;.■:■"■■ mmm ■ ", M i Ma+h a a 4-53 5 37 CORNELL UNIVERSITY LIBRARY CORNELL UNIVERSITY LIBRARY 3 1924 063 308 229 Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924063308229 ELEMENTS GEOMETRY 4 With Exercises for Students, and an Introduction to Modern Geometry. A. SCHUYLEB, LL.D., President of Baldwin University, Author of Higher Arithmetic, Principles of Logic, Complete Algebra, Surveying and Navigation, and Trigonometry and Mensuration. WILSON, HINKLE & CO., 137 Walnut Street 28 Bond Street CINCINNATI NEW YORK GRADED-SCHOOL SERIES. HARVEY'S SPELLER AND READERS. HARVEY'S LANGUAGE COURSE. WHITE'S ARITHMETICS. ECLECTIC GEOGRAPHIES. SCHUYLER'S ALGEBRA AND GEOMETRY. Copyright 1876 by Wilson, Hinkle & Co. ELECTKOTYPED AT ECLECTIC PRESS FP.ANKLIN TYPE FOUNDKY WILSON, HINKLE 4 CO CINCINNATI CINCINNATI PEEFAOE. A new treatise on Geometry, to be of sufficient merit to claim attention, must be both conservative and progressive. It should lay firm hold on the past, embody the present state of the sci- ence, and anticipate future developments. A work claiming to be wholly new might, perhaps with justice, be at once discarded as worthless; while one containing no improvements could not justify its own existence. The geometrical objects, — points, lines, surfaces, solids, and angles, — constitute the subject-matter of the science; the defi- nitions are the tests by which these objects are discriminated and their classification determined ; the axioms are the warrants for the steps taken in the course of demonstration ; the postu- lates justify the assumption of magnitudes having any position, form, and extent. The logical principles which underlie the demonstrations of this volume have be.en carefully -discriminated' and illustrated. The discussion of the axioms and postulates is the result of research, and intent , arid' prolonged' thought. ■ That funda- mental principles have been. r«ache,d ? is. manifest from their underivability, and the simplicity' of 'the deduction of the ordinary so-called axioms from ttreTn as corollaries. Mr. Bain has observed of the principle, If A be greater than B, and B greater than O, much more is A greater than C: " If it can not be deductively inferred from the proper axioms, it will have to be received as a third axiom." Not only can this prin- ciple be inferred (23, 20), but also Mr. Bain's so-called proper axioms (23, 3, 6). (Hi) IV PREFACE. The proposition, " A straight line is the shortest distance from one point to another," has, by some mathematicians, been given as a definition ; by others, as an axiom. This discrepancy raises the question whether it is either. A straight line is not denned by calling it the shortest distance from one point to another; for this does not, as a correct definition would, develop the idea :>f a straight line ; but assuming the idea developed, it affirms of the straight line an additional truth, that any portion of it is the shortest distance from one of the points limiting that por- tion to the other point. But this truth is demonstrable (67), and is, therefore, not an axiom, but a theorem. The discussion of parallels, a matter involved in controversy, requires no additional axiom — an evidence of the sufficiency of the axioms as given, and a justification of the definition of parallels. The theory of limits hg,s been applied to the measurement of the circle, thus securing demonstrations irreproachable in their logical character. When a reference is given, the student should state the prin- ciple involved ; when the query mark (?), which is to be inter- preted "why?" is used, he should give the reason. The discussion of proportion, as properly belonging to Algebra, has been omitted. The references to that subject are to the author's Complete Algebra. Book VIII treats of Modern Geometry, which has within the last century so greatly stimulated the growth of Elementary Ge- ometry, for a time arrested by the analytic method of Descartes. By omitting the Supplementary Sections and the Modern Geometry, Book VIII, those schools ' that wish a moderate course can be accommodated in a simple manner. The Exercises, always interesting to the student, can be taken, one or two each day, while the class is pursuing the next section. Thanks are due to Prof. Warner for invaluable aid. Hoping that the work will receive the candid attention of mathematicians, and the favorable consideration of the public, it is sent forth to accomplish its mission. CONTENTS. Introduction . . . . General Outline Logical Facts and Laws Axioms and Postulates . BOOK I Perpendicular and Oblique Lines Parallel Lines Triangles .... Quadrilaterals Polygons in General Supplementary Propositions . BOOK II Straight Lines and Circles Relative Position of Circles . Measurement of Angles . Constructions .... Inscribed and Circumscribed Polygons Symmetry.— Supplementary BOOK III . . ., Area and Equivalency . ' Proportionality and Similarity Constructions BOOK IV The Circle and Regular Polygons Theory of Limits Measurement of the Circle Maxima and Minima. — Supplementary . 9 9 15 21 27 27 33 43 60 05 71 79 79 93 98 110 123 128 133 133 150 174 191 191 203 207 217 (v) vi CONTENTS. PACE BOOK V 225 Lines and Planes 225 Solid Angles 241 BOOK VI 257 Polyhedrons 257 Prisms 258 Pyramids 273 Similar Polyhedrons 284 Regular Polyhedrons 288 Symmetry. — Supplementary 293 BOOK VII 295 The Cylinder ... 295 The Cone 302 The Sphere 310 Sections and Tangents 310 Spherical Angles and Triangles .... 315 Measurements of the Sphere 329 BOOK VIII— Modern Geometry 347 Transversals 347 Harmonic Proportion 351 Anharmonic Ratio 353 Pole and Polar to the Circle 362 Reciprocal Polars 365 Radical Axes ... .... 366 Centers of Similitude 369 ELEMENTS OF GEOMETET (vii) INTRODUCTION. I. GENERAL OUTLINE. 1. Geometrical Objects. 1. Space is indefinite extension in all directions. 2. A geometric solid is a portion of space having three dimensions, — length, breadth, thickness — and a definite form. In Geometry, the term solid is used for geometric solid. The annexed diagram, which is a representation of a solid, will make the following statements clear : 3. The limits or boundaries of a solid are surfaces. 4. The limits or boundaries of a surface are lines. 5. The limits or extremities of a line are points. 6. The intersection of two surfaces is a line. 7. The" intersection of two lines is a point. 8. The divergence or mutual inclination of lines or sur- faces is an angle. 9. A line can be generated by the movement of a point ; that is, a line is the path traced by a moving point. 10. A surface can be generated by the movement of a line. 11. A solid can be generated by the movement of a surface. 12. An angle can be generated by the revolution of one of two lines or surfaces, from a position of coincidence with the other, about a common point or line. (9) 10 INTRODUCTION. Points, lines, surfaces, and solids, may be defined thus : 13. A point is position without extension. 14. A line is extension in one dimension, — length. 15. A surface is extension in two dimensions, — length and breadth. 16. A solid is extension in three dimensions, — length, breadth, and thickness. 17. Points have position only; but lines, surfaces, solids, and angles have position, magnitude, and form. 18. Geometry is the science of position, magnitude, and form. 2. Points. 1. A point may be regarded, 1st, as a limit of a line; 2d, as the intersection of two lines; or, 3d, without reference to lines, as position in space, without magnitude. 2. A point is designated by a letter, and read by reading the letter. 3. Lines. 1. A line may be regarded, 1st, as a limit of a surface; 2d, as the intersection of two surfaces; 3d, as the path generated by the movement of a point; 4th, as the assem- blage of all the positions of the generating point; or, 5th, without reference to surfaces or points, as extension in one dimension, — length, without breadth or thickness. 2. A line may be designated by letters placed near two of its points, or by a small italic letter near a - B ■ its side, and be read by reading the letters or letter. Thus, the line AB, or the line a. 3. Lines may be classified as elementary and composite. 4. Elementary lines are classified as straight and curved. 5. A straight line is the trace of a point moving contin- uously in the same direction. GENERAL OUTLINE. 11 6. A curved line is the trace of a moving point which continually changes its direction. 7. Composite lines are classified as broken and mixed. 8. A broken line is a series of connected straight lines, called its components, lying in different directions. Thus, ABCD is a broken line, and ^•1JS, BG, CD, are its components. 9. A convex broken line is a broken line no component of which can, if produced, enter the space inclosed by the broken line and the straight line joining c its extremities. Thus, ABODE is a convex broken line. 10. A mixed line is a combination of straight and curved lines. Thus, the last figure below is a mixed line. 11. Summary classification of lines : ' Elementary Lines < ' Straight Curved Composite ( Broken t Mixed 4. Surfaces. 1. A surface may be regarded 1st, as a limit of a solid; 2d, as the path generated by the movement of a line; or, 3d, without reference to solids or lines, as extension in two dimensions, — length and breadth, without thickness. 2. Surfaces may be classified as elementary and composite. 3. Elementary surfaces are classified as plane and curved. 12 INTRODUCTION. 4. A plane surface, or simply a plane, is a surface such that the straight line joining any two of its points lies wholly in the surface. 5. Plane surfaces are classified with reference to their bounding lines, as rectilinear, curvilinear, and mixtilinear. 6. A rectilinear surface, called also a plane polygon, is a surface bounded by straight lines. The bounding lines are the sides, and the broken line forming the entire bound- ary is the perimeter. 7. A polygon is rep- -a Br- resented by drawing its boundary, is designated by letters, and read by reading the letters. Thus, ABC and DEFO. 8. A curvilinear surface is a surface bounded by a curved line. Thus, C, E, are curvilinear surfaces. 9. A mixtilinear surface mixed line. Thus, S is a mixtilinear surface. 10. A surface is curved if any common intersection of it and a plane is a curved line. 11. Composite surfaces are classified as broken and mixed. 12. A broken surface is a surface composed of plane surfaces. Thus, ABC is a broken surface. 13. A mixed surface is a surface partly plane and partly curved. Thus, V-ABC is a mixed surface. is a surface bounded by a GENERAL OUTLINE. 13 5. Solids. 1. A solid may be regarded, 1st, as the path generated by the movement of a plane ; or, 2d, without reference to a plane, as extension in three dimensions, — length, breadth, and thickness. 2. Solids are clas- sified thus : solids bounded by plane sur- faces; solids bounded by curved surfaces ; sol- ids bounded by mixed surfaces. G. Angles. 1. An angle is the divergence of intersecting lines or planes. 2. Angles are classified as ■plaive angles and solid angles. 3. A plane angle is the divergence of two intersecting lines. 4. The diverging lines are called the sides of the angle. 5. The point of intersection is called the vertex of the angle. y Thus, BAC is a plane angle; AB, AC, are a* c its sides; .4 is its vertex. 6. An isolated angle may be read by reading the letter at its vertex ; thus, J. above. Angles having a common vertex are read by reading the three letters designating the sides of each, observing to read the letter at the vertex be- tween the other two. Thus, BAC. CAD, BAD. An angle may be designated by a small italic letter placed within, near the vertex; thus, a, b. 14 INTRODUCTION. 7. Adjacent angles are angles having a common vertex and a common side between them ; thus, BAC, CAD, in the previous figure. 8. Equal angles are angles which can be made to co- incide. 9. A right angle is an angle equal to the adjacent angle formed by producing either side through the s vertex. Thus, BAC is a right angle, if it is equal -*> — j| ° to BAD or to CAE, formed by producing I CA or BA. M 10. An oblique angle is an angle less or greater than a right angle. 11. Oblique angles are classified as acute and obtuse. 12. An acute angle is an oblique angle less than a right angle. 13. An obtuse angle is an oblique angle greater than a right angle. 7. Loci. A locus is the place of all the points having a common property. Loci may be lines or surfaces. 8. Exercises. 1. Give a classification of surfaces, solids, angles. (3, 11)* 2. If the solid represented in the first article retain its length and breadth, and contract in thickness till it reaches its limit, what would it become? 3. If a surface retain its length and contract in breadth till it reaches its limit, what would it become? 4. If a line contract in length till it reaches its limit, what would it become? s The reference is to Article 3, Paragraph 11. See page 11. LOGICAL FACTS AND LAWS. 15 II. LOGICAL FACTS AND LAWS. 9. Definitions. 1. A judgment is the mental decision that a certain re- lation exists between two objects of thought. 2. A proposition is the expression of a judgment. 3. Propositions are classified in reference to form, as cate- gorical, hypotlietieal, disjunctive, and dilemnmtic. 4. A categorical proposition is a proposition not quali- fied by a condition. Thus, An acute angle is an oblique angle. 5. An hypothetical proposition is a proposition consist- ing of an hypothesis — a supposition or fact assumed- — and a conclusion. Thus, If an angle is obtuse, it is oblique. 6. A disjunctive proposition is a proposition expressing an alternative. Thus, An oblique angle is acute or obtuse, 7. A dilemma tic proposition is a combination of an hypothetical and a disjunctive proposition. Thus, If an angle is oblique, it is acute or obtuse. 8. The converge of a categorical or a disjunctive proposition is the proposition obtained by interchanging the subject and predicate. Thus, No right angle is an oblique angle ; con- versely, No oblique angle is a right angle : also, An oblique angle is an acute or an obtuse angle ; conversely, An acute or an obtuse angle is an oblique angle. 9. The converse of an hypothetical or a dilemmaiie proposi- tion is the proposition obtained by making the hypothesis of the original proposition the conclusion, and the conclusion the hypothesis. Thus, If an angle is right, it is not oblique ; conversely. If an angle is oblique, it is not right: also, If an angle is oblique, it is acute or obtuse ; conversely, If an angle is acute or obtuse, it is oblique. 16 INTRODUCTION. 10. Propositions are classified in reference to the relation of the subject and predicate, as analytic and synthetic. 11. An analytic proposition is a proposition in which the predicate expresses what is involved in the mere notion of the subject. Thus, A solid is extended. 12. A synthetic proposition is a proposition in which the predicate expresses something not involved in the mere notion of the subject. Thus, The sum of the two adjacent angles formed by one straight line meeting another is equal to two right angles. 13. Propositions are classified in reference to nature, as definitions, axioms, absurdities, postulates, theorems, problems, lemmas, corollaries, and scholiums. 14. A definition is such a description of an object as distinguishes it from all other objects. A definition of a class of objects includes every thing of that class, and ex- cludes every thing else. An object is defined by affirming it to belong to the class immediately containing it, and by distinguishing it from the rest of the class by stating its essential characteristics. A definition is a categorical, analytic proposition. The converse of a definition is true. 15. An axiom is a fundamental, self-evident truth. An axiom is a synthetic proposition, fundamental and indemonstrable, and is necessarily, universally, and self- evidehtly true. 16. An absurdity is a self-evident falsity. 17. A postulate is a self-evident possibility. It asserts what evidently can be done, but does not tell how. 18. A theorem is a truth which requires proof. 19. A formula is a theorem expressed in algebraic lan- guage. 20. A problem is something proposed for solution. LOGICAL FACTS AND LA WS. 17 21. A corollary is an obvious consequence. 22. A scholium is a note or remark. 23. A lemma is an auxiliary theorem or problem. 24. A demonstration is the proof of a proposition. 25. Demonstration is direct or indirect. 26. Direct demonstration proves a proposition in either of two ways : 1st. By superposition of one figure upon another. 2d. By logical combination of definitions, axioms, and propositions previously demonstrated. 27. Indirect demonstration, called also redvdio ad ab- surdum, proves a proposition true by showing that the supposition that it is false involves a contradiction or an absurdity. 10. Relation of Propositions. 1. Any two propositions are congruent or confiictive. 2. Congruent propositions are those which are compati- ble. Thus, a and b are unequal, and a is less than b. 3. Confiictive propositions are those which are incom- patible. Thus, a and b are equal, and a and b are unequal. 4. Confiictive propositions are contraries or co7itradietories. 5. Contrary propositions are confiictive propositions which are not universally inclusive. Thus, a is equal to b, and a is less than b, are contrary propositions; for they are confiictive, but are not universally inclusive, since another relation, a is greater than b, is possible. 6. Contradictory propositions are confiictive propositions which are universally inclusive. Thus, a and b are equal, and a and b are unequal, are contradictory propositions; for they are confiictive, since o and b can not be both equal and unequal ; and they are universally inclusive, since a and b are either equal or unequal. s. G.-2. 18 INTRODUCTION. 11. General Laws. 1 . Every proposition is either- true or false. For, since truth and falsity are universally inclusive, no other case is possible. 2. A proposition can not be both true and false. For, if so, it would be self-destructive. 3. If a proposition is true, whatever it involves is also true. For the truth of a proposition necessitates the. truth of what it involves. 12. Law of Congruents. The congruity of two propositions is consistent witJi the truih of both, the falsity of both, or ilie truth of one and tlie falsity of the oilier. Thus, a and b are unequal, and a is less than b, are congruents ; and both true, if 6 is greater than a ; both false, if a is equal to b ; one true and the other false, if a is greater than b. 13. Corollary. From mere congruence, the truth or falsity of one proposition does not involve the truth or falsity of another. 11. Law of Conflictives. Two conflictive propositions can not both be true. For, if so, they would mutually destroy each other, and 'both be false. 15. Corollai'ies. 1. If one of two conflictives is true, the other is fake. For, otherwise, we should have two conflictives both true, which is impossible. LOGICAL FACTS AND LAWS. 19 2. Two true propositions can not be confiictives. For, if so, we should have two confiictives both true, which is impossible. 3. All trutJis acid in harmony. For, if two truths conflict with one another, the truth of cither involves the falsity of the other, and each would be both true and false. 4. A proposition is false if it involves the conflictive of a truth. For, if true, what it involves would be true, and we should have two confiictives both true, which is impossible. 16. Law of Contraries. Contrary propositions can not both be true, but may both be false. For, since they are conflictive, both can not be true ; and since they are not universally inclusive, other cases are pos- sible, and both may be false. 17. Corollaries. 1. Hie truth of either of tioo contrary propositions involves Vie falsity of the oilier. For, since they are conflictive, both can not be true. 2. The falsity of either of tivo contrary propositions does not involve tile trutii of the oilier. For, since they are not universally inclusive, other cases are possible, and both may be false. 18. Law of Contradictories. One of two contradictory propositions jiiiwf be true, and- the other false. For, since thev are universally inclusive, no other case is possible, and one must be true ; and, since they are con- flictive, both can not be true. 20 INTRODUCTION. 19. Corollaries. 1. The truth of either of two contradictory propositions in- volves the falsity of the other. For, since they are connectives, both can not be true. 2. The falsity of eitlier of two contradictory propositions in- volves the truth of the other. For, since they are universally inclusive, no other case is possible, and one must be true. 20. Laws of Genera and Species. 1. Inclusion in a speeies is inclusion in the genus. Thus, since both acute and obtuse angles are included as species in the genus oblique angles, if an angle is acute, it is oblique ; or if an angle is obtuse, it is oblique. 2. Inclusion in a species is exclusion from any thing from which the species is excluded. Thus, since acute angles are excluded from obtuse angles, if an angle is acute, it is not obtuse. 3. Inclusion in the genus is not necessarily inclusion in a particular species. Thus, if an angle is oblique, it is not necessarily acute, for it may be obtuse ; neither is it necessarily obtuse, for it may be acute. 4. Inclusion in tlie genus is disjunctive inclusion in the species. Thus, if an angle is oblique, it is acute or obtuse. 5. Exclusion from the genus is exclusion from all the species. Thus, if an angle is not oblique, it is neither acute nor obtuse. 6. Exclusion from a species is not necessarily exclusion from the genus. Thus, if an angle is not acute, it is not necessarily not oblique; for it may be obtuse, and therefore oblique. AXIOMS AND POSTULATES. 21 7. Exclusion from all the species vs exclusion from the genus. Thus, if an angle is neither acute nor obtuse, it is not oblique. 8. Whatever can be predicated, affirmatively or negatively, universally of any class, can- in like manner be predicated of any tiling contained in that class. HI. AXIOMS AND POSTULATES. 21. Preliminaries. 1. Similar magnitudes are magnitudes identical in form. 2. Equivalent magnitudes are magnitudes identical in extent. 3. Equal magnitudes are magnitudes identical in form and extent. 4. Similar magnitudes may not be equivalent, and equiv- alent magnitudes may not be similar ; but equal magnitudes are both similar and equivalent. 5. The test of the equality of two magnitudes is the fact that they can be made to coincide. 6. The greater of two unequal magnitudes is the one which exceeds the other in extent. 7. The less of two unequal magnitudes is the one which is exceeded by the other in extent. 8. The whole of a magnitude is all of it, 9. A part of a magnitude is any amount of it less than the whole. 10. The sum of all the parts taken in any order, and the sum of all the parts taken in any other order, are always equivalent, and, if similar, are equal. 11. A substitute for a magnitude is a magnitude which 22 INTRODUCTION. will sustain to another magnitude the same relation as the given magnitude, if put in its place. 12. The algebraic signs will, in general, be employed in their usual sense ; but the sign (=), which denotes either equality or equivalency, is, in a relation of equality, read, k equal to, or equals; but, in a relation of equivalency, is equivalent to. 22. Axioms. 1. THE AXIOM OF SIMILARITY. Either of two similar magnitudes is a substitute for the other, if only form be considered. 2. THE AXIOM OF EQUIVALENCY. Either of two equivalent magnitudes is a substitute for the other, if only extent be considered. 3. THE AXIOM OF EQUALITY. Either of two equal magnitudes is a substitute for the other, if both form and extent, or either, be considered. 23. Corollaries. 1. If one of two similar magnitudes is similar to a third magnitude, the other is similar to that magnitude. For, if a, b, c, represent three magnitudes, and if (1) a is similar to b, and (2) b is similar to c, then, substituting a for b in (2), or c for b in (1), we have (3) a is similar to c. 2. If one of two similar magnitudes is not similar to a third magnitude, the other is not similar to that magnitude. For, if (1) a is similar to 6, and (2) b is not similar to c, then, substituting a for b in (2), we have (3) a is not similar to c. AXIOMS AND POSTULATES. 23 3. If one of two equivalent magnitudes is equivalent to a third magnitude, the other is equivalent to that magnitude. For, if (1) a — b, and (2) b = c, then, substituting a for 6 in (2), or c for b in (1), we have (3) a = c. 4. If one of two equivalent magnitudes is greater than a third magnitude, the other is greater tlian Uiat magnitude. For, if (1) a=^b, and (2) b > c, then, substituting a for b in (2), we have (3) a > c. 5. If one of two equivalent magnitudes is less than a third magnitude, the other is less than that magnitude. (?) 6. If equivalents be added to equivalents, the sums will be equivalent^ For, let (1) a = b, (2) c = d ; now (3) a -\- c = a -j- c, since the members are identical in extent ; then, substituting b for a, and d for c in the second member of (3), we have (4) a + e = b + d. 7. If equivalents be subtracted from equivalents, the remainders will be equivalent. (?) f 8. If equivalents be multiplied by the same number, the products will be equivalent. (?) 9. If equivalents be divided by the same number, the quotients will be equivalent. (?) 10. Like powers of the numerical values of equivalents, re- ferred to the same unit, are equivalent. (?) 11. Like roots of the numerical values of equivalents, referred to the same unit, are equivalent. (?) 12. If equivalents be added to inequivalents, the sums will be inequivalent, and the resulting inequivalency urill subsist in the same sense. Let (1) a = b, (2) c > d. Let (3) c = d + q. Now (4) a-{-d-{-q^>a-\-d (21, 6). Then, substituting c for d -\- q in the first member of (4), and b for a in the second mem- ber, we have (5) a -\- c > b -f d. 24 INTRODUCTION. 13. If equivalents be subtracted from inequivalents, the re- mainders loill be inequivalent, and the resulting inequivalence/ will subsist in the same sense. (?) 14. If inequivalents be multiplied by the same number, the products will be inequivalent, and the resulting inequivalency will subsist in the same sense. (?) 15. If inequivalents be divided by the same number, the quotients will be inequivalent, and the resulting inequivalency will subsist in the same sense. (?) 16. If inequivalents be subtracted from equivalents, the re- mainders will be inequivalent, and the resulting inequivalency will subsist in a contrary sense. (?)' 17. If equivalents be multiplied by unequal numbers, the products ivill be inequivalent, the greater product corresponding to the greater multiplier. (?) 18. If equivalents be divided by unequal numbers, the quotients will be inequivalent, the greater quotient corresponding to the less divisor. (?) 19. If two inequivahncies subsisting in the same sense be added, member to member, the resulting inequivalency will subsist in the same sense. Let a > b, and c > d. Let a = b + p, and c = d -f- q. Now, b-j-p + d J r q>b + d (21, 6). Then, substituting a for b -f p, and c for d -\- q, we have a -j- c > b + d. 20. If a magnitude is greater than the greater of two magni- tudes, it. is greater than the less. Let a>b, and 6>c; then, a + 6 > b -f c (23, 19). Subtracting b from each member, we have a> c (23, 13). 24. Scholiums. 1. Analogous corollaries can, in like manner, be deduced from the axiom of equality (22, 3). 2. The corollary from the axiom of equality, corresponding AXIOMS AND POSTULATES. 25 to the first from the axiom of equivalency (23, 3), is usually stated as an axiom, thus: Things equal to the same thing are equal to each oilier. 25. Postulates. 1. THE POSTULATE OF POSITION. A magnitude can have any position. 2. THE POSTULATE OP POEM. A magnitude can have any form. 3. THE POSTULATE OP EXTENT. A magnitude can have any extent. 26. Corollaries. 1. A straight line can he drawn from any point, in any di- rection, to any extent. For, a magnitude can have any position and extent. 2. A straight line can be drawn from any point to any other point. For, draw a line in any plane ; move the plane, carrying with it the line, till the line passes through the first point ; revolve the plane about the line as an axis, till it embraces the second point; revolve the line in the plane about the first point as a center, till it passes through the second point. 3. A straight line passing through two fixed points is deter- mined in position. For, if the line should change its position, it could no longer pass through the two points ; hence, the two points determine the position of the line. 4. Two straight lines having two points common are coincident. For, since a straight line passing through two points is determined in position, the two lines are identical in posi- tion, and hence coincident. s. G.— 3. 26 INTRODUCTION. 5. Two straight lines can not intersect in more tlian one point. For, if they intersect in two points, they have two points common, and are therefore coincident. 6. Two straight lines can not inclose a space. For, if they inclose a space, they would have two points common, and therefore be coincident, and the inclosed space would vanish. 7. A finite straight line can i>e produced in either direction to any extent. For, a point can move from either extremity as origin, along the line to the other extremity, and forward, in the same direction indefinitely, thus tracing, beyond the ex- tremity, the prolongation of the line to any extent. 8. A finite straight line can be bisected. For, a j>oint moving from either extremity of the line as origin, along the line toward the other extremity, will at each position divide the line into two parte. The part at first less than the other, will increase continuously, while the other will in like manner diminish, till, at length, the parts become equal, when the line is bisected. 9. A plane angle can be bisected. For, let a straight line, drawn through the vertex in co- incidence with one side of the angle, revolve about the vertex as a center, in the plane of the angle, from that side toward the other. The angle which the revolving line makes with the side left, at first less than the angle "which it makes with the other side, increases continuously, while the other angle in like manner diminishes, till, at length, these angles become equal, when the angle is bisected. BOOK I. I. PERPENDICULAR AND OBLIQUE LINES. 27. Definitions. 1. Perpendicular lines are lines which make a right angle with each other (6, 9). Thus, if R is a right angle, its sides are per- pendicular lines. If one line is perpendicular to another, the angle formed is a right angle, and s- the second line is perpendicular to the first. 2. Oblique lines are lines which make an oblique angle with each other (6, 10). Thus, if A and are oblique - \ angles, their sides are oblique lines. / \ If one line is oblique to another, the angle formed is an oblique angle, and the second line is oblique to the first. 3. The complement of an angle is an angle which, added to the given angle, will make the sum a right angle. 4. The supplement of an angle is an angle which, added to the given angle, will make the sum two right angles. 5. If two straight lines intersect each other, the two angles on the same side of one of the lines and on opposite sides of the other, are called adjacent angles; and the two angles on opposite sides of both lines are called vertical angle.-'. (27) 28 GEOMETRY.— BOOK I. Thus, a and b, b and c, c and d, d and a, are adjacent angles; and a and c, 6 and d, are vertical angles. // d 28. Proposition I. — Theorem. At any point in a straight line an indefinite number of perpendiculars to the line can be drawn. Let C be a point of the line AB. From *•' C, draw, in any plane embracing AB — for : / example, the plane of the paper — any line \/ CD, making with AB the angles ACD, '' DOB, which must be equal or unequal. If these angles are equal, each will be a right angle (6, 9), and the line CD will be perpendicular to AB (27, 1). If these angles are unequal, let CD revolve, in the plane, about C as a center, so that the less angle shall continuously increase, and the greater in like "manner diminish, till the two angles become equal, when each will be a right angle, and CD, in this position represented by CE, will be perpen- dicular to AB. Now, let the plane revolve about AB as an axis, carry- ing with it the perpendicular CE. The angles, BCE, ECA, remaining right angles, the line CE will, in every position throughout the revolution, be perpendicular to AB. But CE takes an indefinite number of positions in the revolu- tion, each of which is the position of a perpendicular to AB. Hence, an indefinite number of perpendiculars to AB can be drawn from the point C. 29. Corollaries. 1. From a point of a line only one perpendicular to Hie line can be drawn in a plane embracing the line. PERPENDICULAR AND OBLIQUE LINES. 29 For, if the line revolve in the plane, either way, from the position in which the adjacent angles are equal, one of the angles would increase, the other diminish, and the angles, no longer equal, would not be right angles; hence, the line would not be a perpendicular. The perpendicular produced across the line is a perpen- dicular on the other side of the line. (?) 2. All right angles are equal. Let ACD, DCB, be right angles; also EGH, HGF. Then will these angles all be equal. For, place the figure ADB on EHF, so that AB shall coincide with EF, and -A q B -E - C with G. Then CD will co- incide with GH, otherwise there would be two perpendic- ulars to AB, drawn from C in a plane embracing AB, which is impossible. Then, ACD = EGH, and DCB = HGF (21, 5). But, ACD = DCB, and EGH= HGF (6, 9). ACD = HGF, and DCB =. EGH (22, 3). 3. The complements of equal angles are equal. Let A, A', respectively, denote two angles ; C, C, their respective complements ; and JR, a right angle. Then, A+C=R, and A' + C = R (27, 3), A + C = A'+C (22, 3; 24, 2). If A = A', then C=C (23, 7). 4. The supplements of equal angles are equal. (?) 30. Proposition II.— Theorem. From a point wiAout a straight line, one perpendicular to the line can be drawn, and only one. 30 GEOMETRY.