 867 The area of a rhombus is 135 sq. in., and one diagonal is 15 in. Find the other diagonal. 868 Find the area of a rhombus if the perimeter is 40 in. and the shorter diagonal 12 in. 869 The sum of the diagonals of a rhombus is 26, and their ratio 5:8. Find the area. [Denote the diagonals by 5 x and 8 x.] 870 The base and the altitude of a triangle are 15 and 12. Find the perimeter of the equivalent rhombus whose altitude is 9. 871 Two homologous sides of two similar polygons are as 3 to 5; the area of the first is 27. Find the area of the second. 872 Find the legs of a right triangle in which the area is 960 sq. ft., and the legs are as 8 to 15. [Denote the legs by 8 x and 15 x.] 873 Compute the area of an equilateral triangle in terms of its altitude. 874 Compute the area of an equilateral triangle whose altitude is 12 in. 875 Find the ratio of two equilateral triangles, if the side of one is equal to the altitude of the other. 876 Find the side of an equilateral triangle equivalent to a square whose side is 15 in. 877 Find the area of a triangle whose sides are 17, 25, 26. 878 Find the area of a triangle, if the perimeter is 192 yd., and the radius of the inscribed circle is 17.5 yd. [Ex. 830.] 212 PLANE GEOMETRY — BOOK IV 879 A square and a rectangle are each enclosed by a chain 100 ft. long, and the rectangle is 20 ft. wide. Compare their areas. 880 Find the dimensions of a rectangle whose area is 540 sq. in., and whose base and altitude are as 3: 5. 881 Find the area of an isosceles trapezoid in which the bases are 66 ft. and 84 ft., and a leg is 41 ft. 882 Find the area of a trapezoid in which the bases are 20 ft. and 34 ft., the legs 15 ft. and 13 ft. 883 Find the area of a trapezium ABCD in which AB = 7, AD = 15, BC = 13, DC = 11, and BD = 20. 884 Find the side of a square equivalent to a trapezium whose sides in order are 6, 25, 8, 35, and whose shortest diagonal is 29. 885 Two similar polygons together contain 195 sq. ft., and the ratio of similitude is 4: 7. Find the area of each. 886 The ratio of two similar polygons is 1: 4. If the shortest side of the first is 5, find the shortest side of the second. 887 The base of a triangle is 12 ft. Find the length of the line parallel to the base, and which bisects the triangle. 888 The side of a square is a. Find the side of a square m times as large. 889 The side of a square is 10 ft. Find the side of a square 5 times as large. 890 Can the altitude of a rhombus be equal to the base? Compute S, R, and r in the following triangles in which the sides are: 891 12, 55, 65; 15, 28, 41; 19, 20, 37. 892 15, 26, 37; 11, 90, 97; 33, 34, 65. 893 17, 25, 26; 17, 25, 28; 20, 51, 65. 894 17, 28, 39; 25, 38, 51; 29, 52, 75. 895 19, 60, 73; 28, 65, 89; 32, 53, 75. 896 25, 39, 40; 29, 35, 48; 25, 29, 36. 897 26, 51, 55; 29, 60, 85; 24, 35, 53. 898 33, 41, 58; 34, 61, 75; 35, 53, 66. 899 35, 78, 97; 37, 91, 96; 39, 41, 50. 900 39, 62, 85; 41, 50, 73; 48, 85, 91. EXERCISES 213 THEOREMS 901 The square on the diagonal of a square is double the given square. 902 The square on a line is four times the square on half the line. 903 The square of either leg of a right triangle is equal to the product of the sum and difference of the other two sides. 904 If one acute angle of a right triangle is double the other, the square of one leg is three times the square of the other. 905 The square on the hypotenuse of a right triangle is equivalent to four timnes the square on the median to the hypotenuse. 906 The square of the altitude of an equilateral triangle is equal to three times the square of half the side. 907 Three times the square of the side of an equilateral triangle is equal to four times the square of the altitude. 908 If two triangles have the same base and their vertices in a line parallel to the base, they are equivalent. 909 A median divides a triangle into two equivalent triangles. 910 The diagonals of a parallelogram divide it into four equivalent triangles. 911 A parallelogram is bisected by any straight line drawn through the mid-point of a diagonal. 912 In the parallelogram ABCD, E, the mid-point of BC, is joined to A and D. Prove the triangles AEB and DEC equivalent. 913 If squares are constructed on two adjacent sides of a rectangle, the rectangle of their diagonals is twice the given rectangle. 914 If squares are constructed on two adjacent sides of a rectangle, the sum of the squares on their diagonals is equivalent to twice the square on the diagonal of the rectangle. 915 A trapezoid is divided into four triangles by its diagonals, two of which are similar, the other two are equivalent. 916 If the mid-points of the adjacent sides of a quadrilateral are joined, the parallelogram thus formed is equivalent to half the quadrilateral. 917 If through the vertices of a quadrilateral parallels are drawn to the diagonals, the parallelogram thus formed is equivalent to twice the quadrilateral. 214 PLANE GEOMETRY-BOOK IV 918 If a line bisects the bases of a trapezoid, it bisects the trapezoid. 919 If similar polygons are constructed on the sides of a right triangle as homologous sides, the polygon on the hypotenuse is equivalent to the sum of the polygons on the legs. 920 If P is any point in the circumference of a circle whose diameter is AB, the sum of the squares on PA and PB is constant. 921 If two equal circles so intersect that the circumference of each passes through the center of the other, the square on their common chord is equivalent to three times the square on the radius. 922 The sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. 923 If the mid-points of two adjacent sides of a parallelogram are joined, the triangle thus formed is one eighth of the parallelogram. 924 If AB, BC, and CA of the triangle ABC are each produced its own length to B', C', and A' respectively, the triangle A'B'C' is seven times the triangle ABC. B* 925 If the sides of a parallelogram ABCD are produced in order, each its own length, to A', B', C', D', A'B'CID' is a parallelogram five times the parallelogram ABCD. 926 The square inscribed in a sector which is one fourth of a circle is five eighths of the square inscribed in the semicircle. [Let q be a side of the square in the sector and s a side of the square in the semicircle whose radius is / s \ r. Then r2 = 2 q2 (1). Also r2= s2 + 2= 5 (2). 4 4 5 S2 5 From (1) and (2), 2 q2= 5 *.. 8 q2= 5 s2; or q2 = 2. 4 8 927 The square inscribed in a semicircle is two fifths of the square inscribed in the circle. 928 If two equivalent triangles have an angle in each equal, the including sides are inversely proportional. 929 If from the mid-point of the base of a triangle lines are drawn parallel to the other two sides, the parallelogram thus formed is equivalent to half the triangle. EXERCISES 215 NOTE. Exercises 930-939 relate to the annexed figure, repeated from ~ 418. 930 The points E, B, F, are in a straight line. 931 AD and CG are parallel. s 932 CE and BK are perpendicular lines. F 933 BC produced bisects FH at Q, and CQ equals the half of AB. 934 The sum of the perpendiculars dropped from E L and F to AC produced is equal to AC. 935 The sum of the angles HCF and KAE is equal to three right angles. 936 If DG, EK, and FH are joined, each of the triangles DBG, EAK, and FCH is equivalent to the triangle ABC. 2 2 2 2 937 BA + BH BC + BK2. 938 FH = AB + 4BC. 2 2 2 939 FH + EK2 = 5AC. 940 If two equivalent triangles are on opposite sides of the same base, the common base bisects the line joining their vertices. 941 If the medians of a triangle ABC intersect in 0, the triangle BOC is one-third the triangle ABC. 942 Four times the sum of the squares of the medians of any triangle is equivalent to three times the sum of the squares of the sides. 943 In a triangle whose base is b and altitude a, prove that the side of a+b the inscribed square is equal to - a+ b 944 In the annexed figure, ABCD is a square; E, F, A i G, and H are the mid-points of the sides. Prove IJKL a \ L square equivalent to one fifth of the square ABCD. 945 The sum of the squares of the four sides of a quadrilateral is equal to the sum of the squares of the diagonals plus four times the square of the line joining B F o the mid-points of the diagonals. 946 In a quadrilateral the lines which join the mid-points of the opposite sides and the line which joins the mid-points of the diagonals meet in a point of bisection. 216 PLANE GEOMETRY-BOOK IV PROPOSITION XI. PROBLEM 423 To construct a square equivalent to the sum of two given squares. DT4. I an K rRtovs e I iDA. iH o o a —re two g squarei es. DATA. H and K are two given squares. REQUIRED. To construct a square equivalent to H + K. SOLUTION Construct the rt. A ABC whose legs are respectively equal to the sides of the squares H and K. Construct the square R whose side is equal to the hypotenuse AC. CONCLUSION. R is the square required. Q. E. F. 2 2 PROOF. AC AB + BC2. "The square on the hypotenuse of a rt. A is equivalent to the sum of the squares on the other two sides." ~ 418.'. R H + K. Q.. D. EXERCISES 947 Construct a square equivalent to the sum of two squares whose sides are 3 and 4. 948 Construct a square equivalent to the sum of two squares whose sides are 8 and 15. 949 Compute the side of a square whose area is equal to the areas of two squares whose sides are 13 in. and 84 in. 950 Compute the side of a square whose area is equal to the areas of two squares whose sides are 20 in. and 99 in. CONSTRUCTIONS 217 PROPOSITION XII. PROBLEM 424 To construct a square equivalent to the difference of two given squares. A,' B x DATA. H and K are two squares, K being the larger. REQUIRED. To construct a square equivalent to K - H. SOLUTION Construct the rt. A ABC in which the leg AB is equal to the side of H, and the hypotenuse AC is equal to the side of K. Construct the square R whose side is equal to BC. CONCLUSION. R is the square required. Q. E. F. PROOF. BC2 AC2 - AB2. "The square on either leg of a right triangle is equivalent to the difference of the squares on the hypotenuse and on the other leg." ~ 419.. Ra KK-H. Q.E.D. EXERCISES 951 Construct a square equivalent to the difference of two squares whose sides are 12 and 13. 952 Construct a square equivalent to the difference of two squares whose sides are 24 and 25. 953 Compute the side of a square whose area is equal to the difference of the areas of two squares whose sides are 35 and 37. 954 Compute the side of a square whose area is equal to the difference of the areas of two squares whose sides are 91 and 109. 218 PLANE GEOMETRY-BOOK IV PROPOSITION XIII. PROBLEM 425 To construct a square equivalent to the sum of any number of given squares. E a ~,/ \~ DATA. a, are sies of gn squares DATA. a, b, c, d, are sides of given squares. REQUIRED. To construct a square equivalent to the sum of the squares on a, b, c, d. SOLUTION Construct the rt. A ABC with legs a and b. Construct the rt. A ACD with legs AC and c. Construct the rt. A ADE with legs AD and d. CONCLUSION. AE is the side of the required square. Q. E. F. PROOF. AE2 _ A-D + d2 AC + 2 +d2 2a + b2 q c2 2- d2. ^a2+b2+c2+d2 ~ 418 Q. E. D. 426 SCHOLIUM. By this problem we can construct a line equal to the square root of any number. Thus, if a = b = c 1, AD = -V3. EXERCISES 955 Construct a line whose length = x/5 in. Compute the side of a square equivalent to the sum of the squares whose sides are: 956 12, 21, 28. 958 16, 25, 33, 39, 52, 56, 60, 63. 957 7, 24, 60, 72. 959 13, 36, 40, 51, 68, 75, 77, 84. NOTE. The answer to each of the last four exercises is a whole number. CONSTRUCTIONS 219 PROPOSITION XIV. PROBLEM 427 To construct a triangle equivalent to a given polygon. A P C D a DATA. ABCDE is a given polygon. REQUIRED. To construct a triangle to ABCDE. SOLUTION Join any two alternate vertices, as A and C. Draw BF 11 to AC, meeting DC produced at F, and join AF. Again. Join the alternate vertices A and D. Draw EG |1 to AD, meeting CD produced at G, and join AG. CONCLUSION. The A AFG is ^ to the polygon ABCDE. Q. E. F. PROOF. The polygon AFDE has one side less than the polygon ABCDE, but is equivalent to it; for the part ACDE is common, and the A AFC isag to the A ABC, Ai having the samteae be(AC) and the same altitude." ~ 407 Again. The A AFG is - to the polygon AFDE; for the part AFD is common, and the A AGD is ~ to the A AED, "' having the same base (AD) and the same altitude." ~ 407 We now have polygon ABCDE = to polygon AFDE = to A AFG. Q. E. D. 220 PLANE GEOMETRY-BOOK IV PROPOSITION XV. PROBLEM 428 To construct a square equivalent to a given parallelogram. 1A1 /, \1 I / l...... -..........-. b h b DATA. P is a given 27, b its base, and h its altitude. REQUIRED. To construct a square equivalent to the U P. SOLUTION Find x, the mean proportional between h and b. ~ 387 Upon x construct the square R. CONCLUSION. R is the square required. Q. E. F. PROOF. h: x = x: b. Const..'. x2=h X b; ~ 328 that is, R = P. ~ 401 Q. E. D. 429 COROLLARY 1. To construct a square equivalent to a given triangle, take for the side of the square the mean proportional between the base and half the altitude of the triangle. 430 COROLLARY 2. To construct a square equivalent to a given polygon, reduce the polygon to an equivalent triangle, and proceed as in ~ 429. EXERCISES 960 Compute the side of a square equivalent to a parallelogram whose base is 32 and altitude 8. 961 Compute the side of a square equivalent to a triangle whose base is 49 and altitude 32. CONSTRUCTIONS 221 PROPOSITION XVI. PROBLEM 431 To construct a square which shall have a given ratio to a given square. m -A- - x I A ^..... B ---~ —x A B C DATA. S is a given square, m: n a given ratio. REQUIRED. To construct a square which shall be to S as m: n. SOLUTION On the line AX take AB = n, and BC = m. On AC as a diameter, describe a semicircle. Draw BD ~ to AC, meeting the Oce at D, and join DA and DC. On DA take DE = to a side of S, and draw EF II to AC, meeting DC at F. On a line equal to DF, construct the square S'. CONCLUSION. S' is the square required. PROOF. The / ADC is a rt. Z... DC: DA = m: n (1). Now DC: DA = DF: DE.. 2 2 - 2.. DC: DA2 = DF: DE2 (2). From (1) and (2), DF2: DE2= mn: n. That is, S':S = m:n. Q. E. F. ~ 277 ~ 366 ~ 340 ~ 338 Ax. 11 Q. E. D. 222 PLANE GEOMETRY -BOOK IV PROPOSITION XVII. PROBLEM 432 To construct a polygon similar to a polygon and having a given ratio to it. given a m ft Q b DATA. P is a given polygon, m: n a given ratio. REQUIRED. To construct a polygon similar to P, and which shall be to P as 1L: n. SOLUTION Draw a line b, such that b2: a2=m n. ~ 431 b, homologous to a, construct the polygon Q ~ 390 On the line similar to P. CONCLUSION. PROOF. Q is the polygon required. Q: P = b2: a2. But b2: a2 = m: n..'. Q: P=m: n. Q. E. F. ~ 416 Const. Ax. 11 Q. E. D. EXERCISES 962 A side of a given polygon is 6 in. Find the homologous side of a similar polygon with twice the area. 963 To construct a right triangle equivalent to a given triangle. 964 To bisect a triangle by drawing a line through the vertex. 965 To trisect a triangle by drawing lines through the vertex. CONSTRUCTIONS 223 PROPOSITION XVIII. PROBLEM 433 To construct a polygon similar to one of two given polygons and equivalent to the other.* a S b DATA. P and Q are two given polygons, a being a side of P. REQUIRED. To construct a polygon similar to P and equivalent to Q. SOLUTION Find m and n, the sides of two squares respectively equivalent to P and Q. ~ 430 Find b, the fourth proportional to m, n, and a. ~ 385 On b, homologous to a, construct the polygon S similar to P. ~ 390 CONCLUSION. S is the polygon required. Q. E. F. PROOF. m: n = a: b. Const... 2: n=a2: b2. ~ 338 But m2 - P, and n2 - Q. Const.... P:Q=a: b2. But P: S a2: b2..'.P:S=P:Q..'. S Q. ~ 416 Ax. 11 Q E. D. EXERCISE 966. On a given base to construct a triangle equivalent to a given triangle. * The solution of this problem is attributed to Pythagoras. 224 PLANE GEOMETRY -BOOK IV MISCELLANEOUS EXERCISES PROBLEMS OF CONSTRUCTION 967 To divide a triangle into two parts in the ratio of 2 to 3. 968 On a given base to construct a right triangle equivalent to a given triangle. 969 On a given base to construct an isosceles triangle equivalent to a given triangle. 970 To construct an equilateral triangle equivalent to a given triangle. 971 To construct a triangle equivalent to the sum of two given triangles. 972 To construct a triangle equivalent to the difference of two given triangles. 973 To find a point within a triangle from which lines drawn to the vertices divide the triangle into three equivalent triangles. 974 To construct a square equivalent to three fifths of a given square. 975 On a given base to construct a parallelogram equivalent to a given triangle. 976 On a given base to construct a parallelogram equivalent to a given parallelogram. 977 To construct an equilateral triangle equivalent to a given square. 978 To transform a trapezoid into an equivalent isosceles trapezoid. 979 To divide a triangle into two equivalent parts A by drawing a line through a given point in a side. [Let P be the point in AB of the A ABC. Draw the median P AM. Join PM and draw AD II to PM. Then PD is the line required.] B DO - 980 To divide a triangle into two equivalent parts by drawing a line parallel to the base. [On the alti- A tude AH as a diagonal construct a square. Take AE equal to a side of this square. Then DEG II to BC divides the A ABC as required.] D 981 To divide a triangle into two equivalent parts H by drawing a line perpendicular to the base. 982 To bisect a quadrilateral by drawing a line through one of its vertices. 983 To bisect a quadrilateral by drawing a line through a given point in one of its sides. BOOK V REGULAR POLYGONS MEASUREMENT OF THE CIRCLE 434 DEFINITION. A regular polygon is a polygon which is equilateral and equiangular. The square and the equilateral triangle are regular polygons. PROPOSITION I. THEOREM 435 An equilateral polygon inscribed in a circle is a regular polygon. A HYPOTHESIS. ABCDE is an equilateral polygon inscribed in a circle. CONCLUSION. ABCDE is a regular polygon. PROOF The arcs AB, BC, CD, etc., are equal. ~ 243.. the arc BCDE = the arc CDEA. Ax. 1.. /A =Z B. ~276 Likewise B = / C= D =ZE. That is, the polygon ABCDE is equiangular, and therefore regular. ~ 434 Q. B. D. 225 226 PLANE GEOMETRY BOOK V PROPOSITION II. THEOREM 436 A circle may be circumscribed about, or inscribed in, any regular polygon. A. A's H -'B HYPOTHESIS. ABCDE is a regular polygon. CONCLUSION. 1 A circle may be circumscribed about ABCDE. PROOF Describe a circumference through A, B, and C, and from its center 0 draw the radii OA, OB, OC. Then Z ABC = / BCD, ~ 434 and Z OBC = Z OCB. ~ 173 Subtracting, Z OBA = Z OCD. Ax. 4 The A OAB and OCD are equal. ~ 162.OA = OD. ~166 That is, the Oce described through A, B, and C passes through D. In the same manner it may be proved that the Oce through B, C, and D also passes through E, and so on for a regular polygon of any number of sides. ~ 236.'.A circle can be circumscribed about ABCDE. 2 A circle may be inscribed in ABCDE. PROOF Draw OH lJ to AB. Then, since the sides of the inscribed polygon are equi REGULAR POLYGONS 227 distant from the center of the circle (~ 248), a circle described from center 0 and with radius OH will be inscribed in the polygon. ~ 236 Q. E. D. DEFINITIONS 437 The center of a regular polygon is the common center of the inscribed and circumscribed circles. 438 The radius of a regular polygon is the radius of the circumscribed circle. 439 The apothem of a regular polygon is the radius of the inscribed circle. 440 The angle at the center of a regular polygon is the angle formed by radii drawn to the end-points of any side. 441 COROLLARY 1. Radii of a regular polygon bisect the angles of the polygon. 442 COROLLARY 2. The angles at the center of a regular polygon are equal, and each is equal to four right angles divided by the number of sides of the polygon. 443 COROLLARY 3. If a circle is circumn- A scribed about a regular polygon, tangents drawn parallel to the sides of the poly- BI a gon form a regular circumscribed polygon o —... whose sides intersect on the radii of the inscribed polygon produced. PROOF. The polygon A'B'C'D'E' is equi- c' H D angular (~ 129). OH is l to C'D' (~ 251), to CD (~ 117), and bisects CD at M (~ 245). Likewise ON bisects DE..-. DM = DN (Ax. 9)..-. DO bisects the L HOK (~ 193). Also D'H = D'K (~ 254)... D'O bisects the Z HOK (~ 193). That is, D' is in OD produced. Again, A OC'D', OD'E', etc., are equal, for they are similar to the equal A OCD, ODE, etc. (~ 351) and have equal altitudes... C'D' = D'E', etc..'. the polygon A'B'C'D'E' is regular (~ 434). 228 PLANE GEOMETRY-BOOK V PROPOSITION III. THEOREM 444 If the circumference of a circle is divided into any number of equal arcs, their chords form a regular inscribed polygon; and the tangents drawn at the points of division form a regular circumscribed polygon. K D Ht EN O YPOTHESIS. The arcs ABBC, C, E, EA, are equal; and FBG, GCH HYPOTHESIS. The arcs AB, BC, CD, DE, EA, are equal; and FBG, GCH, HDK, KEL, LAF, are tangents. CONCLUSION. 1 ABCDE is a regular polygon. PROOF The polygon ABCDE is equilateral. ~ 243.'. it is regular. ~ 435 2 FGHKL is a regular polygon. PROOF The A FAB, GBC, etc., are equal isosceles A. ~~ 281, 167.-. Z F = Z G = Z H = Z K = L, ~ 166 and FG = GH = HK = KL = LF. Ax. 1.'. FGHKL is a regular polygon. ~ 434 Q. E. D. 445 COROLLARY 1. The perimeter of an inscribed polygon is less than the perimeter of an inscribed polygon of double the number of sides. 446 COROLLARY 2. The perimeter of a circumscribed polygon is greater than the perimeter of a circumscribed polygon of double the number of sides. REGULAR POLYGONS 229 PROPOSITION IV. THEOREM 447 Two regular polygons of the same number of sides are similar. Di D /D ECCT P A B l —A B' ---HYPOTHESIS. P and P' are regular polygons, each having n sides. CONCLUSION. P and P' are similar. PROOF Each angle of either polygon = n ~ rt. zs. ~ 216 n That is, the polygons are mutually equiangular. Again, AB = BC = CD, etc., and A'B' = B'C' = C'D', etc. ~ 434 Whence AB: A'B' = BC: B'C' = CD: C'D', etc. That is, the polygons have their homologous sides proportional.. the polygons are similar. ~ 349 Q. E. D. EXERCISES 984 An angle of a regular polygon is the supplement of the angle at the center. 985 If each vertex of a regular inscribed polygon is joined to the midpoints of the adjacent arcs, a regular polygon of double the number of sides will be inscribed. 986 The tangents drawn through the mid-points of the minor arcs intercepted by the sides of a regular circumscribed polygon form a regular circumscribed polygon of double the number of sides. 230 PLANE GEOMETRY-BOOK V PROPOSITION V. THEOREM 448 The perimeters of two regular polygons of the same number of sides are to each other as their radii or as their apothems. Pt D E( \ E E of A B ' H' HYPOTHESIS. P and P' are the perimeters, O and 0' the centers, OA and O'A' the radii, OH and O'H' the apothems, of the two regular polygons. CONCLUSION. P: P' = OA: O'A' = OH: O'H'. PROOF Draw the radii OB and O'B'. Then Z AOB = Z A'O'B', ~ 442 and OA: OB = O'A': O'B'..'. the A OAB and O'A'B' are similar. "If two triangles have an angle of the one equal to an angle of the other, and the including sides proportional, they are similar." ~ 354.. AB: A'B'= OA: O'A' ~ 349 = OH: O'H'. " In two similar triangles, homologous altitudes have the same ratio as any two homologous sides." ~ 359 But P: P'= AB: A'B'. ~ 361.'. P: P' = OA: O'A' = OH: O'H'. Ax. 11 Q. E. D. 449 COROLLARY. The areas of two regular polygons of the same number of sides are to each other as the squares of their radii or as the squares of their apothems. ~ 417 REGULAR POLYGONS 231 PROPOSITION VI. THEOREM 450 If the number of sides of a regular inscribed polygon is doubled indefinitely, the apothem of the polygon approaches the radius of the circle as a limit. HYPOTHESIS. AB is a side, OA the radius, and OH the apothem of a regular inscribed polygon of n sides. CONCLUSION. As n is doubled indefinitely, OH approaches OA as a limit. PROOF OA-OH < AH. ~ 161.'. OA -OH < AB. By doubling n indefinitely, AB can be made less than any assigned quantity... OA- OH can be made less than any assigned quantity..'. OH approaches OA as a limit. ~ 263 Q. E. D. 451 COROLLARY. If the number of sides of a regular inscribed polygon is doubled indefinitely, the square of the apothem approaches the square of the radius of the circle as a limit. For by the theorem OH approaches OA as a limit. Therefore OH2 approaches OA2 as a limit (~ 271). 232 PLANE GEOMETRY-BOOK V PROPOSITION VII. THEOREM 452 The circumference of a circle is less than any enveloping line. D A C HYPOTHESIS. MNO is the circumference of a circle, and ABC any enveloping line. CONCLUSION. MNO is less than ABC. PROOF Of all the lines inclosing the circle, there must be at least one shortest line. Draw the tangent DE. Then DE is less than DBE. Ax. 15.'. ADEC is less than ABC..-. ABC is not the shortest line. In like manner we may show that no other enveloping line is the shortest.. MNO is the shortest. Q. E.D. 453 COROLLARY. The circumference of a circle is less than the perimeter of any circumscribed polygon. EXERCISES 987 The area of a circle is greater than the area of any inscribed polygon. 988 The area of a circle is less than the area of any circumscribed polygon. REGULAR POLYGONS 233 PROPOSITION VIII. THEOREM 454 The circumference of a circle is greater than the perimeter of any inscribed polygon. A HYPOTHESIS. ABC... is a polygon of n sides inscribed in a circle whose circumference is C. Denote the perimeter of the polygon by P. CONCLUSION. C is greater than P. PROOF The sides of the polygon divide the circumference into n arcs. Each arc is greater than the chord which subtends it. Ax. 15.-. C is greater than P. Ax. 3 Q. E. D. EXERCISES 989 The square circumscribed about a circle is twice the square inscribed in the circle. 990 The square inscribed in a circle is twice the square described on the radius. 991 The side of an inscribed equilateral triangle is half the side of the circumscribed equilateral triangle. 992 The area of an inscribed equilateral triangle is one fourth the area of the circumscribed equilateral triangle. 234 PLANE GEOMETRY-BOOK V PROPOSITION IX. THEOREM 455 The circumference of a circle is the limit which the perimeters of regular inscribed and circumscribed polygons approach, if the number of their sides is doubled indefinitely; and the area of the circle is the limit which the areas of these polygons approach. HYPOTHESIS. C is the circumference of a circle whose area i S; P and P' are the perimeters, R and R' the apothems, A and A' the areas, of two similar polygons respectively inscribed in and circumscribed about the circle. CONCLUSION. 1 C is the limit of P and P', if the number of sides of the polygons is doubled indefinitely. PROOF P': P = R': R. ~ 48 By division, P'- P: P' = R'- R: R'. ~ 334 Whence P' - P = p(R' ~ 328 By doubling the number of sides of the polygons indefinitely, R'- R can be made less than any assigned quantity. ~ 450 R'-R Therefore, - can be made less than any assigned quanty. 266 tity. ~ 266 REGULAR POLYGONS 235 Therefore, P'(R R) can be made less than any assigned quantity. ~ 265 Therefore, P'- P can be made less than any assigned quantity. But C is always greater than P and less than P'. ~~ 454, 453 Therefore, C- P and P' -C can each be made less than any assigned quantity. That is, C is the limit of P and P'. ~ 263 2 S is the limit of A and A'. PROOF A': A = R2: R2. ~ 449 By division, A'-A: A' = R'2 - R2: R'2. ~ 334 Whence A' — A A' — ) ~ 328 By doubling the number of sides of the polygons indefinitely, R'2 - R can be made less than any assigned quantity. ~ 451 tR'2_ R[2 Therefore, R can be made less than any assigned R't quantity. ~ 266 Therefore, A'(R ) can be made less than any assigned quantity. ~ 265 Therefore, A'-A can be made less than any assigned quantity. But S is always greater than A and less than A'. Ax. 12 Therefore, S - A and A' - S can each be made less than any assigned quantity. That is, S is the limit of A and A'. ~ 263 Q. E. D. 236 PLANE GEOMETRY -BOOK V PROPOSITION X. THEOREM circumferences have the same ratio as 456 Two their radii. HYPOTHESIS. C and C' are two circumferences whose radii are R and R'. CONCLUSION. C: C' = R: R'. PROOF Inscribe in the circles regular polygons of n sides each, and denote their perimeters by P and P'. Then P:P'= R: R'. ~ 448 By alternation, P: R = P': R'; P P' or - -. R ' ~ 331 If n is doubled indefinitely, P and P' will approach C and C' as their respective limits. ~ 455.. approache s a limit, iR R P' C' and - approaches - as a limit. But P P R R' C C' R R" or C C=R:R'. ~ 268 Proved ~ 264 Q. E. D. REGULAR POLYGONS 237 457 COROLLARY 1. Two circumferences have the same ratio as their diameters. 