— BOOK I. Let G be the point and AB the line. The point and the line are in the same plane, since any plane embracing AB can revolve about AB, as an axis, till it .embraces the point C. At any point of AB, as E, let EF in the plane be per- pendicular to AB, on the same side as C, forming the right angle AEF. Let this right angle move in the plane toward C, the side AE moving in AB, till FE passes through C and becomes CD. Since ADC is one position of the right angle AEF, CD is perpen- dicular to AB. Hence, from a jDoint without a line, one perpendicular to the line can be drawn. No other perpendicular to AB can be drawn from C; for, if CG is another perpendicular, AGC is a right angle. Let this angle move in the plane, AG moving in AB till G coin- cides with D, when GC, having moved, takes the position DH. Since AGC in its position AD H is a right angle, DH is perpendicular to AB. Then DC and DH are both perpen- dicular to AB in a plane embracing AB, which is impossible (29, 1). Hence, the supposition that a perpendicular differ- ing from CD can be drawn from G to AB, which led to this impossibility, is false (15, 4). Therefore, from a point with- out a straight line, only one perpendicular to the line can be drawn. 31. Proposition III.— Theorem. The sum of the two adjacent angles formed by one straight line meeting another, is equal to two right angles. 1. Let CD be perpendicular to AB, and b let R denote a right angle. /■ Then, ACD = R, DCB = E (27, 1). A i, * ACD + DCB --= 2B (23, 6). PERPENDICULAR AND OBLIQUE LINES. 31 2. Let CE be oblique to AB. AGE + ECB = ;1CE — DOE + DCE + ECB, since — DCE and -)- DCE cancel in the second member. But, AGE — DCE = ACD, DCE + ECB = DCB. . ■ . ACE + ECB = ACD + DCB. But, ACD + DCB = 2R, .-. ACE + ECB = 2R. 32. Corollaries. 1. If one of the two adjacent angles formed by one straight line meeting another is a rigid, angle, the other is also a right angle. If not, the sum would not be equal to two right angles. 2. The sum of all the consecutive angles formed by any number of straight lines, drawn from the same point of a straight line on the same side, is equal to two right angles. For, ACF + FCB = 2R. B f. But, ACF = ACD + DCE+ ECF. .-. ACD+ DCE+ ECF+FCB = 2R. A ~ 33. Proposition IV. — Theorem. If two straight lines intersect, the vertical angles are equal. For, a + b = 2R, b + c = 2R (31). a-\-b=b + e. (?) .-. a=r:-c. (?) Again, b is the supplement of a. (?) Also, (Ms the supplement of a, .•. 6 = rf. (?) 31. Corollaries. 1. If two straight lines intersect, the sum of the four angles formed is equal to four right anglf*. For, a + b = 2R, and c + d = 2R (31). a + b + c-\-d = 4R (23, 6). a ' <[ yd 32 GEOMETRY.— BOOK I. 2. The sum of all the angles formed by straight lines meeting at a common point is equal to four right angles. For, let two straight lines intersect at this point. The sum of the angles formed by the two lines will be equal to the sum of the angles formed by the given lines. But the sum of the angles formed by the two lines is equal to four right angles (34, 1). Hence, the sum of the angles formed by the given lines is equal to four right angles. 3. The straight line bisecting an angle bisects Us vertical angle. For, a = d, and b=c (33). If a — b, then a = c, . ■ . d = c. (?) , V \f/ 4. TIi£ bisectors of the tivo pairs of vertical ■ /■» i«^ angles are perpendicular to each otlier. For, e =f. But, a — b, and b = c, . • . a = c. e-\-a —f -\- c, . • . the bisectors are perpendicular. 35. Proposition V.— Theorem. If ilie sum of two adjacent angles is equal to two right angles, tlwir exterior sides are in tlie same straight line. Let ACD + DCB = 2R. Then ACB is a straight line. Let AC be produced to E. A — Then, ACD + DGE = 2R. (?) ACD + DCB = ACD + DCE. .-. DCB = DCE. (?) .-. C£ coincides with CB; other- wise, DCB and DCE would not be equal. But ACE is a straight line ; .• . ACB is a straight line. -B PARALLEL LIKES. II. PARALLEL LINES. 36. Definitions. 1. Parallel lines are lines every-wherc equally distant, Thus, if AB and CD are parallel, the perpendicular distance from E to CD is -i - equal to that from G to AB, or to that from I to CD, and so on. ' f g j 2. A secant of two lines is a line intersecting both ; and, 1st. Interior angles are those which lie within . the two lines. Thus, «, d, g, f. "^~"^<, 2d. Exterior angles are those which lie with- / out the two lines. Thus, b, c, h, e. ~»7j 3d. Interior angles on the same side are those which lie within the two lines, on the same side of the secant. Tims, a and /, J and CA (21, 6) ; but, CA = DB, since AB and CD are parallel (36, 1). Therefore, FA > DB (22, 3). But, DB > OB, .-. FA> GB (23, 20). Hence, the lines AB and FG are not every-where equally distant, and therefore are not parallel. 3. A straight line perpendicular to one of two parallels and lying in the same plane is perpeiulicidar to the other. Let AB and CD be parallel, and EF perpendicular to AB; then will EF be perpendicular to CD. For, in the PARALLEL LINES. 37 plane of AB and EF, let a perpendicular to EF be drawn at E. Then this perpendicular will be parallel to AB (41), and must, therefore, coincide with CD ; otherwise, there would be two parallels to c — AB drawn through the point E, which is A impossible (Cor. 2). Since CD coincides with the perpendicular to EF through E, CD is itself a perpendicular; hence EF, which is perpendicular to AB, is also perpendicular to CD, which is parallel to AB. 4. Two straight lines lying in the same plane and respect- ively parallel to a third straight line lying in that plane are parallel to each other. Let CD and EF, lying in the same plane, G be respectively parallel to AB, lying in that plane. Then will CD and EF be parallel. For, let GH, lying in the same plane, be perpendicular to AB ; then GH will also be perpendicular to CD and to EF (42, 3). Hence, CD and EF, lying in the same plane and perpendicular to a third line, GH, lying in that plane, are parallel. (?) 5. If one of three parallels lying in the same plane move in that plane toward the second, but in such a manner as to con- tinue parallel to the third, it ivill, in one of its positions, coin- cide with the second. Let a, b, c, be such parallels, and let a move in the stated manner toward b, till it c meets 6 in one point. Then, if a and b do not coincide, they will intersect and have but one point common. We should then have two lines through the same point parallel to the same line, which is impossible (42, 2) ; hence a and b coincide. 38 GEOMETR Y.—B 00 K I. 43. Proposition XI.— Theorem. If a straight line intersect tioo straight lines lying in the mine plane, making the sum of the interior angles on tlie same side of the secant equal to two right angles, the two lines are parallel. Let EF intersect AB and CD, lying in the same plane, making EHB -f- FGD = two right angles. Then will AB and CD be parallel. For, _,, Gy E through I, the middle point of GH, draw X/ KL perpendicular to AB. The vertical A . / \ p angles, KIG, HIL, are equal (33). By f hypothesis, IHL + FGD = 2R ; but IGK A- FGD = 2R (31). . • . IHL + FGD = IGK+ FGD ; . ■ . -IHL = IGK (23, 7). Now, conceive the portion of the figure below I to revolve, in the same plane, about J as a center, till IH coincides with IG, and II with G. Since KIG = HIL, IL will fall in IK; hence, L will fall in IK. Since IHL = IGK, HL will fall in GC; hence, L will fall in GC. Since L is in both IK and GC, it must be at their intersection, K. Then ILH and IKG coincide, and hence are equal. But ILH is a right angle by construction ; therefore IKG is a right angle. Hence, AB and CD, both perpendicular to KL and lying in the same plane, are parallel (41). 44. Corollaries. 1. If tlie sum of the exterior angles on the same side is equal to two right angles, the tiao lines are parallel. Let b -\- e = two right angles. But, b = d, a/f and e — g (33). . ' . d -\- g = two right angles ; f/g__ therefore the two lines .ire parallel (43). PARALLEL LIXES. 39 2. //" either two alternate interior angle* are equal, the two lines are parallel. Let (i — g ; then o + d =g + rf. (?) But a-\-d = 2R. (?) .•. gr^^^^A'; therefore the two lines are parallel (43). 3. If either two alternate exterior angle* are equal, the two line* are parallel. Let c = e ; but e = a, and e = g ; (?) . • . a = .F (48, 3) ; but GDF is the supplement of KDG ; therefore KML is the supplement of KDG. Also, KML is the supplement of any angle whose sides are respectively parallel to the sides of KDG, and lie in the same direction or in opposite directions. (?) 50. Exercises. 1. What is the locus (7) of all the points in a plane embracing a line, which are equally distant from that line? 2. Of how many lines does the locus of Ex. 1 consist? 3. Does this locus embrace all the points equally distant from the line? What then is the locus of all the points equally distant from the line? III. TRIANGLES. 51. Definitions and Classification. 1. A triangle is a polygon of three sides (4, 6). i / 2. A right triangle is a triangle having a right angle. .'!. An oblique triangle is a triangle having all its angles oblique. Thus, A, B, are right, triangles ; C, D, E, F, G, are oblique. 4. An acute triangle is an oblique triangle having all its angles acute. 5. An obtuse triangle is an oblique triangle having an obtuse angle. Thus, f, A E, are acute triangles ; F, G, are obtuse. 44 GEOMETRY— BOOK I. G. A scalene triangle is a triangle having no two sides equal. 7. An isosceles triangle is a triangle having at least two sides equal. Thus, A, C, F, are scalene triangles; B, D, E, G, are isosceles. 8. A bi-equilateral triangle is an isosceles triangle having only two equal sides. It is usually called isosceles. 9. A tri-equilateral triangle is an isosceles triangle having its three sides equal. It is usually called equilateral. Thus, B, D, G, are bi-equilateral triangles; E is tri- equilateral. 10. Summary classification .of triangles : ( Scalene ' Right m ( Isosceles — Bi-equilateral ' Scalene Oblique Acute !' Isosceles- Bi-equilateral Tri-equilateral [ Scalene ■ Obtuse X I Isosceles — Bi-equilateral 11. The hypotenuse of a right triangle is the side oppo- site the right angle. 12. The base of a triangle is the side on which it is assumed to stand. TRIANGLES. 45 13. In a bi-equilateral triangle, the side unequal to either of the others is generally regarded as the base. 14. The vertical angle of a triangle is the angle opposite the base. 15. The vertex of a triangle is the vertex of the vertical angle. 16. The altitude of a triangle is the perpendicular from the vertex to the base. 17. A medial line of a triangle is the line drawn from any vertex to the middle point of the opposite side. 18. An exterior angle is the angle which the prolonga- tion of any side of the triangle makes with another side. 19. The adjacent angle of an exterior angle is the angle one of whose sides is prolonged. 20. Opposite interior angles are the two angles not adja- cent to the exterior angle. ° A Thus, ACD is an exterior angle; ACB is the adjacent angle of ACD ; A and B are the opposite interior angles. B 52. Proposition XVI.— Theorem. The sum of the three angles of any triangle is equal to two right angles. Let ABC be a triangle. Then the B / AC and drawing CE parallel to AB c and on the same side of AC, w6 have ACB + BCE + ECD = two right, angles (32, 2). BCE = ABC (16, 2\ ECD = BAC (46, 4). • . ACB -f ABC + BAC= two right angles. 46 GEOMETRY.— BOOK I. 53. Corollaries. 1. An exterior angle is equal to the sum of the opposite in- terior angles. For, ACB + BCD = BAG + ABC + ACB (31), (52). BCD = BAC + ABC (23, 7). 2. An exterior angle is greater than either opposite interior angle. (?) 3. If one angle of a triangle is right, each of the others is acute. (?) 4. If one angle of a triangle is obtuse, each of the others is acute. (?) 5. In a right triangle, the sum of the two acute angles is equal to a right angle, and each acute angle is the complement of the other. (?) 6. In an acute triangle, ilie sum of any tivo angles is greater than a right angle. (?) 7. In an obtuse triangle, ilie sum of the two acute angles is less than a right angle. (?) 8. In any triangle, any angle is the supplement of the sum of the other two. (?) 9. If the sum of two angles of one triangle is equal to the sum of two angles of another, the third angle of ilie one is equal to the third angle of the other. (?) 10. Each angle of an equiangular triangle is one-third of two right angles or two-thirds of one right angle. (?) 54. Proposition XVII. — Theorem. The angle contained by two straight lines^drawn from any point within a triangle to the extremities of one of the sides is greater than the angle conUiiveil by the other sides of the triangle. TRIANGLES. 47 Let DB and DC be straight lines drawn from any point D within the triangle ABC to the ex- tremities of BC. Then, the angle BDC a is greater than the angle BAC. For, BDC > EEC, BEG > BAC (53, 2). Hence, BDC> BAC (23, 20). 55. Proposition XVIII.— Theorem. Two triangles are. equal if two sides and Hie included angle of the one are respectively equal to two sides and the included angle of the oilier. Let ABC and DEF be two triangle?, having the side AB equal to the side DE, the side AC equal to the side DF, and the included angle A equal to the in- cluded angle D. Then the triangle ABC is equal to the triangle DEF. For, let ABC be placed upon DEF, so that the angle A shall coincide with the equal angle D; then, the side AB will coincide with the equal side DE, and the side AC with the equal side DF, the point B with E, C with F, and BC with EF (26, 4). Hence, the tri- angles coincide, and are therefore equal (21, o\ 56. Corollaries. 1. If tivo sides and, the included angle of one triangle are respectiveli/ equal to two sides and the included angle of another, the third sides will be equal, and tlie remaining angles respect- ively equal. For, since the triangles coincide, the like part* coincide, and hence are equal. 48 GEOMETR Y.—B OK I. 2. If two triangles are equal, the equal sides are opposite equal angles, and the equal angles opposite equal sides. 3. If the equal sides of the equal angles of two triangles are reversed in position, the triangles can be made to coincide, if one be turned over. 57. Proposition XIX.— Theorem. .. Two triangles are equal if a side and the adjacent angles of the one are respectively equal to a side and the adjacent angles of the other. Let ABC and DEF be two triangles, having the side BC equal to the side EF, the angle B equal to the angle E, and the angle C equal to the angle F. Then will the triangles be equal. For, let ABC be placed upon DEF, so that BC shall coincide with its equal EF; then, since the angle B is equal to the angle E, BA will fall on ED, and, since the angle G is equal to the angle F, GA will fall on FD ; hence, A, the intersection of BA and GA, will coincide with D, the intersection of ED and FD. Hence, the triangles coincide, and are therefore equal. 58. Corollary. If a side and the adjacent angles of one triangle are respect- ioely equal to a side and the adjacent angles of another, tiie third angles are equal, and the remaining sides respectively equal — the equal sides lying opposite equal angles. For, since the triangles can be made to coincide, the corresponding parts are equal. TRIAXGLES. 49 59. Proposition XX.— Theorem. The angles opposite the equal sides of an isosceles triangle are equal. Let ABC be an isosceles triangle, having the side AC equal to the side AB. Then is the angle B, opposite the side AC, equal to the angle C, opposite the equal side AB. Let AD bisect the angle BAC (26, 9;. The triangles ABD and ACD are equal, b— since AD is common, AB and AC are equal by hypothesis, and the angles BAD and CAD equal (551 Hence, the angles B and C are equal, since they are oppo- site the common side AD (56, 2). But B and C are oppo- site the equal sides, AC, AB, of the isosceles triangle ABC. Hence, in an isosceles triangle, the angles opposite the equal sides are equal. 60. Corollaries. 1. The line bisecting the vertical angle of an isosceles triangle bisects the base at right angles. (?) 2. The line bisecting Hie base of an isosceles tria>igle at right angles bisects the vertical angle. (?) 3. The line joining the verier of an isosceles triangle and the middle point of tlie base bisects Uie vertical angle, and is perpen- dicular to tlie base. (?) 4. Every equilateral triangle is equiangular. (?) 61. Proposition XXI— Theorem. If two angles of a triangle are equal, tlie sidts opposite are equal. S. G.— 5. 50 GEOMETRY— BOOK J. Let ABC be a triangle, having the angle B equal to the angle C; then is the side AC, opposite B, equal to the side AB, opposite G. A Let AD bisect the angle BAC (26, 9). /|\ Then, in the triangles, ABB, ACT), B = C, / j \ and BAD = CAD; .-. ADB = ADC. * * ° Hence, the triangles, ABD, ACD, are equal (57) ; there- fore, AB = AC. (?) 62. Corollary. Every equiangular triangle is equilateral. (?) 63. Proposition XXII.— Theorem. The angle opposite tlie greater of two unequal sides of a tri- angle is greater tlian the angle opposite the less. Let the side AC of the triangle ABC be greater than the side AB. Then, the angle ABC, opposite the side AC, is greater than the angle ACB, opposite the side AB. On AC cut off AD equal to AB, and draw BD. Then, the angle ABD is equal to the angle ADB (59) ; and the angle ADB is greater than the angle DCB (53, 2) ; hence, the angle ABD is greater than the angle ACB. But the angle ABC is greater than the angle ABD (21, 6) ; hence, the angle ABC is greater than the angle ACB. 64. Proposition XXIII.— Theorem. The side opposite the greater of two unequal angles of a tri- angle is greater than the side opposite the less. TRIANGLES. 51 Let the angle jB of the triangle ABC be greater than the angle C; then, the side AC, opposite B, is greater than the side AB, opposite C. Now, either AC = AB, AC < AB, or AC > AB. If AC =AB, B=C (59), which is contrary to the hypothesis ; there- fore, AC is not equal to AB (15, 4). If AC < AB, B < C (63), which is contrary to the hypothesis; .•. -If is not less than AB. Now, neither AC=AB, nor AC AB. 65. Proposition XXIV.— Theorem. Any side of a triangle is less than tfie sum of tJie other sides. Let ABC be a triangle. Produce BC till CD = CA, and draw AD. Since CD = C1I, the angle CiD is , v\ equal to the angle CZL1 (59 \ The \ angle BAD is greater than the angle jb— '; -'# CLID ; hence, the angle BAD is greater than the angle BDA. Therefore, the side BD is greater than the side AB (64) ; .-. AB < 5Z>. BD = BC+ CD; but, CD = C.I ; .-. BD = BC + CA. But, ABc; .-. a > c — b, 6 > c — a. b -\- c > a ; . • . 6 .> a — c, c > a — 6. c-f «>■!); .". c > & — a, a > 6 — c. 67. Proposition XXY. — Theorem. The shortest distance from one point to another is the straight line having these points for Us extremities. 1. Let A and B be the points, AB a straight line having these points for its extremities, ACB a broken line having two components and the same extremities. ACB is a triangle; .-. AB < AG + CB (65). 2. Let AB be a straight line, ACDB a broken line having more than two com- ponents and the same extremities. Then, AB < AC + CD + D.B. For, -4D<^LC + CZ>; (?) .-. AD + DB. 54 GEOMETRY.— BOOK I. 2. Let ABCD be a convex broken line and AmD an enveloping curve. Inscribe in the curve m the' broken line AEFD. Then, J^^^C V.M ABCD < AEFD (68, 1). But, .iE^D < AmD (67, 3) ; .-. ABCD < ilmD. 69. Proposition XXVII.— Theorem. If from a point -without a straight line a perpendicular be drawn to this line and oblique lines be drawn from the same point to different points of tlie line, then : 1. The perpendicular is shorter titan any oblique line. 2. Two oblique lines which meet the given line 'at equal dis- tances from the foot of the perpendicular are equal. 3. Of two oblique lines which meet the given line at unequal distances from the foot of the perpendicular, that which cuts off the greater distance is the greater. 4. Of two oblique lines drawn from the same point wWiout the perpendicidar, in the plane of the perpendicular and line, cutting off on the line equal distances from tlie foot of the per- pendicular, that which intersects the perpendicular is the greater. 1. Let CD be perpendicular to AB, CA an oblique line. Then, CD < CA. Pro- duce CD till DF = CD, and draw A F. The angle ADC is a right angle (27, 1) ; therefore the angle ADF is a right angle (32, 1). The triangles, ADC, ADF, are equal since AD is common, DC = DF by construction, and the included angle ADC = the included angle ADF (29, 2) (55). Then, AC opposite the angle ADC, is equal to AF TMJASGLES. 55 opposite the equal angle ADF. Since CD = DF and CA = AF, CD = |OF and d = $(CA + AF). Since .1CF is a triangle, CF < Ci + .If; • • • W < i (CI + J F) (23, 15). .-. CD DA, and draw EF. Then, the triangles, ECD, EFD, are equal. (?) .-. CE = EF. (?) .-. CF = * (CF + FF). But, CI = £ (CI + AF ). and CE + FF> Ci + AF y 68). . • . i (CE — FF) > A (C4 -f JF), . • . CE > CA. 4. Let (? be a ]>oint in the plane CD B. Draw OB, and GA intersecting the perpendicular in C. Then we have GC + CB > GB. But, CA = C8. .-. (?C + CI > CB, .-. CI > GB. 70. Corollaries. 1. From a given point to a given straight line only two equal straight lines can be drawn. (?) 2. The perpendicular intersecting a straight line at its middle point is the locus of all the points in the plane of Vie line and perpendicular whieJi are equally distant from the extremities of the line. (?) 56 GEOMETRY.— BOOK I. 3. If each of two points of one straight line is equally distant from the extremities of anoilier straight line in the same plane, the first line is perpendicular to tlie second at its middle point. (?) 4. If tlie plane embracing a straight line revolve about this line as an axis, carrying with it the perpendicular at th-e middle point, this perpendicular indefinitely extended will generate a plane of indefinite extent which will be the general locus of all the points equally distant from tlie extremities of tlie line. (?) 71. Scholium. By the distance from a point to a line, we are to under- stand the perpendicular or shortest distance. 72. Proposition XXVIII.— Theorem. Two right triangles are equal in the following cases: 1. If the hypotenuse and a side of the one are respectively equal to the hypotenuse and a side of the otJier. 2. If the hypotenuse and an adjacent angle of the one are respectively equal to the hypotenuse and an adjacent angle of the other. 3. If a side adjacent to the right angle, and the adjacent or opposite acute angle of the one, are respectively equal to a side adjacent to the right angle, and the adjacent or opposite acute angle of tlie otl\er. 1. Let ABC, DEF, be right 4 s triangles having the hypote- y\ nuses, AB, DE, equal, and the sides, AC, DF, equal. Then the triangles are equal. / TXIANGLES. 57 For, place the triangle ABC on DEF so that the side AC shall coincide with its equal DF; then, since the right angles are equal, CB will fall on FE, and the point B will be found somewhere in FE, either at E or at the right or left of E. B can not fall at the right of E; for then AB would be less than DE (69, 3), which is contrary to the hypothesis. Neither can B fall at the left of E, for then AB would be greater than DE, which is contrary to the hypothesis. Hence B must fall at ' E; in which case, the triangles coin- cide, and are therefore equal. 2. Let the hypotenuses, AB, DE, be equal, also the ad- jacent angles, A, D. Then will the triangles be equal. For, B is the complement of A and E of D (53, 5) ; but .4=2), . • . B ~ E (29, 8). Hence, the triangles having the side AB and the adjacent angles, -1, B, of the one, respectively equal to the side DE and the adjacent angles, D, E, of the other, are equal (57"). 3. If BC=EF and B = E, then, since C = F (29, 2), the triangles are equal (57\ If BC = EF and A — D, then B = E (53, 5), and the triangles are equal, as just shown. 73. Proposition XXIX.— Theorem. If two triangles haw tim sides of the one respect tidy equal to two sides of the other, and the included angle of the one greater than the included angle of Vie other, tlie third side of Vie one having tlie greater included angle is greater titan the third side of tlie otiier. Let ABC and DEF be two triangles having AB = DE, 58 GEOMETRY.— BOOK I. AC — DF, and the angle BAG > the angle EDF. Then the side BC > the side EF. For, place DEF on BAC so that it shall take the position ABF, bisect the angle GAF by AG, and draw GF. In the triangles, CAG, FAG, we have AC— AF by hypothesis, AG common, and the angle CAG = FAG by construction; .-. GO = GF. (?) Now, BG+GF>BF; but, GC= GF, BG+GOBF. But, BG+ GC--= BC, and BF = jEF, BC>EF. 74. Proposition XXX. — Theorem. Jf too triangles have two sides of the one respectively equal to two sides of the otlier and tlie third side of the one greater than the third side of tlie otlier, the angle opposite the greater third side is greater than the angle opposite the less. If AB = BE, AC = DF, and BC > EF, then A > D. For, either A = D, A D. If A-=B, BC=EF(oG, 1). If A EF; hence, neither A = D, nor A < D (15, 4) ; .-. A> D. 75. Proposition XXXI.— Theorem. Triangles mutually equilateral are mutually equiangular and equal. TRIAXGL1CS. 59 Let ABC and DEF be two triangles having AB = DE, AC=DF, and BC=EF. Then, .1 = D, B = E, C — .F. For, either .l>A^D, BG>EF; if JL Z>, nor ,1 < D ; .• . J. = D. In like manner, it is proved that B = E, and C = F. Hence, the triangles can be made to coincide, and are therefore equal. 76. Exercises. 1. Prove by means of the annexed diagram, in which DE is parallel to AC, that the sum of the three angles of a triangle is equal to two right angles. 2. The sum of the three straight lines drawn from any point within a triangle to the vertices is greater than one-half the sum of the three sides (65 s ). 3. The sum of the three straight lines drawn from any point within a triangle to the vertices is less than the sum of the three sides (,68). 4. In a right triangle, the middle point of the hypotenuse is equally distant from the vertices of the three angles (61). 5. If A EC = BED, and F is any A other point of CD than E, prove that AE + EB < AF + FB. 6. If one of the acute angles of a right G triangle is double the other, the hypotenuse is double the shortest side. 60 GEOMETRY.— BOOK I. IV. QUADRILATERALS. 77. Definitions and Classification. 1. A quadrilateral is a polygon of four sides. Thus, A, B, C, D, E, F, are quadrilaterals. £ 2. A trapezium is a quadrilateral having no two sides parallel. Thus, J. is a trapezium. 3. A trapezoid is a quadrilateral having at least two sides parallel. Thus, B, C, D, E, F, are trapezoids. 4. A parallelogram is a trapezoid having its opposite sides parallel. Thus, C, D, E, F, are parallelograms. 5. A rhomboid is an oblique parallelogram. Thus, C and D are rhomboids. 6. A rhombus is an equilateral rhomboid. Thus, D is a rhombus. 7. A rectangle is a right parallelogram. Thus, E and F are rectangles. 8. A square is an equilateral rectangle. Thus, F is a square. 9. Two parallel sides of a trapezoid are called bases. 10. The base on which it is supposed to stand is called its lower base; the base opposite, the upper base. 11. The altitude of a trapezoid is the perpendicular dis- tance between its bases. QUADRILATERALS. 61 78. Proposition XXXII— Theorem. Tlie sum of the four angles of any quadrilateral is equal to four right angles. Let R denote a right angle, and -,d draw AC, dividing ABCD into two /£? triangles. Then, \ a + D + d =2R, *" -^° b + B + c = 2R (52). .-. a + 6 -j- 5-r e-J- d + D = 4R. But, a + &=.!, e + d=C; .-. .1 + £-f- C+ D = 4K. 79. Corollary. if ftro angrfcg of o quadrilateral are supplemental, the otlier two angles are supplemental. (?) 80. Proposition XXXIII. — Theorem. The opposite \tngles of any parallelogram are equal. For, the sides of the angles, A, C, A B are parallel and lie in opposite direc- / tions; hence A = (18, 2). In like _ _ manner, it L? proved that B = D. 81. Corollary. If one angle of a parallelogram is a right angle, each of its otlier angles is a right angle. (?) 82. Proposition XXXIV— Theorem. The opposite sides of a parallelogram are equal. 62 GEOMETRY.— BOOK I. For, drawing the diagonal AG, the alternate angles, AGB, GAD, are equal (46, 2); also, BAG, AGD. Then, the triangles, AGB, AGB, /\ 7 are equal (57); ■ -. AB = CD, and / X v / AD = BC (56, 2). 83. Corollaries. 1. A diagonal of a parallelogram divides it into two equal triangles. 2. Two parallelograms are equal if two sides and Hie included angle of the one are respectively equal to two sides and Hie in- cluded angle of the other (21, 5). 3. Two rectangles are equal if Hiey have equal bases and equal altitudes (83, 2). 84. Proposition XXXV.— Theorem. A quadrilateral is a parallelogram if its opposite angles are equal. Let A BCD be a quadrilateral having A i ° its opposite angles equal. By hypothe- / sis, we have. B A = C, B = D; .-. A + B=C+D. But, A + B+C+D = 4R (78). 2(A + B)=4R, .-. A^B = 2R. Hence the sides, AD, BC, are parallel (43). A = C, D = B; .-. A + D=C+B. But, A + D+ C+ B = 4R. 2(A + D)=4R, .-. A + D = 2R. Q VADRILA TERALS. 63 Hence the sides, AB, CD, are parallel (43). Therefore ABCD is :i parallelogram (77, 4). 85. Proposition XXXYL— Theorem. -1 quadrilateral is a parallelogram if its opposite sides are equal. Let ABCD be a quadrilateral having A _ its opposite sides equal. Drawing the / \ / diagonal AC the triangles, ABC, ADC, having AC common, AB = CD and AD = BC by hypothesis, are equal (75). Hence, the angle ACB is equal to the angle CAD (56, 2). .-. AD and BC are parallel (44, 2). In like manner, it is proved that AB and CD are parallel. Hence ABCD is a parallelogram (77, 4). 86. Proposition XXXVII.— Theorem^ A quadrilateral is a parallelogram if two of its opposite sides are equal and parallel. Let ABCD be a quadrilateral having AD, BC, equal and parallel. Drawing / *\ / AC, the angles, ACB, CAD, are equal / ' (46, 2). Then, the triangles, ACB, M ACD, are equal (55). Hence, the angles, BAC, ACD, are equal. (?) Therefore, AB, CD, are parallel (44, 2). Hence ABCD is a parallelogram (77, 4"). 87. Corollary. Tim lines arc parallel if two points of citlier are equally dis- tant from Hie oilier. (?) 64 GEOMETRY.— BOOK I. 88. Proposition XXXVIII .— Theorem. The diagonals of a parallelogram bisect each other. Let ABCD be a parallelogram, AC, A BD, its diagonals intersecting at E. j - >r --''' J AD = BC (82), the angles, ABE, B L^\J C'BE, are equal (46, 2), also, the angles, DAE, BCE. Hence, the triangles, AED, BEC, are equal (57). Therefore, the sides, AE, CE, are equal (56, 2), also, the sides, BE, DE. Hence, the diagonals bisect each other. 89. Corollaries. 1. The diagonals of a rhombus or a square bisect each other at right angles (70, 3). 2. Each diagonal of a rhombus or a square bisects two opposite angles (60, 3). 90. Proposition XXXIX —Theorem. A quadrilateral is a parallelogram if its diagonals bisect each other. Let ABCD be a quadrilateral whose . diagonals bisect each other. Then, / \ „,--''"/ ABCD is a parallelogram. A.--'' B " ' For, AE = CE, DE = BE, by i "" hypothesis, and the angles, AED, BEC, are equal (33). Hence, the triangles, AED, BEC, are equal (55), and the angles, ADB, CBD, are equal (56, 2). Therefore, the sides, AD, CB, are parallel (44, 2). In like manner, it is proved that AB, CD, are parallel. Hence, ABCD is a parallelogram (77, 4). POLYGONS IX GENERAL. Go 91. Corollary. A quadrilateral in a rhombus or a square if its diagonals bisect each other at right angles. (?) 92. Exercises. 1. If two adjacent angles of a quadrilateral are supple- mental, the quadrilateral is a trapezoid. 2. If the angles adjacent to one base of a trapezoid are equal, the angles adjacent to the other base are equal. 3. The diagonals of a rectangle are equal. 4. A parallelogram is a rectangle if its diagonals are equal. 5. If on passing round the perimeter of a square four points be taken equally distant from the vertices left, and these points be joined in order, two and two, the figure formed will be a square. (>. If from a variable point in the base of an isosceles triangle parallels to the equal sides be drawn, a parallelo- gram is formed, whose perimeter is equal to the sum of the equal sides of the triangle. V. POLYGONS IX GENERAL. 93. Definitions and Classification. 1. A plane polygon is a plane figure bounded by straight lines. 2. Polygons are classified in reference to the number of angles, as trigone (triangles), tetragons {quadrilaterals), penta- gons, hexagons, heptagons, octagons, enneagom, decagons, hen- p. G.-6. 66 GEOMETR Y.—B 00 K 1. decagons, dodecagons, decatrigons, decatetragons, decapentagoiis, decahexagons, decaodagons, decaenneagons, icosagons .... The number of sides of a polygon is equal to the number of angles. 3. A pentagon is a polygon of five sides. 4. In like manner, define a liexagon, a heptagon, etc., up to an icosagon. 5. An equilateral polygon is a polygon which has all ite sides equal. 6. An equiangular polygon is a polygon which has all its angles equal. 7. A regular polygon is a polygon which is both equi- lateral and equiangular. 8. Mutually equilateral polygons are polygons which have their corresponding sides equal. 9. Mutually equiangular polygons are polygons which have their corresponding angles equal. 10. Equal polygons are those which can be made to coincide. 11. A convex polygon is a polygon of x x which each interior angle is less than two V. right angles. X - V X \ It is evident that no side of a convex polygon will, when produced, divide the polygon, and that a straight line can not intersect the perimeter in more than two points. 12. A concave polygon is a polygon of which at least one interior angle is greater ,X than two right angles. L_ Such an angle is called re-entrant, and each of its "sides produced through the Vertex will divide POLYGOSS IX GENERAL. 67 the polygon. A straight line may intersect the perimeter of a concave polygon in more than two points. 13. The diagonals of a convex polygon are all interior. 14. A diagonal of a concave polygon maj' be interior, exterior, or partly interior and partly exterior. 15. A polygon is to be regarded convex unless it is stated to be concave. 16. A polygon can be divided into triangles, 1st. By drawing diagonals from the vertex of one of its angles to the vertices of all the angles not adjacent. In this case, the num- ber of triangles is equal to the number of sides of the polygon minus two. -d. By drawing lines from a point within to the vertices of the angles. In this case, the number of triangles is equal to the number of sides. "7 94. Proposition XL. — Theorem. Equal polygons can be divided by diagonals into tlie same number of triangles respectively equal and similarly arranged. For, since these polygons are equal, one may be placed on the other so that they shall coincide. Then, diagonals drawn from the common vertex of any two coincident angles to the common vertices of all the angles not adjacent will divide the coincident polygons into coincident triangles, which are therefore equal and similarly arranged. 68 GEOMETRY.— BOOK I. 95. Proposition XLL— Theorem. Polygons are equal if they can be divided by diagonals into the same number of triangles respectively equal and similarly arranged. For, placing one on the other, the equal parts being sim- ilarly arranged can be made to coincide. Hence, the poly- gons will coincide, and are therefore equal. 96. Proposition XLIL— Theorem. Equal polygons are mutually equilateral and mutually equi- angular. For, since they are equal, they can be made to coincide. Hence, the coincident sides and angles are equal; that is, the polygons are mutually equilateral and mutually equi- angular. 97. Proposition XLIIL— Theorem. Polygons are equal if they are mutually equilateral and mutu- ally equiangular, the equal parte being similarly arranged. For, placing one on the other, the equal parts can be made to coincide. Hence, the polygons can be made to coincide, and are therefore equal. 98. Proposition XLIV.— Theorem. The sum of all the angle* of a polygon U equal to tivo right angles taken as many timr* fc« two as the polygon lias sides. For, let n denote the number of sides, / R one right angle, and .s the sum of the \ angles. Then, by drawing diagonals from y>- y the vertex of one of the angles to the ver- tices of all the angles not adjacent, it is evident that the POLYGOXS IX OESEEAL. 69 polygon will be divided into u — 2 triangles. It is also evident that the sum of all the angles of all the triangles is equal to the sum of all the angles of the polygon. But the sum of all the angles of one triangle is 2R. Hence, the sum of all the angles of the n — 2 triangles, which is the sum of all the angles of the polygon, is equal to n — 2 times 2R, which is 2R (i> — 2). Hence, a = 2R (n — 2). 99. Corollaries. 1 . If the polygon is equiangular, denote each angle by A ; then, since there are as many angles as sides, n = the number of angles, and since each angle is A. of the sum, 4 = 2i?fn— 2) _ n 2. Since, in a triangle, n = 3, in a quadrilateral, n = 4, etc., we have s for all cases, and A for equiangular polygons: In general, * = 2i? (ii - In a triangle, * = 2R, -2), A = -"' v '" ^ n A = fi?. In a quadrilateral, s = 4i?, A= R. In a pentagon, s = 6i?, A = f R, In a hexagon, .< = Si?, A = i R. 100. Proposition XLY— Theorem. If each tide of any polygon be produced in both directions. Hie sum of all the angles, interior and exterior, will be equal to four right angles talrn as many times a.< the polygon has sides. 70 GEOMETRY.— BOOK I. Let R denote one right angle, n, the number of sides or angles, I, the sum of the interior angles, E, the sum of the exterior angles, V, the sum of the exterior angles vertical to the interior, A, the sum of the exterior angles adjacent to the interior. Then, since the sum of the angles about one vertex is 4R, the sum of the angles about the n vertices will be n times 4R, which is 4nR. But I 4- E is equal to this sum. Hence, I -f- E = 4nR. 101. Corollaries. 1. The sum of the exterior angles is 2nR + 4R. For, E = 4ni? — I = 4reii — 2E (ji — 2) = 2ni? + 4R. 2. TAe stw)i of the exterior angles vertical to the interior is 2R(n — 2). For, V=I=2R(n — 2). 3. T/ie sum of the exterior angles adjacent to the interior is &R. For, A = E— V= 2nR + 4R — 2R (n — 2) = 8R. 4. if 6«t one exterior angle adjacent to the interior be talcen at each vertex, their sum will be 4R. (?) 102. Exercises. 1. Prove that the sum of the interior angles of a polygon of n sides is equal to 2nR — 4R, by drawing lines from a point within to the vertices of the polygon. 2. If each side of a polygon be produced in one direction forming one exterior angle at each vertex, prove that the stun of these exterior angles is 4R, by drawing, from any point in the plane of the polygon, lines respectively parallel to the sides. SUPPLEMENTARY PROPOSITIONS. 71 3. What is the greatest number of interior acute angles that any convex polygon can have? 4. Prove that the whole number of diagonals that can be drawn in a polygon of n sides is in (« — 3). 5. Find from the above formula the number of diagonals that can be drawn in a quadrilateral, in a pentagon, in a hexagon, and so on, to an icosagon ; also, in a triangle. 6. What regular polygons of the same kind can be used in making a pavement? (31, 2), (99, 2). 7. What combinations of different regular polygons can be used in making a pavement? VI. SUPPLEMENTARY PROPOSITIONS. 103. Proposition XL VI.— Theorem. The straight line drawn from the middle point of one side of a triangle, parallel to a second side and terminating in the tliird, bisects tiie tliird side, caul is equal to one-half the second. In the triangle ABC, let DE be drawn from D, the middle point of AB, parallel \ to BC and terminating iu AC. Draw DF T . / > parallel to AC; then, in the triangles, /\ \ ADE, DBF, AD = DB, ADE = DBF, 2 J \ DAE = BDF (46, 4) ; hence, the tri- angles, ADE, DBF, are equal (571; .-. AE = DF. But, DF ^ EC, since DECF U a parallelogram (82); .-. AE= EC; hence, DE bisects AC. In the equal triangles, ADE, DBF, DE = BF; (?) but D£= FC; (?) .-. BF=FC; .-. BF = IBC, ,-. DE=-hBC. E 72 , GEOMETRY— BOQK I. 104. Corollaries. 1. The straight line joining the middle points of two sides of a triangle is parallel to the third side and equal to its half. (?) 2. The straight line draivn from the middle point of one of the non-parallel sides of a trapezoid, parallel to the bases, and terminating in the oilier non-parallel side, bisects that side, and 'is equal to one-half tlie sum of the bases. In the trapezoid ABCD, let EF be drawn from E, the middle point of AB, one of the non-parallel sides, parallel to the bases, and terminating in the other non-parallel side. Draw AC intersecting EF in G; then, since E is the middle point of AB, and EG is parallel to BC, G is the middle point of AC, and EG = \BC (103). Since G is the middle point of AC, and GF is parallel to AB, F is the middle point of DC, and GF=\AD. (?) .-. EG+GF=i(BC+AD), .-. EF=i(BC+AD). 3. The straight line joining the middle points of the non- parallel sides of a trapezoid is parallel to the bases, and equal to one-half their sum. (?) 105. Proposition XL VII.— Theorem. Every point in the bisector of an angle is equally distant from the sides of the angle. Let E be any point of AD, the bisector of the angle BAG, and EF, EG, the perpendiculars from E to AB, AC, respectively. The right triangles, AEF, AEG, are equal (72, 2), and EF = EG (56, 2). Hence, E is equally distant from the sides of the angle BAC (71). SUPPLEMENTARY PROPOSITIONS. 73 106. Proposition XLYIIL— Theorem. Ever// point within an angle, not in the bisector, is unequally distant from the sides of the angle. 1. Let BAC be right or acute, AD the p bisector, E a point within the angle, but ^--j; n not in the bisector, EF, EG, perpendiculars to AB, AC, respectively. One of these per- pendiculars will intersect the bisector in H. Let fall HI perpendicular to AC, and draw IE. Since EG is perpendicular- to AC, EI is oblique (30) ; .• . EG' AD the bisector, E, E', E", points AD. For, the chord AD, not passing through the center, and the radii, CA, CD, drawn to the extremities of the chord, form a triangle ACD. Now, AC + CD > AD (65) ; but CD = CB (117, 1) ; AC + CB > AD; but AC+ CB = AB; .-. AB>AD. 122. Proposition VI.— Theorem. The diameter perpendicular to a chord bisects the chord and the arcs subtended by the chord. Let C be a circle, and DE a diam- eter perpendicular to the chord AB. Then, DE bisects AB and the sub- tended arcs, AEB, ADB. For, revolve the semi-circumference DAE about DE as an axis, till it co- incides with the semi-circumference DBE. Since DE is perpendicular to AB, FA will fall on FB, and A, the point common to DAE and FA, will be found in DBE and FB, and hence at their intersection B. Then, JFLrl coincides with FB,„EA with ££, DJ with ZXB. .-. Jf.4 = FB, EA = EB, DA == DJ3. 123. Corollaries. 1. 27*e perpendicular to a chord at its middle point passes through the center of tiie circle and bisects Vie arcs subtended by the cliord. (?) 84 GEOMETRY.— BOOK II. 2. The intersection of the 'perpendiculars to two cliords at their middle points is tlie center of the circle. (?) 3. The straight line which passes through any two of the four points, the center of the circle, the middle point of a chord, the middle points of tlie arcs subtended by the chord, passes through the other points, and is coincident with the diameter perpendic- ular to the cliord. (?) 4. The locus of tlie middle points of a system of parallel chords is tlie diameter perpendicular to these chords, and the extremities of this diameter are the middle points of all arcs subtended by these chords. (?) 5. A diameter perpendicular to a chord bisects tlw angle formed by two lines draivn from any point of the diameter to the extremities of tlie cliord. (?) 6. The distance of a cliord from tlw center is Hie straight line having the center and the middle point of the cliord for its ex- tremities. (?) 124. Proposition YIL— Theorem. In the same circle or in equal circles, equal chords are at equal distances from the center. Let C be the center of a circle, AB, DE, equal chords. Then, the perpendiculars, CF, CG, which meas- ure the respective distances of the chords from the center, are equal. For, these perpendiculars bisect the chords (122); .-. AF = DG. (?) Hence, the right tri- angles, AFC, DGG, are equal (72, 1); .-. CF = CG. In case of equal circles, the demonstration is similar. STRAIGHT LIXKS AXI) CIRCLES. DnE. Then, AB > DE. Take AmG=DnE, and draw AG, CA, CB, CG. Since AmG = DnE, AG = DE (126). The angle ACB > ACG (21, 6). Then, in the triangles, ACB, ACG, AB > AG (73) ; .-. AB > DE. Conversely, let AB > DE, then AmB > DnE; for, if AmB = DnE, AB = DE; (?) if AmB < DnE, AB DE; .-. neither AmB = DnE nor AmB < DnE; . ■ . AmB > DnE. STRAIGHT LIXi;s AXD CIRCLES. 87 128. Scholiums. 1. Each of the arcs considered in the last proposition is supposed to be less than a semi-circumference. 2. If each arc is greater than a semi-circumference, the greater arc is subtended by the less chord, and conversely. 3. If one arc is greater and the other less than a semi- circumference, then, 1st. If the sum of the arcs is less than a circumference, the greater arc is subtended by the greater chord, and conversely. 2d. If the sum of the arcs is equal to a circumference, the arcs are subtended by equal chords, and conversely. 3d. If the sum of the ares is greater than a circumference, the greater arc is subtended by the less chord, and conversely. 129. Proposition XI.— Theorem. 7» tlie same circle or in equal circles, if tuv chords are unequal, tlie less is at tlie greater distance from the center; and conversely, if two chords are at unequal distances from the center, that which is at tlie greater distance is tlie less. Let C be the center of a circle, the chord AB less than the chord DE\ then, CG, perpendicular to AB, is greater than Cfl", perpendicular to DE. If DE is a diameter, it passes through the center; .-. Cff=0; and since AB < DE, AB is not a diameter ; .-. «?>0; .-. CO> CH. 88 GEOMETR Y.—B OK II. If DE is not a diameter, on the less of the two arcs sub- tended by DE take the arc DF equal to the arc AB, draw the chord DF, and CI perpendicular to DF. DF = AB (126), and CI = CG (124). The chord DE intersects CI in some point K. (?) Since DI is perpendicular to CI, DK is oblique to CI (30) ; that is, CK is oblique to IXE; but CH is perpendicular to DE; .-. CK"> CH (69, 1); but CI> CK; .-. CJ> Cfi"; but CG = CI; .-. CG> CH. Conversely, let CG > CH; then, .A.B < DE; for, if .4.B = DE, CG = CH (124) ; if AB > DE, CG < C1T; (?) but both these results are contrary to the hypothesis that CG > CH; .-. neither AB = DE nor AB > DE; . • . AB <: DE. 130. Proposition XII —Theorem. Through any three points not in the same straight line a cir- cumference of a circle can be made to pass. Let A, B, D, be three points not in a straight line. Draw the straight lines, AB, BD, and in the plane of the points erect the perpendiculars, EG, FH, to AB, BD, respectively, at their middle points. These perpendiculars will meet ; for, if not, they are par- allel (40) ; then, AB, perpendicular to EG, would be per- pendicular to FH, parallel to EG (42, 3); but BD is perpendicular to FH. (?) STRAIGHT LIXES ASD CIRCLES. 89 Then, since the three points, -4, B, D, are not in a straight line, we have two straight lines, AB, BD, drawn from the same point B, perpendicular to the same straight line FH, which is impossible (30). Hence, EG and FH must meet in some point C. Now, C being in EG, perpendicular to AB, at its middle point is equally distant from J. and B ; that is, 111 = CB (69, 2). Also, C being in FH, perpendicular to BD, at its middle jioint is equally distant from B and D ; that is, CB = CD. Hence, C is equally distant from the three points, A, B, D. If, therefore, a circumference be described having C for a center and CA for a radius, it will pass through the points, A, B, D. 131. Corollaries. 1. All the circumferences which pass through time points not in the same straight line are coincident. (?) 2. Two circumference* can interact in only two points. (?) 132. Proposition XIII.— Theorem. An indefinite straight line oblique to a radius at its extremity is a secant; a)ui conversely, a secant is oblique to the radius drawn to either point of intersection. Let AB be oblique to the radius CD, at ite extremity D ; then, AB is a secant. For, let CE be perpendicular to AB (30) ; then, CE< CD (69, IV But. CD is a radius; hence, CE is less than a radius; therefore, the point E is within the P, G.— s. 90 GEOMETRY.— BOOK II. circumference. Now, AB having the point D in common with the circumference (115, 4), and the point E within, intersects the circumference in one point D, and hence, in another point F (119), and lies partly within and partly without the circle, and is therefore a secant (115, 14). Conversely, let AB be a secant, and CD a radius drawn to D, one point of intersection of the secant and circumference. Since the secant passes within the circumference, it has points nearer the center than D ; hence, CD is not the perpendicular from C to AB; there- fore, AB is oblique to CD. 133. Proposition XIV— Theorem. A straight line perpendicular to a radius at its extremity is a tangent to the circumference; and conversely, a tangent to a circumference is perpendicular to tlie radius drawn to tiie point of tangency. Let AB be perpendicular to the ra- dius CD, at its extremity D. Then, > AB is a tangent. For, since CD is perpendicular to AB, any straight line drawn from C to any other point of AB, as E, is oblique (30); .-. CE > CD (69, 1). But CD is a radius; hence, CE is greater than a radius; therefore, the point E is without the circle. Hence, AB has only one point in common with the circumference, and is therefore a tangent (115, 12). Conversely, the tangent AB is perpendicular to the radius STRAIGHT LISES ASD CIRCLES. 91 CD drawn to the point of tangeney ; for, if not, it is oblique to CD, and hence enters the circle and intersects the circum- ference in another point, and is therefore a secant. (132), and not a tangent (115, 12), which is contrary to the hy- pothesis ; hence, the tangent AB is not oblique, and is there- fore perpendicular to the radius CD, drawn to the point of tangeney. 134. Corollaries. 1. The indefinite perpendiadar to the tangent at the point of tangeney passes through Hie center of the circle. (?)' 2. Every point of the tangent except, Vie point of tangeney is ivithout Hie circle. (?) 3. At a given point of a circumference only one tangent can he drawn. (?) ■i, If a secant revolve about eittier point of intersection with the circumference, the second point of intersection will move along the circumference ; and when it coincides irith the other, the secant becomes a tangent whicli may, tlierefore, be regarded as a secant whose points of intersection are coincident. 135. Proposition XT. — Theorem. Tivo parallels meeting a circumference intercept equal arcs. 1. "When the parallels are secants, or chords: Let AB and DE be parallel secants or chords, and FG the diameter per- pendicular to them. Then, AF = BF^tmd DF = EF (122). ... AF—DF = BF — EF(i3,7\ Hence, AD = BE. 11 F I Z> ''"" ~~"\£ jl! Xb m\ In 92 GEOMETRY.— BOOK IT. JV"" ~^"\J5 ^t / VR lit 2. When one of the parallels is a secant or a chord and the other a tangent, as AB and HI: Let DE move toward HI, but in j r j- j such a manner as to continue parallel to AB, till it coincides with HI (42, 5). The points D and E will approach and finally meet in i* 7 , J..D will become AF, and J3£, J5F. But in all positions of the moving parallel, AD = BE. Hence, when DE becomes the tangent HI, AF = BF. 3. When the parallels are both tangents, as HI and KL : Let DE move toward HI, and AB toward KL, but in such a manner as to continue parallel, till DE coincides with HI, and AB with KL. The points D and E will finally meet in F, and A and B in G, AD will become GmF, and BE, GnF. But in all positions of the moving parallels, AD =-. BE. Hence, when the parallels become tangents, GmF = GnF. 136. Exercises. 1. What is the locus of all the points in a given plane which are at a given distance from a given point in that plane? Does this locus embrace all the points at the given distance from the given point ? 2. What is the locus of all the points which are at a given distance from the circumference in the plane of a cir- cle? Of how many lines does this locus consist? When would one of these lines reduce to a point? 3. The least chord that can be drawn through a given point in a circle is perpendicular to the diameter through the point (129). RELATIVE POSITION OF CIRCLES. 93 4. The angle contained by two tangents at the extremities of a chord is twice the angle contained by the chord and the diameter drawn from either extremity of the chord (133). II. RELATIVE POSITION OF CIRCLES. 137. Remarks. If two circumferences lie in the same plane, then, 1. They are wholly external to each other when they arc separated by intervening space. 2. They are ta>igent externally when they are external to each other except one common point of tangency. .'i. They intersect when they cross each other. 4. They are tangent internally when one is within the other except one common point of tangency. 5. One is wholly witJtin the other when the former is in- closed by the latter, no point being common. 6. They are concentric when they have a common center. 7. They are coincident when equal and concentric. 138. Proposition XYL— Theorem. If two circumference* intersect in one point, then, 1. They internet in another point. 2. The straight line joining tlieir centers bisects their common chord at right, angles. 3. The distance between, their centers is less than the sum of their radii and greater than their difference. 94 GEOMETRY.— BOOK II. 1. Let the circumferences whose centers are C, C, inter- sect at A. Each has points without and within the other. Let in be a point in the cir- cumference of C without that of C". The circumference of C, starting from m, passing through A, and continuing in the same direction round to A again, passes within the circumference of C" at A; and since it passes through m before reaching A again, it must pass without, and hence intersect the cir- cumference of C in another point B. 2. Draw the common chord AB, and in the plane of the circles draw a perpendicular to AB at its middle point. This perpendicular passes through the centers, C, C (123, 1), and is therefore coincident with the line joining the centers (26, 4) ; hence, the line joining the centers bisects the chord at right angles. 3. Since A and B are different points, and CC passes through the middle point of AB, the point A is not in the line CC ; hence, ACC is a triangle. .-. CC CA — AC. (?) 139. Proposition XVII— Theorem. If two circumferences are tangent to each other externally, tlien, 1. , The straight line joining their centers passes through their common •point of tangency, and is perpendicular to tlieir common tangent at that point. 2. The distance between their centers is equal to the sum of their radii. RELATIVE POSITION OF CIRCLES. 95 1. Let the circles whose centers are C, C, intersect in A •and B. CC is perpendicular to the common chord AB (138, 2), and hence to the common secant EF, which is the common chord produced both ways. \ \ Now, let the circles move so that the centers may move from each other along the prolongations of CC, in such a manner that EF shall remain the common secant through the points of intersection. Since the chord AB is less the greater its distance from the center (129), and is continually bisected by the perpen- dicular CC (138, 2), the points, A, B, approach as the circles move, and finally meet on CC, when EF becomes their common tangent perpendicular to C"C" (115, 12). 2. The straight line, C"C", the distance between the centers when the circumferences become tangent, passes through the common point of tangency (139, 1), and is, therefore, the sum of the radii. 140. Proposition XVIII.— Theorem. If hco circles are wltolly e.iiemal to each oilier, the distance bettveen their centers is greater than the sum of tlieir radii. Let C and C be the cen- ters of two circles which are wholly external to each other. / c _ a\ (b c , Then, CC the distance be- tween their centers, is greater than the sum of their radii. For, CC = CA -f AB + BC, which exceeds the sum of the radii CA + BC by AB. 96 GEOMETRY.— BOOK II. 141. Proposition XIX. — Theorem. If two circumferences are tangent to each oilier internally, then, 1. The indefinite straight line passing through their centers in perpendicular to tlw common tangent at Hie point of tan- gency. 2. The distance between tlieir centers is equal to the difference of tlieir radii. 1. Let the circles whose centers are G, C", be tangent to each other inter- nally at A. EF, tangent to the greater circumference at A, has this point in common with the less circumference, since A is the common point of tangency. Now, EF and the circumference of C have no other point than A common ; for, every point of EF, except A, is without the circumference of G (134, 2), and every point of the circumference of C, except A, is within the circum- ference of C (137, 4) ; hence, EF is tangent to the circum- ference of G' at A (115, 12). The internal perpendicular to EF at A passes through G and C" (134, 1), and is therefore coincident with the line passing through the centers (26, 4). 2. It is evident from the diagram that CG' = CA — G'A. 142. Proposition XX.— Theorem. If one circumference is wholly within another, tJie distance between their centers is less than the difference of their radii. RELATIVE POSITION OF CIRCLES. 97 Let C and C" be the centers of two circles whose re- spective radii are CA and C'B, and let the circumference of C be within that of 0. Then, tt" = CA— C'B — BA, which is less than the difference of the radii by BA. 143. Exercises. Remark. — The first five of the following propositions can he proved by showing that the conditions are inconsistent with any other relation. Thus, in the first, the two circumferences can not be wholly external ; for then the distance between their centers would be greater than the sum of their radii, which is contrary to the hypothesis. In like manner, it can lie proved that no other relation, except that of intersection, is possible; hence, tliev intersect. 1. Two circumferences intersect, if the distance between their centers is less than the sum of their radii and greater than the difference. 2. Two circumferences are tangent externally, if the dis- tance between their centers is equal to the sum of their radii. 3. Two circumferences are wholly external, if the distance between their centers is greater than the sum of their radii. 4. Two circumferences are tangent internally, if the dis- tance between their centers is equal to the difference of their radii. 5. One circumference is wholly within another, if the dis- tance between their centers is less than the difference of their radii. S. G -0. 98 GEOMETRY.— BOOK IT. 6. What is the locus of the centers of all the circles having a given radius, and tangent externally to a given circumference ? 7. What is the locus of the centers of all the circles having a given radius, and tangent internally to a given circumference ? 8. With a given radius, describe a circumference tangent to a given circumference (139). III. MEASUREMENT OF ANGLES. 144. Definitions. 1. Ratio is the relation of two quantities of the same kind expressed by the quotient obtained by dividing the first by the second. 2. A measure of a quantity is a quantity of the same kind which is contained in the given quantity a whole number of times without a remainder. 3. A common measure of two quantities of the same kind is a .measure of each of those quantities. 4. A unit of measure is any measure taken as a unit. 5. A common unit of measure is any common measure taken as a unit. 6. To measure a quantity is to find the ratio of that quantity to the unit of measure. 7. To find the ratio of one quantity to another of the same kind is to measure the first by the second taken as the unit of measure. MEASUREMENT OF ASGLEX. 99 8. The numerical value of a quantity is the ratio of that quantity to the unit of measure. If this ratio is a mixed number, the integral part is the approximate numerical value of the quantity. 9. The numerical ratio of two quantities is the ratio of their numerical values referred to a common unit. 10. Commensurable quantities are quantities which have a common measure. 11. Incommensurable quantities are quantities which have no common measure. 12. An incommensurable ratio is the ratio of two in- commensurable quantities. 13. The approximate numerical value of the ratio of two incommensurable quantities is the ratio of the approxi- mate numerical value of one to the exact or approximate numerical value of the other referred to the same unit. 14. An inscribed angle is an angle whose vertex is in the circumference and whose sides are chords. 15. An angle is inscribed in a segment if its vertex is in the arc of the segment and its sides pass through the ex- tremities of the subtending chord. 16. A central angle is an angle whose vertex is at the center and whose sides are radii. A central angle is said to intercept the arc included between its sides. 17. Similar arcs are those which are intercepted by equal central angles. 18. Similar sectors are those which have equal central angles. 100 GEOMETRY.— BOOK II. 145. Proposition XXI.— Theorem. Tlie ratio of commensurable quantities is equal to their nu- merical ratio. Let a and c be two commensurable quantities, u their common unit of measure, m the numerical value of a, and n of c, referred to their common unit of measure, u. Then, Hence, a u = m, and ~ = n (144, 8). a = mu, and c = nu. a c mn nu m u m „ m ii a n n 146. Proposition XXII.— Problem. To find the approximate numerical value of the ratio ' of hvo incommensurable quantities urithin any required degree of pre- cision. Let a and c be two incommensurable quantities, and let it be required to find the approximate numerical value of their ratio within — . n Divide c into n equal parts, and denote each part by u. Since, by hypothesis, a and c are incommensurable, u will not divide a without a remainder. Suppose that u is con- tained m times in a with a remainder r less than u. Then, — = m + - , . ■ . a = mu -\- r. /> But, — — n, . ■ . c = nu. a mu r _ m 1 r c nu nu n n u MEASUREMENT OF ANGLES. 101 But, v <; », . • . r (6 < 1, .-. 1 n X r u < 1 n a e = m n within 1 n As 5t increases, u diminishes, m increases, and — diminishes. n By making n sufficiently great, — may be made as small as we please, and — will express the approximate numerical value of the ratio of a to e within the required degree of precision. 147. Proposition XXIII— Theorem. Two iiicomm,ensurable ratios are equal if their approximate numerical values within the same degree of precision, however exact, are always equal. Let — and -r be two incommensurable ratios whose nu- e d merical values within the same degree of precision are always equal, each, found as in (146), being — within — • Then, - — — within - e n n , b m . , . 1 and T — — within -• d n n a h I e d 11 By making h, sufficiently great, -- can be made as small as we please. Hence, j is less than any quantity however small ; . • . . =• — ("),.■. - = -=-, e d c d 102 GEOMETRY— BOOK II. 148. Proposition XXIV —Theorem. In the same circle or in equal circles, equal central angles intercept equal arcs of the circumference, and conversely. ' Let ACB, DC'E, be equal central angles in equal circles, C, C. Then, the arcs, AB, DE, are equal. For, place one circle on the other, so that the equal angles coincide. A will coincide with D, and B with E; hence, the arcs, AB, DE, coincide, and are therefore equal. If the equal central 'angles are in the same circle, each of the intercepted arcs is equal to the arc intercepted by an equal central angle in an equal circle ; hence these arcs are equal to each other. Conversely, if the arcs, AB, DE, are equal, the angles, ACB, DC'E, are equal. For, make the equal circles coin- cide, also the equal arcs, AB, DE. Then, since the centers coincide, also, A and D, and B and E, AG coincides with DC, and BC with EC ; hence, the angles, ACB, DC'E, coincide, and are therefore equal. 149. Proposition XXV. — Theorem. In tlie same circle or in equal circles, tlte greater of two un- equal central angles intercepts the greater arc. Let C, C, be centers of equal circles, and ACB > DC'E; then, AB > DE. Place one circle on the other so that AC and DC MEASUREMENT OF ANGLES. 103 shall coincide. Then, since DC'E < ACB, C'E will fall between CA and CB, and E between .1 and B; hence, AB > DE.^ Conversely, if AB > BE, ACB > DC'E. For, place one circle on the other so that AC and DC shall coincide. Then, since AB > Z>£, E will fall between .4 and 5, and EC between AC and 5C; hence, .4C.fi > DC'E. 150. Proposition XXVI.— Theorem. In Vie tame circle or in equal circles, the ratio of two central angles is equal to Hie ratio of tlieir intercepted arcs. 1. When the angles are ,, C n C commensurable: jx A A Let ACB and DC'E be a / /'7|\\ r A ''/ \ central angles of equal eir- w cles, and suppose these angles have a common unit of measure v which, for example, is contained 5 times in -4CB and 3 times in DC'E. Then, 5 is the numerical value of ACB, and 3, of DC'E. .■ ACB DC'E :: o : 3 (145). Let radii be drawn dividing ACB into 5 parts, and DC'E into 3 parts, each equal to u. These radii divide AB into 5 parts, and DE into 3 parts, all equal (14S\ Then, 5 is the numerical value of AB, and 3 of DE. .-. AB : DE :: 5 : 3. Comparing these proportions, we have ACB : DC'E :: AB : DE (Algebra, 317. 9). 104 GEOMETRY.— BOOK II. 2. When the angles are incommensurable: Let ACB and DC'E be in- commensurable central angles of equal circles. Suppose the angle DC'E divided into a number n of equal parts, with- out a remainder. Then, since the angles are incommensura- ble, ACB will contain a certain number m of these parts, with a remainder less than one part. ™ -ACB m ... . 1 ,,.-,. Then, ^^ = _ wiflun - (U6). The radii, separating these parts, divide DE into n equal parts, without a remainder, and AB into m such parts (148), with a remainder less than one part (14:9). AB m „ . 1 — within — n n ali» III ... Then, -plj = — within Since the approximate numerical values of the ratios of the angles and arcs, within the same degree of precision, however exact, are always equal, the ratios themselves are equal (147). A OTl A R •'• ThFe = de'' that is ' ACB : DCE '■'■ AB '■ DE - 151. Corollaries. 1. Assuming a central angle as the unit of measure for angles, and the intercepted arc as the unit of measure for arcs, ilie nu- merical value of any central angle is equal to the numerical value of iti intercepted arc. MEASUREMENT OF ANGLES. 105 Let c be any central angle, a its intercepted arc, u the unit of measure for angles, and u' the unit of measure for arcs, u' being the arc intercepted by u. Then, C - - a -, (150). But — is the numerical value of the central angle c, and — is the numerical value of its intercepted arc a (144, 8). Hence, the numerical value of a central angle is equal to that of its intercepted arc. On this account, a central angle is said to be measured by its intercepted arc. 2. The ratio of any central angle to four right angles is equal to the ratio of its intercepted arc to the circumference. For, four right angles may be regarded as the central angle whose intercepted arc is the circumference. 3. In the name circle or in equal circles, the ratio of two sectors is equal to the ratio of their arcs. The proof is similar to that for Prop. XXYI. 4. The ratio of a sector to the circle is equal to the ratio of the arc of the sector to the circumference. For, the circle may be regarded as a sector whose arc is the circumference. 152. Scholium. Two diameters perpendicular to each other form four equal central angles, and divide the circumference into four equal arcs (14-8), each of which, called a quadrant, is intercepted by a right angle. 106 GEOMETRY.— BOOK II. The unit of measure for angles, which is -fa of a right angle, is called a degree, and intercepts ^ of a quadrant, which is the unit of measure for arcs, and is also called a degree. A degree, whether of angle or arc, is written 1°. A minute is -£-$ of 1°, and is written 1'. A second is -^ of 1', and is written 1". Thus, 10° 30' 20" is read, 10 degrees, 30 minutes, 20 seconds. The number of degrees in an angle is equal to the number of degrees in its measuring arc. Equal central angles in unequal circles intercept arcs of the same number of degrees, though differing in length. 153. Proposition XXVII.— Theorem. An inscribed angle is measured by one-half of its intercepted arc. 1. Let BAD be an inscribed angle, the center being on the side AB. Draw the radius CD. Then, since BCD is an exterior angle to the triangle ACD, CAD + CD A = BCD (53, 1). Since CA, CD, are equal, the triangle A CD is isosceles; . • . CD A = CAD (59) ; . • . WAD = BCD ; .-. BAD = \BCD. But the central angle BCD is measured by the arc BD (151, 1) ; hence, the angle BAD is measured by iBD. MEASUMEMENT OF ANGLES. 107 2. Let. BAD be an inscribed angle, within the angle. Draw the diameter AE. Then, by the first ease, BAE is measured by IBE, and EAD by IED ; .*. .B.i-E -f EAD, or £.41) is meas- ured by IBE + IED, that is, by *(££ + £7T), or by «32). 3. Let BAD be an inscribed angle, the center being without the angle. Draw the diameter AE. Then, by the first case, BAE is measured by \BE, and DAE by WE; .-. BAE— DAE. or BAD is measured by ±BE — 1D.E, that is, by ±{BE—DE^, or by iJSD. the center being 154. Corollaries. 1. All angles inscribed in the same segment are equal. For, each is measured by one-half of the intercejited arc. '2. Ami angle inscribed in a semicircle is a right angle. For, it is measured by one-half of a semi-circumference ; that is, by a quadrant which is the measure of a right angle. 3. Any angle inscribed in a segment greater titan a semi- circle is an aeute angle. (?) 4. -1r angle inscribed in a segment less than a~ semicircle is an obtuse angle. (?) 