458 COROLLARY 2. The ratio of the circumference of a circle to the diameter is constant. PROOF. C:C' =2 R:2 R' (~457)..'. C:2R=C':2R' (~ 331). 459 DEFINITION. The ratio of the circumference of a circle to the diameter is designated by the Greek letter 7r (pi). The value of xr, as shown in ~ 483, is 3.1416 nearly. 460 COROLLARY. C - 2 R = 7r..'. C=2 7rR. PROPOSITION XI. THEOREM 461 The area of a regular polygon is equal to half the product of its apothem and perimeter. b HYPOTHESIS. Q is a regular polygon of n sides, R the apothem, P the perimeter, S the area, and b a side. CONCLUSION. S = R X P. PROOF By drawing all the radii, the polygon is divided into n equal triangles. The altitude of each triangle is R, and the area of each triangle is R x b (~ 406)..-. S = ( R x b)n = 2R (b X n) = R x P. Ax. 13 Q. E. D. 462 DEFINITION. Similar arcs, sectors, and segments are those having equal central angles. 238 PLANE GEOMETRY-BOOK V PROPOSITION XII. THEOREM 463 The area of a circle is equal to half the product of its radius and circumference. HYPOTHESIS. R is the radius of a circle, C the circumference, and S the area. CONCLUSION. S = R X C. PROOF Circumscribe about the circle a regular polygon of n sides, and denote its perimeter by P, and its area by S'. Then S' = R x P. ~461 If n is doubled indefinitely, P approaches C as a limit, ~ 455 and S' approaches S as a limit. ~ 455.*. R x P approaches I R x C as a limit. ~ 267 But S' = R x P. Proved.. S = R x C. ~264 Q. E. D. 464 COROLLARY 1. The area of a circle is 7rR2. PROOF. S = 2 R C= R x 2 rR = 7rR2. 465 COROLLARY 2. The areas of two circles are to each other as the squares of their radii, or as the squares of their diameters. PROOF. S: S' = 7TR2: 7rR12 = R2: R'2 = D2: D2. 466 COROLLARY 3. The area of a sector is equal to half the product of its radius and arc. 467 COROLLARY 4. Similar arcs are to each other as their radii, and similar sectors are to each other as the squares of their radii. CONSTRUCTIONS 239 PROPOSITION XIII. PROBLEM 468 To inscribe a square in a given circle. A - D DATA. O is the center of a given circle. REQUIRED. To inscribe a square in the circle 0. SOLUTION Draw the I diameters AC and BD. Draw AB, BC, CD, and AD. CONCLUSION. ABCD is the square required. Q. E. F. PROOF. The Zs A, B, C, and D are right /s, for each is inscribed in a semicircle; ~ 277 and the sides AB, BC, CD, and AD are equal, for each subtends a quadrant. ~ 243 Therefore, ABCD is a square. ~ 196 Q. E. D. 469 COROLLARY 1. A regular polygon of eight sides can be inscribed in a circle by bisecting the quadrants; and by continuing the process of bisection, regular polygons of 16, 32, 64, 128, etc., sides can be inscribed. 470 COROLLARY 2. A square can be circumscribed about a circle by drawing tangents through the vertices of the inscribed square. ~ 444 240 PLANE GEOMETRY-BOOK V PROPOSITION XIV. PROBLEM 471 To inscribe a regular hexagon in a given circle. 0 A B DATA. O is the center of a given circle. REQUIRED. To inscribe a regular hexagon in the circle 0. SOLUTION Draw a chord AB equal to the radius. CONCLUSION. AB is a side of the hexagon required. Q. E. F. PROOF. Draw OA and OB. The A OAB is equilateral. Const..-. the Z 0 = 60~. ~ 157. the arc AB is one sixth of the Oce. Hence if the chord AB is applied six times, it will divide the Oce into six equal arcs. ~ 243. AB is the side of a regular inscribed hexagon. Q. E. D. 472 COROLLARY 1. An equilateral triangle can be inscribed in a circle by joining the alternate vertices of the regular inscribed hexagon. 473 COROLLARY 2. A regular polygon of twelve sides can be inscribed in a circle by bisecting the arcs of a regular inscribed hexagon; and by continuing the process of bisection, regular polygons of 24, 48, 96, etc., sides can be inscribed. EXERCISES 241 EXERCISES Let R denote the radius, r the apothem, and a the side of a regular polygon. 993 In an inscribed square a = R x/2, r = I R /2, and the area of the square = 2 R2. 994 In an inscribed equilateral triangle a = R \/3, and r = R. [Let ABC be an inscribed equilateral A, and let the radius OD be I to BC at H. Then (1) in the rhombus OBDC, OH = I OD= 1 R... r =1 R. 2a _ a2 2 2 2 2 2 2 (2) B = OB - OH...4BH =40B -40H; orBC =40B 2 a= R2. OD = 3 R2; or a2 = 3 R2. a=R.] 995 The altitude of an inscribed equilateral triangle is equal to I R. 996 The area of an inscribed equilateral triangle is equal to - R2 /3. [Exs. 994, 995.] 997 In an inscribed regular hexagon a = R, and r = ~ R /3. 998 The area of an inscribed regular hexagon is equal to 2 R2 x/. [By drawing all the radii, the hexagon is divided into six equilateral A; and by Ex. 827, the area of each A is I R2 v/3.] 999 If the radius of a circle is 5 feet, compute the side, the apothem, and the area of the inscribed square. 1000 Compute the radius of the circle circumscribed about a square whose side is 16 inches. 1001 If the radius of a circle is 6 feet, compute the side, the apothem, and the area of the inscribed equilateral triangle. 1002 The side of an inscribed equilateral triangle is 10 inches; compute the radius of the circle. 1003 Find the perimeter of a regular hexagon inscribed in a circle whose radius is 9 inches. 1004 Find the apothem of a regular hexagon inscribed in a circle whose radius is 20 inches. 1005 Find the area of a regular hexagon inscribed in a circle whose radius is 10 inches. 1006 Find the radius of the circle circumscribed about a regular hexagon whose apothem is 7 feet. 1007 Find the radius of the circle circumscribed about a regular hexagon whose perimeter is 72 feet. 242 PLANE GEOMETRY-BOOK V PROPOSITION XV. PROBLEM 474 To inscribe a regular decagon in a given circle. B' DATA. 0 is the center of a given circle. REQUIRED. To inscribe a regular decagon in the circle O. SOLUTION Draw any radius AO, divide it internally at C in extreme and mean ratio, ~ 389 and draw a chord AB equal to OC the greater segment. CONCLUSION. AB is the side of the decagon required. Q. E. F. PROOF. Draw BO and BC. By construction, AO: OC = OC: AC, and AB = OC..'. AO: AB = AB: AC. That is, the A OAB and BAC have the Z A common and the including sides proportional..*. the A OAB and BAC are similar. ' If two A have an angle of the one equal to an angle of the other, and the including sides proportional, they are similar." ~ 354 But the A OAB is isosceles. ~ 221. the A BAC is isosceles, and AB = BC = OC. CONSTRUCTIONS 243.-. the A COB is isosceles, and Z 0 = Z n. ~ 173 Whence ext. L m = 2 Z 0. ~158 But Z m = OAB = OBA. ~ 173.. OAB = 2 O, and / OBA = 2 / 0. Also, 0 = Z O. Adding, L OAB + / OBA + Z = 5 Z O =2 rt. As... 0 = 1 of 2 rt. A, or 1 of 4 rt. A..'. the arc AB is 1 of the circumference. Hence, if the chord AB is applied ten times, it will divide the circumference into ten equal arcs. ~ 243. AB is the side of an inscribed regular decagon. Q. E. D. 475 COROLLARY 1. A regular pentagon can be inscribed in a circle by joining the alternate vertices of an inscribed regular decagon. 476 COROLLARY 2. A regular polygon of twenty sides can be inscribed in a circle by bisecting the arcs of an inscribed regular decagon; and by continuing the process of bisection, regular polygons of 40, 80, 160, etc., sides can be inscribed. EXERCISES 1008 In an inscribed regular decagon a = 2 R (/5- 1), and r = R V/10+ 2V. 1009 In an inscribed regular octagon a = R V2 - V2, r = ~ R/2 + /V2, and the area = 2 R2 /2. 1010 In an inscribed regular dodecagon a = R /2- V3, r = R /2 + /3, and the area = 3 R2. 1011 The area of a regular dodecagon is 75 sq. in. Find the radius. 244 PLANE GEOMETRY-BOOK V PROPOSITION XVI. PROBLEM 477 To inscribe a regular pentadecagon in a given circle. c DATA. O is the center of a given circle. REQUIRED. To inscribe a regular pentadecagon in the circle O. SOLUTION Draw a chord AB equal to the radius of the circle. Draw the chord AC equal to the side of the inscribed regular decagon. ~ 474 CONCLUSION. The chord CB is a side of the pentadecagon required. Q. E. F. PROOF. The arc AB is - of the circumference, ~ 471 and the arc AC is y1 of the circumference. Const..'. the arc CB is - 1, or y1- of the circumference... CB is the side of an inscribed regular pentadecagon. ~ 243 Q. E. D. 478 COROLLARY. A regular polygon of thirty sides can be inscribed in a circle by bisecting the arcs of an inscribed regular pentadecagon; and by continuing the process of bisection, regular polygons of 60, 120, 240, etc., sides can be inscribed. NOTE. We have now seen how to inscribe regular polygons of 3, 6, 12, etc.; 4, 8, 16, etc.; 5, 10, 20, etc.; 15, 30, 60, etc., sides. All these constructions were known in Euclid's time. The construction of no other regular polygon, with ruler and compasses, was considered possible until 1796, when Gauss of Germany discovered a method of constructing a regular polygon of 17 sides, and in general any regular polygon of 2" + 1 sides, n being an integer, and 2n + 1 a prime number. PROBLEMS OF COMPUTATION 245 PROPOSITION XVII. PROBLEM 479 Given the side and the radius of an inscribed regular polygon, to find the side of a similar circumscribed polygon. a F,r D DATA. AB is the side of a regular polygon inscribed in a circle 0, whose radius is R. REQUIRED. To find CD, a side of a similar circumscribed polygon. SOLUTION Let CD be drawn through F, the mid-point of the arc AB. ~ 443 Draw OC, OF, and OD. Then OAC and OBD are straight lines. ~ 443 AB and CD are II, for each is I to OF. ~~ 247, 251.'. the A OAB and OCD are similar. ~ 351.'. CD: AB = OF: OE. ~ 349 GOFxAB RxAB Whence CD =OF = E (1). ~ 328 OE OE In the rt. A OEA, OE2 = OA2 - EA2; ~ 370 or OE = VR2 - 4 AB2 =1 V4 R2- AB2. Substituting this value of OE in (1), 2Rx AB CD = - Q.E.F. \/4 R2 - AB' 480 COROLLARY. If the diameter = 1, R= 1 and CD= AB 2 -\ — -- 2 l - AB2 246 PLANE GEOMETRY -BOOK V PROPOSITION XVIII. PROBLEM 481 Given the side and the radius of an inscribed regular polygon, to find the side of an inscribed regular polygon of double the number of sides. C A \B DATA. AB is the side of a regular polygon inscribed in the circle 0, whose radius is R. REQUIRED. To find AC, the side of an inscribed regular polygon of double the number of sides. SOLUTION Draw OA and OC. OC is _ to AB at its mid-point E. ~ 247 = 2 - In the AAOC, A - = AO + CO-2 CO x EO; ~ 373 or AC2 =2R2- 2 R xEO (1). In the rt. A AEO, EO2 = AO2 - -AE2; ~ 370 or EO = R2 - AB2 = 4 R - AB2. Substituting this value of EO in (1), AC2 = 2 R2 - 2 R (-V/4R2 - AB) = 2 R2 - R /4 R2 - AB-.-. AC = 2 R2 - R 4 R2 - AB2. Q. E. F. 482 COROLLARY. If the diameter = 1, R =, and AC = /1 - V1 I AB2 2 PROBLEMS OF COMPUTATION 247 PROPOSITION XIX. PROBLEM 483 To compute the ratio of the circumference of a circle to the diameter. DATA. Let C be the circumference of a circle whose diameter is unity; whence C = 7r (0 460). REQUIRED. To compute the value of 7r. SOLUTION Since the diameter of the circle is 1, the radius is 1, the side of the inscribed square is V /2 (~ 369), and its perimeter is 4(2 V2) = 2.8284271, the side of the circumscribed square is 1, and its perimeter is 4. By using the formulas of ~ 482 and ~ 480, we compute the perimeters of inscribed and circumscribed regular polygons of double the number of sides, and by continuing the process we derive the following values: Number Perimeter of Perimeter of Number Perimeter of Perimeter of of Inscribed Circumscribed of Inscribed Circumscribed Sides. Polygon. Polygon. Sides. Polygon. Polygon. 4 2.8284271 4.0000000 256 3.1415138 3.1417504 8 3.0614675 3.3137085 512 3.1415729 3.1416321 16 3.1214452 3.1825979 1024 3.1415877 3.1416025 32 3.1365485 3.1517249 2048 3.1415914 3.1415951 64 3.1403312 3.1441184 4096 3.1415923 3.1415933 128 3.1412773 3.1422236 8192 3.1415926 3.1415928 From the last two numbers we see that the circumference of the circle whose diameter is unity is greater than 3.1415926 and less than 3.1415928 (~ 455); and since C = r when the diameter = 1, therefore r = 3.1415927 which is correct to the seventh decimal place. Q. E. F. 248 PLANE GEOMETRY-BOOK V For practical purposes we take 7r = 3.1416. 484 HISTORICAL NOTE. The quadrature of the circle, or " to square the circle," is the most celebrated problem in elementary geometry. It means to find the side of a square whose area is equal to the area of a given circle; or, which means the same thing, though it sounds different, to find the numerical value of zr. In 1761 Lambert proved 7r incommensurable. In 1882 Lindemann proved r transcendental, and that the quadrature of the circle, by the use of the straight edge and compasses alone, is impossible. The oldest document known on the subject is the Rhind Papyrus written by Ahmes, an Egyptian priest, probably about 2000 B.C. Ahmes' method of finding the area of a circle consists in squaring eight ninths of the diameter, which makes 7r = 3.1604. Archimedes, born 287 B.C., the most illustrious mathematician of antiquity, by methods similar to those of the text, showed that the value of 7r lies between 31 and 31, or between 3.1428 and 3.1408. Ptolemy, the celebrated astronomer of Alexandria, made or = 3.1417. 355 Metius of Holland, A.D. 1640, found 3 =-, which gives 133' the correct value to 6 decimal places. Soon afterward Romanus and Ludolph van Ceulen, both of Holland, extended the value of 7r, the former to 15, the latter to 35 decimal places. In recent years, by the aid of higher mathematics, Vega computed the value of 7r to 140 decimal places. Clausen and Dan, independent of each other, extended this value to 200 places, their results agreeing exactly. Richter carried the number to 500, and Shanks to 707 decimal places. The following is the correct value of vr to 30 decimal places. 7r = 3.141,592,653,589,793,238,462,643,383,279. EXERCISES 249 485 The following values should be used in numerical computations: 7= 3.1416 V2= 1.4142 -/6 = 2.4495 1.3183 -/3 = 1.7321 -7 = 2.6458 7r -V7 = 1.7725 5 = 2.2361 V/0 = 3.1623 MISCELLANEOUS EXERCISES PROBLEMS OF COMPUTATION 1012 How many degrees in an angle of the following regular polygons: (1) triangle; (2) pentagon; (3) hexagon; (4) octagon; (5) decagon; (6) dodecagon; (7) pentadecagon? 1013 How many degrees in the angle at the center of each of the above polygons? 1014 How many sides has a regular polygon whose angle at the center is 12~? 1015 How many sides has a regular polygon whose angle is 165~ 36'? 1016 If two non-adjacent sides of a regular pentagon are produced to meet, find their included angle. 1017 Find the circumference of a circle whose diameter is 100 feet. 1018 Find the circumference of a circle whose radius is 11 inches. 1019 The circumference of a circle is 50.2656 inches. Find the radius. 1020 In a circle whose radius is 30 in., find the length of the arc subtended by a side of the following inscribed regular polygons: (1) triangle; (2) square; (3) pentagon; (4) hexagon; (5) octagon; (6) decagon. 1021 Find the area of a circle whose radius is 20 feet. 1022 Find the radius of a circle whose area is 25 sq. in. 1023 Find the area of a circle whose circumference is 18 ft. 1024 Find the circumference of a circle whose area is 100 sq. in. 1025 The radius of a circle is 15 in. Find the radius of a circle twice as large; one half as large. 1026 Find the area of a circle inscribed in a square whose side is 40 feet. 250 PLANE GEOMETRY-BOOK V 1027 Find the area of a circle circumscribed about a square whose side is 60 feet. 1028 Find the side of an equilateral triangle circumscribed about a circle whose radius is R. 1029 The radius of a circle is 8. Find the side of the circumscribed equilateral triangle. 1030 The side of the circumscribed equilateral triangle is 12. Find the radius of the circle. 1031 Find the side of a square equivalent to a circle whose radius is 23 inches. 1032 The side of a circumscribed regular hexagon is - R /3. 1033 Find the perimeter of a regular hexagon circumscribed about a circle whose radius is 15 inches. 1034 Find the area of a regular hexagon circumscribed about a circle whose radius is 15 inches. 1035 How many revolutions does a bicycle wheel, 28 in. in diameter, make in going one mile? 1036 A chord is 60 in. long, and the subtended arc is 25 in. high. Find the diameter of the circle. 1037 Find the radius of a circle equivalent to two circles whose radii are 6 ft. and 8 ft. 1038 Find the area of the ring between two concentric circumferences whose radii are 7 in. and 9 in. [Area of ring = rR2 - irr2 = -r (R2 - r2) =32 r.] 1039 How far apart are two parallel chords of a circle whose radius is 14 in., if the chords are respectively a side of an inscribed square and a side of an inscribed equilateral triangle? 1040 Find the radius of a circle whose radius is n times the area of a given circle. 1041 How many degrees in an arc whose length is equal to the radius? 1042 Find the angle subtended at the center of a circle by an arc 3 feet long if the radius is 5 feet. 1043 Find the area of a sector whose radius is 30 and arc 36~. 1044 If a 6-inch pipe fills a cistern in one hour, how long will it take a 2-inch pipe to fill it? 1045 Find the radius of a circle inscribed in a sector which is the fourth part of a circle whose radius is R. EXERCISES 251 THEOREMS 1046 A circle can be inscribed in a polygon if the bisectors of all its angles meet in a point. 1047 The circle described upon the hypotenuse of a right triangle as a diameter is equivalent to the sum of the circles described upon the other two sides as diameters. 1048 An inscribed equilateral triangle is half the regular hexagon inscribed in the same circle. 1049 The area of the inscribed regular hexagon is \ three fourths of the area of the circumscribed regular hexagon. 1050 The area of the six-pointed star formed by producing the sides of a regular hexagon is equal to twice the area of the hexagon. 1051 If two diagonals of a regular pentagon intersect, the greater segment of each is equal to a side of the pentagon. 1052 Two intersecting diagonals of a regular pentagon divide each other in extreme and mean ratio. 1053 Two diagonals of a regular pentagon trisect the angle from whose vertex they are drawn. 1054 The diagonals of a regular pentagon form by their intersections a regular pentagon. 1055 The area of a circular ring is equal to the area of a circle whose diameter is a chord of one circle and a tangent to the other. 1056 An equilateral polygon circumscribed about a circle is regular if it has an odd number of sides. 1057 An equiangular polygon circumscribed about a circle is regular. 1058 An equiangular polygon inscribed in a circle is regular if it has an odd number of sides. 1059 The radius of an escribed circle of an equilateral triangle is three times the radius of the inscribed circle. 1060 If the alternate sides of a regular polygon are produced to meet, their points of intersection are the vertices of a regular polygon. 1061 The apothem of an inscribed regular hexagon is equal to half the side of an inscribed equilateral triangle. 252 PLANE GEOMETRY-BOOK V 1062 The area of the inscribed regular hexagon is the mean proportional between the areas of the inscribed and circumscribed equilateral triangles. 1063 Two diagonals of a regular hexagon, not drawn from the same vertex, are parallel or perpendicular; or, they bisect or trisect each other. 1064 The intersections of the diagonals joining the alternate vertices of a regular hexagon form a regular hexagon whose area is one third that of the given hexagon. 1065 The radius of an inscribed regular polygon is the mean proportional between its apothem and the radius of the similar circumscribed polygon. 1066 The side of a circumscribed regular hexagon is two thirds of the side of an inscribed equilateral triangle. 1067 The perimeter of the inscribed equilateral triangle is three fourths of the perimeter of the circumscribed regular hexagon. 1068 The sum of the squares of the three lines drawn from any point on the circumference to the vertices of an inscribed equilateral triangle is constant. 1069 The difference of the squares of the diagonal and the side of a regular pentagon is equal to the product of the diagonal and the side. 1070 The inscribed regular octagon is equivalent to a rectangle whose dimensions are the sides of the inscribed and the circumscribed squares. 1071 If from any point within a regular polygon of n sides perpendiculars are drawn to the sides, the sum of these perpendiculars is equal to n ti~mes the apothem. 1072 An angle of a regular polygon of n sides is to an angle of a regular polygon of n + 2 sides as n2 - 4 is to n2. 1073 In a regular polygon of n sides the diagonals drawn from the vertex of any angle to the opposite vertices divide the angle into n - 2 equal parts. 1074 If the diameter of a circle is divided into two parts, and on opposite sides of these parts as diameters semicircles are described, the circle is divided by the semicircumferences into two parts which are proportional to the segments of the diameter. 1075 Arcs of equal length on two circumferences subtend angles at the center which are inversely proportional to the radii. EXERCISES 253 PROBLEMS OF CONSTRUCTION 1076 To circumscribe an equilateral triangle about a given circle. 1077 To circumscribe a square about a given circle. 1078 To circumscribe a regular pentagon about a given circle. 1079 To circumscribe a regular hexagon about a given circle. 1080 To circumscribe a regular octagon about a given circle. 1081 To construct a circle equivalent to twice a given circle. SOLUTION. Let R be the radius of the given circle. Construct an isosceles right A with legs equal to R. Call the hypotenuse R'. R' is the radius of the required circle. PROOF. R'2 2 R2 (~ 418)..'. 7rR'2 = 2 7rR2. But 7rR'2 is the area of the required 0, and 2 7rR2 is twice the area of the given 0 (~ 464). 1082 To construct a circle equivalent to the sum of two given circles. 1083 To construct a circle equivalent to the difference of two given circles. 1084 To bisect a circle by a concentric circumference. 1085 To trisect a circle by concentric circumferences. 1086 To construct an equilateral triangle on a given line. 1087 To construct a square on a given line. 1088 To construct a regular pentagon on a given line. 1089 To construct a regular hexagon on a given line. 1090 To construct a regular octagon on a given line. 1091 To construct a regular decagon on a given line. 1092 To construct a regular dodecagon on a given line. 1093 To construct a regular pentadecagon on a given line. 1094 To divide a right angle into five equal parts. 1095 To cut off the corners of a square so as to form a regular octagon. 1096 To cut off the corners of a regular pentagon so as to form a regular decagon. 1097 To construct a circle equivalent to five times a given circle. 1098 To construct a circle equivalent to two thirds of a given circle. 1099 To divide a circle into n equal parts by concentric circumferences. 1100 To construct a circle which shall be to a given circle as m is to n. 254 PLANE GEOMETRY-BOOK V SUPPLEMENTARY MAXIMA AND MINIMA DEFINITIONS 486 A maximum magnitude is the greatest of its class. 487 A minimum magnitude is the least of its class. Thus, the diameter of a circle is the maximum chord, and the perpendicular is the minimum line that can be drawn from a given point to a given straight line. 488 Isoperimetric figures are figures which have equal perimeters. PROPOSITION I. THEOREM 489 Of all triangles having two given sides, that in which these two sides include a right angle is the maximu n. A D B ---— C E F HYPOTHESIS. In the & ABC and DEF, AB = DE, BC = EF, AB is I to BC, and DE is oblique to EF. CONCLUSION. A ABC > A DEF. PROOF Draw DG 1 to EF. Then DE > DG. ~ 182 But AB = DE. Hyp... AB > I)G..'. A ABC > A DEF. ~ 408 Q. E. D. MAXIMA AND MINIMA 255 PROPOSITION II. THEOREM 490 Of all isoperimetric triangles on the same base, the isosceles is the maximum. E, --- /-, / I HYPOTHESIS. In the ABC and DBC, the perimeters are equal, AB = AC, and DB > DC. CONCLUSION. A ABC > A DBC. PROOF On BA produced take AE = AB, join EC, ED, and draw AH and DK ~ to EC. Then the / BCE is a rt. L, for it may be inscribed in a semicircle whose center is A and radius AC. ~ 277 The A AEC is isosceles by const... EH = HC. ~ 178 Now BE BA + AE = BA + AC = BD + DC. Const. and Hyp. But BD + DE > BE. Ax. 15.'. BD + DE > BD + DC. Subtracting BD = BD, Iden. DE > DC. Ax. 5.. EK > KC. ~ 191 That is, K falls between H and C..'. HC > KC. Ax. 12.-. A ABC > A DBC. ~ 408 Q. E. D. 491 COROLLARY. Of all isoperimetric triangles the equilateral is the maximum. 256 PLANE GEOMETRY -BOOK V PROPOSITION III. THEOREM 492 Of all polygons having all the sides given but one, the maximum can be inscribed in a semicircle whose diameter is the undetermined side. D E A F HYPOTHESIS. ABCDEF is the maximum of all polygons whose given sides are AB, BC, CD, DE, EF. CONCLUSION. ABCDEF can be inscribed in a semicircle whose diameter is AF. PROOF Join any vertex C to A and F. Then the Z ACF is a rt. Z; for if not, without changing the length of AC and CF, the area of the triangle ACF can be increased by making the Z ACF a rt. Z, while the area of the remainder of the polygon remains unchanged. ~ 489 But this would increase the area of the polygon, which is contrary to the hypothesis that the polygon is a maximum..*. the / ACF is a rt. L, and the vertex C lies in the semicircumference whose diameter is AF. ~ 277 Likewise every vertex of the polygon lies in the semicircumference..'. the polygon can be inscribed in a semicircle whose diameter is AF. Q. E. D. MAXIMA AND MINIMA 257 PROPOSITION IV. THEOREM 493 Of all polygons having their sides respectively equal, that is the maximum which can be inscribed in a circle. El -F A C XB HYPOTHESIS. The polygons ABCDE and A'B'C'D'E' are mutually equilateral, ABCD can be inscribed in a circle, A'B'C'D'E' cannot be inscribed in a circle. CONCLUSION. ABCDE > A'B'C'D'E'. PROOF Draw the diameter CF, and join FA and FE. On A'E' construct the A F'AE' = A FAE, and join C'F'. Because the polygons FABC and FEDC are inscribed in semicircles,.E. FABC > F'A'B'C', and FEDC > F'E'D'C'. ~ 492 Adding, ABCDEF > A'B'C'D'E'F'. Ax. 3 Subtracting A FAE = A F'A'E', Const. ABCDE > A'B'D'C'E'. Ax. 5 Q. E. D. 258 PLANE GEOMETRY BOOK V PROPOSITION V. THEOREM 494 Of all isoperimetric polygons having the same number of sides, the maximum is equilateral. A B - ----- ----— C HYPOTHESIS. P is the maximum of all polygons having the same perimeter and the same number of sides. CONCLUSION. P is equilateral. PROOF Since P is the maximum, the A ABC must be the maximum.,. AB = AC. "Of all isoperimetric triangles on the same base, the isosceles is the maximum." ~ 490 That is, any two consecutive sides of P are equal..'. P is equilateral. Q. E. D. 495 COROLLARY. Of all isoperimetric polygons having the same number of sides, the maximum is regular. PROOF. The maximum polygon is equilateral (~ 494), can be inscribed in a circle (~ 493), and is therefore regular (~ 435). MAXIMA AND MINIMA 259 PROPOSITION VI. THEOREM 496 Of all isoperimetric regular polygons, that which has the greatest number of sides is the maximum. A K --- — B C HYPOTHESIS. The equilateral triangle ABC and the square S have equal perimeters. CONCLUSION. S > ABC. PROOF Join C to any point K in AB, and construct the isosceles triangle HKC isoperimetric with AKC. Then the quadrilateral HKBC is isoperimetric with the triangle ABC, and therefore isoperimetric with S. Now the A HKC > the A AKC. ~ 490.. the quadrilateral HKBC > the A ABC. Ax. 2 But S > the quadrilateral HKBC. ~ 495.-. S > ABC. In like manner it may be shown that a regular pentagon isoperimetric with S is greater than S, and so on. Q. E. D. 497 COROLLARY. The area of a circle is greater than the area of any polygon of equal perimeter. 