5. Tivo inscribed angles are complementary if Hie half sum. of tlieir intercepted arcs is equal to a quadrant, and con- versely. (?) 6. Two inscribed angles are supplementary if the half sum 108 GEOMETRY.— BOOK II. of their intercepted arcs is equal to a semi-circumference, and conversely. (?) , 7. In the same circle or in equal circles, equal inscribed angles intercept equal arcs, and conversely. (?) 8. In unequal circles, equal inscribed angles intercept arcs of the same number of degrees, and conversely. (?) 155. Proposition XXVIII.— Theorem. An angle formed by a tangent and a chord drawn to tlie point of tangency is measured by one-lialf of the intercepted arc. Let CD be a tangent, and AB a chord drawn to the point of tan- gency. Draw the secant AF, and revolve it about A as a center till the point F coincides with A, when the secant becomes the tangent CD, and the arc BF, the arc BFA. But in all positions of AF, the inscribed angle BAF is measured by \BF; hence, BAD is measured by \BFA. 156. Proposition XXIX— Theorem. An angle formed by two intersecting chords is measured by one-half of the sum of the arcs intercepted by its sides and by the sides of its vertical angle. Let BC, DE, be two chords inter- secting at A. Drawing BE, EAC = EBC + BED. (?) But EBC is measured by \EC, and BED by \BD ; . ■ . EAC is measured by * EC + iBD; that is, by * (EC + BD). MEASUREMENT OF ANGLES. 109 157. Proposition XXX.— Theorem. The angle formed by two secant* intersecting without Hie cir- cumference is measured by one-lialf of Hie difference of the inter- cepted arcs. Let AB, AC, be two secants inter- secting at -1 without the circumference. Drawing BE, BEC = BAC + ABE. .-. BAC = BEC— ABE. But BEC is measured by iBC, and ABE by iDE; .-. BAC is measured by IBC — iDE; that is, by i(BC-DE). 158. Corollaries. 1. The angle formed by a tangent and a secant is measured, by one-half of Vie difference of the intercepted arcs. Let AB be a tangent, and AC a a secant. Drawing the secant AE between AB arid AC, and revolving it about A till it. coincides with AB, the points, E, F, will unite at B, EAC will become BAC, also EC will become BC, aud FD, BD; but iu all positions of AE. EAC is measured by i(_EC — FD) : (?) therefore, BAC is measured by i(BC-BD). 2. 27ie cut^fe formed by two tangents is measured by one-lmif of tlie difference of the intercepted arcs. (?) 159. Exercises. 1. If two circles are tangent externally or internally, and two straight lines be drawn through the point of tangency, 110 GEOMETRY.— BOOK II. the chords of the intercepted arcs are parallel (155), (154, 8), (44, 2). 2. If two circles have a common tangent, the chords which join the points of tangency and the points in which the straight line through the centers meets the circumfer- ences, on the same side of the centers, are parallel. 3. If two circles are tangent internally, and a . chord of the greater is tangent to the less, the line drawn from the point of tangency of the circles to the point of tangency of the chord, bisects the angle formed by the lines drawn from the point of tangency of the circles to the extremities of the chord (155), (44, 4), (135), (153). IV. CONSTRUCTIONS. 160. Proposition XXXI.— Problem. 1. To draw a line equal to a given line. 2. To cut off from the greater of two given lines a part equal to the less. 3. To produce a line tiM the part produced shall be equal to a given line. 1. Let AB be the given line. Place one point of the dividers on A and extend the other to B. Take the dividers thus extended, place one foot at any point C, about which, as a center, describe an arc with the other foot. Draw a radius CD from to any point D in the arc. Then, CD = AB, since each CONSTBUCMOS& 111 is equal to the distance from one foot of the dividers to the other (21, 2). 2. Let AB be the less of the given lines and CD the greater. With a radius equal to AB, found as above, describe about C as a center an arc intersecting CD in E. c ) n Then, CE = AB. ' l: 3. Let it be required to produce CD till DE shall be equal to AB. With a radius equal to AB, describe about D as a center an arc on the side of D opposite C. Pro- \ duce CD by the aid of a ruler till it c D I E meets the arc in E. Then, DE = AB. 161. Proposition XXXII.— Problem. To find the sum or difference of two given lines. Let AB and CD be the given lines. Draw an indefinite line EF. With a radius equal _ to AB, describe about E as a center an arc intersecting EF E — in G ; and about G as a center, with a radius equal to CD, describe two arcs, one inter- secting EF in H, the other in I. From the diagram we have EH=EG + GH, and EI=EG— GI. .-. EH=AB + CD, and EI= AB — CD. 162. Proposition XXXIIL— Problem. To erect a perpendicular to a straight line at a given point in Vie line. 112 OEOMETRY.—BOOK II. With the dividers ^ Let AB be the line and C the point, lay off from C the equal distances, CD, GE. Opening the dividers a little wider, place one point at D, and with the other describe an arc. Then place one point at E, and with the other describe an arc intersecting the first at F. Draw CF, and it will be perpendicular to AB at C (70, 3). 163. Proposition XXXIV— Problem. To let fall a perpendicular upon a line from a point without the line. Let AB be the line and C the point. Place one point of the di- viders at C, extend the other so that when revolved about C it shall inter- sect AB in two points, D, E. With D and E as centers, and a radius greater than the half of DE, describe two arcs inter- secting at F. Draw CF, and it will be the perpendicular required. (?) 164. Proposition XXXV —Problem. To bisect a given finite straight line. Let AB be the straight line. With A and B as centers, and a radius greater than the half of AB, describe arcs intersecting in D and E, and % draw DE. Then, since each of the points, D, E, is equally distant from A and B, DE is per- pendicular to AB at its middle point, and hence, bisects AB. COXSTR VCTIOXS. 1 1 3 165. Proposition XXXVI.— Problem. To construct the supplement of a given angle. Let BAC be the given angle. Produce d CA to D; then, (31), DAB + BAC=- two right angles; .-. DAB is the supplement of BAC. 166. Proposition XXXVIL— Problem. To construct the complement of a given angle. Let BAC be the given angle. Erect j> AD perpendicular to AC (162). Then, DAB + BAC =- a right anale. DAB is the complement of JL4f. 167. Proposition XXX VIII.— Problem. Jb construct an angle equal to a given angle. Let A be the given angle. With the vertex A as a center, and any radius AB, describe an arc intersecting the sides of the angle in the points, B, C, and draw the straight line BC. Draw an indefinite straight line DE. With D as a center, and a radius DE equal to AB, describe the indefinite arc EF. With E as a center, and a radius equal to BC, describe ail arc intersecting the arc EF in F, and draw the straight line DF. Then, AB = DE, AC=DF, BC=EF: .-. A = D (75\ 114 GEOMETRY.— BOOK II. 168. Proposition XXXIX— Problem. To draw through a given point a line parallel to a given line. Let C be the given point and AB the given line. Draw a straight line from C to any point of AB, as D. Construct the angle DGE equal to the angle CDB (167). Then, CE is parallel to AB (44, 2). 169. Proposition XL. — Problem. Given two angles of a triangle, to find the third. Let A and-U be the a given angles. Draw the / indefinite straight line CD, and from any point of this line, as E, draw EF, making the angle DEF = A ; also the line EG, making the angle FEG = B (167). Now, DEF + FEG + GEO = two right angles (32, 2). Hence, GEO is the supplement of DEF + FEG (27, 4). But, DEF = A, FEG = B, GEO is the supplement of A -4- B. Also, the third angle of the triangle is the supplement of A -4- B. .• . GEO = the third angle of the triangle. 170. Proposition XLI. — Problem. To bisect a given angle. Let BAG be the given angle. With A as a center, and a radius less than either side, describe the arc BC; then, i '< i XSTR UCTIONS. 1 1 f> with B and (' as centers, and a radius greater than one- half the distance from B to 0, describe arcs intersecting in D, and draw AD. Then, AD bisects iLlC; for, drawing BD and CD, the triangles, ABD, ACD, are mutually equilateral, and bence mutually equiangular. .-. the angle BAD = the angle CUT); (?) .-. .1/) Insects 7MC 171. Proposition XLIL— Problem. Ghrn two sides of a triangle and tlieir included, angle, to construct the triangle. Let b, c, be the given a sides, and A their in- c eluded angle. Construct. the angle EAF = A. On AF\nyoffAC=h, on AE lay off JLB = e, and draw 5C. Then, J3.4C is the triangle required (55 s ). 172. Proposition XLIII. — Problem. Giivn one tide and tiro angles of a triangle, to construct the triangle. 1. When the - l angles are both / \ adjacent to the -C — -c r. - - <• given side: Let a be the given side, B, C, the adjacent angles. Construct the straight line BC = a, the angle CBA = B. and BCA = C (160, 1\ (167\ Then, ABC is the re- quired triangle. (?) 116 GEOMETRY.— BOOK II. 2. When one of the given angle's is opposite the given side : First, find the third angle (169), then proceed as above. 173. Proposition XLIV — Problem. Given the three sides of a triangle, to construct tJw triangle. Let a, b, e, be the sides. n Draw.BC=a. With C as a center, and a radius equal to b, describe an arc, and with B as a center, and a radius equal to c, describe an arc intersecting the first arc in A. Then, ABC is the required triangle (75). 174. Proposition XLV. — Problem. Given two sides of a triangle and an angle opposite one of them, to construct the triangle. Let a and b be £ the given sides, A / y r / the angle opposite a^— — - a* a. Construct the angle BAC = A, AC = b, and p the perpendicular from C to AB. 1. ft < b. In this case, A is acute ; for, if A were right or obtuse, it would be the greatest angle of the triangle (53, 3, 4), and a would be the greatest side (64) ; but this is contrary to the hypothesis that a < 6. 1st. If a > p, there are two solutions. For, with C as a center and a as a radius, an arc can be described inter- CONSTRUCTIOXS. 117 secting the side of the augle A opposite C iu two points -B and B'; and drawing CB and CB', either triangle ABC, or ^4J3'C having two sides and an angle opposite one of them equal to those given, satisfies the conditions of the problem. 2d. If « = p, there is one solution. For, then B and B' unite at D, and the two triangles ABC and AB'C coincide with the right triangle ADC. 3d. If <( < p, there is no solution. For, a can not reach AD, the construction is impossible, and the triangle im- aginary. 2. a = b. In this case, also, A is acute ; for, if ^1 were right or obtuse, since the angles opposite the equal sides are equal (59), the triangle would have two right angles or two ob- tuse angles, which is impossible. (?) The triangle ABC is real and isosceles, but the triangle AB'C vanishes into the line AC, since B' would coincide with A. 3. a > b. In this case, ^1 may be right or oblique. B and B' fall on opposite sides of A. 1st. If A is right, the triangles, ABC, AB'C are \ equal, and either satisfies n^^ZZT^Z^^' 11 the conditions. 2d. If A is oblique, ABC will satisfy the conditions, but AB'C will not; for, n'~;r" '< b^: if -4 is acute, B'AC is ob- tuse, and if A is obtuse, B'AC U acute. // 118 GEOMETRY.— BOOK II. 175. Proposition XLYL— Problem. Through a given point to draw a tangent to a circumference. 1. If the given point is on the circumference : Let P be the given point, and C the center of the given circle. Draw the radius CP, and erect the per- pendicular PA ; then PA will be • tangent (133). 2. If the given point is without the circumference: Let P be the given point, and C the center of the given circle. Draw PC and bisect it at C. With C" as a center, and CC as a radius, describe a circumference intersecting the given circumference in A, A'. Draw PA, PA', both of which will be tangent to the circumference of the given circle. For, drawing the radii, CA, CA', the angles, PAC, PA'C, are right angles (154, 2) ; hence, PA, PA', are tangents. (?) 3. If the given point is within the circumference, through it no tangent can be drawn. (?) 176. Corollaries. 1. The two tangents drawn from a point without a circle to the circumference are equal, and make equal angles with the line drawn from that point to tlie center. (?) (72, 1). 2. A line drawn through any point of a tangent, maldng with Hie line drawn from that point to the center an angle equal to that made by the tangent, is also a tangent. (?) (176, 1). COXSTRUCTIUXS. 119 3. The line drawn- from the center to the point of intersection of two tangent* bkeett the angle formed by the radii drawn to the points of tangency. (?) 177. Proposition XLYIL— Problem. To draw a common tangent to two given circles. 1. When the common tangent is exterior: Let and t" be the centers of the given circles, the radius of the first being the greater. With C as a center, and a radius CD equal to the difference of the radii of the given cir- cles, describe a circumference. From C draw CD tangent to this circumference (175, 2"). Draw CD and produce it to A in the given circumference, also, C'B parallel to CA. Draw AB and it will be a common exterior tangent. For, CD = CA — C'B, also, CD — CA — DA. . • . C B =-- CA — CD. and DA =- CA — CD. C'B ^ DA; .-. ADC'B is a parallelogram (S6\ But D is a right angle (133") ; " . --1 and B are right angles i_Sl). Hence, AB is a tangent to both circum- ferences. (T) Produce -li? and CC till they meet in P; from P draw PA', making the angle CPA' equal to the angle CP.-1 ; A'B'P is another common exterior tansrent < 170. 2). 120 GEOMETRY— BOOK II. 2. When the common tangent is interior: With C as a center, and a radius equal to the sum of the radii, describe a D circumference. ■ / From C" draw the tangent CD. Draw CD intersecting the given circumference in some point A. Draw C'B parallel to CD and in the opposite direction. Draw AB and it will be a common interior tangent. For, CD = CA + C'B, also, CD = CA + AD. .-. C'B = CD— CA, and AD = CD — CA. .'. C'B = AD; .-. ABC'D is a parallelogram (86). But D is a right angle; (?) .". A and B are right angles ; AB is a common interior tangent. (?) 178. Proposition XL VIII.— Problem. On a given straight line to describe a segment which sliall contain a given angle. Let AB be the given line, and ABD an angle equal to the given angle. Draw BC perpendicular to BD, and EC perpendicular to AB at its middle point, intersecting BC in C. About G as a center, with CB as a radius, describe a circumference ; then, ABF is the segment required. For, BD is a tangent (133), and the angle ABD is CONSTRUCTIONS. 121 measured by \AmB (155). But any angle AFB, inscribed in the segment ABF, is also measured by \AmB, and there- fore equal to the given angle. Any angle inscribed in the segment ABm is equal to ABG, and is therefore the sup- plement of ABD. 179. Corollaries. 1. The locus of the vertices of all the angles ivhose sides pass through two fixed points, and whose magnitude is equal to that of the angle formed by the common chord of tioo equal circum- ferences passing through these points with the interior portion of tlve tangent to either circumference, at either of these points, is tlie exterior ares of tJie circumferences. Let C, C, be the centers of equal circumferences passing through the given fixed points, A, B. The common chord AB of the equal circles subtends equal arcs (126). .-. AEB = AE'B =ABD=ABD' (153), (155). Now, any angle whose sides pass through the points, A, B, respectively, and whose vertex is within either arc, AEB, AE'B, is within some angle whose vertex is in one of these arcs, and is therefore greater than that angle (54). For like reason, any angle whose sides pass through A, B, and whose vertex is without the arcs, AEB, AE'B, is less than the angle AEB. Hence, AEBE'A is the specified locus. 2. The locus of the vertices of all the angles whose sides pass through two fixed points, and whose magnitude is equal to that of the angle formed by the common chord of two equal circum- s. G— 11. 122 GEOMETRY.— BOOK II. ferences passing through these points ivith the exterior portion of the tangent to either circumference, at eitiier of these points, is the interior arcs of the circumferences. (?) 3. The locus of the vertices of all the right angles lohose sides pass tJvrough two fixed points, is the circumference whose diam- eter is the straight line terminated by the fixed points. (?) 180. Exercises. 1. Construct a right triangle; given, first, the hypotenuse and a side ; second, the hypotenuse and an acute angle. 2. Trisect a right angle (53, 10). 3. From two points on the same side or on opposite sides of a line, draw two equal lines which shall meet in the given line. 4. From two points on the same side of a line, draw two lines which shall meet in the given line and make with it equal angles (76, 5). 5. Construct a square, given the difference between its diagonal and side. 6. Draw a tangent to a circumference ; first, parallel to a given line; second, perpendicular to a given line. 7. Determine a point in the prolongation of a diameter such that a tangent drawn from it to the circumference shall be equal to a given line. 8. With a given radius describe a circumference passing through a given point and tangent to a given straight line (50, 1), (117, 5). INSCRIBED AX1) CIRCUMSCRIBED POLYGONS. 123 i). Describe a circumference which shall pass through two 4 given points and have its center in a given straight line. 10. With a given radius describe a circumference tangent to a given circumference and to a given straight line. 11. Describe a circumference tangent to a given straight line at a given point, such that the tangents drawn to it from two given points in the straight line may be parallel. 12. With a given radius describe a circumference tangent to a given straight line, such that the tangents drawn to it from two given points in the straight line may be parallel. 13. To construct a triangle, given the base, the vertical angle, and the point of intersection of the base with the perpendicular to the base from the vertex of the opposite angle (178). 14. To construct a triangle, given the base, the altitude, and the vertical angle. 15. To construct a triangle, given the base, the medial line to the base, and the vertical angle. 16. Through a given point draw a straight line so that the part intercepted by a given circumference shall be equal to a given straight line. V. INSCRIBED AND CIRCUMSCRIBED POLYGONS. 181. Definitions. 1. An inscribed polygon is a polygon whose vertices are all in the circumference. In this ease, the circle is circum- scribed about the polygon. 124 GEOMETRY— BOOK II. 2. A circumscribed polygon is a polygon whose sides are all tangent to the circumference. In this case, the circle is inscribed in the polygon. 3. An inscriptible polygon is a polygon which can be inscribed in a circle. 4. A circumscriptible polygon is a polygon which can be circumscribed about a circle. 182. Proposition XLIX. — Theorem. Any triangle is inscriptible. For, a circumference can be made to pass through the vertices (130). 183. Proposition L.— Theorem. Any triangle is circumscriptible. Let ABD be a triangle. Bisect any two of its angles, as A and B, by straight lines meeting in C. From C let fall the perpendiculars, CE, CF, CO, on the three sides of the tri- angle. Then, these perpendiculars are equal. For, the right triangles, ACE,. ACQ, are equal (72, 2); .-. CE=CG (56, 2). In like manner, it is proved that CE = CF. Hence, the circumference having C for a center and CE for a radius will pass through the points E, F, O. The sides of the triangle are tangents, since each is per- pendicular to a radius at its extremity (133). Hence, the triangle is circumscriptible (181, 4, 2). INSCRIBED AND CIRCUMSCRIBED POLYGONS. 125 184. Proposition LL— Theorem. If a quadrilateral is iuseriptible, its ojiposite angles are sup- plementary; and conversely, if the opposite angles of a quadri- lateral are supplementary, tlie quadrilateral is iuseriptible. Lot the quadrilateral ABCD be in- scribed in a circle. Then, its opposite angles are supplementary. For, the angle A is measured by one-half the arc BCD, and the angle G by one-half the arc DAB (153) ; hence, A -f- C is measured by one-half the sum of the arcs BCD and DAB; that is, by a semi-circumfer- ence ; . ■ . .1 and C are supplementary (161, 6) ; . • . ABC and ADC are supplementary (79). Conversely, if A and C are supplementary, ABCD is iu- seriptible. Describe a circumference through the points, B, A, D, and draw the chord BD. Now, any angle inscribed in the segment BmD is the supplement of the angle A. (?) If the vertex C were within the arc BmD, the angle C would be greater than the supplement of the angle A (JO) ; if the vertex C were without the arc BmD, the angle C would be less than the supplement of the angle .1 ; (?) but both of these results are contrary to the hypothesis ; hence, the vertex C can be neither within nor without the are, and must, therefore, be on the arc BCD ; . • . ABCD is inscriptible. 185. Proposition LII. — Theorem. If a quadrilateral is circumtcriptible, the sum of two opposite sides is eptal to the sum of the other two sides; and conversely, 126 GEOMETRY.— BOOK II. if tiie sum of two opposite sides of a quadrilateral is equal to the sum of the other two sides, Hie quadrilateral is cireum- scriptible. Let the circumscriptible quadrilat- ^ i A eral ABDE be circumscribed about a circle, F, G, H, I, being the points of tangency. Now, AF = AT, BF = BG, DH = DG, EH = EI (176, 1). .-. AF + BF+DH+ EH=AI+BG + DG + EI. .-. AB + DE=AE + BD. Conversely, if AB -\- DE = AE + BD, then, ABDE is circumscriptible. There must be two consecutive angles, for example, A and E, whose sum does not exceed two right angles, since the sum of the four angles is equal to four right angles. Let AC and EC be the bisectors of the angles A and E. The sum of the angles, GAE, CEA, is less than two right angles; (?) therefore, AC and EC intersect in some point C (47). The perpendiculars let fall from C to the three sides, AB, AE, ED, are equal (105). If, therefore, with C as a center, and one of these perpendiculars as a radius, a circumference be described, it will be .tangent to the sides AB, AE, ED. It is now to be proved that this circumference is tangent to the fourth side BD. If BD is not tangent, let BK be tangent ; then, BK will intersect ED, or ED produced in some point as K, since the sum of the angles, A, E, does not exceed two right angles, and since BK, if tangent, must fall between AB IXSCRIBED AXn CIRCUMSCRIBED POLYGOSS. 127 and ED. If BK is tangent, ABKE is a circumscribed quadrilateral ; then, AB + KE = £ Jv + .IE, but .IjB + Z>£ = BD + AE. . ■ . KE — DE = BK — BD, or KD = BK— BD. That is, one side of a triangle is equal to the difference of the other sides, which is impossible (66) ; hence, the sup- position that BD is not a tangent, which led to this im- possibility, is false (15, 4). Therefore, BD is tangent, and ABDE is circumscriptible. 186. Exercises. 1. Circumscribe a circle about a given rectangle. 2. Inscribe a circle in a given rhombus. 3. In a given circle inscribe a triangle whose angles are respectively equal to the angles of a given triangle. 4. About a given circle circumscribe a triangle whose angles are respectively equal to the angles of a given tri- angle. 5. Circumscribe a circle about a triangle, and prove by (153) that the sum of the three angles is equal to two right angles. 6. Prove by (V2T), (153), that the greatest angle of a triangle is opposite the greatest side ; that, in an isosceles triangle, the angles opposite the equal sides are equal; and that an equilateral triangle is equiangular. 7. In a right triangle, the sum of the hypotenuse and diameter of the inscribed circle is equal to the sum of the other sides. 8. The central angles subtended by two opposite sides of a circumscriptible quadrilateral are supplementary. 128 GEOMETRY.— BOOK II. VI. SYMMETRY.— SUPPLEMENTARY. 187. Definitions. 1. Two objects are symmetrical when every point of either has a corresponding point in the other on the opposite side of a point, line, or plane, taken as an object of reference, and at an equal distance from it. 2. A point taken as the object of reference is called the center of symmetry. 3. A line taken as the object of reference is called the axis of symmetry. 4. A plane taken as the object of reference is called the plane of symmetry. 5. Two points are symmetrical with respect to a center when the center bisects the straight line terminated by these points. Thus, P, P', _-■— -''"p' are symmetrical with respect to C as a p center, if C bisects PP'. 6. The distance of either of two symmetrical points from the center of symmetry is called the radius of symmetry. Thus, either CP or CP' is the radius of symmetry. 7. Two figures are symmetrical with respect to a center when every point of either has its symmetrical point in the other ] with respect to that center. Thus, ^ ■--/,» the lines AB, A'B', are symmet- rical with respect to the center C, if every point of either has its symmetrical point in the other, with respect to C as a center. SYMMETR Y.—SUPPLEMENTAR Y. 129 Also, the polygons, ABDE, A'B'D'E', are symmetrical with respect to G as a center, if A. every point in the perimeter of either has its symmetrical point in s{ the perimeter of the other, with respect to C as a center. r " — ' b a'' ' 8. Two points are symmetrical with respect to an axis when the axis bisects at right angles the straight line terminated by these points. Thus, P, P', are symmet- rical with respect to the axis XY, x ~ if XY bisects PP' at right angles. , ; 9. Two figures are symmetrical with respect to an axis when every point of either has its symmetrical point in the other, with respect to that axis. Thus, the lines, AB, A'B', are symmetrical with re- spect to XY, if every point of either has its symmetrical point in the other, with respect to XY as an axis. Also, the triangles, ABC, A'B'C, are symmetrical with respect to the axis XY, if every point of the pe- rimeter of either has its symmetrical point in the perimeter of the other, with respect to XY as an axis.. 10. Two points are symmetrical with respect to a when the plane bisects at right angles the straight line terminated by these points. Thus, P, P', are f symmetrical with respect to XY, if x — XY bisects PP' at right angles. plane 130 GEOMETRY.— BOOK II. ah'- a' 11. Two figures are symmetrical with respect to a plane, when every point of either has its symmetrical point in the other with respect to that plane. Thus, AB, A'B', are symmetrical P""""- 1 - with respect to XY, if every point j'C 1 T of either has its symmetrical point x l ! j / in the other with respect to XY as ,1^^b' a plane of symmetry. Also, the triangles, ABC, A'B'C, are symmetrical with respect to the plane XY, if every point in the perimeter of the one has its symmet- rical point in the perimeter of the other with respect to XY as a plane of symmetry. 12. Symmetrical points and lines in two symmetrical figures are called Jwmobgous. 13. A single figure is symmetrical, when it has a center, axis, or plane, with respect to which every point of its pe- rimeter or surface has its symmetrical point in the perimeter or surface. Thus, ABDEB'D' is symmetrical with respect to the center C, if C bisects every straight line through it terminated by the perimeter. Also, ABCDC'B' is symmetrical with respect to the axis XY, if XY so divides it that the portions ABCD, AB'C'D, are symmetrical with re- spect to XY as the axis of symmetry. 14. The straight line drawn through the center of a single symmetrical figure, and terminated by the perimeter if the *\>. D> B< SYMMETRY.— SUPPLEMENTARY. 131 figure is plane, or by the surface if the figure is solid, is called a diameter. 15. Symmetry is right or oblique according as the axis or plane of symmetry bisects, at right or oblique angles, the lilies joining homologous points. 188. Proposition im.— Theorem. The symmetrical of a straight line with respect to a. center is an equal and parallel straight line. Let AB be a straight line, and C /\ /i B ' the center of symmetry. Draw AC, pi- — -X;- — -L< and produce it till CA' is equal to AC s l~' V , Then, A' is the symmetrical of A. Likewise, find jB' the symmetrical of B, and draw A'B'. The triangles, ACS, A'CB', are equal (55) ; .-. A'B = AB, and the angle A' = the angle A ; hence, A'B' is parallel to AB (44, 2). From P, any point of AB, draw PC, and produce it to P' in A'B'. Then, the triangles, ACP, A'CP', are equal (57) ; .-. CP'=CP; therefore, P' is the symmetrical of P; but P is any point of AB ; hence, every point of AB has its symmetrical in A'B; therefore, A'B is the symmetrical of AB. 189. Corollaries. 1. Tim lines symmetrical with respect to a center lie in oppo- site directions from their symmetrical extremities as origins. 2. Two equal parallel lines are symmetrical loitli respect to a center. 3. If the extremities of one line are respectively the symmet- rieals of the extremities of another line, with respect to Hie same center, the two lines are symmetrical witJi respect to tiiat center. 132 GEOMETRY.— BOOK II. 190. Exercises. 1. The symmetrical of a polygon with respect to a ^center is an equal polygon. 2. A point can be made to coincide with its symmetrical by revolving its radius of symmetry in the same plane through two right angles. 3. How can a figure be made to coincide with its sym- metrical with respect to a center? 4. The symmetrical of a straight line with respect to an axis is an equal straight line. 5. If a straight line is perpendicular or parallel to the axis of symmetry, its symmetrical is perpendicular or par- allel to the axis. 6. If a straight line is oblique to the axis of symmetry, its symmetrical is oblique to the axis, and the two sym- metricals intersect the axis at the same point and make equal angles with it. 7. If two straight lines intersect, their symmetricals with respect to an axis intersect, and the points of intersection are symmetrical. 8. Construct the symmetrical of a polygon with respect to an axis, and prove that it is equal to the given polygon. 9. If a figure is symmetrical with respect to two axes at right angles, it is symmetrical with respect to their in- tersection as a center. BOOK III. I. AREA AND EQUIVALENCY. 191. Definitions. 1. A superficial unit is an assumed unit of measure for surfaces (144, 4). It is usually the square whose side is a linear unit. 2. The area of a surface is its numerical value (144, 8). 3. Equivalent surfaces are those whose areas are equal. 4. The projection of a point upon a line is the foot of the perpendicular drawn from the point to the line. If the point is on the line, it is its own projection. 5. The projection of a finite straight line upon a given line is the distance between the projections of its extremities. Thus, a is the projection of A, B is its own projection, ac is the projection of AC, aB of AB. 192. Proposition I.— Theorem. Rectangles having equal altitudes an- propoHional to thci bases. (133) 134 GEOMETRY.— BOOK III. B Let b, V, denote the bases of two rectangles, R, R', having equal altitudes. 1. Suppose the bases are commensurable, and that the common unit of measure is contained, for example, 5 times in b and 3 times in V. Then, b : V :: 5 : 3 (146). Now, b can be divided into 5 parts and b' into 3 parts, each equal to the common unit. If perpendiculars be erected to the bases at the several points of division, R will be divided into 5 rectangles and R' into 3 rectangles, all equal, since they have equal bases and equal altitudes (83, 3). .-. R : R' :: 5 : 3. (?) .-. R : R :: b : U (Alg., 317, 9). 2. Suppose the bases are incommensurable, and that b' is divided into n equal parts without a remainder. Since b, V, are incom- mensurable, b will contain ' a certain number in of these parts, with a remainder less than one part. a' Then, b_ V - within - (146). n n The perpendiculars to the bases at the points of division divide R! into n rectangles without a remainder, and R into AREA AND EQUIVALENCY. 135 m rectangles with a remainder. These resulting rectangles are all equal, except the remainder, which is less than one of these rectangles. B m „ . 1 — withm — n n ■'■ J? ~ z TT withm ~ (146). .-. ~ = L or B : B' :: 6 : V (147). if o 193. Corollary. Bectangles liaving equal bases are proportional to their alti- tudes. For, the bases may be taken as altitudes and the altitudes 194. Scholiums. 1. The term rectangles, in the proposition and corollary, is the usual abbreviated expression for the areas of rectangles. 2. The rectangle of two lines is the rectangle having one of these lines for its base and the other for its altitude. 3. By the product of two lines is to be understood the product of their numerical values. If the lines are equal, their rectangle is a square, and their product a square number. 195. Proposition II.— Theorem. Any two rectangles are proportional to the products of their bases by their altitudes. 136 GEOMETRY.— BOOK III. Let R, R', denote two rectangles, b, b', their bases, and a, a', their altitudes. ji" Construct a third rectangle, R", whose altitude is a, the altitude of R, and whose base is b', the base of R'. Then, R : R" : : b : b' (192). Also, R" : R' : : a : a' (193). Taking the product of the corresponding terms of these proportions, and omitting R", the common factor of the first couplet, we shall have R : R' : : ah : a'b' (Alg., 317, 17, 13). 196. Proposition III. — Theorem. Tlie area of a rectangle is equal to t/ie product of its base by its altitude. 1. When the base and altitude are commensurable: Let R denote a rectangle, b the numerical value of its base, and a of its altitude, referred to a com- mon linear unit of measure. The base can be divided into 6 parts and the altitude into a parts, each equal to the linear unit. Perpendiculars to the base and altitude at the points of "i~r>T7" — — i — i — i — AREA AND EQUIVALENCY. 137 division divide the rectangle into superficial units, of which there are a rows of b units each. Since there are b superficial units in one row, in a rows there are a times b, or ab superficial units. .-. R = ab. 2. When the base and altitude are incommensurable: Let R, R', denote two rectangles, b, b', their bases, a, a', their altitudes, a, b, being commensurable, a', b', in- commensurable. Then, R : R' : : ab : a'b' (195). But, R = ab, .-. R' = a'b' (Alg., 317, 2). 197. Proposition IV.— Theorem. The area of a parallelogram is equal to tlie product of its base by its altitude. Let P denote the area of the F r f parallelogram ABCD, b its base, and a its altitude. ji¥- Erect the perpendiculars AF, BE, to the base. Produce CD to F, forming the rectangle ABEF, which denote by R. The right triangles, AFD, BEC, are equal (StJ). (72, 1). Taking these triangles, in succession, from the whole figure, we have the parallelogram ABCD equivalent to the rect- angle ABEF, .-. P=R; but R = ab, .-. P = ab. 19S. Corollaries. 1. Ant/ fiw parallelograms are proportional to the products of their bates by their altitudes. R. O. — 12. 138 GEOMETRY.— BOOK III. For, let P, P', denote two parallelograms, b, b', their bases, a, a', their altitudes. Then, P = ab, and P' = a'b'. P : P' ,:: ab : a'b'. 2. Parallelograms liaving equal altitudes are proportional to their bases. For, P : P' :: ab : a'b' (198, 1). If a' = a, P : P' : : b : b' (Ai/J., 317, 13). 3. Parallelograms having equal bases are proportional to tlieir altitudes. For, P : P' :: ab : a'b'. If V = b, P : P' : : a : a'. 4. Parallelograms liaving equal bases and equal altitudes are equivalent. For, P : P' :: a6 : a'b'. If & = 6'anda = a', ab = a'b' ; .-. P = P'. 5. Parallelograms having bases reciprocally proportional to their altitudes are equivalent. For, P : P' : : a6 : a'b'. If b : V : : a' : a. Then, ab - = a'b' ; .-. P=P', 199. Proposition Y. — Theorem. The area of a triangle is equal to one-half the product of iU base by its altitude. AREA AND EQUIVALENCY. 139 Let T denote the area of the triangle ABC, b the nu- merical measure of its base, and a of its altitude. Draw BD parallel B to AC, and CD parallel to AB, forming the parallelogram ABDC, / « and denote the area of this par- -; l allelogram by P. Now, the triangle ABC is one-half the parallelogram ABDC (83, 1); that is, T=\P. But, P = ab; .-. T = M>. 200. Corollaries. 1. A triangle is equivalent to one-half of any parallelogram having an equal- base and an equal altitude. (?) 2. Any two triangles are proportional to the products of Uieir bases by their altitudes. (?) 3. Triangles having equal altitudes are proportional to their bases. (?) 4. Triangles having equal bases are proportional to their altitudes. (?) 5. Triangles having equal bases and equal altitudes are equiv- alent. (?) 6. Triangles having bases inversely proportional to tlieir alti- tudes are equivalent. (?) 201. Proposition YL— Theorem. The area of any eireumscriptible polygon is equal to one-half the product of its perimeter by the radius of the inscribed circle. 140 GEOMETRY.— BOOK III. Let P denote the area of the polygon ABGD ; a, b, e, d, respectively, the sides ; p, the perimeter ; and r, the radius of the inscribed circle. Lines drawn from the center of the inscribed circle to the vertices divide the polygon into triangles having r for their common altitude and a, b, c, d, respect- ively, for their bases. The area of each triangle is equal to one-half the product of its base by its altitude, and the area of the polygon is the sum of the areas of the triangles. P = \ar -\- %br -|- \er -(- \dr, or P — \(a -\-b -\- c -\- d) r = \pr. IS :.x \ / a \ 202. Proposition TIL— Theorem. The area of a trapezoid is equal to one-half tlie product of its altitude by the sum, of its bases. Let T denote the area of the trap- ezoid ABCD, b the numerical measure of its lower base, b' of its upper base, and a of its altitude. The diagonal BD divides the trap- ezoid into two triangles, ABD, BDC, denoted by 7", T". ABD has b for its base and a for its altitude, and BDC has b' for its base and a for its altitude, since a is equal to the perpendicular distance from D to the upper base, or to the upper base produced. Now, T =■ T + T". But, T'=hab, and T"=\ab'. T = iab -f \ah' = ki(b + b'). AREA AND EQUIVALENCY. 203. Corollary. 141 The area of a trapezoid is equal to the product of its altitude ami tlie Hue joining the middle points of its non-parallel sides. For, let I be the line joining the middle points of AB and CD. Then, I = \ (b + b") (104, 3). But, T = $a(b + l')=aX J(& + &')> . • . T = «/. 204. Scholiums. 1. The area of a polygon can be found thus : Draw diagonals from one vertex to the opposite vertices, thus dividing the polygon into triangles; take the diago- nals for bases ; from the vertices draw perpendiculars to the diagonals, for the altitudes; measure the bases and alti- tudes ; then, P — iab + $eb -\- ied. 2. The area of a polygon can also be found thus Draw one diagonal, and perpendicu- lars to this diagonal from the vertices, thus dividing the polygon into triangles and trapezoids; measure the bases and altitudes of the triangles and trapezoids ; then, P = iab + Ac(6 + d) + hie -f Ifg. 205. Proposition Till.— Theorem. The square of tiie sum of two lines is equivalent to the sum of their squares plus tiriee their rectangle. 142 GEOMETRY.— BOOK III. Let each side of the square ABCB be the sum of the lines a and b. The square has a -f- 6 for its side and (a -}- ft) 2 for its area. By joining the common extremities of a and b in the opposite sides of the square by straight lines, the whole square will be divided into two squares and two rectangles. (?) One square has a for its side and a 2 for its area, and the other square has b for its side and b 2 for its area. (?) Each rectangle has a for one side, b for the other, and ah for its area. (?) Then, 2o& = the area of the two rect- angles. .-. (a + b) 2 — a 2 + b 2 + lab. 206. Corollaries. 1. The square of twice a line is four times ike square of tlie line. For, if b = a, the formula of the last article becomes (2a) 2 = a 2 + a 2 + 2a 2 = 4a 2 . This is also evident from the diagram ; for, if b = a, b 2 = a 2 ; each of the rectangles ab becomes a 2 ; and, since the side is 2a, (2a) 2 = a 2 + a 2 + a 2 + a 2 = 4a 2 . 2. The square of three times a line is nine times tlie square of the line. ' (?) 3. The square of n times a line ism 2 times tlie square of the line. (?) AREA AND EQUIVALENCY. 143 207. Proposition IX.— Theorem. Tlie square of the difference of two lines is equivalent to the sum of tlieir squares minus twice their rectangle. Let each side of the square AD be a, one side of each of the rectangles, BD, CF, be a, and the other 6. Then, each side of AC is a — b, and each side of EF is b. Then, AC = (a — i) 2 , AD = a 2 , EF=b*-, BD = ab, CF *= ab. But, .40 = JD + EF — BD — CF. (a — b)° = a" + b" — 2ab. * """25 C ,1 ~iB 208. Proposition X. — Theorem. The rectangle of the sum and difference of two lines is equiv- alent to the difference of their squares. Let each side of the square AB be a, each side of the square CB be 6, the base "of the rectangle AD be a -\- b, the altitude, a — b. Then, CE = FD, since each has a — b for one side and b for the other. Now, AD = AI + FD = AI + CE = AB — OB. But, AD = (a + b) (a — 6), AB = a 2 , f £ = 6 2 . .-. (a + 6) (a— 6) = a 2 —6*. 209. Proposition XI.— Theorem. The square of the hypotenuse of a right triangle is equivalent to Hie sum of the squares of the other sides. 144 GEOMETRY— BOOK III. Let the triangle ABC be right angled at B. Then, AD, the square of AC, is equivalent to AG + CH, the sum of the squares of AB and BC. From J5 draw BK perpendicular to AC, and produce it to L in ED. Then, isTL is parallel to AE (41), and iL is a rectangle. Draw BE and .FC. Since J.BG and ABC are right angles, GBC is a straight line; and for a similar reason ABH is a straight line (35). The triangles, BAE, FAC, are equal ; for, AB is equal to AF, since they are sides of the same square; AE is equal to AC for a similar reason; and the angles BAE and FAC are equal, since each is the sum of the angle BAG and a right angle (55). The rectangle AL is equivalent to twice the triangle BAE; for, it has the same base AE, and, since the vertex B is in LK produced, it has an equal altitude (200, 1). The square AG is equivalent to twice the triangle FAC; for, it has the same base AF, and, since the vertex C is in GB produced, it has an equal altitude. But the triangle BAE is equal to the triangle FAC; hence, the rectangle AL is equivalent to the square AG. In like manner, it can be proved that the rectangle KD is equivalent to the square CH. But the sum of the rectangles AL and KD is the square AD ; hence, the square AD is equivalent to the sum of the squares AG and CH; that is, AC 2 —■ AB* 4- BC 2 . 210. Corollaries. 1. The square of either side adjacent to the right angle is AREA AND EQUIVALENCY. -145 equivalent to the square of the hypotenuse minus the square of Hie oilier side. For, AL<* = AB 2 + BC 2 , .-. \& =*?-*?. Lbc 2 = ac 2 — ab\ 2. The square of Vie hypotenuse h to the square of either of the otlier sides as the hypotenuse is to the projection of tliat side on ilie hypotenuse. For, AD : AL ::,AC : AK (192). Also, AD : KD :: AG : KG But, AD = AC 2 , AL = AB 2 , KD-^BC 2 . AG 2 : AB 2 :: AC : AK. Also, AG 2 : BC 2 :: AC : KG 3. Tlie squares of the sides adjacent to tJie right angle are proportional to Hieir projections on tlie hypotenuse. For, AL : KD : : AK : KC (192). But, AL = AB 2 , KD = BC 2 . AB 2 : BC 2 :: AK : KG 4. Tlie square of a diagonal of a square is double tlie square. Let s be one side of the square, and d the diagonal. Then, d"- = s 2 + s- = 2s 2 (209). 5. Tlie diagonal of a square is to the side as the square root of two is to one. For, d* = 2A .-. d»: * a :: 2 : 1. (?) Hence, d : s :: \ 2 : 1. (?) S. G.-13. 146. GEOMETRY.— BOOK III. 211. Proposition XII.— Theorem. In any triangle, Hie square of a side opposite an acute angle is equivalent to the sum of the squares of tlw other sides, minus twice the, rectangle of one of these sides and tlie projection of the oilier on that side. Let the triangle ABC be acute angled at A ; a, b, c, respectively, the sides opposite the angles A, B, C; m, n, respect- ively, the projections of a and c on b, or on b produced; p, the perpendicular from B to, b, or to b produced. According as the projection of B falls on b, or on b pro- duced, we have m = b — n, or m — n ■ — -b. In either case, by squaring, we have m 2 = b 2 + » 2 — 25n (207). Adding p 2 to both members, we have m 2 + p % = b 2 -\- n 2 -\- p 2 — 2bn. But, m 2 + p 2 = a 2 , . n 2 + p 2 = c 2 (209). a 2 =zb 2 + c 2 — 26m. 212. Proposition XIII— Theorem. In an obtuse triangle, the square of the side opposite the obtuse angle is equivalent to the sum of the squares of Vie other sides, plus twice Hie rectangle of one of these sides and the projection of Hie other on that side. AREA AND EQUIVALENCY. 147 Let the triangle ABC be obtuse angled at A; a, 6, c, respectively, the sides opposite the angles A, B, C; and m, n, respectively, the projections of a and c on 6 and 6 pro- duced; p, the perpendicular from jB to b produced. Now, m = 6 -f- n. m 2 = 6 2 + n 2 + 26)i (205), m"- -f jj 2 = 6 2 + n 2 _+ p 2 + 26»i. a a =6 2 + c 2 + 26». 213. Proposition XIV— Theorem. 1. In any triangle, Hie sum of Hie squares of any tim sides is equivalent to twice tlie square of half the third side, plus twice the square of tlie medial line to tluit side. 2. Tlie difference of Vie squares of any two sides is equivalent to twice the rectangle of Hie third side aiui Hie projection, on Hiat side, of Hie medial line to Hiat side. Let ABC be a triangle ; a, 6, c, -B v respectively, the sides opposite A, / ; \ \ „ B, C; m, the medial line to 6; / ; \»< *\ », the projection of hi on 6. A ~ ~ ~/7 c Then, a 2 = (|)%- Also, c"- = (J- j> m 2 - Also, a 2 — c 2 =26n. + 6)i (212). - 6)i (,211). 148 GEOMETRY.— BOOK III. 214. Corollaries. 1. Tlve sum of the squares of the four sides of any quadri- lateral is equivalent to the sum of the squares of the diagonals, plus four times the square of the line joining the middle points of the diagonals. Let ABCE be a quadrilateral whose sides are denoted as in the diagram; d, d', respectively, the diagonals, AC, BE; m, n, respectively, medial lines from B and E to the diagonal AG; and I the line joining the middle points of the diagonals. Then, a 2 + l"- = %l \ J+ 1m* (213, 1). Also, c 2 + e 2 =2^ |\ 2 +2n 2 . .-. a 2 _j_ 2 _|_ C 2 _j_ e 2 =, 4/ | \ 2 + 2 ( m 2 4- n 2 ). But, m 2 + » 2 = 2/ j V + 2Z 2 . (?) ... a 2 + 6 2 _|_ C 2 + e 2 = 4 ( |J 2 + 4(J-') 2 + 4Z 2 . .-. a 2 + 6 2 + c 2 + e 2 = d 2 + — &) (&P — c )- 150 GEOMETRY— BOOK III. II. PROPORTIONALITY AND SIMILARITY. 216. Definitions. 1. The internal segments of a line are the distances from the extremities of the line to a point on the line. 2. The external segments of a line are the distances from the extremities of the line to a point on the line produced. 3. Similar polygons are mutually equiangular polygons whose corresponding sides are proportional. 4. Homologous points, lines, or angles, in similar poly- gons, are points, lines, or angles similarly situated. 5. The ratio of similitude of two similar polygons is the ratio of two homologous sides. 217. Proposition XV.— Theorem. A parallel to one side of a triangle, intersecting the oilier sides, divides them into proportional internal segments. Let ABC be a triangle, and DE a par- allel to AC. Draw AE and CD. The triangles, BED, AED, having a common vertex E, and their bases, BD, AD, in the same line BA, have a common altitude — the perpendicular from E to BA — and are therefore proportional to their bases (200, 3). .-. BED : AED :: BD : AD. The triangles, BDE, CDE, having a common vertex D, and their bases, BE, CE, in the same line BC, have a common altitude — the perpendicular from D to BG. .-. BDE : CDE :: BE : CE. PROPORTIONALITY AND SIMILARITY. 151 The two triangles, AED, CDE, having a common base DE, and their vertices, A, C, in the line AC, parallel to the base, have equal altitudes — the distances between the parallels at A and C — and are therefore equivalent. Hence, the two proportions have a ratio of the one equal to a ratio of the other, and therefore the remaining ratios are equal. .-. BD : AD :: BE : CE. 218. Corollaries. 1. The two sides of a triangle whicli are divided into internal segments by a line parallel to the third tide are proportional to the corresponding segments. For, BD : AD:: BE : CE (217). Then, BD + AD : BD :: BE + CE : BE, and BD + AD : AD : : BE + CE : CE. Hence, BA : BD:: BC : BE, and BA : AD:: BC : CE. 2. A parallel to one side of a triangle intersecting Hie other sides produced divides them into proportional external segments. Let ABC be a triangle, and let DE, parallel to AC, intersect the sides, BA, BC, produced. Then, BD : AD : : BE : CE (218, 1). Let FG, parallel to AC, intersect AB and CB produced beyond the vertex. The alternate angles, BFO, BAC, are equal (46, 2). 152 GEOMETRY— BOOK III. Revolve FBG, in its own plane, around the vertex B as a center, till it takes the position F'BG' on ABC. Then, since the corresponding angles, BF'G', BAC, are equal, F'G' is parallel to AC (44, 4). .-. BA : BF':: BG : BG' (?) This proportion will be true when the triangle is revolved to its original position, since the revolution will not change the length of the lines. Then, changing BA to AB, and BG to CB, or, AB BF : : CB : BG. + BF BF : : CB + BG AF . BF : : CG : BG. BG, 3. Two lines inter seated by any number of parallels are divided proportionally. Let MN and PQ be met by the parallels, BE, FG, HI, KL... Then will MN and PQ be divided proportionally. Suppose the lines intersect at 0. OF : DF : : OG : EG (218, 1). GI (217). OF : FH:: OG FO : FL :: GO .-. DF-.FH.-.EG and FH: FL :: GI GK (218, 2), by division. GI, (?) GK. (?) If the lines are parallel, they will not intersect ; but then, DF=EG, FH=GI, FL = GK, and the above pro- portions will be true. PROPORTIONALITY AND SIMILARITY. 153 219. Proposition XVI.— Theorem. A straight line which divides two sides of a triangle into pro- portional internal segments is parallel to tlie third side. Let DE divide the sides, AB, CB, of the triangle ABC, so that BD : AD : : BE : CE. Drawing AE, CD, we have BED : AED : : BD : AD (200, 3) : BE : CE. : BE : CE. : BDE : CDE. (?) A-— Also, BDE : CDE But, BD : AD BED : AED In this proportion the antecedents are identical; hence, the consequents, AED, CDE, are equivalent; (?) and, since these triangles have a common hase DE, their altitudes are equal ; that is, their vertices, A, C, points of the line AC, are equally distant from DE, or the prolongations of DE; hence, DE and AC are parallel (87). 220. Corollaries. 1. A straight line which divide* two sides of a triangle into internal segments proportional, to tlwse sides is parallel to the third side. y r> . . a r> . . B q BC- It ' BA : AD Then, BA — AD : AD or, BD : AD BE CE. CE : CE, CE. Hence, DE is parallel to AC (.219). 154 GEOMETRY.— BOOK III. 2. A straight line which divides two sides of a triangle into •proportional external segments is parallel to the third side. In the triangle ABC, let DE meet the sides BA and BC produced, so that we have BD : AD : : BE : CE. Then, AG and DE are parallel (220, 1). Let FG intersect the sides, AB, CB, produced, so that AF : BF : : CG : BG. Then, AF—BF-.BF or, AB : BF CG — BG : BG, CB : BG. Revolving FBG to the position F'BG' on ABC, and writing BA for AB, and BC for CB, we have BA : BF':: BC : BG'. Hence, F'G' is parallel to AG; (?) hence, the angle BF'G' is equal to BAG (46, 4). Revolving the triangle to its original position, the angle BFG is equal to BAC, since it has not changed in the revolution ; therefore, FG is parallel to AC (44, 2). 221. Proposition XYII .— Theorem. Tlie biseetor of any angle of a triangle divides the opposite side into internal segments proportional to the adjacent sides. ^^ Let ABC be a triangle, and BD the bisector of the angle ABC. Draw AE parallel to DB, meeting CB produced in E. PROPORTIONALITY AND SIMILARITY. 155 The angles, BAE, ABD, are equal (46, 2), also, the angles, BEA, CBD (46, 4); but, by hypothesis, the angles, ABD, CBD, are equal; hence, the angles, BAE, BEA, are equal; .-. EB = AB (61). But, CD': AD : : CB : EB (217). CD : AD :: CB : AB (22, 3). 222. Corollary. A straight line drawn from the vertex of any angle of a tri- angle, dividing the opposite side into internal segments propor- tional to the adjacent sides, is the bisector of that angle. If CD : AD : : CB : AB. Then, BD is the bisector of ABC. Produce CB till EB = AB, and draw AE. Now, substituting EB for ifa equal AB in the proportion, we have CD : AD : : CB : EB. Hence, BD and EA are parallel (219) ; hence, the angles, BAE, ABD, are equal (46, 2); also, the angles, BEA, CBD (46, 4) ; and, since EB = AB, the angles, BAE, BEA, are equal (59) ; hence, the angles, ABD, CBD, are equal ; that is, BD is the bisector of the angle ABC. 223. Proposition XYIII — Theorem. The bisector of any exterior angle of a triangle divides the opposite side into external segments proportional to tiie adjacent sides. 156 GEOMETRY.— BOOK III. Let ABC be a triangle, BD the bisector of the exterior angle ABE. Draw AF parallel to DB. The angles, BAF, ABD, are equal (46, 2), also, the angles, BFA, EBB (46, 4) ; but, by hypothesis, the an- gles, ABD, EBD, are equal; hence, the angles, BAF, BFA, are equal; hence, FB = AB (61). But, CD : AD : : CB : FB (218, 1). CD : AD :: CB : AB (22, 3). 224. Corollary. A straight line drawn from the vertex of any angle of a tri- angle, dividing the opposite side into external segments propor- tional to the adjacent sides, is the bisector of the exterior angle. (?) 225. Scholiums. 1. In case the triangle ABC is isosceles, AB and BC being the equal sides, the angles, BAG, BCA, are equal; and since ABE = BAC + BCA, and EBD = \ABE, EBD = BCA ; and hence the bisector BD is parallel to CA (44, 4), and will not intersect it, and the segments, CD, AD, become infinite. 2. If BA > BC, the angle BCA > the angle BAC; .-. DBE < BCA; and DB produced will bisect the ex- terior angle formed by producing AB, and intersect AC produced, dividing it externally as in (223). 226. Proposition XIX.— Theorem. Mutually equiangular triangles are similar. PROPORTIONALITY AND SIMILARITY. 157 In the triangles, ABC, DEF, let A = D, B = E, and C = F. Place JjBC on DEF so that the angles, A, D, shall coincide, and the triangle ABC take the position DBC. Since the angle DBC=E, BC is parallel to EF. (?) Hence, DB : DE : : DC : DF (218, 1). Substituting ^4i? for DB and J.C for DC, we have AB : DE :: AC : DF. In like manner, by placing ABC on DEF so that the equal angles, B, E, shall coincide, we shall have AB : DE BC : EF. These proportions have a common ratio ; hence, the other ratios are equal, and we have the continued proportion, AB : DE :: AC : DF :: BC : EF. .-. the triangles, ABC, DEF, are similar (216, 3). 227. Corollaries. 1. Tiro triangles are similar when two angles of tlie one are respectively equal to two angles of the otlier. (?) 2. In similar triangles, Hie homologous sides lie opposite equal angles. 3. The ratio of similitude of hvo similar triangles is tlie ratio of any ttvo homologous lines. (?) 158 GEOMETRY— BOOK III. 228. Proposition XX —Theorem. Triangles whose corresponding sides are proportional are sim- ilar. In the triangles, ABC, DEF, let AB : DE : : AC : DF : : BC : EF. Take DG = AB, and draw GH parallel to EF. Then, the triangles, DGH, DEF, are mutually equi- angular, and, therefore, sim- ilar (226). .-. DG : BE : : DH : DF : : GH : EF (216, 3). Since AB — DG, the first ratios of the two continued proportions are equal ; hence, the other ratios are equal. AC : DF : : DH : DF, .-. AC = DH. BC : EF : : GH : EF, .-. 5C = GH. Hence, the triangles, ABC, DGH, are mutually equi- lateral, and therefore equal (75) ; but DGH and DEF are similar ; therefore, ABC and DEF are similar. 229. Scholium. In order to establish the similarity of two polygons, it must be proved that they fulfill the two conditions (216, 3) : 1st. They must be mutually equiangular. 2d. Their corresponding sides must be proportional. PROPORTIONALITY AND SIMILARITY. 159 In the case of triangles, either of these conditions involves the other ; hence, to establish the similarity of two triangles, it will suffice to prove either that they are mutually equi- angular, or that their corresponding sides are proportional. But to establish the similarity of polygons having more than three sides, both conditions must be proved to be ful- filled, since, in this case, equality of angles does not involve proportionality of sides; nor does proportionality of sides in- volve equality of angles. 230. Proposition XXI —Theorem. Two triangles liaving an angle of the one equal to an angle of the other, a)id the including sides proportional, are similar. In the triangles, ABC, DEF, let A = D, and AB : DE : : AC : DF. Then, these triangles are similar. Take DG = AB, DH = AC, and draw GH Then, the triangles, ABC, DGH, are equal (55). Substituting DG for AB, and DH for AC, in the pro- portion above, DG : DE : : DH : DF. Hence, GH is parallel to EF (220, 1) ; hence, the tri- angles, DGH, DEF, are equiangular (46, 4), and, there- fore, similar (220). But the triangles, ABC, DGH, are equal; hence, ABC and DEF are similar. 160" GEOMETRY.— BOOK III. 231. Proposition XXII.— Theorem. Triangles having their sides respectively parallel or perpendic- ular are similar. i Let ABC and D.EF be the triangles ; A and D, B and E, C and F, the pairs of angles whose sides are respect- ively parallel or perpendicular. The angles of these pairs are equal or supplemental (48), (49). Then, denoting a right' angle by R, we shall have one of the following cases : 1. A + D=2R, B + E = 2R, C + F = 2R. 2. A = D, B + E=2R, C+ F=2R. A= D, B E, C= F. If the first case is true, the sum of all the angles of the two triangles would be six right angles, which is impossible, since this sum is four right angles (52) ; hence, the first case is false. If the second case is true, the sum of all the angles of the two triangles exceeds four right angles by A -\- D, which is impossible; hence, the second case is false. Therefore, the third case is true ; that is, the triangles are equiangular, and hence, similar (226). PROPORTIONALITY AND SIMILARITY. 161 232. Scholiums. 1. In two triangles whose sides are respectively parallel, the homologous sides are those which are parallel. '2. In two triangles whose sides are respectively perpendic- ular, the homologous sides are those which are perpendicular. 8. In two triangles whose sides are respectively parallel or perpendicular, the homologous or equal angles are those which are opposite homologous sides. 233. Proposition XXIII.— Theorem. Polygons composed of the same number of triangles, respect- ively similar, and similarly placed, are similar. Let ABCDE, AB'C'D'E', be polygons composed of the same number of trian- gles respectively similar ir ^ — ^c and similarly placed. / ~~^\ "' ^-f- 1. The polygons are .i«— — ,— -^d jU- : mutually equiangular. e \^' J p^ X. For, any two corre- e' sponding angles of the polygons are either homologous angles of two similar tri- angles, or sums of two or more such angles. Thus, B = J5', BCD = BCA + A CD = B'C'A' + A'C'U = B'C'D' 2. The corresponding sides of the polygons are propor- tional. For, denoting the corresponding sides and diagonals by like letters, as in the diagrams, accenting those of the s. r, .— i-i. 162 GEOMETRY.— BOOK III. second, we have, from the similar triangles, the continued proportion, a _ b f c g d e a' ~~ V = f ~ c 7 = rf _ d' = e'' Omitting the ratios of the diagonals, we may write a : a' : : b : b' : : c : c' : : d : d! : : e : e'. Hence, the polygons are similar (216, 3). 234. Proposition XXIV —Theorem. Similar polygons can be decomposed into the same number of triangles, respectively similar and similarly placed. Let ABODE, A'B'C'D'E', be two similar polygons, whose homologous sides, angles, and diagonals are de- _ , B / — -*f>' 6 s Tl a / ^- \ noted by like letters, „ ,/ ^-'' \ . Hence, the triangles, ACD, A' CD', are similar. (?) These triangles are also similarly placed in the polygons. In like manner, the remaining triangles of the one poly- gon can be proved to be similar to those similarly placed in the other. 235. Scholiums. 1. Similar polygons can be decomposed into the same number of triangles, respectively similar and similarly placed, by drawing lines from any two homologous points to the vertices, or to the vertices and other homologous points in the perimeters. 2. If the points from which the lines are drawn are with- out the perimeters, the polygons will be respectively the sums of the triangles partly interior and partly exterior, minus the sums of those wholly exterior. 3. The ratio of any two homologous lines is equal to the ratio of any two homologous sides, and. is, therefore, equal to the ratio of similitude of the polygons. 4. Two similar polygons are equal when any two homolo- gous lines are equal. 236. Proposition XXV— Theorem. The perimeters of similar polygons are proportional to any two homologous lines. 164 GEOMETRY.— BOOK III. Let ABODE, A'B'G'D'E', be two similar polygons whose homologous sides are de- noted by like, letters, those of the second be- ing accented, and whose : o i' perimeters are denoted by p, p' , respectively. For, since the polygons are similar, a : a' : : b : V : : e : o' : : d : d' : : e : e'. .-. a + b + c + d + e : a' + V + c' + d' -f e' : : a : a', or, p : p' : : a : a' : : b : b'. . . Hence, the perimeters are proportional to any two homolo- gous lines (235, 3). 237. Proposition XXVI— Theorem. If three or more straight lines passing through a common point intersect two parallels, the corresponding segments of the parallels are proportional. Let AE, BF, CO, DH, passing through the point P, intersect the parallels, AD, EH, on the same side or on opposite sides of P. The triangles, ^IPP, EPF, are similar, also, BPC, FPG, and CPD, GPH. (?) AB : EF | BC : FG j BC : FG) CD : GH) and PB :. PF PB : PF PC : PG PC : PG AB : EF AB : EF : : BC : FG, BC : FG : : CD : GH. BC : FG :: CD : GH. PROPORTIONALITY AND SIMILARITY. 165 238. Proposition XXVII.— Theorem. Three or more non-parallel straight lines dividing two paral- lels proportionally pass through a common' point. Let the non-parallel straight lines, AE, BF, CG, DH, divide the par- allels, AD, EH, so that AB : EF : : BC : EG : : CD : GH. Since these lines are not parallel, any two, as AE, BF, intersect at some point P. Now, it is to be proved that CG and DH pass through the same point. Draw PG; then, PG produced will pass through C; for, if not, let PG produced intersect AD in some other point, as Q. By hypothesis, AB : EF :: BC : FG, .-. BC = AB X FG EF If PG produced passes through Q, we have by (237), AB X FG AB : EF :: BQ : FG, .-. BQ = EF .•. BQ = BC, which is impossible; hence, PG passes through C; that is, CG passes through P. In like manner, it can be proved that DH passes through P. 239. Proposition XXYIIL— Theorem. If from the vertex of tlie right angle of a right triangle a perpendicular be drawn to the hypotenuse, then, 1. The two triangles into which tJie given triangle is thus divided' are similar to the girni triangle and to each other. 166 GEOMETRY.— BOOK III. 2. The perpendicular is a mean proportional between the segimnts into -which it divides the hypotenuse. 3. Each side adjacent to the right angle is a mean propor- tional between the hypotenuse and tlie adjacent segment. Let AHB be a triangle right angled at H; a, h, b, the sides respectively opposite the angles, A, H, B; p, the perpendicular from the vertex of the right angle to the hypotenuse ; m and n, segments of the hypotenuse respect- ively adjacent to a and b. 1. The triangle BDH is similar to the triangle AHB; for, the angle B is common, the right angle BDH is equal to the right angle AHB, and, therefore, the angle BHD is equal to A (53, 9), (226). In like manner, it can be proved that the triangle ADH is similar to the triangle AHB. Therefore, the triangles, BDH, ADH, are similar. 2. In the similar triangles, BDH, ADH, m and p are homologous, since they are respectively opposite the equal angles, BHD and A; p and n are homologous, since they are respectively opposite the equal , angles, B and AHD. m : p : : p : n, . • . p 2 = mn. 3. In the similar triangles, AHB, BDH, h and a are homologous, since they are respectively opposite the equal angles, AHB and BDH; a and m are homologous, since they are respectively opposite the equal angles, A and BHD. . • . h : a : : a : m, . • . a 2 = hm. In like manner, it can be shown that h : b : : b : n, .• . b 2 = hn. PROPORTIONALITY AND SIMILARITY. 167 240. Corollaries. 1. The squares of Hie sides adjacent to the right angle are proportional to the adjacent segments of tlw hypotenuse. ■p, « 3 hm m F01 "' V-Jn-^n' ••• « 2 ^ 2 ::,»:„. 2. The sum of tlw squares of the sides of a right triangle adjacent to the right angle is equivalent to tlie square of Hie hypotenuse. For, a 2 -f 62 = hm + hn = h(m -f ji) =hxh = A 2 . 3. If from any point in tlie circumference of a circle a per- pendicular be drawn to a diameter, and chords from Hie same point to the extremities of Hie diameter, then, 1st. The perpendicular is a mean proportional between tlie segments of the diameter. 2d. Eaeli clwrd is a mean proportional between Hie diameter and the segment adjacent to tiie chord. For, the triangle formed by the diameter and chords is right angled (154, 2) ; hence, the statements of the corollary are identical with those of 239, '2, 3. 3d. The sum of tlie squares of tlie clwrds is equivalent to tlie square of the diameter. (?) 241. Proposition XXIX— Theorem. Triangles having an angle of the om equal to an angle of Hie other are proportional to die products of tlie sides including the equal angles. 168 GEOMETRY.— BOOK III. Let ABC and DEF be two triangles having the angle A equal to the angle D. Take DG = AB, DH = AC, and draw GH; then, the triangles, ABC and DGH, are equal (55). Drawing EH, the two trian- gles, DGH, DEH, having the common vertex H, and their bases, DG, DE, in the same line, have a common altitude — the perpendicular from H to DE. .-. DGH: DEH:: DG : DE (200, 3). In like manner, it can be proved that DEH : DEF : : DH : DF. Taking the product of the corresponding terms of these proportions, omitting DEH, the common factor of the first couplet, we have DGH : DEF :: DG X DH : DEX DF. Substituting ABC for DGH, and AB X AC for DG X DH, we have ABC : DEF :: AB X AC : DEX DF. 242. Proposition XXX.— Theorem. Similar trianjles are proportional to the squares of any two homologous sides. Let ABC, A'B'C, be similar triangles ; A and A', B and B', C and C", pairs of equal angles ; a and a', b and V, c and c', pairs of homologous sides; T and T', respectively, the areas of the triangles. PROPORTIONALITY AND SIMILARITY. 169 Since the triangles are mutually equiangular, we have, by cm), T : T : : oh : a'6' : : 6c : 6V : : ca : c'a. Therefore, T _ ab _ _ be ca a b be c a Y' ~ a7W ~ We ~ eW ~" a' X W = V X c 7 = 7 X «' ' Since the triangles are similar, we have, by (216, 3), iij, i a b c a : a : : b : 6 : : c : c , . • . —. = •;-; = —. a 6 c Substituting — , for r ,, j-. for -r, -7 for — , we have a b b c c a T' a' 2 b' 2 c' 2 .-. T : T' :: a 2 : a'- : : b 2 : b' 2 : : c 2 : c,' 2 . 243. Proposition XXXI.— Theorem. Similar polygom are proportion to the squares of any tico Iwmologous sides. Let ABODE, A'B'C'D'E', be two similar polygons; A and A', B and B' . . . , , , pairs of equal angles; p ?, c /~^ ( ^-'\ a and a', b and 6'. . . , a/ t,f''' t '\ "> .-- ' r' \ c ' pairs of homologous .i-^-.*— - — - z> .i'/--»' — , - d' sides; / and /', g and °\ f d X r" ;j ;' parallel to CB. Then, C and Z> are the points of division required. For, AC : BC :: AE : FE :: m : n. Also, AD : BD : : AE : GE : : m : n. 259. Corollaries. 1. Given the harmonic ratio and Vie extreme harmonic points, to find the conjugate of each of the extreme points. Take, as in the figure of the last article, AE = in, EF=u, EG = n. Draw ED; then, GB, parallel to ED, will give B, the conjugate of A. Draw FB; then, EC, parallel to FB, will give C the conjugate of D. 2. if' any tiiree of four harmonic points be given, ilie foiirtli can he found from the proportion, AC : BC : : AD : BD. 180 GEOMETRY.— BOOK III. 260. Proposition XXXIX.— Problem. To find the hens of all the points whose distances from two given points are in a given ratio. Let A and B be the given points, and m : n the given ratio. Divide AB harmonic- ally in the given ratio at the points, C, D (258). Let P he a point of the required locus. Then, AP : BP : : AC : BG : : AD : BD : : m : n. Hence, PC is the bisector of the angle APB (222), and PD is the bisector of the exterior angle BPE (224). But the bisectors, PC, PD, are perpendicular to each other (34, 3, 4). Hence, P is in s the circumference whose diam- eter is CD (179, 3). 261. Proposition XL.— Problem. On a given line to construct a polygon similar to a given polygon. bI jc' Let it be required l to construct on the / line A'B' a polygon A '\~ similar to the polygon ABCDE. From A draw the diagonals, AG, AD. Construct the angle A'B'C = B, and B'A'C = BAG. Then, the triangle A'B'C is similar to the triangle ABC (227, 1). / D a'4- CONSTRUCTIONS. 181 In like manner, construct the triangle A' CD' similar to ACD, and the triangle A'D'E' similar to ADE. Then, the polygon A'B'G'D'E' is similar to the polygon ABODE (233). 