260 PLANE GEOMETRY — BOOK V PROPOSITION VII. THEOREM 498 Of regular polygons which have the same area, that which has the greatest number of sides has the least perimeter. A A D/ Bt / C HYPOTHESIS. A and B are regular polygons of the same area, A having more sides than B. CONCLUSION. The perimeter of A < the perimeter of B. PROOF Construct the regular polygon C having the same perimeter as A and the same number of sides as B. Then C < A. " Of all isoperimetric regular polygons, that which has the greatest number of sides is the maximum." ~ 496 But A == B. Hyp..'. C < B..H. the perimeter of C < the perimeter of B. But the perimeter of C = th e perieter of A. Const..t. the perimeter of A < the perimeter of B. Q. E. D. 499 COROLLARY. The circumference of a circle is less than the perimeter of any equivalent polygon. EXERCISES 261 EXERCISES THEOREMS 1101 Of all triangles having the same base and the same altitude, the isosceles has the minimum perimeter. 1102 Of all triangles having the same base and the same area, the isosceles has the greatest vertical angle. 1103 Of all triangles having the same base and the same area, the isosceles has the minimum perimeter. 1104 Of all equivalent triangles, the equilateral has the minimum perimeter. 1105 Of all triangles inscribed in a given circle, the equilateral is the maximum. 1106 Of all equivalent parallelograms having equal bases, the rectangle has the minimum perimeter. 1107 Of all equivalent right parallelograms, the square has the minimum perimeter. 1108 Of all polygons of the same number of sides inscribed in a given circle, the regular polygon is the maximum. 1109 Of all triangles having the same base and equal vertical angles, the isosceles is the maximum. PROBLEMS 1110 To divide a given line into two parts so that their product shall be a maximum. 1111 To inscribe an angle in a semicircle so that the sum of its sides shall be a maximum. 1112 To draw a line through a given point within an angle so as to form the minimum triangle. 1113 To find a point in a given line so that the sum of its distances from two given points on the same side of the given line is a minimum. 1114 To inscribe in a given semicircle the maximum rectangle. 1115 To inscribe in a given semicircle the maximum trapezoid. 262 PLANE GEOMETRY-BOOK V SYMMETRY I SYMMETRY WITH RESPECT TO AN AXIS DEFINITIONS 500 Two points are symmetrical with respect to a straight line, called the axis of symmetry, when this axis bisects at right angles the straight line joining the two P points. Thus, P and P' are symmetrical with x y respect to XY, if XY bisects PP' at right angles. 501 Two figures are symmetrical with respect to an axis, when every point in one has its symmetrical point in the other. Thus, the lines AB and A'B' are symmetrical with respect to the axis XY, if every point in either has its symmetrical point in the other, with respect to XY as the axis of _ symmetry. Also, the triangles ABC and A'B'C' are symmetrical with respect to the axis XY, if every B A I I I I X iY B, A l I. Y B',: /C' A A' point in the perimeter of one has its symmetrical point in the perimeter of the other with respect to XY as an axis. 502 A figure is symmetrical with respect to an axis, if the axis divides the figure into two sym- B metrical figures. Thus, ABCDEF is symmetrical z A< | D t with respect to the axis XY, if XY divides ABCDEF into two sym- p metrical figures, ABCD, AFED, with respect to XY. SYMMETRY 263 503 Symmetrical points and lines in two symmetrical figures are called homologous. In all cases, two figures which are symmetrical with respect to an axis, can be made to coincide by rotating either about the axis of symmetry. II SYMMETRY WITH RESPECT TO A CENTER 504 Two points are symmetrical with respect to a third point, called the center of symmetry, when this center bisects the line joining the two points. Thus, P and P' are symmetrical with P I - P respect to the center 0, if O bisects the ~ line PP'. 505 The distance of a point from the center of symmetry is called the radius of symmetry; as, OP, or OP'. A point P can be brought into coincidence with its symmetrical point P' by turning the radius OP about O as a pivot through 180~. 506 Two figures are symmetrical with respect to when every point of either has its sym- A metrical point in the other. C- Thus, AB and A'B' are symmetrical /, -- with respect to the center 0, if every Bpoint in AB has its symmetrical point in A'B'. Likewise the polygons ABCD and A'B'C'D' are sym with respect to the center O. D Any two figures symmetrical:1,, with respect to a center can be -- brought into coincidence by turn- B "'] ing one of them in its own plane about the center of symmetry as a pivot through 180~. a center, B' A' A' imetrical C' B' 'rDt -i5' 264 PLANE GEOMETRY -BOOK V 507 A figure is symmetrical with respect to a center, when every line drawn through the center cuts - -- -—. the perimeter in two points symmetrical with respect to the center. 508 The straight line drawn through the center of a symmetrical figure, and terminated by the perimeter, is called a diameter. PROPOSITION VIII. THEOREM 509 If a figure is symmetrical with respect to two axes perpendicular to each other, it is also symmetrical with respect to their intersection as a center. B A C _ X x Y' HYPOTHESIS. The figure ABCDEFGH is symmetrical with respect to the perpendicular axes XX' and YY' intersecting at O. CONCLUSION. ABCDEFGH is symmetrical with respect to O. PROOF Let P be any point in the perimeter. Draw PLQ parallel to XX', and QKP' parallel to YY'. Join LK, OP, and OP'. Then QK = P'K. ~ 500 But QK = OL. ~ 200.. P'K = OL. Ax. 11.'. P'OLK is a E/. ~ 205 Whence OP' is = and II to LK. ~ 200 SYMMETRY 265 Likewise POKL is a i1, and PO is = and 1I to LK..P. PO = OP', Ax. 11 and POP' is a straight line. Ax. 16 That is, the figure is symmetrical with respect to O as a center. ~ 507 Q. E. D. EXERCISES 1116 A circle is symmetrical with respect to its center, or with respect to any diameter as an axis. 1117 A parallelogram is symmetrical with respect to the intersection of its diagonals as a center. 1118 Every regular polygon of an even number of sides has a center of symmetry. 1119 An isosceles triangle is symmetrical with respect to the altitude upon the base. 1120 An equilateral triangle is symmetrical with respect to all of its altitudes. 1121 The symmetrical of a straight line, with respect to an axis, or with respect to a center, is an equal straight line. 1122 The symmetrical of an angle with respect to a center is an equal angle. 1123 The symmetrical of an angle with respect to an axis is an equal angle. 1124 If two straight lines are symmetrical with respect to a center, they are equal, parallel, and extend in opposite directions. 1125 If two polygons are symmetrical with respect to a center, or with respect to an axis, they are equal. 1126 A trapezium has no center of symmetry. 1127 An isosceles trapezoid has an axis of symmetry. 1128 Each diagonal of a square is an axis of symmetry. 1129 How many axes of symmetry has each of the regular polygons considered in this book? 1130 How many axes of symmetry has a regular polygon of 2 n sides? of 2 n + 1 sides? 266 PLANE GEOMETRY-BOOK V FORMULAS OF MENSURATION SYMBOLS b = base. b' = upper base. C = circumference. D = diameter. h = altitude. m = median. t = angle-bisector. P = perimeter. R = radius of circumscribed circle. r = radius of inscribed circle. S = area. a = the side of an equilateral triangle. a, b, c = the sides of any triangle. s = (a+b+c). FORMULAS PAGE Any triangle, S = b x h.... 197 S = Vs (s-a)(s- b)(s-c).. 208 S - abc 208 4R S =s x r... 209 R=. abc. 177 4 Vs (s - a)(s - b)(s - c) abc 208 R = -. *...... 208 4S FORMULAS OF MENSURATION r = (s-a)(s-b)(s - c) r = 267 PAGE;.. 209 h (on a) = -V(s- a)(s - b)(s- c) a m (on a) = t 2(b2 + c2)- a2 t (on a) = -- cbcs (s - a) b+c..... 176..... 176 Equilateral triangle, a = hV3. a = R V-. h=jV3 h = R..3 s =..A/ 4 S = R2/3 Rectangle, S = b x h Parallelogram, S = b X h Trapezoid, S = h(b + b') Regular polygon,........ 167........ 241........ 208........ 241........ 208 ~~~~~ ~~~~~ ~~~~~ ~~~~~... 241... 195... 196... 198 Circle, Sector, S = rxP. S=2R S =RxC. S = 7rR2. S = 1 R x arc........ 237........ 237........ 238........ 238........ 238 SOLID GEOMETRY BOOK VI STRAIGHT LINES AND PLANES DEFINITIONS 510 Solid Geometry treats of figures whose parts are not all in the same plane. 511 A plane is a surface such that the straight line which joins any two of its points lies wholly in the surface (~ 54). A plane is understood to be indefi- M nite in extent. We may represent a plane by a parallelogram drawn in perspective and lying in the plane; as, the plane MN. 512 A plane is determined by certain conditions which fix its position. 513 Postulate. A plane may be revolved about any line lying in it. Thus, the plane MN may be revolved about the straight line AB as ------ an axis until it passes through any fixed point in space. Hence any number of / planes may pass through, or embrace, a straight line; that is, one straight line does not determine a plane. 268 LINES AND PLANES 269 PROPOSITION I. THEOREM 514 A plane is determined by a straight line and a point without that line. A — M. \ HYPOTHESIS. AB is a straight line and C is a point without AB. CONCLUSION. AB and C determine a plane. PROOF Let the plane MN pass through AB and turn on AB as an axis until it embraces C. Then the plane is determined; for if it now turns either way on AB, it will no longer contain the point C. ~ 512 Q. E. D. 515 COROLLARY 1. A plane is determined by three points not in the same straight line. For let A, B, and C be the three points. Join AB. Then the line AB and the point C determine a plane (~ 514). 516 COROLLARY 2. A plane is determined by two intersecting lines. For let AB and AC be two intersecting lines. Then AB and any point in AC other than A determine a plane (~ 514). 517 COROLLARY 3. A plane is determined by two parallel lines. For let AB and CD be two parallel lines. Then AB and CD are in the same plane (~ 114), and a plane containing AB and any point in CD is determined (~ 514). A — B N 270 SOLID GEOMETRY-BOOK VI PROPOSITION II. THEOREM 518 The intersection of two planes is a straight line. P HYPOTHESIS. MN and PQ are two intersecting planes. CONCLUSION. The intersection of MN and PQ is a straight line. PROOF Let the planes intersect in points A and B. Draw AB. Then AB lies in both planes. ~ 511 Any point without AB cannot be in both planes. ~ 514 Therefore the intersection of the two planes is the straight line AB. Q. E. D. DEFINITIONS 519 A straight line is perpendicular to a plane when it is perpendicular to every straight line in the plane drawn through its foot, that is, the point in which it meets the plane; and the plane is also perpendicular to the line. 520 A straight line and a plane are parallel if they cannot meet, however far they may be produced. 521 A straight line is oblique to a plane when it is neither perpendicular nor parallel to the plane; and the plane is also oblique to the line. 522 Two planes are parallel if they cannot meet, however far they may be produced. LINES AND PLANES 271 PROPOSITION III. THEOREM 523 If a straight line is perpendicular to each of two intersecting lines, it is perpendicular to their plane. AA B B N HYPOTHESIS. ADis perpendicular to DB and DC. CONCLUSION. AD is I to MN, the plane of DB and DC. PROOF Let DF be any other line in the plane MN, and draw CB cutting DF in F. Produce AD to E, making DE = AD, and join A and E to C, F, and B. Since CD and BD are each L to AE at its middle point D, CA = CE, and BA = BE. ~ 186. AACB == A ECB. ~ 172.. ZACF = ECF. ~ 166.'.A ACF = A ECF. ~ 162.-. FA = FE. ~ 166.. FD is to AE. ~ 187 That is, AD is ~ to DF, any line in the plane MN drawn through D..'. AD is L to the plane MN. ~ 519 Q. E. D. 272 SOLID GEOMETRY-BOOK VI PROPOSITION IV. THEOREM 524 All perpendiculars drawn to a straight line at a given point lie in the same plane. A -— | -N HYPOTHESIS. BC, BD, and BE are perpendicular AB. CONCLUSION. BC, BD, and BE lie in the same plane. PROOF Let BD and BE lie in the plane MN. ~ 516 Then AB is I to MN. ~ 523 We are to prove that BC also lies in MN. Let the plane PQ pass through AB and BC, cutting MN in the line BC'. Then AB is I to BC'. ~ 519 But AB is L to BC by hyp. That is, BC and BC' in the plane PQ are L to AB..'. BC and BC' coincide. ~ 102 But BC' lies in MN by const... BC lies in MN. Q. E. D. 525 COROLLARY 1. Through a given point in a straight line only one plane can be drawn perpendicular to the line. 526 COROLLARY 2. Through a given point without a straight line only one plane can be drawn perpendicular to the line. EXERCISE 1131. Why does a three-legged stool rest steadily upon the floor, while a four-legged one frequently does not? LINES AND PLANES 273 PROPOSITION V. THEOREM 527 From a given point only one perpendicular can be drawn to a given plane. M D -D C CAIA N N Fig. 1 Fig. 2 HYPOTHESIS. MN is a given plane, B a given point either in MN (Fig. i), or without MN (Fig. 2), AB I to MN, and A'B any other line to MN. CONCLUSION. A'B is not ~ to MN. PROOF In both cases, pass a plane PQ through AB and A'B intersecting MN in CD. Then in the plane PQ, AB is I to CD. ~ 519.'. A'B is not I to CD. ~~ 102, 113.'. A'B is not L to MN. ~ 519 Q. E. D. 528 COROLLARY. The perpendicular is the shortest line that can be drawn from a point to a plane. For in the plane PQ (Fig. 2), BA < BA' (~ 182). 529 DEFINITION. The distance from a point to a plane is the length of the perpendicular drawn from the point to the plane. EXERCISE 1132. On what line would you measure the distance from a chandelier to the floor? 274 SOLID GEOMETRY-BOOK VI PROPOSITION VI. THEOREM 530 If a perpendicular and oblique lines are drawn from a point to a plane: 1 Oblique lines meeting the plane at equal distances from the foot of the perpendicular are equal. 2 Of oblique lines meeting the plane at unequal distances from the foot of the perpendicular, the more remote is the greater. A D E B N HYPOTHESIS. AB is I to MN, and in MN BD = BC, and BE > BC. CONCLUSION. AD = AC, and AE > AC. PROOF The rt. A ABD and ABC are equal. ~ 163.'. AD=AC. ~166 On BE take BF = BC, and draw AF. Then AE > AF. ~ 189 But AF = AC (~ 163)... AE > AC. Q. E. D. 531 COROLLARY 1. Equal oblique lines drawn from a point in a perpendicular to a plane meet the plane at equal distances from the foot of the perpendicular; and of two unequal oblique lines so drawn, the greater meets the plane at the greater distance from the foot of the perpendicular. LINES AND PLANES 275 532 COROLLARY 2. The locus of a point equidistant from all points in the circumference of a circle is a straight line passing through the center of the circle and perpendicular to its plane. PROPOSITION VII. THEOREM 533 If two lines in a plane are perpendicular to each other, a line joining their intersection to any point in a line drawn perpendicular to both the plane and one of the lines is perpendicular to the other line. A M - / Vr';, HYPOTHESIS. In the plane MN, BE is ~ to CD, AB is L to MN, and EA is a line drawn from E to any point in AB. CONCLUSION. AE is ~ to CD. PROOF Take EC = ED, and join BC, BD, AC, AD. Then BC = BD. ~ 186.. AC = AD. ~530.'. AE is L to CD. ~ 188 Q. E. D. EXERCISES 1133 What is the locus of a point in space equidistant from the extremities of a given straight line? 1134 What is the locus of a line in space perpendicular to a given line at a given point in the line? 276 SOLID GEOMETRY -BOOK VI PROPOSITION VIII. THEOREM 534 If one of two parallel lines is perpendicular to a plane, the other is also perpendicular to the plane. A C HYPOTHESIS. AB is MN HYPOTHESIS. AB is 11 to CD, and AB is i to the plane MN at B. CONCLUSION. CD is I to MN. PROOF Let CD meet MN at D, and draw BD. Then BD is _ to AB (~ 519)..'. BD is _ to CD. ~ 117 Draw EDF in MN _ to BD, and join AD. Then AD is L to EF. ~ 533 Since EF is 1J to BD and AD, it is L to the plane of ABD (~ 523), and.-. 1 to CD. ~ 519 Since CD is L to BD and ED, it is L to MN. ~ 523 Q. E. D. 535 COROLLARY 1. If two straight lines are perpendicular to the same plane. they are parallel. For draw A'B II to CD. Then A'B is 1 to MN (~ 534), and therefore coincides with AB (~ 527)..'. AB is 1I to CD. 536 COROLLARY 2. If two straight lines are parallel to a third straight line, they are parallel to each c other. Let AB and CD be each 11 to EF. M Draw a plane MIN I to EF. Then AB / and CD are each I to MN (~ 534), and.. B L parallel (~ 535). N LINES AND PLANES 277 PROPOSITION IX. THEOREM 537 Any plane containing one of two parallel lines and not the other is parallel to the other. A B C _D N HYPOTHESIS. AB and CD are parallel lines, and the plane MN contains CD but not AB. CONCLUSION. AB and MN are parallel. PROOF AB and CD are in the plane AD (~ 517) which intersects the plane MN in the line CD by hyp. Hence, AB must meet MN in CD, if at all. But AB and CD are parallel and cannot meet. Therefore, AB and the plane MN are parallel. ~ 520 Q. E. D. 538 COROLLARY 1. If two straight lines are in different planes, a plane can be passed through one parallel to the other. Let AB and CD be the lines. Through any A1 / point in CD draw EF II to AB (~ 290). Then CD and EF determine a plane (~ 516) 11 to AB [E ND (~ 537). 539 COROLLARY 2. Through any point in space a plane can be passed parallel to any two lines in space. Let P be the point, a and b the two lines. / Through P draw x and y respectively parallel to a and b. Then x and y determine a plane (~ 516) II to a and b (~ 537). 278 SOLID GEOMETRY-BOOK VI 540 COROLLARY 3. If a straight line is parallel to a plane, it is parallel to the intersection of this plane with any plane passed through the line. Let AB be II to the plane MN, and CD the intersection of MN with any plane passed through AB. Then AB is II to CD. For AB and CD, being in the M plane AD, must meet, if at all, in MN, which I I \ is impossible since AB is II to MN by hyp. N.. AB is II to CD. 541 COROLLARY 4. If a straight line and a plane are parallel, a parallel to the line drawn through a point in the plane lies in the plane. PROPOSITION X. THEOREM 542 Two planes perpendicular to the same straight line are parallel. H N.X K Q HYPOTHESIS. The planes MN and PQ are J to the line HK. CONCLUSION. MN and PQ are parallel. PROOF Should MN and PQ meet, as at X, there would be two planes through X L to HK, which is impossible. ~ 526 Therefore MN and PQ are parallel. ~ 522 Q. E. D. EXERCISE 1135. What is the locus of a point in a given plane equidistant from two given points without the plane? LINES AND PLANES 279 PROPOSITION XI. THEOREM 543 The lines in which a plane intersects parallel planes are parallel. two HYPOTHESIS. AB and CD. CONCLUSION. The plane RS intersects the parallel planes MN and PQ in AB and CD are parallel. PROOF AB and CD cannot meet; for if so, the parallel planes MN and PQ would meet, which is impossible. ~ 522 Since AB and CD are in the same plane RS and cannot meet, they are parallel. ~ 114 Q. E. D. 544 COROLLARY. Parallel lines included between parallel planes are equal. Let AC and BD be II lines included between the II planes MN and PQ. Then the plane of AC and BD (~ 517) intersects MN and PQ in the II lines AB and CD (~ 543)..'. ACDB is a 0 (~ 195), and AC = BD (~ 200). EXERCISE 1136. Find the locus of a point in space equidistant from two given parallel planes. 280 SOLID GEOMETRY -BOOK VI PROPOSITION XII. THEOREM 545 A straight line perpendicular to one of two parallel planes is perpendicular to the other. A! P C D N 'Q HYPOTHESIS. MN and PQ are parallel planes, and AB is I to MN. CONCLUSION. AB is I to PQ. PROOF Through AB pass any two planes intersecting MN in AC and AE, and PQ in BD and BF. Then AC is II to BD, and AE is II to BF. ~ 543 Since AB is J to MN, it is L to AC and AE. ~ 519 Hence AB is _L to BD and BF. ~ 117 Therefore, AB is _ to PQ. ~ 523 Q. E. D. 546 COROLLARY 1. Through a given point without a plane only one plane can be passed parallel to the given plane. For a plane through B II to MN is L to AB (~ 545), and only one such plane can be passed through B (~ 525). 547 COROLLARY 2. If two intersecting straight lines are parallel to a plane, the plane of these lines is parallel to the given plane. Thus, if BD and BF are II to MN, PQ is II to MN. 548 COROLLARY 3. Two parallel planes are everywhere equidistant. For the JI which measure the distances between the planes from any two points (~ 529) are II (~ 535), and.'. equal (~ 544). LINES AND PLANES 281 PROPOSITION XIII. THEOREM 549 If two angles not in the same plane have their sides respectively parallel and extending in the same direction, they are equal, and their planes are parallel. M \ B N, IP A ' xC \~I, I P. _ _ _ _ __ \. / II I "t I 1, 1 II\ Al l -ag V HYPOTHESIS. Angles A and A' lie in the planes MN and PQ, and AB and AC are respectively parallel to A'B' and A'C'. CONCLUSION. Z A = A', and MN is I1 to PQ. PROOF Take AB = A'B', AC = A'C', and draw AA', BB', CC', BC, and B'C'. Then AB' and AC' are L7. ~ 205 Whence BB' and CC' are each equal and II to AA'. ~ 200 Whence BB' and CC' are equal (Ax. 11) and parallel. ~ 536.-. BC' is a D (~ 205), and BC = B'C'. ~ 200.-. the A ABC and A'B 'C' are equal. ~ 172.-. Z A = A'. ~ 166 Since PQ is parallel to AB and AC, ~ 537 PQ is parallel to MN, the plane of AB and AC. ~ 547 Q. E. D. EXERCISE 1137. The four lines in which two parallel planes intersect two other parallel planes are parallel. 282 SOLID GEOMETRY BOOK VI PROPOSITION XIV. THEOREM 550 If two straight lines are cut by three parallel planes, their corresponding intercepted segments are proportional. CONCLUSION. AE:EB = CG:GD. HYPOTHESIS. The lines AB and CD are cut by the parallel planes MN, PQ, RS, in the points A, E, B, and C, G, D CONCLUSION. AE: EB = CG: GD. PROOF Draw AD piercing the plane PQ in F, and draw AC, BD, EF, and FG In the A ABD, EF is II to BD. ~ 543 In the A DAC, FG is II to AC. ~ 543 Whence AE: EB = AF: ED, ~ 340 and CG: GD = AF: FD. ~ 340 Whence AE: EB = CG: GD. Ax. 11 Q. E. D. EXERCISES 1138 If, in the above figure, AE = 3 inches, EB = 5 inches, and CD = 9 inches, find CG and GD. 1139 A line parallel to two intersecting planes is parallel to their intersection. LINES AND PLANES 283 1140 Does the folding of a sheet of paper form a straight edge? 1141 In how many points can a plane intersect a circumference? 1142 Four points, in general, determine how many planes? 1143 Three straight lines passing through a common point determine how many planes? 1144 Four straight lines passing through a common point determine how many planes? 1145 Show how to erect a perpendicular to a plane by the aid of two carpenter's squares. 1146 Erect a perpendicular to a given plane at a given point in the plane. 1147 Draw a perpendicular to a given plane from a given point without the plane. 1148 If two points A and B are equidistant from a plane, and on the same side of it, the line AB is parallel to the plane. 1149 If a plane is perpendicular to a straight line at its middle point, any point in the plane is equidistant from the extremities of the line. 1150 If one of two parallel lines is parallel to a plane, the other is also parallel to the plane. 1151 If two planes are parallel to a third plane, they are parallel to each other. 1152 If two straight lines are not parallel and will not intersect, a plane cannot contain both of them. 1153 Two parallel lines in two intersecting planes are parallel to the line of intersection. 1154 What is the locus of a point at a given distance from a given plane? 1155 What is the locus of a line parallel to a plane and passing through a fixed point? 1156 Find the locus of a point equidistant from three given points not in a straight line. 284 SOLID GEOMETRY -BOOK VI DIEDRAL ANGLES DEFINITIONS 551 A diedral angle is the figure formed by two planes proceeding from a straight line. The intersecting planes are the faces, and their A _ line of intersection is the edge, of the diedral c angle. Thus, the planes AD and AF intersect- G- - H ing in the line AB form a diedral angle. The K planes AD and AF are the faces, and AB is the B- F edge, of the diedral angle. D 552 A diedral angle may be designated by four letters, one in each face and two in the edge, the two in the edge being written between the other two; as, the diedral angle E-AB-C. When there is but one diedral angle at an edge, it may be designated by two letters in its edge; as, AB. 553 The plane angle of a diedral angle is the angle formed by two straight lines, one in each face, and both perpendicular to the edge at the same point; as, the angle HGK. 554 COROLLARY. The plane angle of a diedral angle is constant for all positions of its vertex in the edge. For the sides of any two such plane angles are respectively parallel (~ 115), and therefore the angles are equal (~ 549). 555 Adjacent diedral angles are those which A have a common edge and a common face be- D tween them; as, D-AB-C and D-AB-E. c 556 When one plane meets another making B - the adjacent diedral angles equal, each is a right diedral angle. DIEDRAL ANGLES 285 557 Two planes are perpendicular to each other when they form a right diedral angle. 558 Diedral angles are acute, obtuse, complementary, etc., when their plane angles are acute, obtuse, complementary, etc. PROPOSITION XV. THEOREM 559 Two diedral angles are equal if their plane angles are equal. A A' G -H ----- K 'K1 B --- —- D B- -D C C/ HYPOTHESIS. The plane angles HGK and H'G'K' of the diedral angles AB and A'B' are equal. CONCLUSION. The diedral angles AB and A'B' are equal. PROOF AB and A'B' are L to the planes HGK and H'G'K' respectively. ~ 523 Apply the Z H'G'K' to the equal Z HGK. Then the plane H'G'K' coincides with the plane HGK. ~ 516. A'B' coincides with AB, ~ 527 the plane A'C' coincides with the plane AC, ~ 516 and the plane A'D' coincides with the plane AD. ~ 516. the diedral angles coincide and are equal. ~ 89 Q. E. D. 560 COROLLARY. Two vertical diedral angles are equal. 286 SOLID GEOMETRY -BOOK VI PROPOSITION XVI. THEOREM 561 Two diedral angles have the same ratio as their plane angles. HYPOTHESIS. ABD and A'B'D' are the plane angles of A-BC-D and A'-B'C'-D'. the diedral angles A'-B'C'-D' Z A'B'D' CONCLUSION. -- A. A-BC-D Z ABD CASE 1. When the plane angles are commensurable. PROOF Let Z K, a common measure of the plane angles, be contained m times in / A'B'D' and n times in / ABD. Then A'B'D' m Then Z-BD Z ABD n By passing planes through the several lines of division of the plane angles and the edges of the diedral angles, the diedral angles will be divided into m and n diedral angles, all equal. ~ 559 A'-B'C'-D' m Whence = - -- A-BC-D n. A'-B'C'-D' / A'B'D' A-BC-D ABD Ax. 11 Q. E. D. DIEDRAL ANGLES 287 CASE 2. When the plane angles are incommensurable. K B A BI A'.'E PROOF Since the angles ABD and A'B'D' are not commensurable, let the angles ABD and A'B'E be commensurable, the ZEB'D' being less than Z K. Pass a plane through B'E and B'C'. A —B'C'-E Z A'B'E Then A-B - ZABE Case 1 A-BC-D Z ABD If Z K be indefinitely diminished, the Z EB'D', always less than / K, will become less than any assigned quantity..'. the Z A'B'E approaches the / A'B'D' as a limit, ~ 263 and A'-B'C'-E approaches A'-B'C'-D' as a limit. ~ 263 / A'B'E ^"R^T AA'B'D' Whence ABD approaches ZABD as a limit, ~ 268 / ABD a / ABD and A'-B'C'-E approaches A-B'C- as a limit. ~ 268 A-BC-) A-BC-D A A-BC P -E As'B a But the variable ratios A-B E and A'B are always equal. A-BC-D ABD Case 1 A'-B'C'-D' A'B'D' ~264 A-BC-D L ABD Q. E. D. 562 COROLLARY. A diedral angle is measured by its plane angle. 288 SOLID GEOMETRY-BOOK VI PROPOSITION XVII. THEOREM 563 If a straight line is perpendicular to a plane, any plane passed through the line is perpendicular to the plane. HYPOTHESIS. AB is perpendicular to the plane MN at B, and PQ is any plane passed through AB intersecting MN in SQ. CONCLUSION. PQ is ~ to MN. PROOF Draw BC in MN _ to SQ. Then AB is ~ to SQ and BC..'. the fight angle ABC is the plane angle of the P-SQ-N..-. P-SQ-N is a right diedral angle..-. PQ is ~ to MN. ~ 519 diedral angle ~ 553 ~ 562 ~ 557 Q. E. D. EXERCISES 1157 A plane perpendicular to the edge of a diedral angle is perpendicular to each of its faces. 1158 If two planes intersect, the adjacent diedral angles are supplementary. 1159 If a plane intersects two parallel planes, the alternate interior diedral angles are equal; the corresponding diedral angles are equal. 1160 If the sum of two adjacent diedral angles is equal to two right diedral angles, their exterior faces are in the same plane. DIEDRAL ANGLES 289 PROPOSITION XVIII. THEOREM 564 If two planes are perpendicular to each other, a straight line drawn in one of them perpendicular to their intersection is perpendicular to the other. PA M B C N HYPOTHESIS. SQ is the intersection of the perpendicular planes MN and PQ, and AB in PQ is I to SQ at B. CONCLUSION. AB is I to MN. PROOF Draw BC in MN I to SQ. Then ABC is the plane angle of the right diedral angle P-SQ-N, and therefore is a right angle. ~ 562 Whence AB is I to BS and BC... AB is ~ to MN. ~ 523 Q. E. D. 565 COROLLARY 1. If two planes are perpendicular to each other, a straight line perpendicular to one of them at any point of their intersection lies in the other plane. Otherwise there could be two Is to PQ at B. 566 COROLLARY 2. If two planes are perpendicular to each other, a perpendicular to one of them drawn from any point in the other lies in the other plane. Otherwise there could be two Is to PQ from C. 290 SOLID GEOMETRY -BOOK VI PROPOSITION XIX. THEOREM 567 If two intersecting planes are each perpendicular to a third plane, their intersection is perpendicular to the third plane. M h HYPOTHESIS. The planes HB and KB intersecting in AB are each perpendicular to the plane MN. CONCLUSION. AB is I to MN. PROOF If from any point in AB a _L is drawn to MN, it will lie in both HB and KB (~ 566) and hence coincide with AB, their intersection. ~ 518..AB is l to MN. Q. E.D. 568 COROLLARY 1. A plane perpendicular to each of two intersecting planes is perpendicular to their intersection. 569 COROLLARY 2. If a plane is perpendicular to each of two perpendicular planes, the intersection of any two of these planes is perpendicular to the third plane, and each of the three intersections is perpendicular to the other two. Examine the corner of a cube. DIEDRAL ANGLES 291 PROPOSITION XX. THEOREM 570 Any point in the plane which bisects a diedral angle is equidistant from the faces of the angle. F G HYPOTHESIS. A is any point in the plane EH which bisects the diedral angle G-EB-F, AC and AD are perpendiculars drawn from A to the faces EF and EG. CONCLUSION. AC = AD. PROOF Pass a plane MN through AC and AD intersecting the planes EF, EG, and EH in BC, BD, and BA respectively. The plane MN is I to the planes EF and EG. ~ 563.-. MN is L to EB, the intersection of EF and EG. ~ 568 Whence EB is I to BC, BD, and BA. ~ 519.'. ABC and ABD are the plane angles of the equal diedral angles H-BE-F and H-BE-G. ~ 553.. L ABC= Z ABD. ~ 562.'. rt. A ABC = rt. A ABD. ~ 168.. AC = AD. ~ 166 Q. E. D. 292 SOLID GEOMETRY- BOOK VI PROPOSITION XXI. THEOREM 571 Through a straight line not perpendicular to a plane only one plane can be passed perpendicular to the given plane. A E HYPOTHESIS. AB is any straight line not 1 to the plane MN. CONCLUSION. Only one plane can be passed through AB L to MN. PROOF Through any point E in AB draw EF I to MN, and pass a plane AD through AB and EF. The plane AD is I to the plane MN. ~ 563 If through AB another plane could be passed I to MN, AB would be _L to MN which is contrary to the hypothesis. ~ 567.. through AB only one plane can be passed -_ to MN. Q. E. D. 572 DEFINITION. The projection of a point upon a plane is the foot of the perpendicular drawn from the point to the plane. 573 DEFINITION. The projection of a line upon a plane is the line which contains the / projections of all its points. EXERCISES 1161 A line of constant length is drawn from a fixed point to a plane. Find the locus of the point in which it touches the plane. 1162 What is the locus of a point equidistant from two given points, and also equidistant from two other given points? DIEDRAL ANGLES 293 PROPOSITION XXII. THEOREM 574 The projection of a straight line not perpendicular to a plane upon that plane is a straight line. A E B MS F N HYPOTHESIS. AB is any straight line not - to the plane MN. CONCLUSION. The projection of AB upon MN is a straight line. PROOF Through AB pass a plane I to MN (~ 571), intersecting MN in the straight line CD. ~ 518 All the perpendiculars which can be drawn from AB to the plane MN lie in the plane AD (~ 566) and hence meet the plane MN in CD.. the straight line CD is the projection of AB upon MN. Q. E. D. 575 DEFINITION. The projecting plane is the plane of a line and its projection upon a given plane. Thus, AD is the projecting plane of AB upon MN. The plane MN is called the plane of projection. 576 DEFINITION. The inclination of a line to a plane, or the angle which a line makes with a plane, is the acute angle which it makes with its projection upon the plane. EXERCISE 1163. Is the converse of ~ 574 necessarily true? 294 SOLID GEOMETRY - BOOK VI PROPOSITION XXIII. THEOREM 577 The acute angle which a straight line makes with its projection upon a plane is the least angle which it makes with any line in the plane. A N HYPOTHESIS. BC is the projection of AB upon the plane MN, and BD is any other line in MN. CONCLUSION. Z ABC < Z ABD. PROOF Take BD = BC, and join AD. In the A ABC and ABD, AB is common, BC = BD by const., and AC < AD. ~ 528.-. Z ABC < Z ABD. ~ 171 Q. E. D. EXERCISES 1164 If a line is parallel to a plane, it is parallel to its projection upon the plane. 1165 If a line is equal to its projection upon a plane, it is parallel to the plane. 1166 If two parallel lines intersect a plane, they make equal angles with the plane. 1167 A line and its projection upon a plane determine a second plane perpendicular to the first. DIEDRAL ANGLES 295 PROPOSITION XXIV. THEOREM 578 Only one common perpendicular can be drawn between two straight lines not in the same plane. A G B D HYPOTHESIS. AB and CD are two lines not in the same plane. CONCLUSION. Only one common perpendicular can be drawn between AB and CD. PROOF Through CD pass a plane MN II to AB (~ 538), and through AB pass a plane AF I to MN, intersecting MN in EF. Since EF is II to AB (~ 540), EF is not II to CD (~ 536); hence EF and CD intersect in some point E. In AF draw AE _ to EF. AE is I to MN (~ 564) to AB (~ 117), and.. I. to CD. ~ 519 That is, AE is J_ to AB and CD. Suppose GK a second I between AB and CD. In MN draw KL II to AB. Then GK is I to KL (~ 117) and.. _ to MN. ~ 523 Draw GH I to EF. Then GH is I to MN, ~ 564 We thus have GK and GH each I_ to MN, which is impossible. ~ 527. only one I can be drawn between AB and CD. Q. E. D. 579 COROLLARY. The common perpendicular is the shortest line that can be drawn between two lines. 296 SOLID GEOMETRY -BOOK VI POLYEDRAL ANGLES DEFINITIONS 580 A polyedral angle is the figure formed by three or more planes proceeding from a point. Thus, S-ABCD is a polyedral angle. The common point S is the vertex; the planes SAB, SBC, etc., are the faces; the intersections of the faces, SA, SB, etc., are the edges; and the angles ASB, BSC, etc., are the face angles of the polyedral angle. Two consecu- A C tive faces of a polyedral angle form a diedral angle. 581 The parts of a polyedral angle are its face angles and its diedral angles. 582 A polyedral angle is convex if a section made by a plane cutting all its faces is a convex polygon. 583 A triedral angle is a polyedral angle of three faces; a tetraedral angle is a polyedral angle of four faces, etc. 584 An isosceles triedral angle is a triedral angle having two equal face angles. 585 Triedral angles are rectangular, bi-rectangular, or trirectangular, according as they have one, two, or three right diedral angles. 586 Two polyedral angles are equal if their face and diedral angles are equal respectively and arranged in the same order. 587 Two polyedral angles are sym- s s' metrical if their face and diedral angles are equal respectively and arranged in reverse order; as, S-ABC A l and S'-A'B'C'. B B POLYEDRAL ANGLES 297 Symmetrical polyedral angles cannot, in general, be made to coincide. For example, the right glove will not fit the left hand. 588 Vertical polyedral angles are those which have a common vertex, and in which the edges of one are the prolongations of the edges of the other. PROPOSITION XXV. THEOREM 589 Vertical polyedral angles are symmetrical. 'B' C? A' S CB B HYPOTHESIS. S-ABCD and S-A'B'C'D' are vertical polyedral angles. CONCLUSION. S-ABCD and S-A'B'C'D' are symmetrical. PROOF The face angles ASB, BSC, etc., are equal to the face angles A'SB', B'SC', etc., respectively. ~ 112 The diedral angles SB and SB', being formed by the same two intersecting planes (~ 516), are vertical diedral angles and therefore equal. ~ 560 Likewise the other diedral angles are equal respectively. But the parts of the polyedral angles are arranged in reverse order.. the polyedral angles are symmetrical. ~ 587 Q. E. D. 298 SOLID GEOMETRY -BOOK VI PROPOSITION XXVI. THEOREM 590 The sum of any two face angles of a triedral angle is greater than the third. s A --- C HYPOTHESIS. In the triedral angle S-ABC, let the face angle ASC be greater than either of the face angles ASB or BSC. CONCLUSION. L ASB + Z BSC > / ASC. PROOF In the face ASC draw SD making Z ASD = Z ASB. Take SD = SB, and through B and D pass a plane cutting the triedral angle in the section ABC. The A ASD and ASB are equal. ~ 162.. AB =AD (1). ~166 But BC + AB > AC (2). ~ 160 Subtracting (1) from (2), BC > DC. Ax. 5 In the A BSC and DSC, we have SB = SD by const., SC common, and BC > DC. Proved.-. BSC > L DSC. ~171 By const. Z ASB = / ASD. Adding, / ASB + Z BSC > Z ASC. Ax. 2 Q. E. D. EXERCISE 1168. Find the locus of a point equidistant from two given planes, and also equidistant from two given points. POLYEDRAL ANGLES 299 PROPOSITION XXVII. THEOREM 591 The sum of the face angles of any convex polyedral angle is less than four right angles. S HYPOTHESIS. S-ABCDE is a convex polyedral angle. CONCLUSION. Z ASB + Z BSC + etc., is less than four rt. A. PROOF Cut the edges of the polyedral angle by a plane whose section is the convex polygon ABCDE (~ 582), and join any point 0 within this polygon to all its vertices. We thus have two sets of triangles, the first set having a common vertex S, the second set having a common vertex O. Since each set has the same number of triangles, the sum of all the angles in each set is the same. But the sum of all the base angles of the first set is greater than the sum of all the base angles of the second set, for / SBA + Z SBC > Z ABC, etc. ~ 590 Therefore the sum of the face angles about S is less than the sum of the angles about 0. Ax. 6 But the sum of the angles about O is four rt. A. ~ 107 Therefore the sum of the face angles about S is less than four rt.. Q. E. D. 300 SOLID GEOMETRY -BOOK VI PROPOSITION XXVIII. THEOREM 592 Two triedral angles are equal or symmetrical, if the three face angles of the one are equal respectively to the three face angles of the other. S S' S' D 'DI A X.FU C At:' C C. Ft. A' B B'I BI Fig. 1 Fig. 2 Fig. 3 HYPOTHESIS. In the triedral angles S and S', the angles ASC, ASB, and BSC are equal respectively to the angles A'S'C', A'S'B', and B'S'C'. CONCLUSION. S and S' are equal or symmetrical. PROOF On the edges of the triedral angles, take SA = SB = SC = S'A' = S'B' = S'C', and complete the A ABC and A'B'C'. Then A SAB = A S'A'B'. ~ 162.. AB = A'B'. ~ 166 Likewise BC = B'C', and AC = A'C'... A ABC A'B'C'. ~ 172 From any point D in SA draw DE in the face SAB and DF in the face SAC each 1 to SA. Since SAB and SAC are isosceles A, Const. DE will meet AB in some point E, and DF will meet AC in some point F. Join EF. In S'A' take A'D' = AD. Draw D'E' in the face S'A'B' and D'F' in the face S'A'C' each 1L to S'A', and join E'F'. POLYEDRAL ANGLES 301 Rt. A ADE = rt. A A'D'E'; ~ 169 for / DAE = Z D'A'E', ~ 166 and AD = A'D'. Const..'. AE = A'E', and DE = D'E'. ~ 166 Likewise AF = A'F', and DF = D'F'... A AEF = A'E'F'; ~ 162 for AE = A'E', AF = A'', and Z EAF = Z E'A'F'. Proved.. EF=E'F'. ~166.. A DEF A D'E'F'; ~ 172 for DE = D'E', DF D'F', and EF = E'F'. Proved. Z EDF = E'D'F'. ~ 166 But the s EDF and E'D'F' measure the diedral s SA and S'A' respectively. ~ 562.'. the diedral A SA and S'A' are equal. Likewise the diedral As SB and SC are equal respectively to the diedral /s S'B' and S'C'. Thus far the demonstration applies equally to figures 1 and 2, or 1 and 3. In figures 1 and 2 the equal parts of the triedral angles are arranged in the same order, and therefore the two triedral angles are equal. ~ 586 In figures 1 and 3 the equal parts of the triedral angles are arranged in reverse order, and therefore the two triedral angles are symmetrical. ~ 587 Q. E. D. 593 COROLLARY. If two triedral angles have the face angles of the one equal to the face angles of the other, each to each, the diedral angles of the one are respectively equal to the diedral angles of the other. 302 SOLID GEOMETRY- BOOK VI EXERCISES 1169 If two parallel lines are not perpendicular to a plane, their projections upon that plane are parallel lines. 1170 Are two lines necessarily parallel whose projections upon a plane are parallel? 1171 If a straight line intersects two parallel planes, it makes equal angles with the planes. 1172 When will the projection of a circle upon a plane be a circle? When will the projection be a straight line? 1173 If the projections of n points upon a plane are in a straight line, the n points are all in the same plane. 1174 If a line meets a plane obliquely, with what line in the plane does it make the greatest angle? 1175 A plane can be perpendicular to only one edge of a polyedral angle. 1176 A plane can be perpendicular to only two faces of a polyedral angle. 1177 If two face angles of a triedral angle are equal, the opposite diedral angles are equal. 1178 The three planes bisecting the three diedral angles of a triedral angle intersect in the same line. 1179 If the three face angles of a triedral angle are equal, the three diedral angles are equal. 1180 The three planes passed through the edges of a triedral angle, perpendicular to the opposite faces, intersect in the same line. 1181 Find the locus of a point equidistant from two intersecting planes. 1182 Find the locus of a point equidistant from the vertices of a triangle; of a rectangle. 1183 Find the locus of a point equidistant from the edges of a triedral angle. 1184 Find the locus of a point equidistant from the faces of a triedral angle. [Let AO be the intersection of the three planes bisecting the three diedral angles of the triedral angle (Ex. 1178). Then all points in AO are equidistant from the three faces (~ 570). Any point without AO is without at least two of the angle-bisecting planes, and is therefore not equidistant from the three faces of the triedral angle. Hence AO is the locus required.] BOOK VII POLYEDRONS, CYLINDERS, AND CONES POLYEDRONS DEFINITIONS 594 A polyedron is a solid bounded by planes. The faces of a polyedron are the polygons bounding it. The edges of a polyedron are the intersections of its faces. The vertices of a polyedron are the intersections of its edges. 595 A diagonal of a polyedron is a straight line joining any two of its vertices not in the same face. 596 A convex polyedron is a polyedron every section of which is a convex polygon. All polyedrons considered in this work are convex. Tetraedron Hexaedron Octaedron Dodecaedron Icosaedron 597 A tetraedron is a polyedron of four faces. A hexaedron is a polyedron of six faces. An octaedron is a polyedron of eight faces. A dodecaedron is a polyedron of twelve faces. An icosaedron is a polyedron of twenty faces. 303 304 SOLID GEOMETRY-BOOK VII PRISMS 598 A prism is a polyedron two of whose faces are equal and parallel polygons, and whose other faces are parallelograms. Prisms 599 The bases of a prism are two equal parallel faces; the lateral faces are all the faces except the bases; the lateral edges are the intersections of the lateral faces; the base edges are the intersections of the bases with the lateral faces; and the lateral area is the sum of the areas of the lateral faces. 600 Prisms are triangular, quadrangular, etc., according as their bases are triangles, quadrilaterals, etc. 601 The altitude of a prism is the perpendicular between the planes of its bases. 602 A right prism is a prism whose lateral edges are perpendicular to its bases. An oblique prism is a prism whose lateral edges are oblique to its bases........... 603 A regular prism is a right prism whose bases are regular polygons. | a 604 A truncated prism is a portion of a " prism included between a base and a section formed by a plane oblique to the base and cutting all the lateral edges.:::':ii PRISMS AND PARALLELOPIPEDS 305 605 COROLLARY. The lateral edges of a prism are equal and parallel; the lateral edges of a right prism are equal to its altitude; and the lateral faces of a right prism are rectangles. 606 A parallelopiped is a prism whose bases are parallelograms. Parallelopipeds 607 An oblique parallelopiped is a parallelopiped whose lateral edges are oblique to its bases. 608 A right parallelopiped is a parallelopiped whose lateral edges are perpendicular to its bases. 609 A rectangular parallelopiped is a right parallelopiped whose bases are rectangles. 610 A cube is a parallelopiped all of whose faces are squares. 611 A right section of a prism is a section formed by a plane perpendicular to its.iii i lateral edges. 612 The volume of a solid is the number of units of volume which it contains. In applied Geometry the unit of volume is usually a cube whose edge is some linear - unit; as, a cubic inch, a cubic foot, etc. 613 Two solids having equal volumes are equivalent. 306 SOLID GEOMETRY-BOOK VII PROPOSITION I. THEOREM 614 Sections of a prism made by parallel planes cutting all the lateral edges are equal polygons....... i i!iiiii.. HYPOTHESIS. ABCDE and A'B'C'D'E' are sections of the prism MN, formed by parallel planes cutting all the lateral edges. CONCLUSION. ABCDE and A'B'C'D'E' are equal polygons. PROOF AB is II to A'B', BC is II to B'C', etc. ~ 543.-. Z ABC = Z A'B'C', Z BCD = Z B'C'D', etc. ~ 549 Also, AB = A'B', BC = B'C', etc. ~ 202.'.ABCDE and A'B'C'D'E' are mutually equilateral and equiangular and can be made to coincide..'. ABCDE = A'B'C'D'E'. ~ 89 Q. E. D. 615 COROLLARY 1. Any section of a prism made by a plane parallel to the base is equal to the base. 616 COROLLARY 2. All right sections of a prism are equal. EXERCISES 1185 Any section of a prism made by a plane parallel to a lateral edge is a parallelogram. 1186 In the above figure, what kind of solid is MD'? PRISMS AND PARALLELOPIPEDS 307 PROPOSITION II. THEOREM 617 The lateral area of a prism is equal to the product of a lateral edge by the perimeter of a right section. El A' D'............... L H B C HYPOTHESIS. AD' is a prism, S its lateral area, E a lateral edge, and P the perimeter of the right section FK. CONCLUSION. S= E x P. PROOF The face AB' is a parallelogram, whose base is AA' and altitude FG. ~ 611.-. area ABf'= AA'x FG = E x FG. ~ 401 Likewise area BC' = BB' x GH = E x GH, etc. ~ 605 Adding, S = E(FG + GH + HK + etc.)..-. S=ExP. Q. E. D. 618 COROLLARY. The lateral area of a right prism is equal to the product of the altitude by the perimeter of the base. EXERCISES 1187 Find the lateral area of a prism, if the lateral edge is 18 inches and the perimeter of a right section is 32 inches. 1188 Find the lateral area of a right prism, if the altitude is 21 inches and the perimeter of the base is 38 inches. 308 SOLID GEOMETRY-BOOK VII PROPOSITION III. THEOREM 619 Two prisms are equal, if the three faces including a triedral angle of the one are respectively equal to the three faces including a triedral angle of the other, and are similarly placed.:;.;i ijjiii i iiilijij.iijiii;:. J Jc ) K r H K' aHt Bg Ce E'-.... A B A' B' HYPOTHESIS. In the prisms AJ and A'J', the faces AD, AG, and AK are respectively equal to the faces A'D', A'G', and A'K', and similarly placed. CONCLUSION. The prisms AJ and A'J' are equal. PROOF The triedral angles A and A' are equal, ~ 592 and the upper bases FJ and F'J' are equal. ~ 599 Apply the triedral angle A to the triedral angle A'. The faces AD, AG, and AK coincide respectively with the faces A'D', A'G', and A'K'; and the points K, F, and G coincide respectively with the points K', F', and G'... the upper bases FJ and F'J' coincide. ~ 515.'. the prisms coincide and are equal. Q. E. D. 620 COROLLARY 1. Two truncated prisms are equal, if the three faces including a triedral angle of the one are respectively equal to the three faces including a triedral angle of the other, and are similarly placed. 621 COROLLARY 2. Two right prisms are equal, if they have equal bases and equal altitudes. PRISMS AND PARALLELOPIPEDS 309 PROPOSITION IV. THEOREM 622 An oblique prism is equivalent to a right prism whose base is equal to a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. J H K'n R E PROOF The altitude of FJ' is equal to FF'. ~ 605 AA' FF'. Hyp. Subtract FA' = FA', and AF = A'F'. Ax. 4 Likewise BG = B'G', etc. Also, AB = A'B', and FG = F'G'. ~ 200.'. the faces AG and A'G' are equal, for they are mutually equilateral and equiangular. ~ 122 Likewise the facesn o A K' ar e equ al. Also, the faces AD and A'D' are equal. ~ 599.-. the truncated prisms AJ and A'J' are equal, ~ 620 and if to each we add the truncated p rism F D', we have AD' = FJ'. Ax. 1 we have AD' E J'. Ax. 1 Q. E. D. 310 SOLID GEOMETRY-BOOK VII PROPOSITION V. THEOREM 623 The opposite lateral faces of a parallelopiped are equal and parallel. I 1 D --- A B HYPOTHESIS. AC and EG are the bases of the parallelopiped AG. CONCLUSION. The opposite faces AF and DG are equal and parallel. PROOF AB is equal and parallel to DC, ~~ 200, 195 and AE is equal and parallel to DH... / EAB- Z HDC. ~549.'. AF = DG, ~ 207 and AF is II to DG. ~ 549 Likewise the faces AH and BG are equal and parallel. Q. E. D. 624 COROLLARY. Any two opposite faces of a parallelopiped may be taken as bases. EXERCISE 1189. The diagonals of a parallelopiped bisect one another. For the section formed by passing a plane through X f two diagonally opposite edges, as AB and HG, is a parallelogram whose diagonals AG and BH bisect each other in 0. Likewise BH and CE bisect each A, other in 0, and so on for any two diagonals. The point 0 is called the center of the parallelo- B C piped. PRISMS AND PARALLELOPIPEDS 311 PROPOSITION VI. THEOREM 625 The plane passed through two diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms. _ _ -------- Em G L -: ""B:-:::^::=::=jB HYPOTHESIS. The plane EACG passes through the opposite edges EA and CG of the parallelopiped EC. CONCLUSION. The prisms ABC-F and ADC-H are equivalent. PROOF Draw the right section JKLM, intersecting the plane AEGC in JL. Since the faces AF and DG are parallel, ~ 623 the lines JK and'ML are parallel. ~ 543 Likewise the lines JM and KL are parallel..-. JKLM is a parallelogram, ~ 195 and the triangles JKL and JML are equal. ~ 201 The prism ABC-F is equivalent to a right prism whose base is JKL and whose altitude is AE, and the prism ADC-H is equivalent to a right prism whose base is JML and whose altitude is-AE. ~ 622 But these two right prisms are equal. ~ 621 Therefore, the prisms ABC-F and ADC-H are equivalent. Ax. 11 Q. E. D. 312 SOLID GEOMETRY-BOOK VII PROPOSITION VII. THEOREM 626 Two rectangular parallelopipeds having equal bases are to each other as their altitudes. pr P A. A' HYPOTHESIS. AB and A'B' are the altitudes of the rectangular parallelopipeds P and P' whose bases are equal. P' A'B' CONCLUSION. = AB P AB CASE 1. When the altitudes are commensurable. PROOF Let k, a common measure of A'B' and AB, be contained m times in A'B' and n times in AB. A'B' m Then A.B. AB n By passing planes through the several points of division in A'B' and AB and perpendicular to A'B' and AB, P' will be divided into m parallelopipeds, and P into n parallelopipeds, all equal. ~ 621 P, m Whence _. P n P' AVB' ~ A1-B =l. Ax. 11 1 AB PRISMS AND PARALLELOPIPEDS 313 CASE 2. When the altitudes are incommensurable. pr P' A "' "A' PROOF Since A'B' and AB are not commensurable, let A'C and AB be commensurable, CB' being less than k. Pass a plane through C perpendicular to A'B', and denote the parallelopiped A'C by Q. Then - = AC Case 1 P AB If k be indefinitely diminished, CB', always less than k, will become less than any assigned quantity.. Q approaches P' as a limit, and A'C approaches A'B' as a limit; ~ 263 whence Q approaches - as a limit, P P A'C A'B' and A- approaches AB as a limit. ~ 268 AB AB But the variable ratios 9 and 'C are always equal. Case 1 P AB P' A '' PA t-R ~264 P AB Q. E. D. 627 DEFINITION. The dimensions of a rectangular parallelopiped are the three edges which meet at the same vertex. 628 COROLLARY. Two rectangular parallelopipeds which have two dimensions in common are to each other as their third dimensions. 314 SOLID GEOMETRY —BOOK VII PROPOSITION VIII. THEOREM 629 Two rectangular parallelopipeds having equal altitudes are to each other as their bases. HYPOTHESIS. P and P' are two rectangular parallelopipeds whose three dimensions are a) b, c, and a, b', c', respectively. CONCLUSION. P b c P' b' x c' PROOF Let Q be a rectangular parallelopiped having two dimensions a and b in common with P, and two dimensions a and c' in common with P'. Then P - Q Cr' Q b and,= b P bf ~ 628 Multiplying these two ber, we have equations together, member by memP bxc P' b'xc' Q. E. D. 630 COROLLARY. Two rectangular parallelopipeds which have one dimension in common are to each other as the products of their other two dimensions. PRISMS AND PARALLELOPIPEDS 315 PROPOSITION IX. THEOREM 631 Two rectangular parallelopipeds are to each other as the products of their three dimensions. HYPOTHESIS. P and P' are two rectangular parallelopipeds whose three dimensions are a, b, c, and a', b', c', respectively. P axbxc CONCLUSION. P = a xx c P' a' x b' x c' PROOF Let Q be a rectangular parallelopiped having two dimensions a and b in common with P, and the third dimension c' in common with P'. Then ~ 628 Q cn, and Q a xb ~630 P' a'x b' Multiplying these two equations together, member by member, we have P aXbXc P a X b x c P a' x b' x c' Q. E. D. EXERCISE 1190. Find the ratio of two rectangular parallelopipeds whose dimensions are 5, 6, 8, and 12, 15, 16, respectively. 316 SOLID GEOMETRY-BOOK VII PROPOSITION X. THEOREM 632 The volume of a rectangular parallelopiped is equal to the product of its three dimensions. HYPOTHESIS. P is a rectangular parallelopiped whose three dimensions are a, b, and c. CONCLUSION. The volume of P = a x b x c. PROOF Let U be the unit of volume. Then P- x b XcaXb c. ~ 631 U 1xIx1 But P = the volume of P. ~ 612 U Therefore, the volume of P = a X b x c. Q. E. D. 633 COROLLARY 1. The volume of a cube is equal to the cube of its edge. 634 COROLLARY 2. The volume of a rectangular parallelopiped is equal to the product of its base by its altitude. 635 SCHOLIUM. If the linear unit is an exact divisor of the three dimensions of a rectangular parallelopiped, the truth of Prop. X is made evident by the annexed figure. Thus, if the three dimensions contain the linear unit 3, 4, 5, times respectively, the solid can be divided into 3 x 4 x 5, or 60 parallelopipeds, each equal to the unit of volume. PRISMS AND PARALLELOPIPEDS 317 PROPOSITION XI. THEOREM 636 The volume of any parallelopiped is equal to the product of its base by its altitude. D F —F G 1 C c.. -.. HYPOTHESIS. P is any parallelopiped whose base is B and whose altitude is H. CONCLUSION. The volume of P = B x H. PROOF Produce CD and all the edges of P II to CD, and on CD produced take EF = CD. Through E and F pass planes _L to EF, cutting all the edges of P produced, thus forming the oblique parallelopiped P' whose bases are rectangles. Produce FG and all the edges of P' II to FG, and on FG produced take HK = FG. Through H and K pass planes 1I to HK, cutting all the edges of P' produced, thus forming the rectangular parallelopiped P". P, P', and P" have the common altitude H. ~ 548 B = B', and B' = B". ~ 402.'. B _ B". Ax. 11 Also, P P', and P' P P". ~ 622.. P P". Ax. 11 But the volume of P" = B" X H. ~ 634 Substituting P for P", and B for B", we have the volume of P = B x H. Q.E. D. 318 SOLID GEOMETRY-BOOK VII PROPOSITION XII. THEOREM 637 The volume of a triangular prism is equal to the product of its base by its altitude. F' El I ise eit HYPOTHESIS. CDE-D' is any triangular prism whose volume is V, base B, and altitude H. CONCLUSION. V = B x H. PROOF Construct the parallelopiped CDEF-D'. CDE-D' 1 ~ CDEF-D'. ~ 625 But CDEF-D' = CDEF x H. ~ 636.'