262. Proposition XLI. — Problem. Given the ratio of similitude of tiro similar polygons, and one of tlie polygons, to construct tlie other. Let m : n be the ratio of similitude, and ABCDE the given polygon. Take any point P for the center of simili- tude. Draw from P lines through the vertices, A, B, C . . . li _- — " / iriS, On PA produced lay off P.-i' equal to the fourth propor- tional to vi, n, and PA (253). From A' draw A'B' parallel to AB, till it meets PB produced, and from B' draw B'C parallel to BC, till it meets PC produced, and so on. Then, A'B'G'D'E' is sim- ilar to ABCDE, and the ratio of similitude is m : n. For, the polygons are equiangular, since their sides are respectively parallel; and the corresponding sides are pro- portional, since the ratio of any two homologous sides is m : n, the ratio of similitude. (?) 263. Proposition XIII.— Problem. To construct a triangle equivalent to a given polygon. 182 GEOMETRY.— BOOK III. Let ABODE be the given polygon. Take any three consecutive vertices, as A, B, C, and draw the diagonal AC. From B draw BF parallel to AG till it meets the pro- longation of DC in F, and draw AF. The triangles, AFC, ABC, are equiva- lent, for they have a common base AC, and equal altitudes, since their vertices, F, B, lie in the line BF, parallel to AC. To each of these equivalent triangles add the quadrilateral, ACDE, and we shall have the quadrilateral AFDE equiva- lent to the pentagon ABCDE. In like manner, we find the triangle AQD equivalent to the triangle AED. To each of these- equivalent triangles add the triangle AFD, and we shall have the triangle AFG equivalent to the quad- rilateral AFDE, and hence, to the pentagon ABCDE. In like manner, we may find a triangle equivalent to any given polygon, whatever be the number of sides. 264. Proposition XLIIL— Problem. To construct a square equivalent to a given 'parallelogram or to a given triangle. 1. Let AC be the parallelogram, a its altitude, b its Find a mean proportional x between a and b (255). CONSTR UCTIONS. 183 Then, a : x : : x : b; .-. x 2 = ah (197). 2. Let ABC be the triangle, a ft§ altitude, b its base. Find a mean proportional x between a and \b. Then, a : x :: x : \b; .-. a; 2 = ia& (199). 265. Corollary. To construct a square equivalent to any given polygon. Construct a triangle equivalent to the given polygon (263), then, a square equivalent to the given triangle (264, 2). 266. Proposition XLIV — Problem. To construct a square equivalent to tlie sum of two or more given squares, or to Hie difference of two given squares. 1. Let /, in, ii, be the £ sides of the given squares. ^ / Take AB = I, and per- ™ — D< ^ pendicular to AB draw n "'\ BC = m, and draw AC. u *"" Then, AC 2 = l 2 + m 2 (209). Perpendicular to AC, draw CD = n, also draw AD. Then, AD' = AC' + »- = P + ™ 2 + » 2 - 2. Let ? and ))i be the sides of two given .a squares. Take AB = hi, and draw JSC ;, perpendicular to ^1-B. With ^4 as center and I as radius, describe an arc cutting BC in C Then, BC' = I 2 — m- (210, 1). 'i 184 GEOMETRY.— BOOK III. 267. Corollary. To construct a square equivalent to tlie sum of any number of given polygons, or to tlie difference of two given polygons. Construct squares respectively equivalent to the given polygons (265), then, a square equivalent to the sum or difference of these . squares (266). 268. Proposition XLY— Problem. On a given line to construct a rectangle equivalent to a given rectangle. Let R be the given rectangle, a its attitude, b its base, and V the given line. Find a fourth proportional, a', to b' , b, and a (253). Then, a' is the altitude of the required rectangle R'. For, V : b : : a : a', . • . a'b' = ab. But, R=a'V, R = ab, .-. R' = R. 269. Proposition XLVL— Problem. Given the area of a rectangle and the sum of two adjacent sides, to construct tlie rectangle. Let s 2 be the given area, AB the sum of two adjacent sides of the required rectangle. On AB as a diameter con- struct a semi-circumference. At A erect a perpendicular to AB equal to s, and at its extremity draw CD parallel to AB. At D, one point of the intersection of this parallel with the circumference, let CONSTRUCTIONS. 185 Ml the perpendicular DE on AB. Then, AE and EB are the sides of the rectangle. For, AE x EB = s 2 (240, 3, 1st). 270. Proposition XLVIL— Problem. Given the area of a rectangle and tlw difference of two adja- cent sides, to construct Hie rectangle. Let s 2 be the given area, AB the difference of two ad- jacent sides of the required rectangle. On JjB as a diameter con- struct a circumference. At A erect a perpendicular AC to AB, equal to s, and from its extremity C draw the secant CE through the center. Then, CE and CD are the sides of the rectangle. For, CE X CD = s 2 (24-8). 271. Proposition XLVIII. — Problem. To find two straight lines having €ie same ratio as tJiat of the areas of two given polygons. Construct squares equivalent to the given polygons (265). Construct a right angle ABC. From B lay off BA and BC, respectively equal to the sides of the squares. Let fall the perpendicular BD. Then, AD, DC, are the required lines. For. AB' : BC 2 : : AD : DC (210, 3). -ir,. 18G GEOMETRY.— BOOK III. 272. Proposition XLIX— Problem. To find a square which shall be to a given square in the ratio of two given straight lines. Let s 2 be the given square, m » and m : n the given ratio. On an indefinite straight line lay off AB == m and BC= n. On AC as a diameter describe a semi-circumference. At B erect BD perpendicular to AC, cutting the circumference in D, and draw DA and DC. On DC lay off DE = s, and draw EF parallel to CA. Then, DF is the side of the quired square. For, DF : DE : DA DC (218, !)• DF 2 DE 2 :: DA 2 DC 2 . (?) Now, DA 2 DC 2 :: AB BC (210, 3). But, AB = ; m and BC = = n. , * , DA 2 DC 2 :: m n. DF 2 DE 2 :: m n. ■(?) DF 2 s 2 :: m n. (?) 273. Proposition L.— Problem. A polygon being given, to construct a similar polygon whose area shall be in a given ratio to Hie area of the given polygon. Let P be the given poly- gon, one side of which is a, and m : n the given ratio. Find a line a! such that a' 2 : a 2 : : m : n (272). CONSTRUCTIONS. 187 On a' as a side homologous to a, construct the polygon P' similar to P (261). Then, P' is the required polygon. For, P' : P : : «' 2 a 2 (2*3) But, a' 2 : a 2 : : m n. .-. P' P : : m 11. (?) 274. Proposition LI.— Problem. 2b construct a "polygon similar to a given polygon P, mid equivalent to a given polygon Q. Find m, n, sides of squares, such that m 2 = P, n, 2 = Q (265). Let a be a side of P; then, find a fourth propor- tional, a', to »i, n, and a, and on a' as a side homologous to a, construct a polygon P' similar to P. Then will P' be the polygon required. For, »i But, But, Ml J n a : : a" a ,J -. P Q ■■ : mi 2 n 2 . P Q ■■ : a 2 a' 2 . (?) P P' : : a 2 a' 2 (243). P Q = : P P'i .-. P' = Q- (?) 188 GEOMETRY.— BOOK III. 275. Proposition LII.— Problem. To prove that a side of a square and its diagonal are in- commensurable. The side CB is contained in the diagonal once, with a remainder AE, which is to be compared with CB, or with its equal AB. Since ,„f AB is perpendicular to the radius CB, / \ AB is tangent to the arc EBF; and, since AF is a secant drawn from the same point A, and AE its external segment, we have AE : AB : : AB : AF (248). Therefore, instead of applying AE to AB, we can apply AB to AF. But AB is contained in AF twice, with a re- mainder AE, which is again to be applied to AB, instead of which we may again apply AB to AF, and so on in- definitely; hence, the process will never terminate, and AB and AC have no common measure. Hence, the side of a square and the diagonal are incommensurable. 276. Exercises. 1. Draw a straight line at random and divide it into 4 parts proportional to 2, 4, 3, 5 (251). 2. Divide a given straight line into '3 equal parts (252). 3. Draw three lines and find their fourth proportional (253). 4. Draw two lines and find their third proportional (254). 5. Given 2 in. : % : : x : 3 in., to construct x (255). CONSTR UCTIOKS. 189 6. Given x = l / 10, y — ~\ ■'!, to construct a; and y (255). 7. Divide a line 2 in. long, internally and externally in extreme and mean ratio (256). 8. Divide a line 3 in. long, harmonically in the ratio of 5 to 3 (258). 9. Construct the locus of all the points whose distances from the extremities of a line 2 in. long are in the ratio of 2 : 1 (260). 10. Draw a pentagon and construct a similar pentagon such that each side of the latter shall be to the homologous side of the former as 3 to 2 (262). 11. Construct a square equivalent to a given pentagon (265). 12. Construct a square equivalent to the sum of the squares whose sides are respectively 3 in., 5 in., 6 in. (266, 1). 13. Construct a square equivalent to the difference of the squares whose sides are respectively 5 in. and 3 in." (266, 2). 14. Construct a square equivalent to~the sum of a given rectangle and a given pentagon (267). 15. On a line 4 in. long construct a rectangle equivalent to the rectangle whose adjacent sides are 2 in. and 5 in., respectively (268). 16. Construct a rectangle whose area shall be 4 sq. in., and the sum of two of whose adjacent sides shall be 6 in. (269) ; also, a rectangle having the same area, the difference of two of whose adjacent sides is 3 in. (270). 190 GEOMETRY.— BOOK III. 17. Given a pentagon, to construct a similar pentagon whose area shall be to that of the given pentagon as 5 to 3 (273). 18. Construct a quadrilateral similar to a given quadri- lateral and equivalent to a given pentagon (274). 19. Given x-\-y = s, x 2 -\-y 2 = h 2 , to construct x and y. 20. Divide a given line into three segments, a, b, c, such that a : b : : m : n and b : c : : p : q (251). 21. Construct the locus of a point whose distances from two given parallel straight lines are in a given ratio. 22. Construct the locus of a point whose distances from two given non-parallel straight lines are in a given ratio. 23. Through a given point draw a straight line so that the distances from two given points to this line shall be in a given ratio. 24. Construct the locus of a point which divides all the chords of a given circle into internal or external segments whose rectangle is constant. BOOK IV. I. THE CIRCLE AND REGULAR POLYGONS. 277. Proposition I.— Theorem. An inscriptible equilateral polygon is regular. Let an equilateral polygon of n sides be inscribed in a circle. The equal sides are equal chords, and therefore divide the circumference into n equal arcs (126). Each angle is measured by i (w — 2) of the equal arcs (153) ; hence, the polygon is equiangular, and, since it is also equilateral, it is regular (93, 7). 278. Corollary. A regular polygon of any number of sides is possible. For, a straight line equal in length to any circumference can be divided into any number of equal parts (252) ; hence, the circumference can be divided into any number of equal arcs. The chords of these equal arcs are equal (126) ; therefore, these chords are the sides of an inscribed equilateral polygon ; but an inscribed equilateral polygon is regular (277). Hence, a regular polygon of any number of sides is possible. (191) 192 GEOMETRY.— BOOK IV. 279. Proposition II— Theorem. An iiucriptible equiangular polygon of an odd number of sides is regular. Let the equiangular polygon ABODE, of an odd number of sides, be inscribed in a circle. The equal inscribed angles intercept equal arcs (154, 7). If each of these arcs be subtracted from the whole cir- cumference, the remaining arcs will be equal ; hence, the arc ABC = the arc BCD. Subtracting the common arc BC, we have the arc AB = the arc CD; that is, the first of any three consecutive arcs is equal to the third. . • . arc AB = arc CD — arc EA = arc BC — arc DE. Likewise, for any odd number of sides, the arcs are equal. Hence, the sides of the polygon, which are the chords of these equal arcs, are equal (126). Therefore, the polygon is equilateral, and, since it is also equiangular, it is regular (93, 7). 280. Proposition III. — Theorem. A eircumscriptible equiangular polygon is regular. Let ABDEFC be an equiangular polygon circumscribed about a circle ; I and K, points of tangency. Draw CA, CI, CB. The angle CAI = the angle CBI (176, 1). The angles, AIC, BIC, are equal ; (?) hence, the angles, ACI, BCI, are equal. (?) Therefore, the triangles, AIC, BIC, are equal ; (?) .-. AI -= BI; (?) .-. AB = 2BI. THE CIRCLE AND REGULAR POLYGONS. 193 Likewise, BD = 2BK; but, BI = BK; .-. AB ■ In like manner, it can be proved that BD = DE Hence, the polygon is regular. (?) BD. 281. Proposition IV.— Theorem. A ciremnscriptible equilateral polygon of an odd number of sides is regular. Let ABDEF be an equilateral polygon of an odd number of sides circumscribed about a circle. Draw CA, CB, CD. The triangles, ABC, BDC, are equal (55) ; hence, the angles, BAC, BDC, are equal. But, BAF=2BAC, BDE=2BDC; (?) In like manner, it can be proved that the first of any three consecutive angles is equal to the third. .-. A = D = F=B = E. Hence, the polygon is regular. (?) BAF=BDE. Bisect AB, BD, 282. Proposition V.— Theorem Any regular polygon is inscriptible. Let ABDEFG be a regular polygon by the perpendiculars, IC, AT. The intersection C of these perpendiculars is equally distant from A, B, D (70, 2). Let ABKC revolve about CK as an axis. Then, since the angles at A' are right, and KB = KD, KB will coincide with KD. Since the angles, B, D, are equal, S». G.— 17. 194 GEOMETRY.— BOOK IV. also the sides, BA, BE (93, 7), BA will coincide with BE, and hence, CA with CE. Therefore, the circumference described about C as a center, with CA as a radius, will pass through the vertices, A, B, D, E. In like manner, it can be proved that this circumference will pass through the vertices, F, G. Hence, the polygon is inscriptible. 283. Proposition VI.— Theorem. By the last propo- Any regular polygon is circtimscriptible. Let ABBEFG be a regular polygon, sition, any regular polygon is inscript- ible, and, being inscribed, its equal sides are equal chords of the circle, and are therefore equally distant from the center. If, then, a circumference be described about this center, with a radius equal to its distance from the sides, it will be tangent to the sides, and the polygon will be circumscribed about the circle. 284. Definitions and Corollaries. 1. The center of a regular polygon is the common center of the inscribed and circumscribed circles. 2. The radius of a regular polygon is the radius of the circumscribed circle. 3. The apothem of a regular polygon is the radius of the inscribed circle. 4. The central angle of a regular polygon is the angle included by the two radii drawn to the extremities of the same side. THE CIRCLE AND REGULAR POLYGONS. 195 5. The straight lines joining Hie center and vertices of a regu- lar polygon bisect Hie angles of the polygon. (?) 6. The central angles of a regular polygon are all equal, and each is equal to four right angles divided by the number of sides, and is the supplement of any angle of Vie polygon. (?) 7. A regular circumscribed polygon divides Hie circumference intxi equal pads at tJie points of tangency. (?) 8. J. regular inscribed polygon being given, a regular inscribed polygon of double the number of sides can be constructed by bi- secting tlm arcs subtended by the sides, and drawing clwrds to tlie arcs thus formed. (?) 285. Proposition VII.— Theorem. Regular polygons of tlie same number of sides are similar. The polygons are mutually equiangular (99, 1), and their homologous sides are proportional; (?) hence, the polygons are similar. 286. Corollaries. 1. The perimeters of regular polygons of the same number of sides are proportional to tlieir radii or to tlieir apothems (236). 2. The areas of regular polygons of the same number of sides are propmiional to the squares of tlieir radii, or to Hie squares of their apothems ("Hi, 1). 287. Proposition VIII— Problem. A regular inscribed polygon being given, to circumscribe a similar polygon about the circle, and conversely. 196 GEOMETRY— BOOK IV. 1. To construct the circumscribed polygon. 1st method. Let ABCD ... be a regular inscribed polygon. Draw tangents at the vertices, A, B, C, . . . intersecting in the points P, Q, R,... Then, PQB ... is a regular cir- cumscribed polygon similar to ABCD . . . For, these polygons have the same number of sides, since PQR . . . has a side for each angle of ABC. . . The triangles, APB, BQC, CRD, . . . are equal and isosceles ; for, AB, BC, CD, . . . are equal, since they are sides of the regular inscribed polygon; and the angles, PAB, PBA, QBC, . . . are equal, since each is formed by a tangent and one of the equal chords, AB, BC, CD, . . . drawn to the point of tangency, and therefore measured by one-half of one of the equal arcs, AB, BC, CD... Hence, the angles, P, Q, R, . . . are equal, also the sides, AP, PB, BQ,... (58) ; .-. PB + BQ = QC+ CR = RD + DS=... or, PQ= QR = RS=... Hence, the circumscribed polygon PQR ... is both equi- lateral and equiangular, and therefore regular (93, 7), and consequently similar to ABC (285). 2d method. It is evident from the construction in the first method, that if the circumference be divided into equal parts, the tangents drawn at the points of division will form a regular circumscribed polygon. Hence, bisect the arcs, AB, BC, . . . and draw the tan- THE CIRCLE AND REGULAR POLYGONS. 197 gents, PQ, QR,... Then, PQR . . . will be the regular circumscribed polygon required ; for, since the arcs, AB, BC, . . . are equal, their halves, AK, KB, BL, . . . are equal ; hence, also, KB -\- BL, LC -+- CM, ... or KL, LM, . . . are equal. 2. To construct the inscribed polygon. Id metliod. Draw chords joining the consecutive points of tangency of the sides of the circumscribed polygon, as in the first figure. Then, since the ares, AB, BC, . . . are equal (284, 7), the chords, AB, BC,... are equal, and ABC. . . is the regular inscribed polygon required. 2d method. Draw lines joining the center with the vertices of the circumscribed polygon, and draw chords joining the consecutive points in which these lines intersect the circum- ference, as in the second figure. Then, ABCD ... is the inscribed polygon required. For, the lines, OP, OQ, . . . bisect the equal angles of the circumscribed polygon {2Si, 5) ; hence, in the triangles, OPQ, OQR, . . . the angles, POQ, QOR, ... are equal (53, 9) ; therefore, the arcs, AB, BC,... axe equal (148) ; hence, the chords, AB, BC, . . . are equal (126) ; therefore, ABCD ... is the regular- inscribed polygon required (2??). In the second construction, the sides of the two polygons are respectively parallel, since they are perpendicular to the radii drawn to the points of tangency (133), (123, 3). 288. Corollaries. 1. If chords be drawn from the vertices of a regular inscribed polygon to flic adjaee)it points of tangency of €\e sides of the similar circumscribed polygon, in case Vie sides are paraUd, a 198 GEOMETRY.— BOOK IV. regular inscribed polygon will be formed of double the number of sides, its perimeter ivill be greater than the perimeter of the given inscribed polygon, and its area greater than the area of that polygon. (?) 2. If tangents be drawn at the vertices of a regular inscribed pohjgon to the sides of the similar circumscribed polygon, in case the sides are parallel, a regular circumscribed polygon will be formed of double tlie number of sides, its perimeter wiM be less than the perimeter of the given circumscribed polygon, and its area less than the area of that polygon. (?) 3. The perimeter of any regular inscribed polygon is less than the circumference (68), and its area is less than the area of the circle (21, 7). 4. The perimeter of any regular circumscribed polygon is greater than the circumference, and its area is greater than the area of the circle. The perimeter is greater than the circumference ; for, let successive regular circumscribed polygons be constructed, each having twice as many sides as the next preceding. The perimeter of each is less than that of the preceding, and approaches more nearly to an equality with the cir- cumference, which is therefore less than any of these perimeters. The area is greater than the area of the circle (21, 6). 289. Proposition IX— Problem. To inscribe a square in a given circle. Draw two diameters at right angles, and join their ex- tremities by chords. The figure thus formed is a square (148), (126), (154, 2). THE CIRCLE AND REGULAR POLYGONS. 199 290. Corollaries. 1. By continually bisecting the arcs and drawing chords, we can construct regular inscribed polygons of 8, 16, 32, . . . sides. 2. Regular circumscribed polygons of 4, 8, 16, . . . sides can be consb'ucted by 287. 291. Proposition X.— Problem. To inscribe a regular hexagon in a given circle. Let ABDEFG be a regular inscribed hexagon. Draw the diameters, AE, BF. Each of the angles, ACB, ABC, BAG, is measured by one of the equal divisions of the circumference (151), (153) ; hence, the triangle ABC is equiangular, and there- fore equilateral. Therefore, the side of a regular inscribed hexagon is equal to the radius. Hence, to inscribe a regular hexagon in a given circle, apply the radius, as a chord, six times to the circumference. 292. Corollaries. 1. To inscribe an equilateral triangle in a given circle, draw chords joining tlie alternate vertices of tlie regular inscribed hexagon. (?) 2. To inscribe a regular dodecagon in a given circle, bisect the arcs subtended by Vie sides of tJie regular inscribed hexagon, and draw chords joining Hie points of bisection with the adjacent vertices of the hexagon. o. In like manner, regular inscribed polygons of 24, 48, 96, . . . sides can he constructed. 4. Regular circumscribed polygons of 3, 6, 12, . . . sides can be constructed by 287. 200 GEOMETRY.— BOOK IV. 293. Proposition XL — Problem. To inscribe a regular decagon in a given circle. Suppose the construction made, and that each of the arcs, ABD, BDE, is five-tenths of the cir- cumference. Then, AD, BE, are diam- eters, and intersect at the center C. Draw BF so that AF shall be two- tenths of the circumference. The angles, ABC, BAC, AGB, are equal, since each is measured by two- tenths of the circumference (153), (156). Hence, the triangles, ABC, ABG, are isosceles (61) and similar (227, 1). .-. BG = AB, and AC : AB : : AB : AG. The angles, CBG, BCG, are equal, since each is measured by one-tenth of the circumference (153), (151, 1). .-. CG = BG, but BG = AB, .-. CG = AB. Substituting CG for AB in the proportion, we have AC : CG ■ : CO : AG. Hence, the radius AC is divided in extreme and mean ratio at the point G (250, 1). Since AB = CG, we have the following construction : Divide the radius in extreme and mean ratio (256) ; then apply the greater segment, as a chord, ten times to the circumference. 294. Corollaries. 1. By continually bisecting tlie arcs and drawing chords, we can construct regular inscribed polygons of 20, 40, 80, THE CIRCLE AND REGULAR POLYGONS. 201 2. By joining tJie alternate vertices of the decagon, ire shaU have a regular inscribed pentagon. 3. Since | — tV = xV> ^ le clwrd of the difference of one- sixth and one-tenili of the circumference is the side of a regular ■inscribed decapentagon, which can be constructed by applying this chord fifteen times to the circumference. The decapentagon being constructed, we can construct inscribed polygons of 30, 60, 120, .. . sides. 4. Regular circumscribed polygons of 5, 10, 20, . . . also of 15, 30, 60, . . . sides, can be constnicted by 287. 295. Proposition XII.— Theorem. Tlie area of a regular polygon is equal to tlw product of its perimeter by one-half of its apothem. Let s denote one side of the polygon, n the number of sides, p the perimeter, r the apothem, and P the area. The radii of the polygon divide it into 7i triangles, each of which has s for its base, r for its altitude, and sX|r for its area; hence, the area of the n triangles is ns X \r. But the area of the n triangles is the area of the polygon, and ns = p. - • . P=j>Xjr. 296. Exercises. 1. An equiangular quadrilateral is inscriptible. 2. An equilateral quadrilateral is circumscriptible. 3. Is an inscribed equiangular polygon necessarily regular? 4. Is a circumscribed equilateral polygon necessarily reg- ular? 202 GEOMETRY.— BOOK IV. 5. If R denote the radius of a regular inscribed polygon, s one side, r the apothem, A one angle, and C the central angle, prove, 1st, That in case of a regular inscribed triangle, s = R VS, r = j s R, .4 = 60°, G = 120°. 2d. In case of an inscribed square, s = R\/2, r = \RV\ .4 = 90°, 0=90°. 3d. In case of a regular inscribed hexagon, s = R, r = \R\/\ ,4 = 120°, C=60°. 1/5 i 4th. In case of a regular inscribed decagon, s = R , -j r = Ji?VlO + 2l/5, ^ = 144°, C=36°. 6. If « 8 denote one side of a regular inscribed triangle, s 4 one side of an inscribed square, prove that s a : * 4 :: ]/3 : l/2. 7. The area of the regular inscribed hexagon is three- fourths of the area of the regular circumscribed hexagon. 8. The area of a regular inscribed hexagon is a mean proportional between the areas of the regular inscribed and circumscribed triangles. 9. The area of a regular inscribed dodecagon is equal to three times the square of the radius. 10. If a circumscriptible quadrilateral has two opposite sides parallel, the line passing through the center, parallel to the parallel sides and terminated by the other sides, is one-fourth of the perimeter. THEORY OF LIMITS. 203 n. THEORY OF LIMITS. 297. Preliminaries. 1. A variable is a quantity which admits of an indefinite number of successive values. 2. The limit of a variable is the constant which the varia- ble approaches indefinitely. 3. A variable never readies its limit. For, if a variable reach its limit, it could not approach the limit indefinitely. 4. The difference between a variable and its limit is a variable whose limit is zero; and conversely, if the limit of the difference between a variable and a constant is zero, the constant is the limit of the variable. For, since the variable approaches its limit indefinitely, the difference approaches zero indefinitely; and if the difference approaches zero indefinitely, the varia- ble approaches the constant indefinitely. 5. Let a point move from A toward JS : in the first second, one-half the distance from A to B, that is, to P; m the next second, -i -P ? J? one-half the distance from P to B, that is, to P'; and so on indefinitely. Then, the distance from .4 to the moving point is an increasing variable which indefinitely approaches the constant AB, as its limit, without ever reaching it; and the distance from the moving point to B is a decreasing variable which indefinitely approaches the constant zero, as its limit, without ever reaching it. Let I be the symbol for limit; then, Ix denotes the limit of x. 204 GEOMETRY.— BOOK IV. 298. Proposition XIII.— Theorem. If two variables are always equal, and each approaclies a limit, these limits are equal. Let x, y, be two variables, always equal ; a, b, their re- spective limits. Both variables are increasing or both decreasing. (?) 1st. Suppose the variables increasing. If a and b are not equal, one, as a, is the greater. Let d = a — 6. Since x approaches a indefinitely, x may be made to differ from a by a quantity less than d. . ■ . d^> a — x, .'. a — b > a — x, . • . a; > b. Since the limit of y is b, and y is increasing, y <^ b, . • . x > y, which is contrary to the hypothesis ; hence, a and b can not be unequal ; . ■. a = b, or h = ly. 2d. Suppose the variables decreasing. If a and 6 are not equal, one, as a, is the greater. Let d = a — b. Since y approaches b indefinitely, y may be made to differ from b by a quantity less than d. ••■ y — b a, . ■ . a; > y, which is contrary to the hypothesis ; hence, a and b can not be unequal ; . • . a = b, or Ix = &/. 299. Proposition XIV.— Theorem. if i/ie limit of a variable is zero, the limit of the product of the variable by a constant is zero. Let c be a constant, and x a variable whose limit is zero. As x approaches zero indefinitely, ex decreases indefinitely, and can be made to differ from zero by a quantity less than any assigned quantity; .•. lex =z= (297, 2). THEORY OF LIMITS. 205 300. Proposition XV.— Theorem. The limit of the algebrak sum of two or more variables is the algebraic sum of tiieir limits. Let x, y, z, be variables, a, b, c, their respective limits, and u, v, w, the variable differences between x, y, z and a, b, c. Then, x = a =p u, y = b =p v, z = c =p w, ac- cording as the variables are increasing or decreasing. ■•■ i(a; + y + »)=i( (, + H« + M+» + i») (298). But, k = a, ly =b, h ■=. c. lu=0, lv=Q, lw=0 (297,4). .-. l(a + b + c +B + «+tii) = 8+ii + c (297, 2). .-. l(x-\-y + z)=a + b + c = Ix + ly + h. 301. Proposition XVI.— Theorem. T/ie limit of the product of a constant and a variable is Hie product of the constant and Hie limit of the variable. Let c be a constant, and x a variable whose limit is a, and u the variable difference between x and a. Then, x = a =F «, . • . cr = ca =P e«. .-. 7ex = Z (ea =F e»)- Since k = a, Z« = (297,4), .-. leu = (299). Z (ca =F cm) = ca, . • . ?e,v = ca = c/.r. 302. Proposition XVII.— Theorem. if tte limit of each of two or more variables is zero, the limit of their product is zero. 206 GEOMETRY.— BOOK IV. Let the limit of each of the variables, x, y, z, . . . be zero. Since each of the variables, x, y, z, . . . approaches zero in- definitely, their product, xyz... , approaches zero indefinitely. .-. hyz... = (297,' 2). 303. Proposition XVIII— Theorem. The limit of Ae product of any number of variables is the product of tlieir limits. Let x, y, z, . . . be variables, a, b, c, . . . their respective limits, and u, v, w, . . . the variable differences between x, y, z, . . . and their respective limits. x = a =p u, y = b =p v, z = c + u... xyz . . . = ahe . . , =p terms whose limits are zero. . • . Ixyz . . . = abc ... = lx .ly .lz. . . 304. Corollaries. 1. Tlie limit of any power of a variable in the same power of its limit. For, lx" = I (x . x . x . . .) == Ix .lx .lx . . . = (lx) n 2. The limit of any root of a variable is the same root of its limit. For, i (Vxy = (i Vxy i but, i (p^y = h. (iV / x)" = fe; .-. iVx = \/k. 3. The limit of the quotient of two variables is the quotient of their limits. Let z = — , . • . yz — x, . • . lyz = lx, 777 i iidu 7 3j LZC ~~ ' ~iy' " y ~iy" MEASUREMENT OF THE CIRCLE. 207 305. Proposition XIX— Theorem. If two variables are always in a constant ratio, and each approaches a limit, tliese limit* are in tlie same ratio. Let x and y be two variables whose constant ratio is r, and whose limits are respectively a and b. Then, — = r, . • . x = ry. lx = Iry; but, Ix = a, and, since ly = b, , . , a lx try = ro; . • . a = rb, . • . 7- = r, . • . =- — r. ■ b ly in. MEASUREMENT OF THE CIRCLE. 306. Proposition XX.— Theorem. If the number of sides of a regular inscribed polygon be in- creased indefinitely. Hie apotlum loill be an increasing variable whose limit is the radius. Let r = OA — the radius, r' = OD = the apothem. s = AB = one side. In the triangle AOD, we have r'< r (69, 1), r — r' < is (66). By increasing the number of sides indefinitely, s can be made less than any assigned quantity ; for a stronger reason, ■^ s can be made less than any assigned quantity ; and for a still stronger reason, r — r' can be made less than any as- signed quantity. .-. l(r — r')=0, .-. lr' = r (297,4). 208 GEOMETRY.— BOOK IV. 307. Proposition XXI.— Theorem. If the number of sides of regular inscribed and circumscribed polygons be increased indefinitely, then, 1. The circumference is the common limit of the perimeter y of tlie polygons. 2. The area of the circle is the common limit of Hie areas of the polygons. 1. Let p, p', respectively, denote the perimeters of similar circumscribed and inscribed regular polygons, r, r', re- spectively, their apothems, and c the circumference of the circle. Then, p : p' : : r : r' (286, 1). p — p' : p : : r — r' : r; . • . p — p' = — (r — r'). ... l(p-p')=l£(r-r')=l£. l(r-f). (?) But, I (r — r') = (306); .-. l^.l(r — r') = 0. l(p — p') = 0. .'. p and p' approach each other indefinitely; but, p > c and p' < c (288, 4, 3). p — c -< p ■ — p', and c — p' <^p — p'. .-. I (p — c) = 0, and I (c — p') = 0. lp = c, and ?p' =e (297, 4). 2. Let P, P', respectively, denote the areas of two similar circumscribed and inscribed regular polygons, p, p', their MEASUREMENT OF THE CIRCLE. 209 perimeters, r, ?•', tlieir apothems, and C the area of the circle. P=%pr, and P' = ^p'r' (295). P~P' =-}{pr —p'r'). But, y : jp' ::■»•: r', . ■ . p'r — pr' = 0. P — P' = I Q>r + /»• — pr' — p'r'). ■■■ P-l" = : k(.P +p')r-i{p+p')r'. .-. P-P' = $ { p + p') (r-O- But, Z (r — /) =0, . • . l[i(p + p') (r — »•')] = 0. . . 1{P — P')—0, .-. P and P' approach each other indefinitely; but P>C, and P'< ( ' (288,4,3). P—C