. CDE-D' = CDEF x H.... CDE-D' = CDE x H; ~ 201 or V = B x H. Q. E. D. EXERCISES 1191 The square of a diagonal of a rectangular parallelopiped is equal to the sum of the squares of its three dimensions. 1192 The sum of the squares of the four diagonals of any parallelopiped is equal to the sum of the squares of its twelve edges. 1193 Any straight line drawn through the center of a parallelopiped (Ex. 1180) and terminated by its faces is bisected at the center. PRISMS AND PARALLELOPIPEDS 319 PROPOSITION XIII. THEOREM 638 The volume of any prism is equal to the product of its base by its altitude. Df EA' /' A B HYPOTHESIS. AC' is any prism whose volume is V, base B, and altitude H. CONCLUSION. V = B x H. PROOF By passing planes through a lateral edge AA' and the diagonals of the base; AD, AC, etc., the given prism will be divided into triangular prisms, all having the common altitude H. The volume of each triangular prism is equal to the product of its base by its altitude. ~ 637 The volume of the given prism is equal to the sum of the volumes of the triangular prisms. Ax. 13 But the sum of the volumes of the triangular prisms is equal to the sum of their bases multiplied by their common altitude; or ABCDE x H. That is, V = B x H. Q. E. D. 639 COROLLARY. Prisms are to each other as the products of their bases by their altitudes; prisms having equivalent bases are to each other as their altitudes; prisms having equal altitudes are to each other as their bases; prisms having equivalent bases and equal altitudes are equivalent. 320 SOLID GEOMETRY-BOOK VII PROBLEMS OF COMPUTATION 1194 The edge of a cube is 8 inches. Find its volume and entire surface. 1195 The dimensions of a rectangular parallelopiped are 5, 8, and 12. Find its volume and entire surface. 1196 The height of a right triangular prism is 30 inches, and the base edges are 7, 15, and 20 inches. Find the volume and entire surface. 1197 Find the edge of a cube whose volume is equal to the volume of a rectangular parallelopiped whose edges are 6, 16, and 18 inches. 1198 Find the volume and the lateral area of a regular hexagonal prism, if the altitude is 14 feet, and a side of the hexagon is 1 foot. 1199 How many bushels will a rectangular bin hold whose inside dimensions are 4, 5, and 6 feet, allowing 14 cu. ft. to a bushel? 1200 How many gallons of water will a rectangular cistern hold whose inside dimensions are 2, 3, and 5 feet, allowing 7) gallons to a cubic foot? 1201 The lateral edge of a prism is 36 inches, and the perimeter of its right section is 42 inches. Find the lateral surface. 1202 Find the volume of a right prism whose altitude is 28 inches, and whose base contains 6 square feet. 1203 The volume of a rectangular parallelopiped is 120 cubic feet, and the dimensions of its base are 3 and 4 feet. Find the altitude. 1204 Show that the diagonal of a cube is equal to its edge multiplied by 3. 1205 Find the edge of a cube whose entire surface is one square yard and fifty-four square inches. 1206 Find the edge of a cube whose diagonal is 5V3. 1207 Find the edge of a cube whose entire surface and volume are expressed by the same number. 1208 Find the volume of a cube whose entire surface is 1014 square inches. 1209 Find the volume and entire surface of a cube in which the diagonal of a face is a. 1210 Find the entire surface, the volume, and a diagonal of a rectangular parallelopiped whose dimensions are a, b, c. PYRAMIDS 321 PYRAMIDS DEFINITIONS 640 A pyramid is a polyedron one of whose faces, called the base, is a polygon, and whose other faces are triangles having a common vertex. 641 The lateral faces of a pyramid are all the faces except the base. s 642 The lateral area of a l i pyramid is the sum of the i"!l I i: areas of the lateral faces. 643 The lateral edges of a _.... pyramid are the intersections of its lateral faces. 644 The vertex of a pyramid is the common vertex of its lateral faces. 645 The altitude of a pyramid is the perpendicular drawn from the vertex to the plane of the base. 646 A pyramid is triangular, quadrangular, etc., according as its base is a triangle, quadrilateral, etc. A triangular pyramid is called a tetraedron. It has four triangular faces any one of which may be taken for the base. 647 A regular pyramid is a pyramid whose base is a regular polygon, and whose altitude meets the base at its center. 648 The axis of a regular pyramid is the straight line drawn from the vertex to the center of the base.. 322 SOLID GEOMETRY-BOOK VII The following corollaries are immediately inferred from the definition of a regular pyramid: 649 COROLLARY 1. The lateral edges of a regular pyramid are equal. ~ 530 650 COROLLARY 2. The lateral faces of a regular pyramid are equal isosceles triangles. ~ 172 651 COROLLARY 3. The altitudes of the lateral faces of a regular pyramid are equal. 652 The slant height of a regular pyramid is the altitude of any lateral face. 653 A truncated pyramid is the portion iil of a pyramid included between its base - '|ii and a section formed by a plane cutting all its lateral edges. 654 A frustum of a pyramid is a truncated pyramid in which the cutting section is parallel to the base. ii.iii,...... The lower base of the frustum is the base of the pyramid, and the upper base ll of the frustum is the parallel section. -| — The lateral faces of a frustum of a pyramid are trapezoids....... 655 The altitude of a frustum of a pyramid is the perpendicular between the planes of the bases. 656 The lateral area of a frustum of a pyramid is the sum of the areas of the lateral faces. The lateral faces of a frustum of a regular pyramid are equal isosceles trapezoids. 657 The slant height of a frustum of a regular pyramid is the altitude of any lateral face. PYRAMIDS 323 PROPOSITION XIV. THEOREM 658 The lateral area of a regular pyramid is equal to half the product of its slant height by the perimeter of its base. HYPOTHESIS. V-ABCDE is a regular pyramid whose lateral area is S, its slant height L, and the perimeter of its base P. CONCLUSION. S= L x P. PROOF The lateral faces of V-ABCDE are equal isosceles triangles. ~ 650 The altitude of each A is L. ~ 652 The area of each A is 1 L x its base. ~ 406.'. the sum of the areas of all the A is ~ L x the sum of the bases. But the sum of the areas of all the A is S, and the sum of the bases is P...S= L x P. 659 COROLLARY. The lateral area of a frustum of a regular pyramid is equal to half the sum of the perimeters of its bases multiplied by the slant height. ~ 411 For the lateral faces. are equal isosceles trapezoids (~ 656). Q. E. D. E, A' - Df/ AAA B R C 324 SOLID GEOMETRY-BOOK VII PROPOSITION XV. THEOREM 660 If a pyramid is cut by a plane parallel to the base: 1 The lateral edges and the altitude are divided proportionally. 2 The section is a polygon similar to the base. 0 o A Atl B HYPOTHESIS. The pyramid O-ABCD is cut by a plane parallel to its base, intersecting the lateral edges in a, b, c, d, and the altitude OK in k. rN Ob Ok CONCLUSION. 1 Ob OA OB OK PROOF abcd is parallel to ABCD by hypothesis... ab is II to AB, be is II to BC *, and ak is 1I to AK. ~543 Oa Ob Ok 3 OA OB OK 2 The polygons abed and ABCD are similar. PROOF Since the corresponding sides of the two polygons are parallel, ~ 543 Z abc = ABC, Z bcd = Z BCD, etc. ~ 549 PYRAMIDS 325 Again. Since the A Oab, Obc, etc., are similar respectively to the A OAB, OBC, etc., ~ 351 ab _(Ob\ be (c) cd etc. AB ~ OB) BC K OC) CD'.'. the polygons abcd and ABCD are similar. ~ 349 Q. E. D. 661 COROLLARY 1. Any section of a pyramid parallel to the base is to the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. For Oc: OK = Oa: OA = ab: AB; ~ 359 2 2 -2 2 whence Ok2: OK2= ab:AB. ~ 338 But abcd: ABCD = ab2: AX. ~ 416. abed: ABCD = k2: OK2. Ax. 11 662 COROLLARY 2. If two pyramids having equal altitudes are cut by planes parallel to the bases, and at equal distances from the vertices, the sections have the same ratio as the bases. -2 ~-2 For abcd: ABCD = Okc2 OK2, ~ 661 and ab'c': A'B'C' = k: O'K'2. ~ 661 But Ok = O'k', and OK = O'K' by hypothesis..'. abed: ABCD = a'b'c': A'B'C'; Ax. 11 or abcd: a'b'c' = ABCD: A'B'C'. ~ 331 663 COROLLARY 3. If two pyramids have equal altitudes and equivalent bases, sections made by planes parallel to the bases, and at equal distances from the vertices, are equivalent. For abcd: a'b'c' = ABCD: A'B'C'. ~ 662 Since ABCD - A'B'C', abcd = a'b'c'. 326 SOLID GEOMETRY -BOOK VII PROPOSITION XVI. THEOREM 664 The volume of a triangular pyramid is the limit of the sum of the volumes of a series of inscribed or circumscribed prisms of equal altitude, when the number of prisms is indefinitely increased. L S H B HYPOTHESIS. S-ABC is a triangular pyramid. CONCLUSION. The volume of S-ABC is the limit of the sum of the volumes of a series of inscribed or circumscribed prisms of equal altitude, when the number of prisms is indefinitely increased. PROOF Divide the altitude AL into any number of equal parts, and denote one of them by h. Through the points of division pass planes parallel to the base, cutting the pyramid in the sections DEF, GHK, etc. Upon the triangles DEF, GHK, etc., as upper bases, construct prisms whose lateral edges are parallel to SA, and whose altitudes are each equal to 7h. A series of prisms, A-DEF, D-GHK, etc., are thuO inscribed in the pyramid. Upon the triangles ABC, DEF, etc., as lower bases, construct prisms whose lateral edges are parallel to SA, and whose altitudes are each equal to h. PYRAMIDS 327 A series of prisms, D-ABC, G-DEF, etc., are thus circumscribed about the pyramid. The inscribed prism A-DEF is equal to the circumscribed prism G-DEF, for each has the same base DEF and the same altitude h. ~ 639 Likewise each inscribed prism is equal to the circumscribed prism next above it. Hence the difference between the sum of the inscribed prisms and the sum of the circumscribed prisms is the prism D-ABC. Now the pyramid is greater than the sum of the inscribed prisms and less than the sum of the circumscribed prisms. Ax. 12 Hence the difference between the volume of the pyramid and the volume of either series of prisms is less than the prism D-ABC. By indefinitely increasing the number of equal parts into which the altitude AL is divided, the volume of the prism D-ABC can be made less than any assigned quantity. Hence the difference between the volume of the pyramid and the volume of either series of prisms can be made less than any assigned quantity. Therefore the volume of the pyramid is the limit of the sum of the volumes of either series of prisms when indefinitely increased. ~ 263 Q. E. D. EXERCISES 1211 The altitude of a regular triangular pyramid is 18 inches, and each side of the base is 10 inches. Find the side of a section made by a plane parallel to the base and 8 inches from the vertex. 1212 The altitude of a regular quadrangular pyramid is 15 feet, and each side of the base is 9 feet. Find the area of a section made by a plane parallel to the base and 5 feet from the vertex. 328 SOLID GEOMETRY-BOOK VII PROPOSITION XVII. THEOREM 665 Two triangular pyramids having equivalent bases and equal altitudes are equivalent. L S S/ A -A,' C' B B' HYPOTHESIS. S-ABC and S'-A'B'C' are two triangular pyramids having equivalent bases in the same plane, and a common altitude AL. CONCLUSION. S-ABC == S'-A'B'C'. PROOF Divide the altitude AL into any number of equal parts, and denote one of them by h. Through the points of division pass planes parallel to the base, cutting the pyramids in the sections DEF, GHK, etc., D'E'F', G'H'K', etc. On these triangles as upper bases inscribe prisms, having h as their altitudes. Since the corresponding sections are equivalent, ~ 663 the corresponding prisms are equivalent. ~ 639 Hence the sum of the volumes of the prisms inscribed in S-ABC is equal to the sum of the volumes of the prisms inscribed in S'-A'B'C'. Ax. 1 Denote these sums by V and V' respectively. Then V= V'. By indefinitely increasing the number of equal parts into which AL is divided, V and V', always equal, approach as their limits the pyramids S-ABC and S'-A'B'C' respectively. ~ 664 Therefore S-ABC = S'-A'B'C'. ~ 264 Q. E. D. PYRAMIDS 329 PROPOSITION XVIII. THEOREM 666 The volume of a triangular pyramid is equal to one third the product of its base by its altitude. D F D D E E B B B HYPOTHESIS. E-ABC is a triangular pyramid, V its volume, B its base, and H its altitude. CONCLUSION. V= B x H. PROOF On the base ABC construct the prism ABC-DEF, whose lateral edges are equal and parallel to EB. The plane EAC divides the prism into the given pyramid E-ABC and the quadrangular pyramid E-ACFD. The plane DEC divides the quadrangular pyramid E-ACFD into the triangular pyramids E-DCF and E-DCA, which have the same altitude and equal bases. ~ 201.-. E-DCF - E-DCA. ~ 665 The pyramid E-DCF is the same as the pyramid C-DEF. ~ 646 But C-DEF - E-ABC. ~ 665 Hence the prism is divided into three equivalent pyramids. Ax. 11.'. the pyramid E-ABC is one third of the prism. But the volume of the prism = B x H. ~ 637. V= Bx H. Q. E. D. 330 SOLID GEOMETRY-BOOK VII PROPOSITION XIX. THEOREM 667 The volume of any pyramid is equal to one third the product of its base by its altitude. HYPOTHESIS. S-ABCDE is any pyramid, V its volume, B its base, and H its altitude. CONCLUSION. V= 1 B X H. PROOF Planes passed through a lateral edge SA and the diagonals of the base, AD, AC, etc., divide the pyramid into triangular pyramids, having H for their common altitude. The base of the given pyramid is equal to the sum of the bases of the triangular pyramids. Ax. 12 The volume of the given pyramid is equal to the sum of the volumes of the triangular pyramids, which is one third the sum of their bases multiplied by their common altitude. ~ 666.'V=.- BXH..~~.V=~~x3 HQ. E. D. 668 COROLLARY 1. The volumes of two pyramids are to each other as the products of their bases and altitudes. 669 COROLLARY 2. Pyramids having equivalent bases are to each other as their altitudes; pyramids having equal altitudes are to each other as their bases. PYRAMIDS 331 670 COROLLARY 3. Pyramids having equivalent bases and equal altitudes are equivalent. 671 SCHOLIUM. The volume of any polyedron may be found by dividing it into pyramids, and computing their volumes separately. PROBLEMS OF COMPUTATION 1213 The altitude of a pyramid is 9 feet, and the area of its base is 216 square feet. Find the area of a section parallel to the base and 3 feet from the vertex. 1214 The slant height of a regular triangular pyramid is 12 feet, and the base edge is 5 feet. Find the lateral area. 1215 The slant height of a regular pentagonal pyramid is 21 feet, and the base edge is 8 feet. Find the lateral area. 1216 Find the total surface of a regular pyramid whose slant height is 10 inches, and whose base is a square with side 3 inches. 1217 Find the volume of a regular pyramid whose altitude is 7, and whose base is a square with side 3. 1218 Find the volume of a regular pyramid whose altitude is 20, and whose base is a triangle with side 6. 1219 Find the volume of a pyramid whose altitude is 13, and whose base is a rectangle with sides 3 and 4. 1220 Find the volume of a triangular pyramid, if the altitude is 60, and the sides of the base are 11, 13, and 20. 1221 Find the altitude of a triangular pyramid, if the volume is 3780, and the sides of the base are 12, 17, and 25. 1222 Find the lateral area of a regular quadrangular pyramid whose altitude is 24, and whose base edge is 14. 1223 Find the lateral area of a regular hexagonal pyramid whose altitude is 12, and whose base edge is 2. 1224 A regular pyramid 35 feet high stands on a square base each side of which is 24 feet. Find its lateral edge. 1225 Each face of a triangular pyramid is an equilateral triangle whose side is 4. Find the volume and total surface. 332 SOLID GEOMETRY-BOOK VII PROPOSITION XX. THEOREM 672 A frustum of a triangular pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and the mean proportional between the bases of the frustum. Q / '@^K HYPOTHESIS., ABC-DEF is a frustum of a triangular pyramid whose lower base is B, upper base B', and altitude H. CONCLUSION. ABC-DEF = three pyramids whose common altitude is H, and whose bases are B, B', and VB x B'. PROOF Planes EAC and EDC divide the frustum into three triangular pyramids, E-ABC, C-DEF, and E-DCA. Denote these pyramids by P, Q, and R respectively. P and Q have the common altitude H, and B and B' for their bases respectively; hence it remains to prove that R is equivalent to a pyramid whose altitude. is H, and whose base is VB X B'. The pyramid P is the same as the pyramid C-AEB, and the pyramid R is the same as the pyramid C-AED. ~ 646.~. P: R= A AEB: AAED. ~ 669 PYRAMIDS 333 The triangles AEB and AED have the same altitude, the altitude of the trapezoid DABE. Hence AB: DE = A AEB: A AED. ~ 409.'. P:R= AB: DE (1). Ax. 11 The pyramid R is the pyramid E-DCA, and the pyramid Q is the same as the pyramid E-DCF. ~ 646... R: Q =A DCA: A DCF. ~ 669 The triangles DCA and DCF have the same altitude, the altitude of the trapezoid DACF. Hence AC: DF = A DCA: A DCF. ~ 409.. R: Q = AC: DF (2). Ax. 11 The triangles ABC and DEF are similar. ~ 660 Hence AB: DE = AC: DF (3). ~ 349 From (2) and (3), R:Q = AB: DE (4). Ax. 11 From (1) and (4), P R = R.: Q. Ax. 11 Whence R = P x Q. ~ 329 ButP= ~ H x B, and Q=~ Hx B'. ~ 666.'. R = V- H x B x 3 H x B' = H x VB/ x B'. That is, R is equivalent to a pyramid whose altitude is H, and whose base is VB x B'. Q. E. D. -673 COROLLARY. Denoting the volume of a frustum of a triangular pyramid by V, the lower base by B, the upper base by B', and the altitude by H, we have V= H x B+ 1 H x B' + H x V/B x B' = - H x (B + B'+ V/B x B'). EXERCISE 1226. Find the volume of a frustum of a triangular pyramid whose altitude is 12 inches, and whose bases are 25 square inches and 16 square inches respectively. 334 SOLID GEOMETRY-BOOK VII PROPOSITION XXI. THEOREM 674 The volume of a frustum of any pyramid is equal to the sum of the volumes of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and the mean proportional between the bases of the frustum. s T A A C N HYPOTHESIS. ABCD-EFGH is a frustum of any pyramid S-ABCD, V its volume, B the lower base, B' the upper base, and H the altitude. CONCLUSION. V = - H (B + B' + x/B x B'). PROOF Let T-MNO be a triangular pyramid whose altitude is equal to the altitude of S-ABCD, and whose base MNO is equivalent to ABCD and in the same plane. Then S-ABCD = T-MNO (1) ~ 670 Let the plane EFGH cut T-MNO in the section PQR. Then EFGH == PQR. ~ 663 Hence S-EFGH o T-PQR (2). ~ 670 Subtracting (2) from (1), ABCD-EFGH I MNO-PQR. Ax. 4 The volume of MNO-PQR = 1 H (B + B' + V/B x B'). ~ 673.-. V = H (B + B' + V/B X B'). Q. E. D. PYRAMIDS 335 PROPOSITION XXII. THEOREM 675 The volumes of two tetraedrons, having a triedral angle of the one equal to a triedral angle of the other, are to each other as the products of the three edges of the equal triedral angles. D G~~ a~G E B E HYPOTHESIS. V and V' are the volumes of the two tetraedrons A-BCD and A-EFG whose triedral angles A and A are equal. CONCLUSION. AE x AC x AD V' AE x AF x AG PROOF Apply the tetraedron A-EFG to A-BCD so that the equal triedral angles A and A shall coincide. Draw DH and GK I to the plane ABC, and let the plane of DH and GK intersect ABC in AKH. Since DH and GK are the altitudes of the pyramids D-ABC and G-AEF respectively, V ABC x DH ABC DH ~ X (1). ~ 668 V' AEF x GK AEF GK Now ABC AB x AC AEF AE X AF' DH AD and DGK A ~349 GK AG Substituting these values in (1), V AB x AC x AD V' AE x AF xAG Q. E. D. 336 SOLID GEOMETRY -BOOK VII PROPOSITION XXXIII. THEOREM 676 A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism, and whose vertices are the three vertices of the inclined section. B B B B Fig. 1 F F D D ' EE A C., C QtA B Fig. 2 HYPOTHESIS. ABC-DEF is a truncated triangular prism, whose base is ABC, and whose inclined section is DEF. CONCLUSION. ABC-DEF is equivalent to the sum of the three pyramids E-ABC, D-ABC, and F-ABC (Fig. i). PROOF Denote these pyramids by P, Q, and R respectively. Planes EAC and EDC divide the prism into the pyramids E-ABC, E-DAC, and E-DFC (Fig. 2). Denote these pyramids by P', Q', and R' respectively. We are to prove that P, Q, and R are equivalent to P', Q', and R' respectively. P and P' are equivalent for they are identical. Q is the same as B-ADC. ~ 646 PYRAMIDS 337 But Q' B-ADC. ~ 665 For the two pyramids have the same base ADC, and equal altitudes, since EB is II to the plane ADC. ~ 537..Q Q'. Ax. 11 R is the same as B-AFC. ~ 646 But R' B-AFC. ~ 665 For the two pyramids have equal altitudes, since EB is parallel to the plane of their bases; ~ 537 and they have equivalent bases, because the A DFC and AFC have the same base CF and equivalent altitudes, since AD is II to CF. ~ 605.-.R R'. Ax. 11.. ABC-DEF P' + Q' + R' % P + Q + R. Q. E. D. 677 COROLLARY 1. The volume of a truncated right triangular prism is equal to the product of its base by one third the sum of its lateral edges. For the lateral edges AD, BE, and: CF are F ~ to the base ABC (~ 602)..-. they are the D / altitudes of the three pyramids whose sum is equivalent to the truncated prism (~ 676). A Denote the volume of the truncated prism B by V, and its base ABC by B. ThenV=BxAD +B x ~BE +B x ~CF =B x ~(AD + BE + CF). 678 COROLLARY 2. The volume of any truncated triangular prism is equal to the product of its right section by one third the sum of its lateral edges. For the right section DEF divides the trun- K cated prism into two truncated right prisms the G volumes of which are F DEF x 1 (AD + BE + CF), A and DEF x (DG + EH + FK) (~ 677). B Adding, volume of ABC-GHK = DEF x (AG + BH + CK). 338 SOLID GEOMETRY -BOOK VII SIMILAR POLYEDRONS DEFINITIONS 679 Similar polyedrons are polyedrons which have the same number of faces respectively similar and similarly placed, and which have their corresponding polyedral angles equal. 680 Homologous faces, lines, and angles of similar polyedrons are faces, lines, and angles similarly situated. PROPOSITION XXIV. THEOREM 681 Two similar polyedrons may be divided into the same number of tetraedrons, similar each to each, and similarly placed. L F P K A -— D L! IK I B' C' B HYPOTHESIS. Let P and P' denote the similar polyedrons ABCDE-G and A'B'C'D'E'-G'. CONCLUSION. P and P' may be divided into the same number of tetraedrons, similar each to each, and similarly placed. PROOF Let B and B' be any two homologous triedral angles, and through the extremities of their edges, A, G, C, and A', G', C', respectively, pass planes. SIMILAR POLYEDRONS 339 In the tetraedrons B-AGC and B'-A'G'C', the faces GAB, GBC, ABC, are similar to the faces G'A'B, G'B'C', A'B'C', respectively. AG _ AB_> AC _ BC>__ GC ~ 349 * *A'G' A'B ' A'C' 'C' G'C'..the faces GAC and G'A'C' are similar, ~ 355 and the homologous triedral angles of these two tetraedrons are equal. ~ 592..the tetraedrons B-AGC and B'-A'G'C' are similar. ~ 679 If the tetraedrons B-AGC and B'-A'G'C' are removed, the remaining polyedrons are similar; for their corresponding new faces are similar, and their corresponding changed polyedral angles are equal since they have been equally reduced. Ax. 4 In like manner other corresponding similar tetraedrons may be removed until the polyedrons are divided into the same number of tetraedrons, similar each to each, and similarly placed. Q. E. D. 682 COROLLARY 1. The homologous edges of similar polyedrons are proportional. ~ 349 683 COROLLARY 2. Any two homologous lines in two similar polyedrons have the same ratio as any two homologous edges. ~ 349 684 COROLLARY 3. Any two homologous faces of two similar polyedrons have the same ratio as the squares of any two homologous edges. ~ 417 685 COROLLARY 4. The entire surfaces of two similar polyedrons have the same ratio as the squares of any two homologous edges. ~ 336 340 SOLID GEOMETRY —BOOK VII PROPOSITION XXV. THEOREM 686 The volumes of two similar tetraedrons are to each other as the cubes of their homologous edges. o ot HYPOTHESIS. O-ABC and O'-A'B'C' are similar tetraedrons, V and V' their volumes. CONCLUSION. V_= OA3 PROOF The triedral angles 0 and 0' are equal. ~ 679.V OA x OB x OC V' O'A' x O'B' x O'C O'A' O'B' O'C' OB OC OA Now B --- = ~ 682 O 'BF O'C'- O'A" V OA OA OA _OT __ _x OA 0A3 Q.E.D. 'V' 0'A' xO'A O'A' o'A'A3 EXERCISE 1227. If in the above figure OA = 5, and O'A' = 3, find 1 The ratio of V to V'. 2 The ratio of the entire surfaces of V and Vt. SIMILAR POLYEDRONS 341 PROPOSITION XXVI. THEOREM 687 The volumes of two similar polyedrons are to each other as the cubes of any two homologous edges. K L Le A B C BI C' HYPOTHESIS. P and P' are similar polyedrons, V and V' their volumes, GB and G'B' any two homologous edges. CONCLUSION. V: V' = GB3: G3B8. PROOF Separate P and P' into tetraedrons, similar each to each, and similarly placed. ~ 681 Let T, T1, T2,..., T', T'1, T'2,..., denote the volumes of these similar tetraedrons respectively. The Gn B T G T2 = -- T - GB3 ~ 686 T' G-~B,3 T G'-B,3' T,- G 3 T Tj _T2 Whence = T= T2.. Ax. 11 Tl T'l T'2 Hence T = T, + T2 +... GB336 T' T'+T'I+T'2+.- G'B''3 V' GB'3 342 SOLID GEOMETRY- BOOK VII REGULAR POLYEDRONS 688 DEFINITION. A regular polyedron is a polyedron whose faces are equal regular polygons, and whose polyedral angles are equal. PROPOSITION XXVII. THEOREM 689 Only five regular convex polyedrons are possible. PROOF. A convex polyedral angle must have not less than three faces, and the sum of its face angles must be less than 360~. ~ 591 1 Each angle of an equilateral triangle is 60~. The sum of three, four, or five such angles is less than 360~, but the sum of six such angles is 360~. Hence three, four, or five equilateral triangles may form A convex polyedral angle, but six or more such triangles cannot form a solid angle. ~ 591 Therefore only three regular convex polyedrons are possible whose surfaces are composed of triangles. 2 Each angle of a square is 90~. The sum of three such angles is less than 360~, but the sum of four such angles is 360~. Therefore only one regular convex polyedron is possible whose surface is composed of squares. 3 Each angle of a regular pentagon is 108~. The sum of three such angles is less than 360~, but the sum of four such angles is greater than 360~. Therefore only one regular convex polyedron is possible whose surface is composed of pentagons. 4 Each angle of a regular hexagon is 120~. The sum of three such angles is 360~. Hence a convex polyedral angle cannot be formed by regular polygons of six or more sides. Therefore only five regular convex polyedrons are possible. Q. E. D. REGULAR POLYEDRONS 343 690 The five regular polyedrons are the tetraedron, the hexaedron, the octaedron, the dodecaedron, and the icosaedron. They may be constructed as follows: Draw the following figures on thin cardboard. Cut out the figures, and cut half through the cardboard on the dotted lines. Fold along the dotted lines, and fasten the edges together by pasting strips of paper along them. I I I I i I - - - - - /\ I>\ II ~ r, \, \ \, \, \ I I I I, I I I I I I I I I I, 8, \ I 'I, I, I —, \, I I I\ I I,, I 11 I \ I, I I 1," I I I I I - I I I - - - - Tetraedron Hexaedron Octaedron Dodecaedron Icosaedron 344 SOLID GEOMETRY-BOOK VII CYLINDERS DEFINITIONS 691 A cylindrical surface is a curved......... surface generated by a moving straight".". il.i. line which always remains parallel to,i!|i|| I its original position and which constantly touches a fixed curved line. The generatrix is the moving straight line. The directrix is the fixed curved line. An element of the cylindrical surface..........i is the generatrix in any position. NOTE. The generatrix is usually considered indefinite in extent. The directrix may be any curve whatever, but closed curves, usually circles, are the only ones considered in Elementary Geometry. 692 A cylinder is a solid bounded __ by a cylindrical surface and two parallel plane surfaces. The lateral surface of a cylinder is i i its cylindrical surface. The bases of li III a cylinder are its two plane surfaces. I 693 COROLLARY. All elements of a cylinder are equal. ll 694 The altitude of a cylinder is _ the perpendicular between the planes.......ii..iiiiii.... of its bases. 695 A section of a cylinder is the figure formed by a plane intersecting the cylinder. A right section of a cylinder is a section perpendicular to the elements. 696 A right cylinder is a cylinder whose elements are perpendicular to its bases. An oblique cylinder is a cylinder whose elements are oblique to its bases. CYLINDERS 345 697 A circular cylinder is a cylinder whose bases. are circles. 698 A cylinder of revolution is a cylinder generated by the revolution of a rectangle about one side as an axis. Such a cylinder is a right circular cylinder. 699 Similar cylinders of revolution are cylinders generated by the revolution of similar rectangles 'about their homologous sides as axes. 700 A tangent line to a cylinder is an indefinite straight line which touches the lateral surface in one point only. 701 A tangent plane to a cylinder is an indefinite plane which contains an element of the cylinder without cutting its surface. The element contained by a tangent plane is called the element of contact. 702 A prism is inscribed in a cylinder when its lateral edges are elements of the cylinder and its bases are inscribed in the bases of the cylinder. 703 A prism is circumscribed about a cylinder when its lateral faces are tangent to the cylinder and its bases are circumscribed about the bases of the cylinder. 346 SOLID GEOMETRY -BOOK VII PROPOSITION XXVIII. THEOREM 704 Every section of a cylinder made by a plane passing through an element is a parallelogram. D HYPOTHESIS. ABCD is a section of the cylinder BD made by a plane passing through the element AB. CONCLUSION. ABCD is a parallelogram. PROOF Let DC' be the element through D. Then DC' is 11 to AB..'. DC' lies in the plane AC. ~ 691 ~~ 515, 517 Since DC' is common to the plane AC and the cylindrical surface, it must be their intersection and coincide with DC..'. DC is parallel to AB. Also, BC is parallel to AD..'. ABCD is a parallelogram. ~ 543 ~ 195 Q. E. D. 705 COROLLARY. Every section of a right cylinder made by a plane passing through an element is a rectangle. CYLINDERS 347 PROPOSITION XXIX. THEOREM 706 The bases of a cylinder are equal. B- F -- HYPOTHESIS. ABC and DEF are the bases of the cylinder AF. CONCLUSION. ABC = DEF. PROOF Through any element AD pass any two planes forming the sections AE and AF. Join BC and EF. Then AE and AF are l. ~ 704.'. BE and CF are equal and parallel. ~ 536.. BF is a /, and BC = EF, AB = DE, AC = DF. ~ 200.-. A ABC =A DEF. ~ 172 Superpose the bases, making the equal A coincide. Since A, B, C, are any points in the perimeter of the lower base, every point in the perimeter of the lower base will fall in the perimeter of the upper base..'. the bases will coincide and are equal. Q. E. D. 707 COROLLARY 1. Any two parallel sections cutting all the elements of a cylinder are equal. For these sections are the bases of a cylinder. 708 COROLLARY 2. Any section of a cylinder parallel to the base is equal to the base. 348 SOLID GEOMETRY -BOOK VII PROPOSITION XXX. THEOREM 709 If a prism whose base is a regular polygon is inscribed in or circumscribed about a circular cylinder, and if the number of sides of the base of the prism is indefinitely increased, The lateral area of the prism approaches the lateral area of the cylinder as a limit. The volume of the prism approaches the volume of the cylinder as a limit. A right section of the prism approaches a right section of the cylinder as a limit. PROOF If the number of lateral faces of either prism is indefinitely increased, its bases will approach the bases of the cylinder as their limits. Therefore, ~ 455 The lateral area of the prism approaches the lateral area of the cylinder as a limit. The volume of the prism approaches the volume of the cylinder as a limit. A right section of the prism approaches a right section of the cylinder as a limit. Q. E. D. CYLINDERS 349 PROPOSITION XXXI. THEOREM 710 The lateral area of a circular cylinder is equal to the product of the perimeter of a right section by an element. K O.....T H...........Ste..A HYPOTHESIS. AK is a circular cylinder, S the lateral area, P the perimeter of a right section, and E an element. CONCLUSION. S= P x E. PROOF Inscribe in the cylinder a prism whose base is a regular polygon, and let S' be its lateral area, and P' the perimeter of a right section. Then S' = P' x E. ~617 If the number of lateral faces of the prism is indefinitely increased, P' approaches P as a limit. ~ 709. P' x E approaches P X E as a limit. ~ 267 Also, S' approaches S as a limit. ~ 709 But S' = P' x E. Proved S =P x E. ~ 264 Q. E. D. 711 COROLLARY 1. The lateral area of a cylinder of revolution is equal to the product of the circumference of its base by its altitude. 350 SOLID GEOMETRY-BOOK VII 712 COROLLARY 2. Let S denote the lateral area, T the total area, H the altitude, and IR the radius of a cylinder of revolution. Then S = 2 wrR x H, and T = 2 rR x H + 2 rRP = 2 7rR (H + R). PROPOSITION XXXII. THEOREM 713 The volume of a circular cylinder is equal to the product of its base by its altitude. K A HYPOTHESIS. AK is a circular cylinder, V the vo'ume, B the base, and H the altitude. CONCLUSION. V = B x H. PROOF Inscribe in the cylinder a prism whose base is a regular polygon, and let V' be its volume, and B' its base. Then V'= B' x H. ~ 638 If the number of lateral faces of the prism is indefinitely increased, B' approaches B as a limit. ~ 455.. B' x H approaches B X H as a limit. ~ 267 Also, V' approaches V as a limit. ~ 709 But V' = B' x H. Proved.'.V=BxH. ~264 Q. E. D. 714 COROLLARY. Let V denote the volume, H the altitude, and R the radius of a cylinder of revolution. Then V = 7rR2 x H. CYLINDERS 351 PROPOSITION XXXIII. THEOREM 715 The lateral areas, or the total areas, of two similar cylinders of revolution are to each other as the squares of their altitudes, or as the squares of their radii; and their volumes are to each other as the cubes of their altitudes, or as the cubes of their radii. H HI HYPOTHESIS. S and S' are the lateral areas, T and T' the total areas, V and V' the volumes, H and H' the altitudes, R and R' the radii of two similar cylinders of revolution. CONCLUSION. S: S' = T: T = H2: H'2 = R2: R'2, and V: V' = H3: H'3 = R3: R'3. PROOF H R H~R H' R' H' ~+ 349,336 H' R' H' + iR2' S 2T-H RH R 1 H2 R2 S' 2 rR'IH -t R' ' H'2 - R,2' andT 2 7rR (H + R) R H + R H2 R2 T' 2 7R' (H' + R) R' H'+R' H'2 R'2 V 7TR2H R2 xH H3 H R3 — = ~ -X - ~~ 714 V' TrrI'H' R'2 H' H'3 R '3 Q. E. D. 352 SOLID GEOMETRY - BOOK VII PROBLEMS OF COMPUTATION 1228 Find the lateral area of a cylinder of revolution whose altitude is 12 inches, and the diameter of whose base is 4 inches. SOLUTION. S = 2 7rR x H =487r = 150.80 sq. in. 1229 Find the total area of a cylinder of revolution whose altitude is 3 feet, and the diameter of whose base is 18 inches. 1230 Find the volume of a right circular cylinder whose altitude is 6 inches, and the diameter of whose base is 2 inches. 1231 A stand pipe is 12 feet in diameter and 48 feet high. How many gallons of water will it hold, estimating 7- gallons to a cubic foot? 1232 How many square feet of sheet iron in a smokestack 52 feet high and 6 feet in diameter? 1233 A cylindrical cistern is 6 feet deep and 6 feet in diameter. How long will it take to fill it if 200 cubic inches of water flow into it per minute? 1234 How many cubic yards of earth are removed in constructing a tunnel 135 yards long, a.right section of which is a semicircle with a radius of 15 feet? 1235 A cylindrical oil tank is 28 feet long and 6 feet in diameter. How many gallons of oil does it hold, estimating 71 gallons to a cubic foot? 1236 Find the volume generated by a rectangle, whose dimensions are 4 and 7 inches respectively, in revolving (a) about its longer side as an axis; (b) about its shorter side as an axis. 1237 The altitudes of two similar cylinders of revolution are as 3: 5. What is the ratio of their volumes? of their total areas? 1238 The lateral area of a cylinder of revolution is 108 sq. ft., and its radius 3 ft. Find the lateral area of a similar cylinder whose radius is 5 ft. 1239 The altitudes of two similar cylinders of revolution are as 2: 3. If the volume of the first is 16, find the volume of the second. 1240 The volumes of two similar cylinders of revolution are as 125: 216. Find the ratio of their lateral areas. 1241 How much water will flow through a pipe in one minute, if it flows 21 feet per second, and the interior diameter of the pipe is 6 in.? CONES 353 CONES DEFINITIONS 716 A conical surface is a surface generated by a moving straight line which constantly touches a fixed curve and passes through a fixed point not in the plane of the curve. The moving straight line is called the generatrix; the fixed curve, the directrix; and the fixed point, the vertex. Any straight line in the surface which corresponds to some position of the generatrix is called an element of the surface. When the generatrix is of indefinite length,: the conical surface described consists of two portions called nappes, the upper nappe and the lower nappe. A Conical Surface It is customary, however, when we speak of a cone to refer to one nappe only. 717 A cone is a solid bounded by a conical plane cutting all its elements. The conical surface is called the lateral surface of the cone, and the plane surface is called the base of the cone. The vertex of a cone is the vertex of the conical surface. The altitude of a cone is the perpendicular: drawn from the vertex to the plane of the base. surface and a 718 A circular cone is a cone whose base A Cone is a circle. The axis of a circular cone is the straight line which joins the vertex and the center of the base. The axis of a cone is not necessarily its altitude. 354 SOLID GEOMETRY -BOOK VII 719 A right circular cone is a circular cone whose axis is perpendicular to its base. An oblique circular cone is a circular cone whose axis is oblique to its base. Right Cone Oblique Cone Cone of Revolution 720 A right circular cone is also called a cone of revolution, because it may be generated by the revolution of a right triangle about one of its legs as an axis. The hypotenuse generates the lateral surface, and the remaining leg generates the base of the cone. The hypotenuse in any position is an element of the surface, and any element is the slant height of the cone. 721 COROLLARY. All elements of a right circular cone are equal. 722 Similar cones of revolution are cones generated by the revolution of similar right triangles about homologous legs as axes. 723 A tangent line to a cone is an indefinite straight line which touches the lateral surface in one point only. 724 A tangent plane to a cone is an indefinite plane which contains an element of the cone without cutting its surface. The element contained by a tangent plane is called the element of contact. CONES 355. 725 A pyramid is inscribed in a cone when its lateral edges are elements of the cone and its base is inscribed in the base of the cone. 726 A pyramid is circumscribed about a cone when its lateral faces are tangent to the cone and its base is circumscribed about the base of the cone. Circumscribed Pyramid Inscribed Pyramid Frustum of a Cone 727 A truncated cone is the portion of a cone included between its base and a section formed by a plane cutting all its elements. 728 A frustum of a cone is a truncated cone in which the cutting section is parallel to the base. The lower base of the frustum is the base of the cone, the upper base of the frustum is the parallel section, and the altitude of the frustum is the perpendicular drawn between its bases. 729 The lateral surface of a frustum of a cone is the portion of the lateral surface of the cone included between the bases of the frustum. 730 The slant height of a frustum of a cone of revolution is the portion of any element of the cone included between the bases. 356 SOLID GEOMETRY -BOOK VII PROPOSITION XXXIV. THEOREM 731 Every section of a cone made by a plane passing through its vertex is a triangle. HP EI. isa a D ct HYPOTHESIS. S-ABC is a cone, and SBD is any section formed by a plane passing through the vertex S. CONCLUSION. The section SBD is a triangle. PROOF Draw the straight lines SB and SD. These straight lines lie in the lateral surface, ~ 716 and also in the plane SBD. ~ 511 That is, they are the lines in which the plane SBD cuts the lateral surface..'. the plane SBD cuts the lateral surface in the straight lines SB and SD. Also, BD is a straight line. ~ 518.*. the section SBD is a triangle. ~ 134 Q. E. D. EXERCISE 1242. What kind ot triangle is the section made by a plane passing through the vertex of a cone of revolution? CONES 357 PROPOSITION XXXV. THEOREM 732 Every section of a circular cone made by a plane parallel to the base is a circle............. iiii,......,....... s a""_ I a il dX i6 0 DC HYPOTHESIS. The section abd of the circular cone S-ABD is parallel to the base ABD. CONCLUSION. The section abd is a circle. PROOF Let O be the center of the base, and let o be the point in which the axis SO pierces the parallel section. Through SO and any two elements SB and SC, pass planes cutting the base in the radii OB and OC, and the parallel section in the straight lines ob and oc. ~ 518 Since ob is II to OB, and oc is II to OC, ~ 543 the A Sob and SOB are similar; likewise A Soc and SOC. ~ 351 Whence ob = (So) - o ~349 OB SO) OC Since OB = OC, ob = oc. ~ 221. the section abd is a circle. ~ 220 Q. E. D. 733 COROLLARY. The axis of a circular cone passes through the center of every section parallel to the base. 358 SOLID GEOMETRY-BOOK VII PROPOSITION XXXVI. THEOREM 734 If a pyramid whose base is a regular polygon is inscribed in or circumscribed about a circular cone, and if the number of sides of the base of the-pyramid is indefinitely increased, The lateral area of the pyramid approaches the lateral area of the cone as a limit. The volume of the pyramid approaches the volume of the cone as a limit. Any section of the pyramid approaches the section of the cone made by the same plane as a limit. PROOF If the number of lateral faces of either pyramid is indefinitely increased, its base approaches the base of the cone as a limit. ~ 455 Therefore, the lateral area of the pyramid, its volume, and any section, approach as their respective limits the lateral area of the cone, its volume, and the section made by the same plane. Q. E. D. 735 COROLLARY. The above theorem remains true if for "pyramid" and "cone" we substitute "frustum of a pyramid" and "frustum of a cone" respectively. CONES 359 PROPOSITION XXXVII. THEOREM 736 The lateral area of a cone of revolution is equal to half the product of the slant height by the circumference of the base. HYPOTHESIS. S is the lateral area of a cone of revolution, C the circumference of its base, and L its slant height. CONCLUSION. S= C x L. PROOF Circumscribe about the cone a regular pyramid, and denote the perimeter of its base by P, and its lateral area by S'. Then S'= P x L. ~658 If the number of lateral faces of the pyramid is indefinitely increased, P approaches C as a limit. ~ 455. P. ~ P x L approaches - C x L as a limit. ~~ 268, 267 Also, S' approaches S as a limit. ~ 734 But S'= 1 P x L. Proved.-. S= CxL. ~264 Q. E. D. 737 COROLLARY. If S is the lateral area, T the total area, L the slant height, R the radius of the base, of a cone of revolution, we have S = -(2 7R x L) = 7RL, and T = 7rRL + ~rR'= 7rR(L + R). 360 SOLID GEOMETRY- BOOK VII PROPOSITION XXXVIII. THEOREM 738 The volume of a circular cone is equal to one third the product of its base by its altitude. HYPOTHESIS. V is the volume of a circular cone, B its base, and H its altitude. CONCLUSION. V = B x H. PROOF Inscribe in the cone a pyramid whose base is a regular polygon, and denote its volume by V', and its base by B'. Then V' = B' x H. ~ 667 If the number of lateral faces of the pyramid is indefinitely increased, B' approaches B as a limit. ~ 455. B' x H approaches I B x H as a limit. ~~ 268, 267 Also, V' approaches V as a limit. ~ 734 But V'= I B' x H. Proved.. V = B x H. ~ 264 Q. E. D. 739 COROLLARY. If we denote the radius of the base of a circular cone by R, then B = -R2. ~ 464.. V= rR2 X H. 3 CONES 361 PROPOSITION XXXIX. THEOREM 740 The lateral areas, or the total areas, of two similar cones of revolution are to each other as the squares of their altitudes, as the squares of their radii, or as the squares of their slant heights; and their volumes are to each other as the cubes of their altitudes, as the cubes of their radii, or as the cubes of their slant heights.!LL, HYPOTHESIS. S and S are the lateral areas of two similar cones of revolution, T and T' their total areas, V and V' their volumes, H and H' their altitudes, R and R' their radii, L and L' their slant heights. CONCLUSION. S:S' = T:T' = H2: H'2= R2: R'2= L2:L'2, and V:V'= H3: H'3 = R3:R'3 = L3:L'3. PROOF H _I; L L+R H R' L- L~+R ~~349, 336 H' R' L' L! + R' S 7rRL R L R2 L2 H2 S' 7rR '' Rt' L'- R'2 L'2 Hr2 a T _ rR(L + R) R L + R R2 L2 H2 x ~ 737 T' 7r R'(L' + R') R' L' + R 2' R' L'2 H12 V _ 7rR 2H R2 H R3 H3 L3 V' rR'2H '1 R H' R '3 H'3 L'3 Q. E. D. 362 SOLID GEOMETRY- BOOK VII PROPOSITION XL. THEOREM 741 The lateral area of a frustum of a cone of revolution is equal to half the sum of the circumferences of its bases multiplied by its slant height. HYPOTHESIS. S is the lateral area, C and C' are the circumferences of the bases, R and R' their radii, and L is the slant height. CONCLUSION. S = ~ (C + C') x L. PROOF Circumscribe about the frustum of the cone a frustum of a regular pyramid, and denote its lateral area by S', the perimeters of its bases by P and P'. Its slant height is L. ~ 730 Then S'= 2 (P+ P') x L. ~ 659 If the number of lateral faces of the frustum of the pyramid is indefinitely increased, P + P' approaches C + C' as a limit. ~ 455.. ~ (P + P') x L approaches 2 (C + C') x L as a limit. ~~ 268, 267 Also, S' approaches S as a limit. ~ 735 But S' = ~ (P + P') x L. Proved SS = (C + C') x L. ~264 Q. E. D. 742 COROLLARY. The lateral area of a frustum of a cone of revolution is equal to the circumference of a section equidistant from its bases multiplied by its slant height. CONES 363 PROPOSITION XLI. THEOREM 743 The volume of a frustum of a circular cone is equivalent to the sum of the volumes of three cones whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and the mean proportional between the bases of the frustum. HYPOTHESIS. V is the volume of a frustum of a circular cone, B its lower base, B' its upper base, and H its altitude. CONCLUSION. V=1 H (B + B' + /B x B'). PROOF Let V' denote the volume of an inscribed frustum of a pyramid whose bases are regular polygons, b its lower base, and b' its upper base. The altitude of the frustum is H. ~ 548 Then V'= I (b + b'+ x b'). ~ 674 If the number of lateral faces of the inscribed frustum is indefinitely increased, b approaches B as a limit, and b' approaches B' as a limit. ~ 455.*. b x b' approaches B x B' as a limit. ~ 272.'. Vb x b' approaches VB x B' as a limit. ~ 271.'. ~ H (b +b' + Vb x b') approaches H (B + B' + Y/B x B') as a limit. ~~ 268, 267, 272 Also, V' approaches V as a limit. ~ 735 But V' + H (b + b' b b') Proved.-. V = ' H (B + B + /VB x B'). ~ 264 Q. E. D. 744 COROLLARY. If R and R' are the radii of the bases of a frustum of a circular cone, then B = 7rR2, B' = -r '2. ~ 464 ~Y VB x B'= -/YR2 X 7rR' = 7rRR'.... V =- rH (R2 + R12 + RR'). 364 SOLID GEOMETRY-BOOK VII PROBLEMS OF COMPUTATION 1243 Compute the lateral area of a cone of revolution of which the radius of the base is 3 inches and the altitude is 4 inches. 1244 Compute the total area of a right circular cone of which the radius of the base is 7 feet and the slant height is 23 feet. 1245 Find the volume of a cone of revolution, the radius of its base being 48 inches, and its slant height being 73 inches. 1246 Find the volume of a circular cone whose altitude is 10 feet, and the diameter of whose base is 10 feet. 1247 The radii of the bases of a frustum of a cone of revolution are 6 inches and 4 inches, and the slant height is 5 inches. Find the lateral area. 1248 The diameters of the bases of a frustum of a circular cone are 18 feet and 12 feet, and the altitude is 16 feet. Find the volume. 1249 The radii of the bases of a frustum of a right circular cone are 7 feet and 4 feet, and its altitude is 5 feet. Find the altitude of the cone from which the frustum is cut off. 1250 The altitudes of two similar cones of revolution are as 2:1. What is the ratio of their total areas? of their volumes? SOLUTION. T: T' = H2: H'2 = (2)2: (1)' = 4: 1. V:V'= H3:H13=(2)3: (1)3 =8:1. 1251 The volumes of two similar cones are 216 cu. ft. and 512 cu. ft. What is the height of the first, if the height of the second is 12 ft.? 1252 The altitude of a cone is 36 inches. How far from the vertex must it be cut by a plane so that the frustum shall be equivalent to half the cone? 1253 How many square inches of tin are required to make a funnel, if the slant height is 9 inches, and the diameters of its bases are 6 inches and 1 inch, allowing i inch for seam? 1254 The volumes of two similar cones of revolution are as 27:125. Compare their convex surfaces. 1255 Of two cylinders of revolution, the altitude of one is three times the altitude of the other. Compare their convex surfaces and their volumes. 1256 Find the edge of a cube equivalent to a right circular cylinder whose diameter is 6 ft. and whose altitude is 10 ft. BOOK VIII THE SPHERE DEFINITIONS 745 A sphere is a solid bounded by a surface all points of which are equidistant from a point within called the center of the sphere. A sphere may ---- be generated by the revolution of a semicircle about its diameter as an axis. t is a straight line drawn from - the center to aly point of the..... equa. surface. The diameter of a sphere is a straight line drawn through the center and terminated by the surface. 747 A line or a plane is tangent to a sphere when it has one point only in common with the surface of the sphere. Two spheres are tangent to each other when their surfaces have one point only in common. 748 COROLLARY. All radii of a sphere are equal. All diameters of a sphere are equal. Spheres having equal radii are equal, and conversely. A point is within a sphere, on its surface, or without a sphere, when its distance from the center is less than, equal to, or greater than, the radius. 365 366 SOLID GEOMETRY —BOOK VIII PROPOSITION I. THEOREM 749 Every section of a sphere made by a plane is a circle..m...... II. '0:ii'ii i jiiiziiiii: HYPOTHESIS. O is the center of a sphere, and ABC is any section made by a plane. CONCLUSION. The section ABC is a circle. PROOF Draw OD I to the section, and join O to any two points in the perimeter of the section, as A and B. Then OA =OB..-. DA = DB. ~ 748 That is, all points in the perimeter of the section ABC are equidistant from the point D..'. the section ABC is a circle. Q. E. D. 750 DEFINITION. A circle of a sphere is any section of the sphere made by a plane. 751 COROLLARY 1. The line joining the center of a sphere and the center of a circle of the sphere is perpendicular to the plane of the circle. 752 COROLLARY 2. Circles of a sphere equidistant from the center of the sphere are equal, and conversely. 753 COROLLARY 3. Of two circles unequally distant from the center of a sphere, the more remote is the smaller, and conversely. THE SPHERE 367 DEFINITIONS 754 A great circle of a sphere is a circle passing through the center of the sphere. A small circle of a sphere is a circle of the sphere not passing through the center of the sphere. 755 The axis of a circle of a sphere is the diameter of the sphere perpendicular to the circle. The poles of a circle are the extremities of its axis. 756 The distance between two points on the surface of a sphere is the length of the minor arc of the great circle joining them. 757 COROLLARY 1. The center of a great circle coincides with the center of the sphere. 758 COROLLARY 2. All great circles of a sphere are equal. For their radii are radii of the sphere. ~ 757 759 COROLLARY 3. Any two great circles of a sphere bisect each other. For since they have the same center, they have the same diameter. 760 COROLLARY 4. A great circle bisects the sphere. For the two parts may be made to coincide. ~ 745 761 COROLLARY 5. An arc of a circle may be drawn through any three points on the surface of a sphere. For the three points determine a plane (~ 515) which cuts the sphere in a circle. ~ 749 762 COROLLARY 6. An arc of a great circle may be drawn through any two points on the surface of a sphere. For the two points on the surface and the center of the sphere determine a great circle. ~ 515 The extremities of a diameter and the center of the sphere, being in a straight line, do not determine a circle. 763 COROLLARY 7. If the planes of two great circles are perpendicular, each circle passes through the poles of the other. 368 SOLID GEOMETRY-BOOK VIII PROPOSITION II. THEOREM 764 All points in the circumference of a circle of a sphere are equidistant from either of its poles. HYPOTHESIS. P and P' are the poles of the circle ABC, and A, B, and C are any three points in the circumference. CONCLUSION. The arcs PA, PB, and PC are equal. Also, the arcs P'A, P'B, and P'C are equal. PROOF The straight lines PA, PB, and PC are equal. ~ 530.-. the arcs PA, PB, and PC are equal. ~ 243 In like manner it may be proved that the arcs P'A, P'B, and P'C are equal. Q. E. D. 765 DEFINITIONS. The polar distance of a circle of a sphere is the distance measured on the surface of the sphere from a point in the circumference of the circle to its nearer pole. A quadrant in Spherical Geometry is one fourth the circumference of a great circle. 766 COROLLARY. The polar distance of a great circle is a quadrant. For let EDF and PDP' be great circles. Then O is their common center, and the right angle POD is measured by the quadrant PD. THE SPHERE 369 PROPOSITION III. THEOREM 767 If, on the surface of a sphere, a point is at the distance of a quadrant from each of two other points, not the extremities of a diameter, it is a pole of the great circle passing through those points. P --- --- -A B n D A 9; zC HYPOTHESIS. ABCD is a great circle, PA and PB are quadrants, A and B are not the extremities of a diameter. CONCLUSION. P is a pole of the great circle ABCD. PROOF Draw the radii OP, OA, and OB. Since PA and PB are quadrants, -Hyp. the angles POA and POB are right angles. ~ 274.. PO is 1 to the plane ABCD. ~ 523.'. P is a pole of the great circle ABCD. ~ 755 Q. E. D. 768 SCHOLIUM. The preceding propositions enable us to draw arcs of small and great circles upon the surface of a material sphere. For this purpose we use compasses with curved arms, called spherical compasses. 370 SOLID GEOMETRY-BOOK VIII PROPOSITION IV. PROBLEM 769 To find the diameter of a material sphere. P p C' A 0 a o A' — Of DATA. APP' is a material sphere. REQUIRED. To find the diameter of APP'. SOLUTION From any point P on the surface as a pole describe any circumference ABC on the sphere. ~ 768 Take three points, A, B, C, in this circumference, and with the compasses measure the chords AB, AC, and BC. Construct the A A'B'C' whose sides are respectively equal to AB, AC, and BC, and circumscribe about it a 0 whose center is 0'. The radius A'O' is equal to the radius AO of the ( ABC. Construct the rt. A apo whose hypotenuse ap is equal to the chord AP, and a side ao equal to A'O'. Draw ap' _L to ap, meeting po produced in p'. CONCLUSION. pp' POP', the diameter required. Q. E. F. PROOF. The rt. A apo and APO are equal. ~ 178. the rt. A app' and APP' are equal. ~ 169..pp'= PP'. Q.E. D. 770 SCHOLIUM. The chord of a quadrant (h) is readily computed when the radius (r) of the sphere is known. For h = rV2. r THE SPHERE 371 PROPOSITION V. THEOREM 771 A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere. HYPOTHESIS. O is the center of a sphere, OP a radius, MN a plane perpendicular to OP at P. CONCLUSION. MN is tangent to the sphere at P. PROOF Let K be any point, except P, in MN, and draw OK. Then OK > OP. ~ 528 -Hence K is without the sphere. ~ 748 That is, every point in MN, except P, is without the sphere..'. MN is tangent to the sphere at P. Q. E. D. 772 COROLLARY 1. A plane tangent to a sphere is perpendicular to the radius drawn to the point of contact. 773 COROLLARY 2. A straight line tangent to a circle of a sphere lies in the plane tangent to the sphere at the point of contact. 774 COROLLARY 3. Any straight line in a tangent plane drawn through the point of contact is tangent to the sphere at that point. 