e square* of their radii. Let r, r', respectively, be the radii of two circles, c, c', their circumferences, ( ', C, their areas, p. p', the pe- rimeters of similar regular inscribed polygons, and P, P', their areas. Then, £=-,, and J, = £ (286, 1. 21. s <^ —IS. / r' P' 210 GEOMETR Y.—BOOK I V. In these relations, which are true whatever be the number of sides, r, r', are constants, while p, p', P, P', are varia- bles. If the number of sides be indefinitely increased, the perimeters approach the circumferences as their limits, and the areas of the polygons approach the areas of the circles as their limits (307). That is, Ip = c, lp' = e', IP = C, IP' = C". ••• ? = ^» and c r = 7i (305) - 309. Corollaries. 1. The circumferences of two circles are proportional to their diameters, and their areas are proportional to the squares of their diameters, or to the squares of their circumferences. Let d, d', be the diameters. ~, c r 2r d Then, _=_=_=_. C _ r 2 __ 4r 2 d 2 c 2 Also, c ,— /2 — 4/2 _ d , 2 — e , 2 • 2. The ratio of a circumference of a circle to its diameter, or to its radius, is constant ; that is, the same for all circles. -p, c d c c' c c' c c' ' c 7- d" •"' d~ I" "'■ 2r = 2?' '"' r~ /* The constant ratio of the circumference to the diameter is denoted by -. Thus, -=- = *, . • . c = 7td, . • . c = 2711*. MEASUREMENT OF THE CIRCLE. 3. Similar arcs are proportional to their radii, and sectors are proportional to the squares of their radii. Let a, a', denote the ares; », »', the sectors ; r, r', the radii ; A, A\ the central angles; e, c', the circumferences; C, C, the ureas of the circles ; and R, a right angle. Then, 211 similar 1. But, But, 2. But, But, a A .a' 7=4R' and 7 A = A' (1U, 17), 4R 061,2). ? =7 (308), a a' 'I a c a a s C a' <■' C C a c a c and = £ (151, 4). -r, , (308), s s C" 310. Proposition XXIII.— Theorem. T/ic area of a circle is equal to the product of its circumfer- ence by one-half of its radius. Let r denote the radius of the circle ; c, its circumference ; C. its area ; r, the apothem of the circumscribed polygon ; ;>, its perimeter ; P, its area. Then, P = p X ir (295). 212 GEOMETRY.— BOOK IV. In this equation, which is true whatever be the number of sides, P, p, are variables, and r is a constant. If the number of sides be indefinitely increased, the perimeter ap- proaches the circumference as its limit, and the area of the polygon approaches the area of the circle as its limit (307). That is, Ip = c, and IP = C. C=«Xlr (301). 311. Corollaries. 1. The area of a cirde ii equal to Hie square of its radius multiplied by -. For, C=cXlr, and c = 2-r (309,2). 2. Tlie area of a sector is equal to the product of its arc by one-half of its radius. For, t=l (^1,4), ,. £=J-£f;- (?) But, C = c X ir, ••• « = a X %r. (?) 312. Proposition XXIV.— Theorem. Given tlie perimeter of a regular circumscribed polygon, and tlie perimeter of a similar inscribed polygon, to compute the perimeters of regular circumscribed and inscribed polygons of double Hie number of sides. Let AB be a side of the given circumscribed polygon ; CD, parallel to AB, a side of the similar inscribed polygon. MEASUREMENT OF THE CIRCLE. 213 Then, E, the point of tangency, is the middle point of the arc CD (135, '2), and the chord CE is a side of the regular inscribed polygon of double the number of J l ^,--- sides (288, 1). ^.iri- Draw tangents at C and „ \ D, intersecting AB at F \ and G. Then, FG is a side of the regular circum- ° scribed polygon of double the number of sides (288, 2). Let p, p', denote the perimeters of the given circumscribed and inscribed polygons ; p d , p' d , the perimeters of the regular circumscribed and inscribed polygons of double the number of sides. Since (LI is the radius of the circumscribed polygon, P — 0A — 0A (OKfi n p' ~ '60 ~0E (-bb ' lj - Since OF bisects the angle AOE (176, 3"), we have OE ~ FE ( ^" a ' • * p'~ FE This proportion taken by composition gives ■ p+p' _ AF + FE p' ~ FE Dividing by 2 and observing that AF-\- FE= AE, and that 2FE = FG, we have P + p' _ AE 2p' FG But AE is one-half of a side of the given circumscribed polygon whose perimeter is p; (?"> FG i± one side of the 214 GEOMETRY.— BOOK IV. circumscribed polygon of double the number of sides whose perimeter is p d . Hence, FO is contained the same number of times in p d as AE is contained in p. 1±- FG P+P' P . AE P_. Pd' Pi = AE FO P+P' Pd X Pd — p- 2p Xp' p +p The triangles, EFI, CEH, are similar, since the angles, I, H, are right angles ; (?) and the alternate angles, FEI, ECU, are equal. (?) EL EF cm ce' Now, EF, EI, respect-' ively, are halves of sides of polygons of the same number of sides whose perimeters are p d , p' d . Hence, EF is contained the same number of times in p d as EI is contained in p' d . Pd _ p± EF ~~ ET El 'EF' Pi. Pd Also, CH is one-half of a side of the given inscribed polygon whose perimeter is p', and CE is one side of the inscribed polygon of double the number of sides whose perimeter is p' d . Hence, CE is contained the same number of times in p' d as CH is contained in p'. CM _p' CE ~ p' d Pd CE P . CH' MEASUREMENT OF THE CIRCLE. 215 PJ P d P.< 1 P' i».C 313. Proposition XXV —Problem. To compute the ratio of the circumference of a circle to its diameter. Taking a circle whose diameter is 1, let p it p s , p 1B , . . . denote the perimeters of the regular circumscribed polygons of 4, 8, 16, . . . sides, and p\, p' g , p\ 5 , ■ ■ ■ the perimeters of the regular inscribed polygons of 4, 8, 16, . . . sides. Each side of the circumscribed square is 1, and each side of the inscribed square is A 1 2. (?) C Pi = 4. [p't = 4X|l 2 = 2.8284 271. i»* — -P* \ p\ Pi + P'i -= 3.3137 = 3.0614 085. U's 1 ^4 X P S 675. miug this process, we find Pis = 3.1825979, P'lS = 3.1214452. Pzi = 3.1517249, P'S2 = 3.1365485. Psi 2*128 PlS 6 Ps 1 2 = 3.1441184, 3.1422236, 3.1417504, 3.1416321, P'si P 128 P'loS J>' 5 .2 = 3.1403312. = 3.1412773. = 3.1415138. = 3.1415729, /'l024 PlSiS 1*409 6 ^819 2 = 3.1416025, 3.1415951, 3.1415933, 3.1415928, P'l024 P 2 04 8 P 4096 J* 8192 = 3.1415877. = 3.1415914 = 3.1415923. = 3.1415926. 216 GEOMETRY.— BOOK IV. The circumference of the circle whose diameter is 1 is less than 3.1415928, and greater than 3.1415926 (288, 3, 4), and is therefore 3.1415927, which is correct to the seventh decimal place. .-. - = | = 3 - 14 * 5927 = 3.1415927 (309,2). For ordinary purposes, let n = 3.1416. The above method of finding the ratio of the circumfer- ence to the diameter is called the method of perimeters. 314. Formulas for the Circle. 1. r = $d. 7. c = 2-r. *' = £■ 8. e = -d. 3. r = > /v 9. e = 2 V^G. 4. rf^ 2r. 10. G = w 2 . 5. d = C -- 11. G = l~d\ 6. d = 2\^r- r 2 12. C=~- 4, 315. Exercises. 1. In circles of different radii, central angles subtended by arcs of equal length are inversely proportional to the radii. 2. Divide a given circle into a given number of equal parts by concentric circumferences. MAXIMA AND MINIMA.— SUPPLEMENTARY. 217 3. Construct two circles whose radii are proportional to two given lines, and the sum of whose areas is equal to the area of a given circle. 4. In a given equilateral triangle, inscribe three circles, each tangent to two sides of the triangle and to the other two circles, and find their radii in terms of a side of the triangle. 5. Find the radii of three equal circles tangent to each other, and including between them one acre of land. 6. In a given circle inscribe three equal circles tangent to the given circle and to each other, and determine the radii of these equal circles in terms of the radius of the given circle. 7. If B, r, respectively denote the radius and apothem of a regular polygon, R', r', the radius and apothem of the isoperimetric polygon of double the number of sides, prove that These formulas furnish a method of finding the ratio of the circumference to the diameter, called the metlwd of iso- perimeters. IV. MAXIMA AND MINIMA. - SUPPLEMENTARY. 316. Definitions and Illustrations. 1. A maximum magnitude is the greatest of a class. 2. A minimum magnitude is the least of a class. Thus, the diameter is the maximum chord of a circle. The perpendicular is the minimum line drawn from a given point to a given straight line. S. Q.—W. 218 GEOMETRY.— BOOK IV. 317. Proposition XXVI— Theorem. Of all triangles having two sides respectively equal, that in which these sides include a right angle is the maximum. Let BAC, DAC, be two triangles having the sides, BA, AC, respectively equal to the sides, DA, AC, and the angle BAC right, while the angle DAC is oblique; then, the area of BAC is greater than the area of DAC. For, let fall the perpendicular DE upon the base or the base produced. Then, DA>DE; (?) but BA = DA, .-. %BAX AOiDExAC But, BAC =iBAx AC, DAC- BAC> DAC. BA > DE. (23, 14). ■\DEXAC (199). 318. Proposition XXVII— Theorem. Of all triangles having equal areas and equal bases, that lohich is isosceles has the minimum perimeter. Let ABC, ADC, be two triangles j. having equal areas and equal coincident bases, and let AB, BC, be equal, and AD, DC, unequal. Since the areas are equal and the bases equal, the altitudes are equal. Since the altitudes are equal, the ver- tices, B, D, being on the same side of the base, are in a line parallel to the base. Draw CF perpendicular to BD, and produce it to E in the prolongation of AB, and draw DE. MAXIMA AND MINIMA.— SUPPLEMENTARY. 219 The angles, FBC, BCA, BAC, EBF, are equal (46, 2), (59), (46, 4). The right triangles, BFC, BFE, are equal (57). .-. BC=BE, FC=FE; (?) .-. DC = BE (69, 2). Now, AE / coincident buses, AB, BC, being \ /.-'"' ""^O './ equal, AD, DC, unequal. Then, area ABC > area ADC. The vertices, B, D, being on the same side of AC, D must be on LM, the parallel to AC through B, above LM, or below LM. If D is on LM, at any point E, the triangles, ABC, A EC, having equal altitudes and equal bases, have equal areas. Then, AB + BC + CA < AE + EC -f CA (318), which is contrary to the hypothesis ; .*. D can not be on LM. 220 GEOMETRY.— BOOK IV. If D is above LM, at any point F, EC < EF + FC. .-. AE+EC< AE+EF-JrFC,orAE+EC p, or c < p. 222 GEOMETRY.— BOOK IV. If c = p, then, O E (322), which is contrary to the hypothesis ; therefore, c is not equal to p. If c > p, G is greater than if c =p, since Coc e 2 (309, 1). But if e =p, then, C > 2? ; hence, if c > p, we have, for a stronger reason, C > -E, which is contrary to the hypoth- esis ; therefore, e is not greater than p. Now, neither c — p nor c > p ; . • - c < p. 324. Proposition XXXI— Theorem. Of all polygons having their sides respectively equal, that which is inscriptible in a circle is the maximum. Let P, P' , be polygons whose sides are a, b, c, d, e, respectively, P being in- scriptible in a circle, and P' not inscriptible. Then, P > P'. Let a circle C be cir- cumscribed about P, and on the sides, a, b, c, d, e, of the polygon P', let exterior circular segments, equal to those on the corresponding sides of P, be constructed. The circle C and the irregular figure F are isoperimetric ; hence, C > F (322). From each figure, subtracting its circular segments, we have P > P'. 325. Proposition XXXII— Theorem. Of all isoperimetric polygons having the same number of sides, that which is regular is the maximum. MAXIMA AND MINIMA.— .SUPPLEMENTARY. 223 Let P be the maximum polygon having the given number of sides. Then, P is equilateral. For, draw the diagonal AC. Then, the triangle ^iPC must be a maximum, in order that the polygon be a maximum. But ABC is a maximum when AB = BC (320). Hence, any two consecutive sides of P are equal, and therefore P is equilateral. Since P is the maximum polygon having given sides, it is inseriptible in a circle (324), and therefore regular (277). 326. Proposition XXXIII.— Theorem. Of all polygons having equal areas and the same number of sides, the regular polygon lias the minimum perimeter. Let P be a regular polygon, and P' an irregular polygon, having equal areas and. the same number of sides. Let ' p be the perimeter of P, and p' the perimeter of P'. Then, p < p'. For, p — p, p > p', or p P' (325\ which is contrary to the hypothesis; hence, p is not equal to p'. If p > p', P is greater than if p = p', since the area of P varies as the square of one side (24:3), and hence, as the square of its perimeter (236). But if p =-p', then, P > P'; hence, if p > p', we have, for a stronger reason, P > P', which is contrary to the hypothesis ; hence, p is not greater than p'. Now, neither p = p' nor p > p' ; . " . p < p' ■ 224 GEOMETRY— BOOK IV. 327. Proposition XXXIV.— Theorem. Of any two isoperimetric regular polygons, that which has the greater number of sides has the greater area. Let S be a square and T a regular triangle having an equal perimeter. Take any point P in one side of T, and consider it as dividing that side into two sides. Then, the triangle may be regarded as an irregular quadrilateral, while the square is regular. .-. $> T (325). Also, it can be proved that a regular pentagon is greater than a square having an equal perimeter, and so on. 328. Proposition XXXV.— Theorem. Of any two regular polygons having equal areas, that which has tlie greater number of sides has the- less perimeter. Let P and P' be two regular polygons having equal areas, P having a greater number of sides than P'. Let p be the perimeter cf P, and p' the perimeter of P'. Then, p < p'. For, p =p', p > p', or p < p'. If p = p', then, P > P' (327), which is contrary to the hypothesis; hence, p is not equal to p'. If p > p' , P is greater than if p = p' , since P oc p 2 . Therefore, if p > p', we have P > P', which is contrary to the hypothesis; hence, p is not greater than p'. Now, neither p = p' nor p > p', . • . p < p'. BOOK Y. I. LINES AND PLANES. 329. Definitions. 1. A plane is a surface such that a straight line joining any two of its points lies wholly in the surface. A plane, though indefinite in extent, is usually represented by a par- allelogram. 2. The foot of a line is the point in which it meets a plane. 3. A straight line is perpendicular to a plane, if it is per- pendicular to every straight line of the plane passing through its foot. In this case, the plane is perpendicular to the line. 4. The distance from a point to a plane is the perpen- dicular from the point to the plane. 5. A line is parallel to a plane, if all of its points are equally distant from the plane. Li this case, the plane is parallel to the line. 6. A line is oblique to a plane, if it is neither perpendic- ular nor parallel to the plane. In this case, the plane is oblique to the line. 7. Two planes are parallel, if all the points of either are equally distant from the other. 226 GEOMETRY.— BOOK V. 8. The projection of a point on a plane is the foot of the perpendicular from the point to the plane. 9. The projection of a line on a plane is the locus of the projections of all its points. 10. The angle which a line makes with a plane is the angle which it makes with its projection on the plane. This angle is called the inclination of the line to the plane. 11. A plane is determined by lines or points, if no other plane can embrace these lines or points without being coin- cident with the first. 12. The intersection of two planes is the locus of all the points common to the two planes. 330. Proposition I. — Theorem. An indefinite number of planes can embrace the same straigM line. In any plane draw a straight line, and move the plane till the line drawn in it coincides with the given line. The plane revolving about the line as an axis takes an indefinite number of positions, each of which is the position of a plane embracing the line. 331. Proposition II.— Theorem. A plane embracing a straigM line and a point without that line, is determined in position. For, let any plane embracing the line revolve about the line as an axis, till it embraces the point. Now, if the plane revolve either way about the line as an axis, it will cease to embrace the point; hence, any LINES AND PLANES. Ill other plane embracing the line and point must coincide with the first plane, which is therefore determined (329, 11). 332. Corollaries. 1. A plane embracing three points not in the same straight line is determined in position. For, the plane embraces the line joining any two of these points (329, 1), and the third point which is without that line (331). '2. A plane embracing two intersecting straight lines is de- termined in position. For, this plane embraces one of the lines and any point of the other without the first. 3. A plane embracing two parallels is determined in position. For, this plane embraces one of the parallels and any point of the other which is without the first. 4. The intersection of two planes is a straight line. For, if any three points of the intersection were not in the same straight line, the two planes embracing them would coincide (332, 1), and therefore would not intersect. 5. A line embraced by each of two planes is their intersection. For, if not, the two planes embrace their intersection which is a straight line, and the points of the given line which are without the intersection, which is impossible (331). 333. Proposition III. — Theorem. From any point wffliout a planp, one perpendicular to ilie plane can be drawn, and only one. 228 GEOMETRY.— BOOK V. Let P be the point and MN the plane. There can not be two minimum lines drawn from P to MN; for, if PA and PB are mini- mum lines, they are equal, the triangle APB is isosceles, and PC drawn from P to the middle point of AB is per- pendicular to AB (60, 3) ; therefore, PA and PB are not perpendicular to AB (30) ; hence, PC is less than either PA or PB (69, 1), which are, there- fore, not minimum lines ; consequently, there can be only cine minimum line from P to MN. Let PF be that mini- mum line ; then, PF is perpendicular to any line, EF, drawn in the plane through its foot. For, if not, the per- pendicular from P to EF would be less than PF; then, PF would not be the minimum line, which is contrary to the hypothesis ; hence, PF is perpendicular to any and, con- sequently, to every line of the plane drawn through its foot, and therefore perpendicular to the plane (329, 3). Since the minimum line and perpendicular are identical, and since there can be only one minimum line, there can be only one perpendicular. 334. Corollary. At any point in a plane, one perpendicular to the plane can he erected, and only one. Let F be the point in the plane MN. Let P'F' be the perpendicular from P' to the plane M'N (333). Now, let M'N' be brought into coincidence with MN, so that F' shall coincide with F, F'P' taking the position FP, which is therefore perpendic- LINES AND PLANES. 229 ular to MN at the point F, since it is perpendicular to the coincident plane. Only one perpendicular to MN can be erected at F ; for, let FA be any other line through F, and let the plane PFA intersect MN in FB. Since PF is perpendicular to MN, tt is perpendicular to FB (329, 3) ; hence, FA is not per- pendicular to FB (29, 1), and therefore not perpendicular to the plane (329, 3). 335. Proposition IV.— Theorem. If from a point without a plane a perpendicular and oblique lines be drawn, 1, Oblique lines which meet the plane at equal distances from the foot of Hie perpendicular are equal. 2. Of two oblique lines ivhich meet the plane at unequal dis- tances from the foot of the perpendicidar, tiie one which meets it at tlie greater distance is {tie greater. Let P be the point; MN, the plane ; PF, the perpendicular ; PA, PB, PC, oblique lines. Then, 1. If FA = FB, PA = PB (56, 1). 2. If FC>FA, PC>PA; for, on FC take FD = FA, and draw PD. Then, PD = PA; but, PC>PD (69, 3); .-. PC>PA. 336. Corollaries. 1. Equal oblique lines from a point to a plane meet the plane at equal distances from the foot of Hie perpendicular from that point, and make equal angles wWi Hie perpendicular and also xoidi tlie plane. (?) 230 GEOMETRY.— BOOK V. 2. Of two unequal oblique lines from a point to a plane, the greater meets the plane at Hie greater distance from the foot of tlie perpendicular, and makes the greater angle with the perpen- dicular and ike less with the plane. (?) 3. Equal straight lines from' a point to a plane meet the plane in the circumference of a circle whose center is the foot of the perpendicular from tJiat point. (?) Hence, to draw a perpendicular from a point to a plane, draw any oblique line from the point to the plane ; revolve this line about the point, tracing the circumference of a circle in the plane ; draw a line from the point to the center of the circle ; this line will be the perpendicular to the plane and the axis of the circle. 4. The locus of all Hie points equally distant from the extremi- ties of a line is Hie plane perpendicular to the line at its middle point. (?) 337. Proposition Y. — Theorem. A straight line perpendicular to each of two straight lines at their intersection is perpendicular to their plane. Let PF be perpendicular to FA, FB, at their intersection F; then, PF is perpendicular to their plane MN. For, let FC be any other straight line through F in the plane MN, and AGB any straight line in- tersecting FA, FC, FB, in A, C, B. Produce PF till FP' = FP, and draw PA, PB, PC, P'A, P'B, P'C. Now, AP^AP', BP=BP' (69, 2); therefore, the triangles, APB, AP'B, are equal (75). Let the triangle AP'B revolve about AB till it coincides with APB; then, LINES AND PLANES. 231 P'C, PC, coinciding, are equal ; therefore, FC is perpen- dicular to PP' (60, 3), or PF is perpendicular to FC. Hence, PF is perpendicular to any, that is, to every line of MX through its foot, and consequently is perpendicular to the plane MX. 338. Corollaries. 1. At any point in a straight line, one jikuie can be perpen- dicular to that line, and only one. For, the plane of the perpendiculars, FA, FB, to PP', is perpendicular to PP' (337). Any other plane through F can not contain both A and B (,332, 1), and must therefore intersect either PA or PB, say PB, in some other point D. Then, since PFB is a right angle, PFD is not a right angle ; hence, PF is not perpendicular to FD, and therefore not perpendicular to the plane containing FD (329, 3). 2. If one of two perpendiculars revolve about the other as an axis, maintaining its perpendicularity, it will generate a plane perpendicular to die axis, and this plane will be tJie locus of all Hie perpendiculars to Hie axis at the intersection of the two per- pendiculars (329, 3), (338, 1). 3. Through any point without a straight line, one plane can be passed perpendicular to that line, and only one. For, the plane generated by the revolution of the per- pendicular from the point to the line, about the line as an axis, contains the point and is perpendicular to the line (338, 2). If any other plane containing the given point intersect the line at the same point as the first, it is not perpendic- ular to the line (338, 1) ; and if it intersect the line at any 232 GEOMETRY.— BOOK V. other point, the line joining the given point and the point of intersection is not perpendicular to the line (30) ; hence, the plane containing it is not perpendicular to the line (329, 3). 339. Proposition VI.— Theorem. If from tlis foot of a perpendicular to a plane a straight line be drawn at right angles to any line of the plane, the line drawn from its intersection with the line of the plane to any point of the perpendicular, is perpendicular to the line of the plane. Let PF be perpendicular to the plane MN, FC a perpendicular from the foot of PF to any line AB of MN; then, CP is perpendicular to AB. For, take CA = GB, and draw FA, FB, PA, PB ; then, FA = FB (69, 2) ; . ■ . PA -— PB f 335, 1) ; hence, PC is perpendicular to AB (70, 2). 340. Proposition VII. — Theorem. A straight line and a plane are parallel if they can not meet, tlwugh both be produced indefinitely; and conversely. For, if they are not parallel, all the points of the line would not be equally distant from the plane (329, 5) ; hence, the line would approach the plane in one direction or the other, and if sufficiently produced, would meet the plane, which is contrary to the hypothesis. Conversely, if a line and plane are parallel, they can not meet. For, if they could meet, all the points of the line would not be equally distant from the plane, which is con- trary to the definition (329, 5). LINES AND PLANES. 233 341. Proposition VIII.— Theorem. Two planrs are parallel if tiiey can not meet, though both be produced indefinite!)/; and ' eonvensely. For. if the planes are not parallel, all 'the points of one would not be equally distant from the other ; hence, the planes would appioaeh, and if sufficiently produced, would meet, which is contrary to the hypothesis. Conversely, two parallel planes can not meet. For, if they could meet, all the points of one would not be equally distant from the other, which is contrary to the definition (329, 7). 342. Corollary. A line in one of two parallel planes is parallel to Hie otlier plane. For, since the planes can not meet (341), any line in one can not meet the other, and is therefore parallel to it (340). 343. Proposition IX.— Theorem. Either of hvo parallel lines is parallel to every plane embracing the bther. Let AB, CD, be parallel lines, and MX any plane embracing CD ; then, AB is parallel to MX. I _i i_ / For, the parallels, AB, CD, are in / Y the same plane ABDC (37), (332, 3). The plane ABDC intersects MX in CD (332, 5) : hence, if AB meet MX, it must meet it in CD (329, 12). But AB can not meet CD (38) ; therefore, AB can not meet MX; hence, AB is parallel to MX (340). 234 GEOMETRY.— BOOK V. 344. Corollaries. 1. Through any straight line a plane can be passed parallel to any oilier straight line. For, the plane of the first line and a parallel to the second through any point of the first, is parallel to the second line (343). 2. Through any point a plane can be passed parallel to any two straight lines. For, the plane of the lines through the point, respectively parallel to the given lines, is parallel to each of the given lines (343). 345. Proposition X.— Theorem. Planes perpendicular to the satne straight line are parallel. For, if the planes are not parallel, they would meet; then, there would be two planes through a point of their intersection perpendicular to the same straight line, which is impossible (338, 3) ; hence, the planes are parallel. 346. Proposition XI.^-Theorem. The intersections of two parallel planes ivith a third plane are parallel. For, if these intersections are not parallel, they would meet (39) ; hence, the planes embracing them would meet, which is impossible (341). 347. Corollary. Parallel lines intercepted between parallel planes are equal. For, the plane of the lines intersects the planes in parallel LINES AND PLANES. 235 lines (346) which, with the given parallel lines, form a parallelogram (77, -i) whose opposite sides are equal (82). Hence, the lines arc equal. 34S. Proposition XII. — Theorem. A straight line perpendicular to one of two parallel planes is perpendicular to the other. Let MX, PQ, be parallel planes, and m, , let AB be perpendicular to PQ ; then, , AB is perpendicular to MX. [_ Let two planes embracing AB intersect MX, PQ, in AC, BE, and AD, BF; 2 their, AC, BE, are parallel, also AD, j BF (346) ; but AB is perpendicular to L PQ, hence, to BE and BF (329, 3), hence, to AC and AD (42, 3), hence, to MX (337). 349. Corollary. Through any point imtkoid a plane, one plane can be passed parallel to the gicen plane, and onbj one. Let AB be a perpendicular from the point A to the plane PQ (333), and MX a plane through A perpendicular to AB (338, 1) ; then, MX is parallel to PQ (345). Now, any other plane through A is not perpendicular to AB (338, 1), and therefore not parallel to PQ (348). 350. Proposition XIII.— Theorem. If one of two parallels is perpendiadar to a plane, tlie other is perpendicular to the plane. 236 GEOMETRY.— BOOK V. Let AB and CD be parallel, and AB perpendicular to MN; then, CD is perpendicular to MN. For, BD in the plane MN is perpen- dicular to AB (329, 3), and therefore nn- to CD (42, 3). Then, ED in the plane A / / J/2V, perpendicular to J3Z), is perpen- / V dicular to ^4.1) (339), hence, to the plane of AD and BD (337), which is the plane of the parallels, and therefore to CD (329, 3). Hence, CD per- pendicular to BD and ED at their intersection is perpen- dicular to their plane MN (337). 351. Corollaries. 1, Tivo lines perpendicular to the same plane are parallel. For, a parallel to one of the perpendiculars through the foot of the other is perpendicular to the plane (350), and hence coincident with the given perpendicular at that point, since only one perpendicular can be erected to a plane at the same point (334). • 2. Two straight lines parallel to a third straight line are par- allel to each oilier. For, the two lines are perpendicular to any plane perpen- dicular to the third line (350), and hence are parallel to each other (351, 1). 352. Proposition XIT.— Theorem. All the intersections of a plane with planes embracing a line parallel to the plane are parallel to tliat line and to one another. LIXES AXD PLAXES. 237 Let CD, EF, GH, be the intersections of the plane MN with the planes AD, AF, AH, em- bracing the line AB parallel to MX; then, CD, EF, GH, are parallel to AB and to one another. For, since AB is parallel to MX, it can not meet CD, which lies in the plane MX (340) ; and since AB and CD are in the same plane AD, and can not meet, they are parallel (40). For like reason, AB is parallel to EF and to GH. Hence, CD, EF, GH, are parallel to one another (351, 2). 353. Corollaries. 1. A parallel to a line, tlirough any point of a plane parallel to the line, lies in tiie plane. For, a parallel to AB through any point C of MX must coincide with CD, the intersection of MX with the plane embracing AB and the point C (352); otherwise there would be two lines through the point C parallel to AB, which is impossible (42, 2). 2. The locus of any straight line through a given point par- allel to a given plane is the plane through the point parallel to that plane. For, the intersection of the given plane with a plane em- bracing the line is parallel to the line (352), and to the plane through the point parallel to the given plane (342) ; hence, the line lies in that plane (353, 1). 3. If each of two intersecting straight lines is parallel to a given plane, the plane of Hiese lines is parallel to the given plane (353, 2). 238 GEOMETRY.— BOOK V. 354. Proposition XY.— Theorem. Two angles not in the same plane, having their sides respect- ively parallel and in the same direction, are equal, and their planes parallel. Let the angles, A, D, in the planes, M l ^ 7 UN, PQ, have their sides, AB, AC, / /f\ / respectively parallel to DE, DF, and '- f—j i — (v in the same direction. p rtl / 7 The planes of AB, DE, and AC, DF, I j/K I intersect in AD, and the plane BF, /-° J / parallel to AD, intersects MN, PQ, in BC, EF, and AE, AF, in BE, CF. Then, BE, CF, are parallel to AD and to each other (352) ; hence, AE, AF, are parallelograms (77, 4); .-. AB = DE, AC = DF, BE=AD, CF^AD (82); .-. BE=CF; .-. BF is a parallelogram (86); .•. BC = EF. Hence, the tri- angles, ABC, DEF, are equal (75); .'. the angles, BAC, EDF, are equal (56, 2). Since AB, AC, are respectively parallel to DE, DF, they are parallel to PQ, the plane of DE, DF (343); hence, MN, the plane of AB, AC, is parallel to PQ (353, 3). 355. Corollaries. 1. The angles formed by tlie intersections of two intersecting planes with parallel planes cutting the intersection of the two planes, are equal. For, MN, PQ, being parallel, AB, DE, are parallel; also, AC, DF (346); therefore the angles, BAC, EDF, are equal (354). 2. The triangles formed by joining the corresponding extremi- LINES AND PLANES. 239 ties of three. equal parallel straight lines not in the same plane, are equal, and their planes parallel. For, if AD, BE, CF, are equal and parallel, AE, AF, BF, are parallelograms ; therefore, AB, DE, are equal and parallel ; also, AC, DF, and BC, EF ; hence, the triangle?, ABC, DEF, are equal (75), and their planes are parallel (354). 356. Proposition XVI.— Theorem. The corresponding segments of straight lines witersected by parallel planes are proportional. Let the straight lines, AB, CD, be intersected by the parallel planes, MX, PQ, RS, in the points, A, E, B, and C, F, D. Then, AE : EB :: CF : FD. For, let G be the point in which AD pierces PQ; then, EG, FG, the inter- sections of the planes, ABD, ACD, with PQ, are respectively parallel to BD, AC (346). ■ \ \ \ 7 \ \ > / 1 \\ — 's Then, f JJ5 : EB :: AG: GD, \ \AG : GD:: CF : FD, \ AE :EB::CF: FD. 357. Exercises. 1. Do two lines that are not parallel always meet? 2. Designate any three points, and show how a plane can be conceived to embrace them. 3. Why does a three-legged stool stand firm on a floor, when a four-legged chair may not? 240 GEOMETRY.— BOOK V. 4. Show how to find a perpendicular to the floor from a given point in the ceiling (336, 3). 5. Find a point in a plane equally distant from three given points not in the plane. 6. Through a given point pass a plane parallel to a given plane (353, 3). 7. Through a given point pass a plane perpendicular to a given straight line. 8. Through the point in which a given line pierces a given plane draw a line in that plane making a given angle with the given line. 9. Through a given point draw a straight line which shall intersect two given straight lines not in the same plane. 10. Through a ' given point draw, to a given plane, a straight line at a given inclination to the plane and parallel to another given plane. 11. Find the locus of the points in a given plane equally distant from two given points out of the plane. 12. Draw, from one of two given straight lines not in the same plane to the other, a straight line making a given angle with the first. 13. Through a given point draw a straight line intersect- ing the circumference of a circle and a given straight line not in the plane of the circle. 14. To determine that point in a given straight line which is equidistant from two given points not in the same plane with the given line. SOLID ANGLES. 241 15. Through a given point draw to a given plane a straight line of given length parallel to another given plane. 16. Find the locus of the points equidistant from three given points. 17. From the first to the second of three non-parallel straight lines not in the same plane draw a line parallel to the third. 18. Through a given point of a straight line parallel to a given plane draw a straight line of a given length ter- minating in the plane and making a given angle with the given line. 19. To find a point in one line at a given distance from another line not in the same plane. 20. Through a given point in a plane draw a straight line in the plane which shall be at a given distance from a given point without the plane. II. SOLID ANGLES. 358. Definitions. 1. A solid angle is the divergence of two planes inter- secting in a common line, or of three or more planes inter- secting in a common point. 2. Solid angles are classified as dihedral and polyhedral. 3. A dihedral angle is the divergence of two intersecting planes. 4. The faces of a dihedral angle are the intersecting plane?. S. G.-21. 242 GEOMETRY.— BOOK V. 5. The edge of a dihedral angle is the intersection of its faces. 6. The plane angle of a dihedral angle is the plane angle formed by two straight lines, one in each face, perpendicular to the edge at the same point. Thus, the divergence of the planes, AE, AF, is a dihedral angle ; AE, AF, are its faces; AB is its edge; GIH is its plane angle, if GI, HI, in the faces, are per- pendicular to the edge AB at the same point J. 7. A dihedral angle is expressed by four letters, one in each face and two in the edge between the other two; thus, CABD, or HIBE. If, however, the dihedral angle is isolated, it may be expressed by two letters in its edge; thus, jLB. 8. A dihedral angle may be generated by the revolution of one of two planes about a common straight line from a position of coincidence with the other. 9. Adjacent dihedral angles are those which have a common edge and a common face between them. 10. If the faces of a dihedral angle be produced through the edge, the angles on opposite sides of one plane and on the same side of the other are adjacent; the angles on oppo- site sides of both planes are vertical. 11. Dihedral angles are classified as right and oblique, and the oblique as acute and obtuse, as the analogous plane angles (6), the faces of the dihedral angle taking the place of the sides of the plane angle, and the edge, of the vertex. 12. Two planes are perpendicular to each other, or oblique, according as their dihedral angle is right or oblique. SOLID ANGLES. 243 13. When two planes are cut by a third plane in different lines, the dihedral angles formed take the same names as the analogous plane angles formed by one straight line intersect- ing two others (36, 2). 14. The angle of two straight lines not in the same plane is equal to the angle formed by two straight lines through the same point, respectively parallel to the given lines. 15. A polyhedral angle is the divergence of three or more planes meeting in a common point. 