775 COROLLARY 4. The plane of any two straight lines tangent to a sphere at the same point is tangent to the sphere at that point. 372 SOLID GEOMETRY-BOOK VIII DEFINITIONS 776 A sphere is inscribed in a polyedron when all the faces of the polyedron are tangent to the sphere. 777 A sphere is circumscribed about a polyedron when all the vertices of the polyedron lie in the surface of the sphere. PROPOSITION VI. THEOREM 778 A sphere may be inscribed in any tetraedron. D B HYPOTHESIS. D-ABC is-any tetraedron. CONCLUSION. A sphere may be inscribed in D-ABC. PROOF Let the planes OAB, OAC, and OBC bisect the diedral angles at the edges AB, AC, and BC respectively. Then every point in the plane OAB is equidistant from the faces ABC and ABD; every point in the plane OAC is equidistant from the faces ABC and ADC; and every point in the plane OBC is equidistant from the faces ABC and DBC. ~ 570 Hence 0, the common intersection of these three planes, is equidistant from the four faces of the tetraedron, and is therefore the center of an inscribed sphere. ~ 776 Q. E. D. 779 COROLLARY. The six planes bisecting the six diedral angles of a tetraedron intersect in the same point. THE SPHERE 373 PROPOSITION VII. THEOREM 780 A sphere may be circumscribed about any tetraedron. D F~ G — -- A -— +- C B HYPOTHESIS. D-ABC is any tetraedron. CONCLUSION. A sphere may be circumscribed about D-ABC. PROOF Let H and K be the centers of the circles circumscribed about the faces ABC and DBC respectively. Draw HF I to the plane ABC, and KG 1. to the plane DBC. Join E the middle point of BC to H and K. Then HE and KE are each.L to BC. ~ 188..the plane HEK is _ to BC. ~ 523..the plane HEK is 1. to the planes ABC and DBC. ~ 563.-. HF and KG lie in the plane HEK. ~ 565. HF and KG intersect in some point 0. Since all points in HF are equidistant from A, B, C, ~ 532 and all points in KG are equidistant from D, B, C, ~ 532 0 is equidistant from D, A, B, C, and is therefore the center of a sphere circumscribed about D-ABC. ~ 777 Q. E. D. 781 COROLLARY 1. The four perpendiculars erected at the centers of the circles circumscribing the four faces of a tetraedron meet in a point. 782 COROLLARY 2. The six planes perpendicular to the six edges of a tetraedron at their middle points intersect in a point. 374 SOLID GEOMETRY-BOOK VIII SPHERICAL ANGLES. DEFINITIONS 783 The angle of two intersecting curves is the angle formed by the tangents to the curves at their point of intersection. 784 A spherical angle is the angle formed by the intersection of two arcs of great circles of a sphere. PROPOSITION VIII. THEOREM 785 A spherical angle is measured by the arc of the great circle described from its vertex as a pole and included between its sides, produced if necessary. P HYPOTHESIS. PAP' and PBP' are great circles to which PS and PT are HYPOTHESIS. P and AB i and PBP' arc of a great circles to whose pole is PS and PT are tangents at P, and AB i3 an arc of a great circle whose pole is P. CONCLUSION. The spherical Z APB is measured by the arc AB. PROOF Draw the radii OA and OB. PS is I to PO (~ 251), and OA is ~ to PO. ~ 274. PS and OA being in the same plane are parallel. ~ 115 Likewise PT and OB are parallel..'.ZSPT=- AOB. ~549 But L AOB is measured by the arc AB. ~ 274.. SPT is measured by the arc AB.. Z APB is measured by the arc AB. ~ 783 Q. E. D. THE SPHERE 375 786 COROLLARY. A spherical angle is equal to the plane angle of the diedral angle formed by the planes of the two circles. SPHERICAL POLYGONS. DEFINITIONS 787 A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. The sides of a spherical polygon are the bounding arcs; the vertices are the points in which the sides intersect; the angles are the angles formed by the sides. Since the sides of a spherical polygon are arcs, they are usually measured in degrees, minutes, and seconds. 788 The planes of the sides of a spherical'polygon form by their intersections a / polyedral angle whose vertex is the center of the sphere, whose face angles are meas- ~ — ured by the sides of the polygon, and whose diedral angles are equal to the angles of the polygon. From these relations of polyedral angles and spherical polygons, we have the following 789 COROLLARY. From any property of polyedral angles we may infer an analogous property of spherical polygons. 790 A diagonal of a spherical polygon is the arc of a great circle joining two vertices not consecutive. 791 A convex spherical polygon is a spherical polygon whose corresponding polyedral angle is convex. A spherical polygon is considered convex unless otherwise stated. 792 A spherical triangle is a spherical polygon of three sides. Like a plane triangle, it may be right, obtuse, acute, scalene, isosceles, equilateral, or equiangular. 376 SOLID GEOMETRY —BOOK VIII PROPOSITION IX. THEOREM 793 Any side of a spherical triangle is less than the sum of the other two sides. HYPOTHESIS. AC is the longest side of the spherical A ABC. CONCLUSION. AC < AB + BC. PROOF. Draw the radii OA, OB, OC. Then Z AOC < Z BOA + Z BOC... AC < AB + BC. ~ 590 ~ 788 Q. E. D. PROPOSITION X. THEOREM 794 The sum of the sides of any spherical polygon is less than the circumference of a great circle. HYPOTHESIS. ABCD is a spherical polygon. CONCLUSION. AB + BC + CD + AD < 360~. PROOF. Draw the radii OA, OB, OC, OD. Then Z AOB + Z BOC + / COD + Z AOD < 360~..'. AB + BC + CD + AD < 360~. ~ 591 ~ 788 Q. E. D. THE SPHERE 377 PROPOSITION XI. THEOREM 795 The shortest line that can be drawn between two points on the surface of a sphere is the arc of a great circle, less than a semicircumference, joining them. HYPOTHESIS. A and B are two points on the surface of a sphere, AB the arc of a great circle less than a semicircumference, and ACDEB any other line on the surface of the sphere. CONCLUSION. AB is less than ACDEB. PROOF Let D be any point in ACDEB, and draw the great circle arcs AD and DB. Then arc AD + arc DB > arc AB. ~ 793 Likewise if great circle arcs are drawn from A and D to any point in ACD, and from D and B to any point in DEB, their sum will be greater than AD + DB. By repeating this process indefinitely, the sum of the great circle arcs joining A and B becomes an increasing variable approaching the line ACDEB as a limit, and therefore always less than ACDEB. ~ 263.-. AB is less than ACDEB. Q. E D. 378 SOLID GEOMETRY -BOOK VIII PROPOSITION XII. PROBLEM 796 Through a given point to draw an arc of a great circle perpendicular to a given arc of a great circle. P p DATA. BAC is an arc of a great circle whose pole is P, and E is any point on the surface of the sphere. REQUIRED. To draw an arc of a great circle through E I to BAC. SOLUTION From E as a pole describe an arc of a great circle cutting BAC in D. From D as a pole describe the great circle arc PEA, and it is the arc required. Q. E. F. PROOF. Since the arc PD is a quadrant, ~ 766 the angle PAD is a right angle. ~ 785..the arc PA is 1L to the arc BAC. ~ 783 Q. E. D. EXERCISES 1257 The locus of a point equidistant from the extremities of a given arc of a great circle is the arc of a great circle perpendicular to the given are at its middle point. 1258 A great circle arc drawn through the pole of a great circle is perpendicular to the circumference of the great circle. 1259 A great circle arc perpendicular to the circumference of a great circle passes through the pole of the great circle. 1260 From a given point not the pole, only one great circle arc can be drawn perpendicular to an arc of a great circle. Two cases. 1261 The perpendicular great circle arc is the shortest arc that can be drawn from a point to a given arc of a great circle. THE SPHERE 379 POLAR TRIANGLES 797 DEFINITION. The polar triangle of a spherical triangle is a spherical triangle the poles of whose sides are respectively the vertices of the given triangle. PROPOSITION XIII. THEOREM 798 If one triangle is the polar of a second, the second is the polar of the first. At HYPOTHE. A'B'C' is the polar t e of ABC. HYPOTHESIS. A'B'C is the polar triangle of ABC. CONCLUSION. ABC is the polar triangle of A'B'C'. PROOF Since B is the pole of the arc A'C', and C is the pole of the arc A'B', ~ 797 A' is at a quadrant's distance from B and C. ~ 766.'. A' is the pole of the arc BC. ~ 767 Likewise B' is the pole of the arc AC, and C' is the pole of the arc AB.. ABC is the polar triangle of A'B'C'. ~ 797 Q. E. D. 799 SCHOLIUM. The great circles described from A, B, and C as poles divide the surface of the sphere into eight spherical triangles. Of these, the two polar triangles are those whose corresponding vertices A and A' lie on the same side of BC and B'C'; likewise for the other corresponding vertices. 380 SOLID GEOMETRY-BOOK VIII PROPOSITION XIV. THEOREM 800 In two polar triangles, each angle of the one is the supplement of the side opposite to it in the other. Al HYPOTHESIS. ABC and A'B'C' are polar triangles. CONCLUSION. Z A is the supplement of B'C'. PROOF Produce AB and AC to meet B'C' in D and E respectively. Since B' is the pole of AC, and C' is the pole of AB, ~ 766 B'E and DC' are quadrants... B'E + DC' = B'C' 4- DE = 180~.. DE is the supplement of B'C'. But DE measures the Z A. ~ 785.. Z A is the supplement of B'C'. Likewise Z A' is the supplement of BC, etc. Q. E.D. EXERCISES 1262 The locus of a point equidistant from the sides of a spherical angle is the arc of a great circle bisecting the angle. 1263 The sides of a spherical triangle are 70~, 80~, and 90~. Find the angles of its polar triangle. 1264 The angles of a polar triangle are 60~, 70~, and 110~. Find the sides of its polar triangle. 1265 A sphere whose diameter is 10 in. is cut by a plane 4 in. from the center. Find the diameter of the small circle. 1266 Four points not in the same plane determine a sphere. THE SPHERE 381 PROPOSITION XV. THEOREM 801 The sum of the angles of a spherical triangle is greater than two, and less than six, right angles. HYPOTHESIS. ABC is a spherical triangle, and a', b', c', are the sides of the polar triangle A'B'C'. CONCLUSION. L A + L B + L C > 180~ and < 5400. PROOF Z A = 180~ - a', Z B = 180~ - b, and Z C = 180~ - c'. ~ 800 Adding, A + B+Z C = 540~- (a'+ b + c') (1). Ax. 1.*.Z A+ B +ZC<540~. But a' + b' + c'< 360~ (2). ~ 794 Subtracting (2) from (1), Z A + Z B + Z C > 180~. Ax. 6 Q. E.D. 802 COROLLARY 1. A spherical triangle may have two, or three, right angles; also two, or three, obtuse angles. 803 DEFINITIONS. A bi-rectangular spherical triangle is a spherical triangle having two right angles. A tri-rectangular spherical triangle is a spherical triangle having three right angles. 804 COROLLARY 2. In a bi-rectangular spherical triangle, the sides opposite the right angles are quadrants. For since AB and AC both pass through the pole of BC (~ 763), A is the pole of BC..-. AB and AC are quadrants (~ 766). A B C 382 SOLID GEOMETRY -BOOK VIII 805 COROLLARY 3. In a tri-rectangular spherical triangle each side is a quadrant. 806 COROLLARY 4. Three planes passed through the center of a sphere, each perpendicular to the other two, divide the surface of the sphere into eight equal tri-rectangular triangles. SYMMETRICAL SPHERICAL POLYGONS 807 DEFINITION. Symmetrical spherical polygons are spherical polygons whose successive sides and angles are equal, each to each, but arranged in reverse order........ A.A....'.. A.-....!;,_ -.....-:........................;.................,Thus, the spherical triangles ABC and A'B'C' are mutually equilateral and equiangular, but their corresponding parts are arranged in reverse order. In general, two symmetrical spherical polygons cannot be made to coincide and are not equal. If, however, two symmetrical tri- A A' angles are isosceles, they are equal. For if the symmetrical triangles ABC and A'B'C' are isosceles, AC = A'B', and AB = A'C'... the c C, triangles can be made to coincide. B B Hence the following 808 COROLLARY. Two symmetrical isosceles spherical triangles are equal. THE SPHERE 383 PROPOSITION XVI. THEOREM 809 Two triangles on the same sphere or equal spheres are equal, 1 If two sides and the included angle of the one are equal respectively to two sides and the included angle of the other; 2 If two angles and the included side of the one are equal respectively to two angles and the included side of the other; 3 If the three sides of the one are equal respectively to the three sides of the other; Provided that the equal parts in each case are arranged in the same order. PROOF In cases 1 and 2, the A, like plane A, can be made to coincide, and are therefore equal. ~~ 162, 167 In case 3, the face angles of the corresponding triedral angles at the center of the sphere are respectively equal. ~ 241.'. the corresponding diedral A are equal. ~ 593.'. the zA of the spherical A are respectively equal. ~ 789.'. the triangles are equal. Case 1 Q. E. D. 384 SOLID GEOMETRY -BOOK VIII PROPOSITION XVII. THEOREM 810 Two triangles on the same sphere or equal spheres are symmetrical, 1 If two sides and the included angle of the one are equal respectively to two sides and the included angle of the other; 2 If two angles and the included side of the one are equal respectively to two angles and the included side of the other; 3 If the three sides of the one are equal respectively to the three sides of the other; Provided that the equal parts in each case are arranged in reverse order........................ PROOF In each case, A and B are the given A. Construct the A C symmetrical with the A B. Then A A and C are equal. ~ 809 But A C and B are symmetrical by const..'. the A A, which is equal to the A C, is symmetrical with the A B. Q. E. D. THE SPHERE 385 PROPOSITION XVIII. THEOREM 811 If two triangles on the same sphere or equal spheres are mutually equiangular, they are mutually equilateral, and are either equal or symmetrical. Fig. 1 Fig. 2 Fig. 3 HYPOTHESIS. A and A' are mutually equiangular spherical triangles. CONCLUSION. A A and A' are mutually equilateral, and are either equal or symmetrical. PROOF Let P and P' be the respective polar A of A and A'. Since A A and A' are mutually equiangular, A P and P' are mutually equilateral. ~ 800.'. A P and P' are mutually equiangular. ~~ 809, 810, Case 3.'. A A and A' are mutually equilateral, ~ 800 for they are the polar A of P and P'. ~ 798 Therefore A A and A' are equal if their equal parts are arranged in the same order (Figs. 1 and 2), ~ 809, Case 3 and they are symmetrical if their equal parts are arranged in reverse order (Figs. 1 and 3). ~ 810, Case 3 Q. E. D. EXERCISES 1267 What spherical triangle is its own polar triangle? 1268 A spherical surface can be passed through four points in the same plane if the four points are concyclic. How many solutions? 386 SOLID GEOMETRY — BOOK VIII PROPOSITION XIX. THEOREM 812 Two symmetrical spherical triangles are equivalent. i- / \ HYPOTHESIS. ABC and A'B'C' are symmetrical spherical A. CONCLUSION. A ABC A A'B'C'. PROOF Let P and P' be the poles of the small circles passing through A, B, C, and A', B, C', respectively. The arcs AB, AC, and BC are equal to the arcs A'B', A 'C', and B'C' respectively. ~ 807.'. the chords of the arcs AB, AC, and BC are equal to the chords of the arcs A'B', A'C', and B'C' respectively. ~ 243. the plane A formed by these chords are equal. ~ 172.'. the circles ABC and A'B'C' are equal. ~ 299 Draw the great circle arcs PA, PB, PC, and P'A', P'B', P'C'. These great circle arcs are all equal. ~ 764 Hence the A PAB and P'A'B' are isosceles and symmetrical. ~ 810, Case 3.. A PAB - A P'A'B'. ~ 808 Likewise A PAC = A P'A'C', and A PBC = A P'B'C'. Adding, A ABC A'B'CC'. Ax. 1 Should the poles fall without the triangles, one of the equations must be subtracted from the sum of the other two. Q. E. D. THE SPHERE 387 PROPOSITION XX. THEOREM 813 In an isosceles spherical triangle, the angles opposite the equal sides are equal; conversely, if two angles of a spherical triangle are equal, the sides opposite are equal, and the triangle is isosceles. Fig. 1 Fig. 2 HYPOTHESIS. In the spherical A ABC, AB = AC (Fig. I). CONCLUSION. Z B = L C. PROOF Let the arc AD bisect the Z A. Then A ADB and ADC are symmetrical. ~ 810, Case 1.-. Z B= Z C. ~807 CONVERSELY HYPOTHESIS. In the spherical A ABC, Z B = Z C (Fig. 2). CONCLUSION. AC = AB. PROOF Let A'B'C' be the polar A of ABC. Since Z B = Z C, A'C' = A'B'. ~ 800.. B'= Z C'. ~ 813.-. AC = AB. ~ 800 Q. E. D. 388 SOLID GEOMETRY-BOOK VIII PROPOSITION XXI. THEOREM 814 If two angles of a spherical triangle are unequal, the sides opposite are unequal, and the greater side is opposite the greater angle; conversely, if two sides of a spherical triangle are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side. _ _ _...............D HYPOTHESIS. In the spherical A ABC, Z B > Z C. CONCLUSION. AC > AB. PROOF Draw the great circle arc BD, making Z DBC = Z C. Then DB = DC. ~ 813 Since AD + DB > AB, ~ 793 AD + DC > AB; or AC > AB. CONVERSELY HYPOTHESIS. In the spherical A ABC, AC > AB. CONCLUSION. Z B > Z C. PROOF If L B = L C, AC = AB. ~ 813 If Z B < Z C, AC < AB. ~ 814 Since both these conclusions are contrary to the hypothesis, ZB > ZC. Q. E. D. EXERCISES 389 THEOREMS 1269 Two vertical spherical angles are equal. 1270 If a spherical triangle is equilateral, it is also equiangular. 1271 If a spherical triangle is equiangular, it is also equilateral. 1272 Parallel circles of a sphere have the same poles. 1273 If one of two polar triangles is isosceles, the other is also isosceles. 1274 The arcs bisecting the base angles of an isosceles spherical triangle are equal. 1275 If two sides of a spherical triangle are quadrants, the third side measures the opposite angle. 1276 The polar triangle of a bi-rectangular triangle is bi-rectangular. 1277 A tri-rectangular triangle and its polar are identical. 1278 The three perpendicular bisectors of the sides of a spherical triangle meet in a point. 1279 The three medians of a spherical triangle meet in a point. 1280 The three bisectors of the angles of a spherical triangle meet in a point. 1281 The intersection of two spherical surfaces is the circumference of a circle. 1282 Two symmetrical spherical polygons are equivalent. 1283 An exterior angle of a spherical triangle is less than the sum of the two opposite interior angles. 1284 The sum of the angles of a spherical pentagon is greater than six right angles. 1285 A radius of a sphere perpendicular to the plane of a circle of the sphere passes through the center of the circle. 1286 Each side of a spherical triangle is greater than the difference of the other two sides. 1287 If a right spherical triangle has one side greater than a quadrant, it has a second side greater than a quadrant. 1288 If one of two great circles passes through the pole of the other, their circumferences intersect at right angles. 390 SOLID GEOMETRY-BOOK VIII PROBLEMS OF CONSTRUCTION 1289 To bisect an arc of a great circle. 1290 To bisect a spherical angle. 1291 To inscribe a circle in a spherical triangle. 1292 To circumscribe a circle about a spherical triangle. 1293 At a given point in a great circle to construct a spherical angle equal to a given spherical angle. 1294 To construct a spherical triangle having given two sides and the included angle. 1295 To construct a spherical triangle having given two angles and the included side. 1296 To construct a spherical triangle having given the three sides. 1297 To construct a spherical triangle having given the three angles. 1298 To draw the circumference of a circle through any three points on the surface of a sphere. 1299 Pass a plane through a given point within a sphere so that the section shall be the least circle; the largest circle. 1300 Pass a spherical surface through four given points. 1301 Through a given point on the surface of a sphere pass a plane tangent to the sphere. 1302 Through a given straight line without a sphere pass a plane tangent to the sphere. 1303 To draw a great circle through a given point on a sphere tangent to a given small circle of the sphere. 1304 Pass a spherical surface through two given points and having its center in a given line. With a given radius to construct a spherical surface: 1305 Passing through three given points. 1306 Tangent to two given spheres and passing through a given point. 1307 Tangent to three given spheres. 1308 Tangent to three given planes. EXERCISES 391 PROBLEMS IN LOCI 1309 Find the locus of a point at a given distance from a fixed point. 1310 Find the locus of a line tangent to a given sphere at a given point in the surface of the sphere. 1311 Find the locus of a line drawn from a given point without a given sphere tangent to the sphere. 1312 Find the locus of a point at a given distance from the surface of a given sphere. 1313 Find the locus of the center of a given sphere whose surface passes through a fixed point. 1314 Find the locus of a point on the surface of a given sphere equidistant from three fixed points on the surface of the sphere. 1315 Find the locus of a point on the surface of a given sphere equidistant from two fixed points on the surface of the sphere. 1316 Find the locus of the center of a given sphere whose surface is tangent to a fixed line. 1317 Find the locus of the center of a sphere whose surface passes through the vertices of a given triangle. 1318 Find the locus of the center of a sphere whose surface passes through two given points. 1319 Find the locus of the center of a given sphere tangent to two given spheres. 1320 Find the locus of the center of a sphere tangent to three given planes. 1321 Find the locus of the center of a given sphere tangent to two given intersecting planes. 1322 Find the locus of a sphere tangent to the lateral faces of a given regular prism. 1323 Find the locus of a sphere tangent to the lateral faces of a given regular pyramid. 1324 Find the locus of the middle point of a straight line drawn from a fixed point without to the surface of a given sphere. 1325 Find the locus of the center of a section of a given sphere whose plane passes through a given straight line without the sphere. 392 SOLID GEOMETRY -BOOK VIII MENSURATION OF THE SPHERE SPHERICAL SURFACES DEFINITIONS 815 A zone is a portion of the surface of a sphere included between two parallel planes. The bases of a zone are the circumferences of the sections formed by the parallel planes. The altitude of a zone is the perpendicular between the parallel planes. If one of the parallel planes is tangent to the sphere, the zone formed is called a zone of one base. A C E _ _ DV F............., A,,.,,,...................................... Let ACDB be a semicircle, CE and DF perpendiculars to AB the diameter. If the semicircle is revolved about AB as an axis, the semicircumference ACB will generate the surface of a sphere, the arc CD will generate a zone whose altitude is EF and whose bases are the circumferences generated by the points C and D. The arc AC will generate a zone of one base. 816 A lune is a portion of the surface of a sphere bounded by two semicircumferences of great circles; as, PHP 'K. 817 The angle of a lune is the angle of its bounding arcs. 818 COROLLARY. Two lunes on the same sphere or equal spheres having equal angles are equal. For they can be made to coincide. THE SPHERE 393 PROPOSITION XXII. THEOREM 819 Two lunes on the same sphere or equal spheres are to each other as their angles. _ _ P............................. HYPOTHESIS. Denote the lunes PBP'C and PCP'D by L and L', and the angles of the lunes by A and A'. CONCLUSION. L: L' = A: A'. PROOF Let the Z m, a common measure of A and A', be contained n times in A and n' times in A'. Then A: A' = n: n' (1). By drawing great circles through PP' and the several points of angular division, L will be divided into n lunes, and L' into n' lunes, all equal. ~ 818..L: L' =n:n' (2). From (1) and (2), L: L'= A: A'. Ax. 11 If the angles A and A' are incommensurable, the proof is the same as in ~ 561, Case 2. Q. E. D. 820 COROLLARY. A lune is to the surface of the sphere as the angle of the lune is to four right angles. 821 DEFINITION. A spherical degree is the ninetieth part of a tri-rectangular triangle. Hence the surface of a sphere (~806) contains 720 spherical degrees. 394 SOLID GEOMETRY-BOOK VIII If the lune (L) and the surface of the sphere (S) are expressed in spherical degrees, we have L: 720 = A: 360. ~ 820 Whence L = 2 A. Hence the following 822 COROLLARY. The area of a lune in spherical degrees is numerically expressed by twice its angle. PROPOSITION XXIII. THEOREM 823 If two arcs of great circles intersect on the surface of a hemisphere, the sum of two opposite triangles thus formed is equivalent to a lune whose angle is the angle formed by the given arcs. I m ll I I I B $BI HYPOTHESIS. ACA' and BCB' are semicircumferences of great circles intersecting at C on the hemisphere C-ABA'B', and the arcs ACA' and BCB' produced intersect in C'. CONCLUSION. A CA'B' + A CAB lune CAC'B. PROOF The semicircumference BCB'= the semicircumference C'BC. Take the arc BC from each, and CB' = C'B. Ax. 4 Likewise CA' = C'A, and A'B' = AB..-. A CA'B' and C'AB are symmetrical. ~ 810, Case 3.'. A CA'B' = A C'AB. ~ 812 Add the A CAB to each member, and A CAIB'+ A CAB = lune CAC'B. Q. E. D. THE SPHERE 395 824 DEFINITION. The spherical excess of a spherical triangle is the difference between the sum of its angles and two right angles. Thus, if A, B, C, are the angles of a spherical triangle, and E is its spherical excess, we have E= A + B + C - 180~. PROPOSITION XXIV. THEOREM 825 The area of a spherical triangle in spherical degrees is numerically equal to the spherical excess of the triangle..-. d..._-, —_-_............,, j,, _,........................\:: _=_._.....__. _...__..._ __. _ B._ B' HYPOTHESIS. Denote the values of the angles of the spherical triangle ABC by A, B, C, and its spherical excess by E. CONCLUSION. Area of A ABC in spherical degrees = E. PROOF Complete the Oces of which AB, AC, BC, are arcs. Then A ABC + A 'BC lune ABA'C = 2 A, ~ 822 A ABC + A AB'C lune BAB'C = 2 B, and A ABC + a A 'B'C = lne CAC'B = 2 C. ~ 823 Adding, observing that the sum of the triangles exceeds the surface of the hemisphere (360 spherical degrees) by twice the triangle ABC, we have in spherical degrees, 2 A ABC + 360 = 2 (A + B + C)... A ABC = A + B + C-180 = E. ~ 824 Q. E. D. 396 SOLID GEOMETRY-BOOK VIII 826. DEFINITION. The spherical excess of a polygon is the difference between the sum of its angles, and two right angles taken as many times as the polygon has sides less two. Thus if K is the sum of the angles of a spherical polygon of n sides, and E is its spherical excess, we have E = K - (n - 2) 180~. PROPOSITION XXV. THEOREM 827 The area of a spherical polygon in spherical degrees is numerically equal to the spherical excess of the polygon..........._.._....A is E. CONCLUSION. The area of ABCDE in spherical degrees = E. PROOF From any vertex A draw diagonals dividing the polygon into spherical triangles. The area of each triangle in spherical degrees is numerically equal to the spherical excess of the triangle. ~ 825 Hence the area of the polygon in spherical degrees is numerically equal to the sum of the spherical excesses of the triangles, which is the spherical excess of the polygon. ~ 826.'. the area of ABCDE in spherical degrees = E. Q. E. U. THE SPHERE 397 PROPOSITION XXVI. THEOREM 828 The area of the surface generated by the revolution of a straight line about an axis in the same plane is equal to the product of the projection of the line on the axis by the circumference whose radius is the perpendicular to the line erected at its middle point and terminated by the axis. B A -E X -- C': PG HYPOTHESIS. AB revolves about XY as an axis, CD is the projection of AB on XY, E is the middle point of AB, EG is I to AB, EF is I to XY, and AH is _L to BD. CONCLUSION. The area AB = CD x 2 7rEG. PROOF If AB and XY are not parallel and do not meet, AB generates the surface of a frustum of a cone of revolution. Hence the area AB = AB x 2 7rEF (1). ~ 742 The A EGF and ABH are similar. ~ 357 Whence AB: AH = EG: EF. ~ 349.-.AB x EF = AH x EG. ~ 328 But AH = CD. ~ 200.. AB X 2 7rEF = CD x 2rEG (2). From (1) and (2), area AB = CD x 2 7rEG. If AB is II to XY, AB generates the lateral surface of a cylinder of revolution, and area AB = CD x 2 7rEG. ~ 711 If AB meets XY, AB generates the lateral surface of a cone of revolution, and the conclusion follows from ~ 736. Q. E. D. 398 SOLID GEOMETRY -BOOK VIII PROPOSITION XXVII. THEOREM 829 The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle. B - C A E eF HYPOTHESIS. S is the area of the surface and R the radius of a sphere generated by the semicircle ABD revolving about the diameter AD as an axis. CONCLUSION. S = AD x 2 7rR. PROOF Inscribe in the semicircle the half of a regular polygon of an even number of sides, as ABCD, denote its apothem by a, and the area of the surface generated by its perimeter by S'. Draw BE and CF perpendicular to AD. Then area AB = AE x 2 7ra, area BC = EF x 2 7ra, and area CD = FD x 2 7ra. ~ 828 Adding, S' = AD x 27ra. Ax. 1 By doubling indefinitely the number of sides of the semipolygon, a approaches R as a limit. ~ 450 Whence AD x 2 7ra approaches AD x 2 IrR as a limit. ~ 267 Also, S' approaches S as a limit. ~ 455 But S' = AD x 2 rwa. Proved.'.S =AD x 27R. ~ 264 Q. E. D. THE SPHERE 399 830 COROLLARY 1. The surface of a sphere is equivalent to four great circles. For S = AD x 2 T2R = 2 R X 2 7rR = 4 rR2. 831 COROLLARY 2. The areas of the surfaces of two spheres are to each other as the squares of their radii, or as the squares of their diameters. 832 COROLLARY 3. The area of a zone is equal to the product of its altitude by the circumference of a great circle. For the surface generated by any arc BC is a zone the area of which by the proof of the theorem equals EF x 2 7rR, in which EF is the altitude of the zone (~ 815), and 2,rR is the circumference of a great circle. 833 COROLLARY 4. Zones on the same sphere or equal spheres are to each other as their altitudes. For denote the areas of two zones by Z and Z', and their altitudes by H and H'. Then, Z: Z'= H x 2 rR: H' x 2 rR = H:H'. 834 COROLLARY 5. A zone of one base is equivalent to the circle whose radius is the chord of the generating arc. For zone AB = AE x 2 7rR = rAE x AD = rAB2. ~ 365, Case 3 835 COROLLARY 6. The area of a lune = -72A 90 For a spherical degree = 4R2 ~~ 821, 830 720 1 and the area of a lune = 2 A spherical degrees ~ 822 A 4 7rR2 7rR2A 720 90 836 COROLLARY 7. The area of a spherical polygon =-7rR2. 180 For area of spherical polygon = E spherical degrees ~ 827 4 7R2 7R2E =Ex R -= 720 180 400 SOLID GEOMETRY -BOOK VIII SPHERICAL VOLUMES DEFINITIONS 837 A spherical segment is a portion of a sphere included between two parallel planes. The bases of a spherical segment are the sections of the sphere made by the parallel planes. The altitude of a spherical segment is the perpendicular between the bases. If one of the parallel planes is tangent to the sphere, the segment formed is called a spherical segment of one base. A F B Let ACDB be a semicircle, CE and DF perpendicular to AB the diameter. If the semicircle is revolved about AB as an axis, the solid generated by the figure CEFD is a spherical segment, whose altitude is EF, and whose bases are the circles generated by CE and DF. The solid generated by the figure BDF is a spherical segment of one base. If DF is a radius, the spherical segment generated by BDF is half a sphere. 838 A spherical sector is the portion of a sphere generated by any sector of a semicircle when the semicircle is revolved about its diameter as an axis. The base of a spherical sector is the zone generated by the arc of the circular sector. THE SPHERE 401 If the base of a spherical sector is a zone of one base, the spherical sector is called a spherical cone. A AB Thus, when the semicircle ACDB revolves about AB as an axis, the solid generated by the circular sector COD is a spherical sector, whose base is the zone generated by the arc CD. The solid generated by the sector AOC is a spherical cone. 839 A spherical pyramid is the portion of a sphere bounded by a spherical polygon and the planes of the great circles forming the sides of the polygon. The vertex of a spherical pyramid is the center of the sphere. The base of a spherical pyramid is the spherical polygon which forms a part of its boundary. Thus, O-ABCD is a spherical pyramid, whose vertex is 0, and whose base is the spherical polygon ABCD. E H 0 F 840 A spherical wedge, or ungula, is the portion of a sphere bounded by a lune and two great semicircles. Thus, EOFHK is a spherical wedge. 402 SOLID GEOMETRY -BOOK VIII PROPOSITION XXVIII. THEOREM 841 The volume of a sphere is equal to the product of the area of its surface by one third of its radius. HYPOTHESIS. 0 is the center of a sphere whose radius is R, surface S, and volume V. CONCLUSION. V = S x ~ R. PROOF Circumscribe about the sphere any polyedron; denote its surface by S' and its volume by V'. Straight lines drawn from O to each of the vertices of the polyedron divide the polyedron into as many pyramids as it has faces, whose bases are the faces of the polyedron, whose common vertex is 0, and whose common altitude is R. The volume of each of these pyramids is equal to the area of its base multiplied by ~ R. ~ 667 Therefore, the combined volume of all these pyramids, which is the volume of the circumscribed polyedron, is equal to the area of the surface of the polyedron multiplied by ER. That is, V' = S' x ~ R, no matter how many faces the circumscribed polyedron may have. By passing planes tangent to the sphere at the points in which the lateral edges of the pyramids pierce the surface of the sphere, a part of the polyedron will be cut away, and a new THE SPHERE 403 polyedron, having more faces than the first, will be circumscribed about the sphere, whose volume will be less than the volume of the first polyedron. Ax. 12 By joining O to each of the vertices of this new polyedron, the polyedron will be divided into a new series of pyramids, greater in number than the first series, but whose combined volume is likewise equal to the area of the surface of the new polyedron multiplied by ~ R. By repeating indefinitely this process of circumscribing new polyedrons about the sphere, the volume of each will be less than the volume of the preceding one (Ax. 12), and hence V' will be a decreasing variable approaching V as a limit. Since in the equation V' = S' x I R, V' is a decreasing variable, and R is a constant, therefore S' is a decreasing variable approaching S as a limit. Since S' approaches S as a limit, S' x ~R approaches S x 3R as a limit. ~~ 267, 268 Also, V' approaches V as a limit. Proved But V' = S' x R. Proved.'.V=Sx 3R. ~264 Q. E. D. 842 COROLLARY 1. If V denotes the volume, R the radius, and D the diameter of a sphere, V = 4, rR3 = 7rD3. 843 COROLLARY 2. The volumes of two spheres are to each other as the cubes of their radii. For V: V' = 4 rR3: 4 7rR'3 = R3: R'3. 844 COROLLARY 3. The volume of a spherical pyramid is equal to the product of the area of its base by one third of the radius of the sphere. 404 SOLID GEOMETRY-BOOK VIII 845 COROLLARY 4. The volume of a spherical sector is equal to the product of the area of its base by one third of the radius of the sphere. Let R denote the radius of a sphere, C the circumference of a great circle, V the volume of a spherical sector, H the altitude of the zone forming the base of the spherical sector, and Z the surface of the zone. Then C = 2 rR, Z = 2 wrRH, and V = riR2H. PROPOSITION XXIX. THEOREM 846 The volume of a spherical segment is equal to half the product of its altitude by the sum of its bases, plus the volume of a sphere whose diameter is the altitude of the segment. A D F HYPOTHESIS. O is the center and AB the diameter of the semicircle ACDB, CE and DF are I to AB. Denote CE by r, DF by r', EF by h, OF by a, OE by b, the radius OC by R, and the volume of the segment generated by the figure EFDC by V. CONCLUSION. V = h (7rr2 + 7rr'2) + q1 7rh3. PROOF The volume generated by EFDC, as the semicircle rotates about AB as an axis, is equal to the spherical sector generated by OCD plus the cone generated by ODF minus the cone generated by OCE. THE SPHERE 405 The volume of the sector generated by OCD = ] 7rR2h. ~ 845 The volume of the cone generated by ODF = 7rr '2a. ~ 738 The volume of the cone generated by OCE = 7r rb. ~ 738.*. V =2 R2h + ~ rr'a - a rb. But a - b = h, a2 = R2 - r2, and b2 = R2 - -2.. '. V = R [2 R2 (a - b) + (R2 - a2)- (R2 - b2)b] = r [2 R2 (a - b) + R2 (a - b) - (a3 - b3)] = 7rh[3 R2 - (a2 + ab + b2)] (). But h2 = (a - b)2 = a2 - 2 ab + b2. Whence a2 + ab + b2 2 (a2)(a2 - 2 ab + b2 ab + ) = 3 (a2 + b2) - - = ] (R2 - r2 + R r' 2) - =3 R2 - 3 (2 + r' 2)_ 2 Substituting this value in (1), v= 3 (r2 + r2) +^ h] = h (rr2 + 7rr 2) + wh3. Q. E. D. 847 COROLLARY. If the spherical segment has but one base, r' = 0, and we have V = ~ 7rr2h + 7rh3. 406 SOLID GEOMETRY-BOOK VIII EXERCISES 1326 The volume of a sphere is equal to two thirds of the volume of the circumscribed cylinder. 1327 If a cone has the same base and altitude as the cylinder circumscribed about a sphere, prove that cone: sphere: cylinder = 1: 2: 3. 1328 The radius of a sphere is 10 in. Find the radius of a small circle whose distance from the center of the sphere is 8 in. 1329 The radii of two equal intersecting spheres are 5 in., and the distance between the centers of the spheres is 8 in. Find the circumference of their intersection. 1330 Find the radius of a circle of a sphere whose polar distance is 60~, if the radius of the sphere is 34 in. 1331 The diameter of a sphere is 96 in. Compute the length of a tangent drawn to the sphere from a point 25 in. from the sphere. Compute the area of the surface of a sphere 1332 Whose radius is 10 in. 1333 Whose diameter is 8 in. 1334 Whose circumference is 37.6992 feet. 1335 Whose radius is 7r. 1336 Whose volume is one cubic yard. Compute the radius of a sphere 1337 Whose circumference is 12 in. 1338 Whose surface is one square yard. 1339 Whose volume is one cubic foot. 1340 If the area of a great circle is 1256.64 square feet. 1341 Some of the planetoids are only 10 miles in diameter. How many acres in the surface of such a sphere? 1342 The radii of two spheres are as 2:3. Compare their surfaces. 1343 The surface of one sphere is twice the surface of another. Compare their radii. 1344 The diameter of the base of a right circular cone is equal to the slant height. Compare its total area to the area of the surface of the inscribed sphere. EXERCISES 407 1345 The surface of a sphere is equivalent to the lateral surface of the circumscribed cylinder. 1346 The surface of a sphere is two thirds of the surface of the circumscribed cylinder. 1347 Find the area of a spherical triangle whose angles are 75~, 85~, 105~, the radius of the sphere being 12 inches. 1348 The angles of a spherical triangle are 80~, 90~, 130~. The area of the triangle is what part of the surface of the sphere? 1349 The angles of a spherical triangle are 84~, 96~, 112~. If the radius of the sphere is 9 inches, find the area of a symmetrical triangle. 1350 The sides of a spherical triangle are 47~, 56~, 87~. Find the area of the plane triangle, if the diameter of the sphere is 14 inches. 1351 Find the area of a spherical quadrilateral whose angles are 78~, 107~, 125~, 140~, the radius of the sphere being 10 inches. 1352 Find the area of a spherical pentagon whose angles are 120~, 127~, 130~, 133~, 150~, the surface of the sphere being 148 square feet. 1353 Find the area of a spherical hexagon whose angles are 99~, 138~, 148~, 157~, 158~, 164~, the area of a great circle of the sphere being 35 sq. ft. 1354 Find the area of a zone whose altitude is 3 inches, if the radius of the sphere is 16 inches. 1355 Find the area of a zone of one base, the chord of whose generating arc is 4 inches. 1356 If the radius of a sphere is 4 inches, find the altitude of a zone whose area is that of a great circle. 1357 If the radius of a sphere is 10 inches, find the altitude of a zone whose area is one third the area of the surface of the sphere. 1358 Find the area of a lune whose angle is 20~, the surface of the sphere being 180 square inches. 1359 Find the area of a lune whose angle is 75~, the radius of the sphere being 14 inches. 1360 What is the angle of a lune on a sphere of 15-inch radius, if the area of the lune is equal to the area of the surface of a sphere of 5-inch radius? 408 SOLID GEOMETRY —BOOK VIII Compute the volume of a sphere 1361 Whose radius is 3 inches. 1362 Whose diameter is 12 inches. 1363 Whose circumference is 100 feet. 1364 Whose surface is 1000 square feet. 1365 Find the diameter of a sphere whose circumference and surface are numerically equal. 1366 Find the diameter of a sphere whose circumference and volume are numerically equal. 1367 Find the diameter of a sphere whose surface and volume are numerically equal. 1368 The radii of two spheres are 3 and 5. What is the ratio of their volumnes? 1369 The volumes of two spheres are 8 and 27. What is the ratio of their radii? 1370 One sphere has twice the volume of a second. Compare their radii. 1371 If a sphere 2 inches in diameter weighs 1 ounce, what is the diameter of a sphere of the same material weighing 125 ounces? 1372 The radii of two spheres of lead are as 2 to 1. If the first weighs 216 ounces, how much does the second weigh? 1373 Find the radius of a sphere equivalent to two spheres whose radii are 3 inches and 5 inches. 1374 How many bullets ~ of an inch in diameter can be made from a cubic foot of lead? 1375 What is the diameter of an iron ball which when dropped into a cylinder of 12-inch diameter causes the water in the cylinder to rise one inch? 1376 The exterior diameter of a hollow sphere is 4 inches in diameter. If the sphere will hold 14.1372 cubic inches of water, find the thickness of the shell. 1377 How many cubic inches of lead are required to make a spherical shell ~ inch thick, if the exterior diameter of the shell is 14 inches? 1378 Find the volume of a sphere inscribed in a cylinder whose altitude is 18 inches. EXERCISES 409 1379 The volume of a sphere is to the volume of the circumscribed cube as or is to 6. 1380 The volume of a sphere is to the volume of the inscribed cube as r is to i /3. 1381 The volume of a sphere is 120. Find the volume of the circumscribed cube. 1382 The diameter cf a sphere is 3 inches. Find the volume of the inscribed cube. 1383 The diameters of two spheres of the same material are a and b respectively. If the first weighs n pounds, find the weight of the second. 1384 The base of a circular cone is equal to a great circle of a sphere, and the altitude of the cone is equal to the diameter of the sphere. Compare their volumes. 1385 Find the volume of a spherical wedge whose angle is 60~, if the radius of the sphere is 5 inches. 1386 The volume of a spherical wedge is 7 cubic feet, and the volume of the sphere is 63 cubic feet. Find the angle of the wedge. 1387 What part of a sphere is a spherical pyramid whose base is a trirectangular triangle? 1388 If the volume of a sphere is 214 cubic inches, what is the volume of a triangular spherical pyramid, the angles of whose base are 80~, 104~, and 116~? 1389 Find the volume of a spherical sector whose altitude is 7 inches, if the radius of the sphere is 27 inches. 1390 Find the volume of a spherical sector, if the area of its curved surface is 5, and the radius of the sphere is 2. 1391 The radii of the bases of a spherical segment are 7 and 9, and the altitude is 5. Find the volume. 1392 Find the volume of a spherical cone, if the radius of its base is 8, and the radius of the sphere is 10. 1393 If h is the distance of a light from a sphere whose radius is R, show that the surface illuminated is 2 rR2h R+h 410 SOLID GEOMETRY —BOOK VIII In the following exercises the earth is considered a perfect sphere, and 8000 miles in diameter, unless otherwise stated. 1394 If a man was 1000 miles above the earth, how much of its surface could he see? 1395 How far above the earth must a man be to see one fifth of its surface? 1396 A man standing on the sea-shore can see the water of the ocean how far away, if his eye is 6 feet above the level of the sea? 1397 With the eye at the level of the sea, the top of a mast of a ship 200 feet high is just visible on the horizon. What is the distance of the ship from the observer? 1398 Compute the surface and volume of the earth. 1399 The diameter of the sun is about 108.5 times the diameter of the earth. Compare their surfaces and volumes. 1400 The diameter of Jupiter is 11 times that of the earth. Compare their volumes. 1401 The planet Mars has two moons, Deimos and Phobos, whose diameters are estimated to be 7 miles and 6 miles respectively. Compute their surfaces and volumes. 1402 The diameter of the moon is 2163 miles, and the diameter of the earth is 7920 miles. Show that the volume of the moon is almost exactly T of the volume of the earth. 1403 Taking the earth's diameter as unity, the diameters of the other major planets are roughly as follows: Mercury, I; Venus, 1; Mars, ~; Jupiter, 11; Saturn, 9; Uranus, 4; Neptune, 41. Compare the volume of Jupiter with the combined volumes of all the other major planets. 1404 Assuming that the orbit of the earth around the sun is a circle whose radius is 92,800,000 miles, and assuming that the velocity of the earth in its orbit is uniform, and assuming that the earth completes one revolution around the sun in 3651 days, compute the velocity of the earth in its orbit per second. FORMULAS OF MENSURATION SYMBOLS B = lower base P = perimeter of lower base B' = upper base P' = perimeter of upper base C = circumference of lower base P = perimeter of right section C' = circumference of upper base R = radius of lower base C"t = circumference of mid-section R' = radius of upper base E = lateral edge S = lateral area H = altitude T = total area L = slant height V = volume a, b, c, = dimensions of a parallelopiped FORMULAS Right Prism, Any Prism, Any Prism, Rectangular Parallelopiped, Any Parallelopiped, Regular Pyramid, Any Pyramid, Frustum of Regular Pyramid, Frustum of any Pyramid, Cylinder of Revolution, Any Cylinder, Cone of Revolution, Any Cone, S=HxP S = H x P S = E x P" V=B x H V= a x b x c V=B xH S = - L x P V= B x H S = (P + P) x L V = - H(B + B' + /B x B') S =C2 x H S 2 7rR x 11 T = 2 7rR (H + R) V= 7rR2 x H V=B x H S = C xL S =7rRL T = rR (L + R) V= 1 irR2 x H V= B xH 411 PAGE 307 307 319 316 317 323 330 323 334 349 350 350 350 350 359 359 359 360 360 412 FORMULAS OF MENSURATION Frustum of Cone of Revolution, S = - (C + C') x L S = L x C" V= i 7rH (R2+ R12 +RR') Frustum of any Cone, V = 1 H (B + B' + v/B x B') SPHERICAL AREAS AND VOLUMES PAGE 362 362 363 363 A = no. of degrees in angle B = base D = diameter E = spherical excess Spherical Areas Sphere, Zone, Lune, Spherical polygon, Spherical Volumes Sphere, Spherical pyramid, Spherical sector, Spherical segment, H = altitude r and r' = radii of bases S = area of surface V = volume S =4 rR2 S =2 7rRH r R2A 90 rR2E 180 V= S x R V = trR3 V= 17D3 V=B x IR V= 2 7rR2H V = ~ H (r2 + rr'2) + rH3 399 399 399 399 402 403 403 403 404 404 v/2= 1.4142 3 = 1.7321 v = 2.2361 6= 2.4495 7i = 2.6458 Vi0 = 3.1623 / = 0.7071 -2 = 1.2599 3 = 1.4422 APPROXIMATE VALUES = 1.5874 - =1.7100 - = 1.8171 //7= 1.9129 /9 = 2.0801 /10 = 2.1544 = 0.7937 r = 3.1416 r2 = 9.8696 TrS = 31.0063 Vr = 1.7725 1 = 0.3183 7r I = 0.5642 Vr 4 ir = 4.1888 INDEX OF DEFINITIONS [Figures refer to pages. ] Absurdity, 1. Alternation, 142. Altitude, of cone, 353. of cylinder, 344. of frustum of pyramid, 322. of parallelogram, 54. of prism, 304. of pyramid, 321. of spherical segment, 400. of trapezoid, 54. of triangle, 32. of zone, 392. Analysis, of theorems, 70. of problems, 125. Angle, 7. acute, 8. at center of regular polygon, 227. central, 80. complement of, 9. diedral, 284. exterior of polygon, 30. inscribed in circle, 80. oblique, 8. obtuse, 9. of lune, 392. of two curves, 374. polyedral, 296. reflex, 9. right, 8. round, 8. sides of, 7. spherical, 374. straight, 8. supplement of, 9. tetraedral, 296. triedral, 296. Angle, vertex of, 7. Angles, adjacent, 8. alternate exterior, 23. alternate interior, 23. complementary, 9. corresponding, 23. exterior, 23. homologous, 36. interior, 23. of polygon, 30. supplementary, 10. supplementary adjacent, 17. vertical, 18. Antecedents, 140. Apothem, 227. Arc, 78. Area, 192. Axiom, 1. Axis, of circular cone, 353. of circle of sphere, 367. of regular pyramid, 321. Base, of isosceles triangle, 31. of pyramid, 321. of spherical pyramid, 401. of spherical sector, 400. of triangle, 31. Bases, of cylinder, 344. of frustum of cone, 355. of frustum of pyramid, 322. of parallelogram, 54. of prism, 304. of spherical segment, 400. of trapezoid, 53. of zone, 392. Bisector of angle of triangle, 32. 413 414 INDEX OF DEFINITIONS Center, of circle, 78. of regular polygon, 227. of sphere, 366. Chord, 78. Circle, 78. Circles, concentric, 80. escribed, 122. tangent, 79. Circumference, 78. Commensurable, 91. Complement, 9. Composition, 143. and division, 145. Conclusion, 2. Concyclic points, 80. Cone, 353. altitude of, 353. base of, 353. circular, 353. lateral surface of, 353. oblique, 354. of revolution, 354. right, 354. slant height of, 354. truncated, 355. vertex of, 353. Conical surface, 353. directrix of, 353. element of, 353. generatrix of, 353. nappes of, 353. vertex of, 353. Consequents, 140. Constant, 91. Continued proportion, 140. Converse of theorem, 2. Corollary, 1. Cube, 305. Cylinder, 344. altitude of, 344. bases of, 344. circular, 345. lateral surface of, 344. oblique, 344. of revolution, 345. Cylinder, right, 344. right circular, 345. right section of, 344. section of, 344. Cylindrical surface, 344. directrix of, 344. element of, 344., generatrix of, 344. Decagon, 62. Definition, 1. Degree, 11. spherical, 393. Demonstration, 1. direct, 2. indirect, 2. Diagonal of polygon, 30. Diameter, of circle, 78. of sphere, 365. Diedral angle, 284. acute, 285. edge of, 284. faces of, 284. plane angle of, 284. right, 284. I)iedral angles, adjacent, 284. complementary, 285. obtuse, 285. Dimensions, 4. Distance, from point to line, 47. from point to plane, 273. on surface of sphere, 367. Division, 144. external, 151. internal, 151. Dodecaedron, 303. Dodecagon, 62. Edge of diedral angle, 284. Edges, of polyedral angle, 296. of polyedron, 303. Element, of conical surface, 353. of cylindrical surface, 345. Equal figures, 10. Equivalent figures, 10. INDEX OF DEFINITIONS 415 Equivalent surfaces, 192. Excent( rs of triangle, 122. Extension, 3. Extreme and mean ratio, 187. Extremes, 140. Faces, of diedral angle, 284. of polyedral angle, 296. of polyedron, 303. Figure, curvilinear, 7. geometric, 7. plane, 7. rectilinear, 10. Foot of perpendicular, 10. Fourth proportional, 140. Frustum of cone, 355. altitude of, 355. bases of, 355. lateral surface of, 355. slant height of, 355. Frustum of pyramid, 322. altitude of, 322. bases of, 322. lateral area of, 322. slant height of, 322. Generatrix of conical surface, 353. of cylindrical surface, 344. Geometry, 3, 7. fundamental concepts in, 5. plane, 11. solid, 268. Harmonic division, 151. Heptagon, 62. Hexaedron, 303. Hexagon, 62. Hypotenuse, 31. Hypothesis, 2. Icosaedron, 303. Inclination of line to plane, 293. Incommensurable, 91. Inversion, 143. Isoperimetric figures, 254. Legs, of isosceles triangle, 31. of right triangle, 31. of trapezoid, 53. Lemma, 1. Limit, 91. Line, 5. broken, 6. curved, 6. distance from one to another, 47. distance of point from, 47. mixed, 6. oblique to plane, 270. of centers, 79. parallel to plane, 270. perpendicular to plane, 270. straight, 6. Lines, homologous, 36. oblique, 10. parallel, 20. perpendicular, 10. Locus, 126. Lune, 392. angle of, 392. Magnitude, 3. Magnitudes, equal, 10. equivalent, 10. similar, 10. Major arc of circle, 78. Maximum, 254. Mean proportional, 140. Means, 140. Median, of trapezoid, 53. of triangle, 32. Minimum, 254. Minor arc of circle, 78. Nappes of cone, 353. Octaedron, 303. Octagon, 62. Parallel, lines, 20. planes, 270. 416 INDEX OF DEFINITIONS Parallelogram, 53. oblique, 53. right, 53. Parallelopiped, 305. dimensions of, 313. oblique, 305. rectangular, 305. right, 305. Pentadecagon, 62. Pentagon, 62. Perimeter, 30. Perpendicular, 10. lines, 10. planes, 285. Plane, 7. determined, 268. oblique to line, 270. of projection, 293. parallel to line, 270. perpendicular to line, 270. projecting, 293. Planes, parallel, 270. perpendicular, 285. Point, 5. of contact, 79. of tangency, 79. Polar, distance of circle, 368. triangle, 379. Pole, 367. Poles of circle of sphere, 367. Polyedral angle, 296. convex, 296. edges of, 296. face angles of, 296. faces of, 296. parts of, 296. vertex of, 296. Polyedral angles, equal, 296. symmetrical, 296. vertical, 297. Polyedron, 303. convex, 303. diagonal of, 303. edges of, 303. faces of, 303. Polyedron, regular, 342. vertices of, 303. Polyedrons, similar, 338. Polygon, 30. adjacent angles of, 30. angles of, 30. circumscribed, 80. concave, 62. convex, 62. diagonal of, 30. equiangular, 62. equilateral, 62. exterior angle of, 30. inscribed, 80. perimeter of, 30. regular, 225. sides of, 30. spherical, 375. Polygons, mutually equiangular, 36. mutually equilateral, 36. Postulate, 1. Prism, 304. altitude of, 304. base edges of, 304. bases of, 304. circumscribed about cylinder, 345. inscribed in cylinder, 345. lateral area of, 304. lateral edges of, 304. lateral faces of, 304. oblique, 304. quadrangular, 304. regular, 304. right, 304. right section of, 305. triangular, 304. truncated, 304. Problem, 1. Projection, of line upon line, 166. of line upon plane, 292. of point upon plane, 292. Proportion, 140. continued, 140. Proposition, 1. Pyramid, 321. INDEX OF DEFINITIONS 417 Pyramid, altitude of, 321. axis of, 321. base of, 321. circumscribed about cone, 355. inscribed in cone, 355. lateral area of, 321. lateral edges of, 321. lateral faces of, 321. quadrangular, 321. regular, 321. slant height of, 322. triangular, 321. truncated, 322. vertex of, 321. Quadrant, 78. spherical, 368. Quadrilateral, 62. Radius of circle, 78. of regular polygon, 227. of sphere, 365. Ratio, 91. of similitude, 154. Rectangle, 53. Rhomboid, 53. Rhombus, 53. Right section of prism, 305. Scholium, 1. Secant, 79. Sector of circle, 79. Segment, of circle, 78. of sphere, 400. Semicircle, 79. Semicircumference, 78. Sides, of angle, 7. of polygon, 30. of spherical polygon, 375. Similar, arcs, 237. cones of revolution, 354. cylinders of revolution, 345. polyedrons, 338. polygons, 154. sectors, 237. segments, 237. Slant height of regular pyramid, 322. Solid, 5. Solution, 1. Space, 3. Sphere, 365. center of, 365. circle of, 366. circumscribed about polyedron, 372. diameter of, 365. distance between two points on surface of, 367. great circle of, 367. inscribed in polyedron, 372. radius of, 365. small circle of, 367. Spherical angle, 374. Spherical cone, 401. Spherical excess, of triangle, 375. of polygon, 396. Spherical polygon, 375. angles of, 375. convex, 375. diagonal of, 375. sides of, 375. vertices of, 375. Spherical polygons, symmetrical, 382. Spherical pyramid, 401. base of, 401. vertex of, 401. Spherical sector, 400. base of, 400. Spherical segment, 400. altitude of, 400. bases of, 400. of one base, 400. Spherical triangle, 375. bi-rectangular, 381. polar, 379. tri-rectangular, 381. Spherical wedge, 401. Square, 53. Supplement, 9. Surface, 5. 418 INDEX OF DEFINITIONS Surface, curved, 7. plane, 7. Symmetry, 262. Synthesis, 70. Tangent, common internal, 79. common external, 79. Tangent line, to circle, 79. to cone, 354. to cylinder, 345. to sphere, 365. Tangent plane, to cone, 354. to cylinder, 345. to sphere, 365. Tangent spheres, 365. Terms of a proportion, 140. Tetraedral angle, 296. Tetraedron, 303. Theorem 1, converse of, 2. opposite of, 2. Third proportional, 140. Transversal, 23. Trapezium, 53. Trapezoid, 53. isosceles, 53. Triangle, 62. acute, 31. altitude of, 32. base of, 31. base angles of, 31. centroid of, 65. circumcenter of, 65. equiangular, 31. equilateral, 31. incenter of, 65. Triangle, isosceles, 31. legs of, 31. legs of isosceles, 31. median of, 32. oblique, 30. obtuse, 31. orthocenter of, 65. right, 30. scalene, 31. spherical, 375. vertex of, 31. vertical angle of, 31. Triedral angle, 296. bi-rectangular, 296. isosceles, 296. rectangular, 296. tri-rectangular, 296. Unit, of surface, 192. of volume, 305. Ungula, 401. Variable, 91. Vertex, of angle, 7. of cone, 353. of polyedral angle, 296. of pyramid, 321. of triangle, 31. Volume of solid, 305. Wedge, spherical, 401. Zone, 392. altitude of, 392. bases of, 392. of one base, 392.