16. The faces of a polyhedral angle are the portions of the bounding planes limited by their intersections. 17. The edges of a polyhedral angle are the intersections of its faces. 18. The vertex of a polyhedral angle is the intersection of its edges. 19. The facial angles of a polyhedral angle are the plane angles formed by consecutive edges. 20. The dihedral angles of a r polyhedral angle are the dihedral / \ angles formed by consecutive faces. / / \ \ Thus, V-ABCD is a polyhedral A --"T~ ~"^\ angle ; VAB, YBC, . . . are its faces ; "\l^^~^ Til. VB, ... its edges; T r is its * vertex: AVE, BVC,... are its f.ieiil angles; A VBC, BVCD, . . . its dihedral angles. 21. Polyhedral angles are classified as trihedral, qitadri- hedral, . . . according to the number of faces. 22. Trihedral angles are tretangtdar, bireetangular, or tri- rcetangular, according as they have one, two, or three right dihedral angles. 244 GEOMETRY.— BOOK V. 23. Trihedral angles are scalene, isosceles, or equilateral, according as the facial angles are all unequal, two equal, or three equal. 24. Two trihedral angles are supplementary if the facial angles of one are respectively the supplements of the dihe- dral angles of the other. 25. A polyhedral angle is convex if the polygon formed by the intersections of a plane with all its faces is convex (93, 11). 26. Two polyhedral angles are symmetrical if they have the same number of faces, and the successive dihedral and facial angles respectively equal, but arranged in a reverse order. 359. Proposition XVII. — Theorem. All the plane angles of a dihedral angle are equal. For, since the corresponding sides of these plane angles are in the same plane, and perpendicular to the same line, the edge, they are parallel (41) ; hence, these plane angles are equal (354). 360. Corollary. Two dihedral angles are equal if their plane angles are equal. For, bringing their edges and plane angles into coinci- dence, the dihedral angles will coincide, and hence are equal (332, 2), (21, 5). 361. Proposition XVIII— Theorem. Two dihedral angles are proportional to their plane angles. 1. If the plane angles of the dihedral angles are com- mensurable, suppose lines to be drawn through their vertices SOLID ANGLES. 245 in their planes, dividing them into partial plane angles, each equal to the common unit of measure. The planes embracing the edges and the several lines of division will divide the dihedral angles into equal partial dihedral angles, since their plane angles are equal (360). Each of the given dihedral angles having one of the partial dihedral angles as a unit of measure, has the same numerical value as its plane angle, since it contains its unit of measure the same number of times. The dihedral angles are proportional to their numerical values (145) ; the plane angles are also proportional to their numerical values; but the numerical values of the dihedral angles are equal to the numerical values of their corresponding plane angles ; hence, the dihedral angles are proportional to their plane angles. 2. If the plane angles of the dihedral angles are incom- mensurable, the proposition can be proved by the method employed iu 150, 2. 362. Corollaries. 1. The plane angle of a dihedral angle is the measure of the dihedral angle. (?) 2. If the fiier* of a dihedral angle be produced Uiraugh the edge, tlie adjacent angles are supplemental and tlie vertical angles are equal. (?) 3. Two dihedral angles are equal in the following eases : 1st. If tlieir faces are respectively parallel, and lie in the same direction or in opposite directions from tlieir edges (18). 2d. If their edges are parallel, and the faces of one respect- ively meet, at right angles, the furs of a supplemental angle of Vie other. (?) 246 GEOMETRY.— BOOK V. 4. Two diJiedral angles are supplemental in tlie following cases : 1st. If two of their faces are parallel and lie in the same direction, and the other faces are parallel and lie in opposite directions from their edges. 2d. If their edges are parallel, and the faces of one respect- ively meet, at right angles, the faces of the other (49). 363. Proposition XIX.— Theorem. If a straight line is perpendicular to a plane, every plane embracing the line is perpendicular to tlud plane. Let AF be perpendicular to the plane MN; then, any plane PQ, embracing AF, is perpendicular to MN. For, at F draw, in the plane MN, ■» FB, perpendicular to the intersection / FQ. L Since AF is perpendicular to MN, it is perpendicular to FQ and FB. (?) Hence, the angle AFB is the plane angle (358, 6), and consequently, the measure of the dihedral angle of MN and PQ (362, 1) ; and since AFB is a right angle, the dihedral angle is a right angle; that is, PQ is perpendicular to MN. 364. Corollaries. 1. If three straight lines are perpendicular to each other at a common point, each is perpendicular to the plane of the oilier two, and the three planes are perpendicular to each other. (?) 2. A plane embracing a given line can be passed perpendic- ular to a given plane. For, the plane embracing both the given line and a b / 'N SOLID ANGLES. 247 perpendicular from any one of its points to the given plane, is perpendicular to that plane (363). 365. Proposition XX.— Theorem. If two planes are perpendicular to vaeli otlier, a straight line in one, perpendicular to their intersection, u perpendicular to Hie otlier. Let the planes, MX, PQ, be perpen- dicular to each otlier ; then, AF in PQ, ' perpendicular to their intersection FQ, r is perpendicular to MX. -W/- For, if FB in MX is perpendicular > to FQ, the angle AFB is right, since it L is the measure of the right dihedral angle of the two planes ; hence, AF, perpendicular to FQ and FB at their intersection, is perpendicular to then- plane MX (337). 366. Corollaries. t 1. If two plmies are perpendicular to each other, a straight line through any point of one perpendicular to the oilier lies in the first. For, a straight line in the first plane, through that point, perpendicular to the intersection, is the perpendicular to the second plane (365), (333), (334). 2. If a straight line is oblique or parallel to a plane, only one plane embracing tlie line can be perpoidicular to tlie plane. For, one such plane can be drawn (364, 2) ; but only one, since this plane must contain any perpendicular from the line to the given plane (366, 1), and only one plane can embrace, two intersecting lines (332y 2). 248 GEOMETRY.— BOOK V. 3. The projection of a straight line on a plane is a straight line. For, the plane embracing this line, perpendicular to the given plane, contains the perpendiculars from all the points of the line to the plane (366, 1) ; and its intersection with the given plane, which is a straight line (332, 4), is the locus of the projections of all the points of the line upon the plane, and hence the projection of the line. The plane which embraces a given straight line, and is perpendicular to a given plane, is called the projecting plane of the line. 4. If each of two intersecting planes is perpendicular to a third plane, their intersection is perpendicular to that plane. For, the perpendicular to the third plane, through any point of the intersection of the first and second, is in each of these planes (366, 1), and, hence, is their intersection. Corollary 4 may be stated thus : A plane perpendicular to each of two intersecting planes is perpendicular to their inter- section. 367. Proposition XXI.— Theorem. Every point in a plane bisecting a dihedral angle is equally distant from the faces of that angle. Let P be any point of the plane NQ, bisecting the di- hedral angle, MNAR; then, the perpendiculars, PE, PF, from P to the faces, MN, NR, are equal. The plane PEAF, of PE and PF, is perpendicular to the planes, MN, NR (363), and hence to their intersection SOLID ANGLES. 249 NA (366, 4); .-. NAE, NAP, NAF, are right angles (329, 3) ; hence, the plane angles, PAE, PAF, are re- spectively the measures of the dihedral angles, MXAQ, QXAR, which by hypothesis are equal; therefore their measures, PAE, PAF, are equal; hence the right triangles, PEA, PFA, are equal (72, 2); .-. PE = PF. 368. Proposition XXII— Theorem. T]ie angle hicluded by two perpendiculars drawn from any point within a dUwdral angle to its faces, or Hie angle included by two lines drawn from any point, respectively parallel to Uiese peipendkidars, in Hie same direction or in opposite directions, is the supplement of the dihedral angle. Let PE, PF, be two per- pendiculars drawn from the point P to the faces, MX, XQ, of the dihedral angle, MXAQ. The plane of these per- pendiculars is perpendicular to the faces (363), and, consequently, to the edge XA (366, 4) ; hence its intersections, AE, AF, with the faces, are perpendicular to the edge (329, 3) ; therefore EAF is the plane angle of the dihedral angle (358, 6), and hence, its measure (362, 1). Since the angles, E, F, of the quadrilateral PEAF are right, P is the supplement of ^1 (79). The angle included by two lines drawn from any point, respectively parallel to these perpendiculars, in the same direction or in opposite directions, is equal to the angle included by the perpendiculars, and heuce is the supplement of the given dihedral angle. 250 GEOMETRY.— BOOK V. 369. Proposition XXIII —Theorem. The perpendiculars drawn from any point within a trihedral angle to its faces, or tlie lines drawn from any point respectively parallel to these perpendiculars, in the same direction or in opposite directions, are the edges of a trihedral angle which is tlie supplement of the given trihedral angle. For, the facial angles formed by the perpendiculars are respectively the supplements of the dihedral angles of the given trihedral angle (368), (358, 24). The lines drawn from any point respectively parallel to these perpendiculars, in the same direction or in opposite directions, form facial angles respectively equal to those formed by the perpendiculars, and hence are the supple- ments of the dihedral angles of the given trihedral angle (358, 24). 370. Proposition XXIV.— Theorem. The sum of any two facial angles of a trihedral angle is greater than tlie third. 1. If either of two facial angles is greater than the third, their sum is greater than the third. 2. If each of the facial angles, AVB, BVC, of the trihedral angle, V-ABC, is less than the facial angle AVC, then, AVB + BVOAVG. For, draw VD in the face AVC, making the angle AVD equal to the angle AVB. Through any points, A, C, of the edges, VA, VC, draw AC cutting VD in D; take VB = VD, and draw AB, BC. SOLID ANGLES. 251 The triangles, AVB, AVD, are equal (55); .-. AB = AD (56, 2). But, AB + BC> AC (65) ; . • . AB + BC — AB > AC— AD ■ . • . BC> DC ; . • . the angle B VC > the angle D VC (74) ; .-. AVB+BVOAVC. 371. Proposition XXV— Theorem. The Him of the three faeial angles of a trihedral angle h greater than zero and less than four right angles, but may have any value between tJiese limit*. Let V-ABC be a trihedral angle. % 1. Since each of the facial angles is greater than 0, their sum is greater than 0. 2. Join any three points, ^4, B, C, in the edges, thus forming the triangle ABC. The sum of the angles of the three triangles whose common vertex is V, is equal to six right angles (52) ; and since the sum of the angles ^AB, VAC, is greater than the angle BAC (370), and similarly for the angles about B and C, the sum of the angles at the bases of the same triangles is greater than the sum of the angles of the triangle ABC, that is, greater than two right angles; hence, the sum of the angles at the vertex V, which is the sum of the faeial angles of the trihedral angle, is less than four right angles. This sum may have any value between 0° and 360°. (?) 252 GEOMETRY— BOOK V. 372. Corollary. The sum of tlie three dihedral angles of a trihedral angle is greater than two and less than six right angles, but may have any value between these limits. For, the sum of the dihedral angles of the given trihedral angle, plus the sum of the facial angles of its supplemental trihedral angle, is equal to six right angles (358, 24) ; but the sum of these facial angles is less than four right angles and greater than ; hence, the sum of the dihedral angles of the given trihedral angle is greater than two and less than six right angles. This sum may have any value between 180° and 540°. (?) \ 373. Proposition XXYL— Theorem. Two trihedral angles, having the Hiree facial angles of the one respectively equal to the three facial angles of the other, are equal if the equal facial angles are arranged in the same order; but are symmetrical and equivalent if the equal facial angles ' are arranged in a reverse order. Let the trihedral angles, S, S', have the facial angles, ASB, A'S'B', equal; also, ASC, A'S'C, and BSC, B'S'C. Then will the dihedral angles, SA, S'A', be equal; also, SB, S'B', and SC, S'C. Take any point A in the edge SA, and perpendicular to SA draw AB, AC, in the faces, ASB, ASC; respectively. SOLID AXGLES. 253 Take S'A' — SA, and perpendicular to S'A' draw J/J3', iT', in the faces, A'S'^B', -4'S'C', respectively. The two triangles, ASB, A'S'B', are equal (57); there- fjre, AB = A'B', BS=B'S' (56, 2). For like reason, AG=A'C, and SC=S'C. The 'two triangles, BSC, B'S'C, are equal (55); there- fore, BC = #C. Hence the triangles, ABC, A'B'C, are equal (75); therefore the_angles, BAC, B'A'C, are equal; but these plane angles are the measures of the dihedral angles, SA, S'A' (362, 1), which are therefore equal. In like manner, it can be proved that the dihedral angles, SB, S'B', are equal; also, SC, S'C. If the equal facial angles are arranged in the same order, as in the central and left-hand figures, the trihedral angles can be made to coincide, and are therefore equal; but if the equal facial angles are arranged in a reverse order, as in the central and right-hand figures, the trihedral angles are symmetrical and equivalent. A familiar example of symmetrical equivalency is found in the two hands. 374. Exercises. 1. Pass a plane perpendicular to a given plane so as to embrace a given straight line. 2. Find the locus of points in space which are equallv distant from two given points. 3. The angle which a straight line makes with its pro- jection upon a plane is the least angle which it makes with any line of that plane. 4. A plane perpendicular to a line parallel to a plane is perpendicular to the plane. 254 GEOMETRY.— BOOK V. 5. Find the locus, of the points which are equally distant from two given straight lines in the same plane. 6. The three planes bisecting the three dihedral angles of a trihedral angle intersect in the same straight line. 7. Find the locus of the points which are equally distant from three given planes. 8. The three planes embracing the edges of a trihedral angle, and respectively perpendicular to the opposite faces, intersect in the same straight line. 9. Find the locus of the points which are equally distant from three given straight lines in the same plane. 10. The three planes embracing the edges of a trihedral angle and the bisectors of the opposite facial angles respect- ively, intersect in the same straight line. 11. The three planes embracing the bisectors of the facial angles of a trihedral angle, and respectively perpendicular to the faces, intersect in the same straight line. 12. Find the locus of the points which are equally distant from the three edges of a trihedral angle. 13. Through a given point of a plane draw, in the plane, a straight line perpendicular to a given straight line in space. 14. Cut a given quadrahedral angle by a plane so that the section shall be a parallelogram. 15. Prove that a common perpendicular to any two lines in space can be drawn intersecting both ; that only one such 'common perpendicular can be drawn ; and that the common perpendicular is the shortest distance from one of these lines to the other. SOLID ANGLES. 255 16. The sum of the facial angles of any convex polyhe- dral angle is less than four right angles. 17. Two isosceles trihedral angles are equal if the three facial angles of the one are respectively equal to the three facial angles of the other. 18. The dihedral angles opposite the equal facial angles of an isosceles trihedral angle are equal. 19. A trihedral angle is isosceles if two of its dihedral angles are equal. 20. Two trihedral angles, having two facial angles and the included dihedral angle of one equal to two facial angles and the included dihedral angle of the other, are equal, or symmetrical and equivalent. 21. Two trihedral angles, having a facial angle and the two adjacent dihedral angles of one equal to a facial angle and the two adjacent dihedral angles of the other, are equal, or symmetrical and equivalent. 22. In two trihedral angles, having two facial angles of one respectively equal to two facial angles of the other, and the included dihedral angles unequal, the third facial angle belonging to the trihedral angle having the greater included dihedral angle is the greater. 23. In two trihedral angles, having two fecial angles of one respectively equal to two fecial angles of the other, and the third facial angles unequal, the dihedral angle opposite the greater third fecial angle is the greater. 24. Two .trihedral angles, having the three dihedral angles of one respectively equal to the three dihedral angles of the other, are equal if the equal dihedral angles are arranged in 256 GEOMETRY.— BOOK V. the same order, but are symmetrical and equivalent if the equal dihedral angles are arranged in a reverse order. 25. All trirectangular trihedral angles are equal. 26. If the edges of any polyhedral angle be produced through the vertex, an equivalent polyhedral angle will be formed symmetrical with the given polyhedral angle. 27. In how many ways may three planes intersect one another ? Draw illustrations, and show what kind of angles the planes form in each case. 28. Determine a point in a given plane such that the sum of its distances from two given points on the same side of the plane shall be a minimum. 29. Determine a point in a given plane such that the difference of its distances from two given points on opposite sides of the plane shall be a maximum. 30. Draw, in a given plane, through a given point in the plane, a straight line perpendicular to a given line not in that plane. BOOK VI. POLYHEDRONS. 375. General Definitions. 1. A polyhedron is a solid bounded by polygons. 2. The faces of a polyhedron are the bounding polygons. 3. The edges of a polyhedron are the intersections of its faces. 4. The vertices of a polyhedron are the intersections of its edges. 5. A diagonal of a polyhedron is a straight line joining any two vertices not in the same face. 6. A section of a polyhedron is the polygon formed by the intersection of a plane with three or more faces. 7. A convex polyhedron is a polyhedron every section of which is a convex polygon. 8. A solid unit, called also a unit of volume, is a given polyhedron assumed as the unit of measure for solids. 9. The volume of a polyhedron is its numerical value (144, 8). S. G.-22. ('257) 258 GEOMETRY.— BOOK VI. 10. Similar polyhedrons are polyhedrons identical in form. 11. Equivalent polyhedrons are polyhedrons identical in volume. 12. Equal polyhedrons are polyhedrons identical in form and volume. 13. A tetrahedron is a polyhedron of four faces ; a penta- hedron is a polyhedron of five faces; a heqaliedron, is a polyhedron of six faces Let all the polyhedrons, up to the icosahedron, be defined (93, 2). I. PRISMS. 376. Definitions and Classification. 1. A prism is a polyhedron two of whose faces are equal and parallel polygons having their homologous sides parallel, and whose other faces are parallelograms having homologous sides of the equal polygons for bases. 2. The bases of a prism are the equal parallel faces. 3. The lateral faces of a prism are all the faces except the bases. 4. The lateral surface of a prism is the sum of its lateral feces. 5. The lateral edges of a prism are the intersections of its lateral faces. 6. The basal edges of a prism are the intersections of the bases with the lateral faces. PRISMS. 259 7. The altitude of a prism is the perpendicular distance from one base to the plane of the other. Thus, ABCDE-F is a prism ; ABODE, FGHIJ, are its bases ; AG, BH, ... . are J its lateral faces, whose sum is the lateral /■ surface ; AF, BG, . . . are its lateral edges ; I AB, BC,... FG, GH,... are its basal / edges; KL, perpendicular to the bases, -^xf is the altitude. 2> 'b 8. A right prism is a prism whose lateral edges are perpendicular to its bases. Any lateral edge of a right prism is equal to the altitude. (?) 9. An oblique prism is a prism whose lateral edges are oblique to its bases. Any lateral edge of an oblique prism is greater than the altitude. (?) 10. A regular prism is a right prism whose bases are regular polygons. 11. Prisms are triangular, quadrangular, pentangular, . . . according as their bases are triangles, quadrilaterals, penta- gons, . . . 12. A truncated prism is the portion of a prism included between either base and a section cutting all the lateral edges. 13. A right section of a prism is a section perpendicular to its lateral edges. 14. A parallelepiped is a prism whose bases are paral- lelograms. 15. An oblique parallelopiped is a parallelopiped whose lateral edges are oblique to its bases. 260 GEOMETRY.— BOOK VI. 16. A right parallelopiped is a parallelopiped whose lateral edges are perpendicular to its bases. Hence, its lateral faces are rectangles. (?) 17. A rectangular parallelopiped is a right parallelo- piped whose bases are rectangles. 18. The dimensions of a rectangular parallelopiped are three edges having a common vertex. 19. A cube is a rectangular parallelopiped all of whose faces are squares. / / / / / 20. A cube whose edge is a linear unit is usually taken as the unit of volume. 377. Proposition I. — Theorem. Parallel sectixms of a prism are equal polygons. Let CF, GJ, be parallel sections of the prism AB; then, CF and GJ are equal. For, CD and GH are paraUel (?) and equal ; (?) also, DE and HI, . . . .-. CDE = GHI,... (?) Hence, the sections are mutually equi- lateral and equiangular, and therefore equal. PRISMS. 378. Corollary. 261 Any section of a j)ris»i parallel to Hie base is equal to the ht»e. (?) 379. Proposition II.— Theorem. The lateral area of a prism is equal to the product of a lateral edge by the perimeter of a right section. Let CF be a right section of the prism AB. The lateral edges are all equal. (?) Let I denote the lateral area ; e, a lateral edge ; p, the perimeter of the right section. The face A I is a parallelogram, whose base is AG and altitude DE. AT = AGX DE; also, HJ=AGXEF,... AI + HJ +... = AG X (DE+ EF + ...), .-. l = eXp- Remark. — Let b denote the area of each base, and s the entire surface. Then, i = «Xf + * 380. Corollary. Tlie lateral area of a right prism is equal to tlie perimeter of Hie base multiplied by the altitude. (?) 381. Proposition III— Theorem. Two prisms are equal if the three faces including a trihedral angle of the one are respectively equal to the three corresponding faces including a trihedral angle of the oilier. 262 GEOMETRY.— BOOK VI. Let the three faces, AD, AG, AJ, including the tri- hedral angle A of the prism AI, be respectively equal to the three faces, A'D' , A'G', A' J', including the trihedral angle A' of- the prism AT, and similarly placed. The trihedral an- gles, A, A', are equal, (?) and can be made to coincide ; then, the base AD will coincide with A'D', the face AG with A'G', and AJ with A'J' ; hence, FG will coincide with F'G', FJ with F'J'. There- fore, the planes of the upper bases coincide, (?) also, the planes of the faces, BH, B'H' ; hence, GH, G'H', coincide, and therefore the faces, BH, B'H'. In like manner, it can be shown that the remaining faces respectively coincide ; hence, the prisms coincide and are therefore equal. 382. Corollaries. 1. Two truncated prisms are equal if the three faces including a trUiedral angle of the one are respectively equal to tlw three faces including a trihedral angle of the other, and are similarly placed. The proof is the same as for the last proposition. 2. Two right prisms Iiaving equal bases and equal altitudes are equal. If the faces are not similarly placed, the prisms can be made to coincide if one be inverted. PRISMS. 263 383. Proposition IV —Theorem. Any oblique prism is equivalent to a right prism whose bases are equal to right sections of the oblique prism, and whose alti- tude is equal to a lateral edge of the oblique pi-ism. Let AD' be an oblique prism ; FI, a right section. Complete the right prism FI', making its edges equal to those of the oblique prism. The truncated prisms, A I, AT, are equal. (?) To each add the truncated prism FD'. Then the oblique prism AD' is equivalent to the right prism FI'. 384. Proposition V— Theorem. Any two opposite faces of a paraUelopiped are equal and parallel. Let AG be a paraUelopiped. AT J^~ The bases, AC, EG, are equal / «> j — V and parallel. (?) AE, DH, are equal and parallel, (?) so, also, are AB, DC; hence the angles, EAB, HDC, are equal, and their planes parallel. (?) Therefore the opposite faces, AF, DG, are equal (83, 2) and parallel. Li like manner, it can be proved that AH and BG are equal and parallel. Any two opposite faces of a paraUelopiped can. be taken for bases, since they are equal and parallel parallelograms. ■ / A \ 264 GEOMETRY.— BOOK VI. 385. Proposition VI.— Theorem. The four diagonals of a parallelopiped bisect each other. Any two diagonals are the diagonals of a parallelogram, (?) and hence bisect each other. Any three diagonals, there- fore, pass through the middle point of the fourth ; hence, all the diagonals pass through the same point, which is called the center. 386. Proposition VII— Theorem. The sum of the squares of the four diagonals of a paraUeb- piped is equivalent to the sum of tlie squares of tlie twelve edges. From the diagram of the last proposition we have AE 2 + GO 2 + AG 2 + EG 2 , (?) AG 2 + CE 2 BH 2 + DF 2 == BE 2 DH 2 + BD 2 + FH 2 . Adding, substituting AB + BC + CD + DA for AG 2 + BD 2 , and EF 2 + EG 2 + GH 2 + HE 2 for EG 2 + FH 2 , we have AG 2 + CE~ 2 +BH 2 +DF 2 = AE 2 + BF 2 + CG 2 + DlX 2 + AB 2 + BG 2 + CD 2 + DA 2 + EF 2 + FG 2 + G~H 2 + HE 2 . _ 387. Corollary. The square of a diagonal of a rectangular parallelopiped is equivalent to the sum of the squares of three edges meeting at a common vertex. (?) PHISMS. 265 Remark. — A parallelopiped can be constructed upon any three given straight lines not in the same plane, terminating in the same point, by passing a plane through the extremity of each parallel to the plane of the other two. These planes, together with the planes of the given lines, will by their mutual intersections determine the faces of the parallelopiped. 388. Proposition VIII— Theorem. The plane embracing two diagonally opposite edges of a par- allelopiped divides it into tivo equivalent triangular prisms. The plane AG, embracing the diagonally opposite edges, AE, CG, divides the parallelopiped DF into two equivalent triangular prisms w. -_( — / -„~C / onal of the parallelogram LJ, and \ / ,--'' \ / hence divides it into two equal tri- -^ ■» angles, UK, ILK. The prism ABC-F is equivalent to the right prism whose base is UK and whose altitude is AE. (?) The prism ADC-LT is equivalent to the right prism whose base is ILK and whose altitude is AE. But the two right prisms are equal ; (?) hence, the prisms, ABC-F, ADG-H, are equivalent. 389. Proposition IX.— Theorem. Two rectangular parattelopipeds having equal bases are pro- portional to tlieir altitudes. s. G.— 23. 266 GEOMETRY.— BOOK VI. Let P, P', be two rectangular parallelopipeds having equal bases, and let a, a', be their altitudes. 1. If a, a', are com- mensurable, let their com- mon measure m be con- tained, for example, 5 times in a and 3 times in a'. .-. a : a' :: 5 : 3. \ p ■ \ \ \ \ \ ( \ \ \ \ \ \"K i \ i\ v \ If sections parallel to the bases divide the alti- tudes, a, a', into 5 and 3 equal parts respectively, P will be divided into 5 equal parallelopipeds, and P' into 3 equal parallelopipeds, each equal to each of those in P. (?) .-. P : P' :: 5 : 3. .-. P : P' :: a : a'. 2. If a, a', are incommen- surable, divide a' into any number of equal parts, and apply one part to a as many times as possible. The section through the last point of division of a divides P into two parallelopipeds, Q, Q', whose altitudes are x, x', the altitude x being commensurable with a', and x' being less than one of the equal divisions of a'. ■ §- = ^- m x — h- i i i * i i - \ X 1 1 1 \ Now, if the number of parts into which a' is divided be PBISMS. 267 indefinitely increased, each part will indefinitely diminish, x will approach a as its limit, x' will approach 0, and Q Q P will approach P. Hence, the limit of -—, is p-, , and the limit of '-; is — , a a (?) P P' (?) 390. Corollary. Two rectangular parallelopipeds having two dimensions of one respectively equal to two dimensions of Vie oilier are proportional to their third dimensions. (?) 391. Proposition X.— Theorem. Two rectangular parallelopipeds having equal altitudes are proportional to their bases. Let P, P', be two rectangular parallelopipeds whose dimensions are a, c, d, and a, c', d', respectively, and whose bases are b, V. \ • N i a 1 *\ Let Q be a rectangular parallelopiped having two dimen- sions, a, d, the same as P, and two dimensions, a, c', the same as P'. Then, P : Q : : e : c', and Q : P' : : d : d'. (?) 268 GEOMETRY.— BOOK VI. Taking the product of the corresponding terms of these proportions, and omitting the common factor Q from the first couplet, P : P' :: cXd : c' Xd'. But, b = c X d, and V = c' X d'. (?) P : P' :: b : b'. 392. Corollary. Two rectangular parallelopipeds having a dimension of one equal to a dimension of the other are proportional to the products of their other two dimensions. (?) 393. Proposition XI.— Theorem. Any two rectangular parallelopipeds are proportional to Vie products of their thxee dimensions. \ - i\ 1 <* 1 k \ Let P, P', be two rectangular parallelopipeds whose dimensions are a, c, d, and a', c', d', respectively, and Q a rectangular parallelopiped having two dimensions, c, d, the same as P, and the third dimension, a', the same as P'. Then, P : Q : : a : a', and Q : P' ; : c X d : c' X d'. (?) P : P' : : a X c X d : a' X c' X d'. PRISMS. 394. Proposition XII.— Theorem. 269 Tlie volume of a rectangular parallelopiped is equal to the product of its iJiree dimensions. \ \ \ \ \ \ X \ \ \P\ \ \\ \ \ \ \ \ ^\ \ H x\ d \ \ P' d\ \ 1. If the dimensions of the parallelopiped are commensura- ble, let P be its volume ; a, c, d, its dimensions ; and the cube, whose edge is a common measure of a, c, d, the unit of volume. The edge of this cube, taken as the linear unit, is con- tained in the dimensions, respectively, a, c, d times. Sections parallel to the faces, dividing the edges into parts each equal to the linear unit, divide the parallelopiped into cubes each equal to u, the unit of volume. These cubes are arranged in a strata, each containing d rows of c cubes each, and hence, c X d cubes ; therefore, a strata contain a X c X d cubes, or units of volume. .-. P = a X c X d. 2. If the dimensions of the parallelopiped are incom- mensurable, let P' be its volume, a', c', d', its dimensions. Then, P : P' :: aXcXd : a' Xe? X z - 396. Proposition XIII.— Theorem. The volume of any parallelopiped is equal to the product of its base by its altitude. Let P denote the volume ; b, the base ABCD; a, the alti- tude HR of the par- allelopiped ABCD- F, having all its faces oblique. The oblique paral- lelopiped ABCD-F is equivalent to the right parallelopiped HIJN-L, whose base is the right section HIJN of the oblique parallelopiped, and whose altitude HQ is a lateral edge of the oblique parallel- opiped. (?) The right parallelopiped HIJN-L, regarded as an oblique parallelopiped whose base is IJKL and lateral edge III, is equivalent to the right parallelopiped IOPL-R, whose base is the right section IOPL and whose altitude is the lateral edge IH. But this right parallelopiped is a rectangular parallelo- PRISMS. 271 ILGH=EFGH= ABCD piped, (?) whose base OPQR = b, and whose altitude HR = a. Now, ABCD-F = JKMX-L ■■= OPQR-L P = bxa. bXa. P = b X a. 397. Proposition XIV.— Theorem. T/ie volume of any prism is equal to Hie product of its base by Us altitude. 1. If the base is a triangle, let P denote the volume, b the base, a the altitude. The parallelopiped AH, con- structed on the edges, AB, AC, AE, of the triangular prism ABC-E, has 2P for its volume, 26 for its base, and a for its altitude. (?) .-. 2P=2Z>X«; (?) 2. If the base is a polygon having more than three sides, let P denote the volume, b the base, a the altitude. Planes embracing a common lat- eral edge and its opposite lateral .,-/ -""?, edges respectively, divide the prism \ ,---'■' ° into triangular prisms having a J ^ for their common altitude. De- noting the volumes of these triangular prisms by P', P", . . . and their bases by V, b", ... we have P'+P"+...=b'X«+b"Xa + ... = (b'+b"+...)Xa. But, P' + P"+... = P, b'+b"+... = b. P=bXa. 272 GEOMETRY.— BOOK VI. 398. Corollaries. 1. Any two prisms are proportional to the products of their bases by their attitudes. Let P, P', denote the volumes; b, b', the bases; a, a', the altitudes of any two prisms. Then, P = &X«, P' = b'Xa'; .-. p = ^-J • 2. Two prisms having equivalent bases are proportional to their altitudes. „.„,,, P b X « P a For, if 6 =6, F = 6 __ 7 ; . . p = -,- 3. Two prisms having equal altitudes are proportional to tlieir v ■* , P bXa . P b For, ifa=a, p = ft -^- ; . . p = gj • 399. Exercises. 1. The three edges of a rectangular parallelopiped are Z, ?», n;* find its diagonal, surface, and volume. 2. The surface of a cube is s ; find its edge, diagonal, and volume. 3. The volume of a triangular prism is equal to the product of the area of a lateral face by one-half the per- pendicular from the opposite edge to that face. 4. Of all quadrangular prisms having equivalent surfaces, the cube has the maximum volume. PYRAMIDS. 273 II. PYRAMIDS. 400. Definitions. 1. A pyramid is a polyhedron one of whose feces is a polygon, and whose other faces are triangles having a common vertex and the sides of the polygon for bases. 2. The base of a pyramid is the face whose sides are the bases of the triangles having a common vertex. 3. The lateral faces of a pyramid are all the faces except the base. 4. The lateral surface of a pyramid is the sum of its lateral faces. 5. The lateral edges of a pyramid are the intersections of its lateral faces. 6. The basal edges of a pyramid are the intersections of its base with its lateral faces. 7. The vertex of a pyramid is the common vertex of its lateral faces. 8. The altitude of a pyramid is the perpendicular from its vertex to the plane of its base. Thus, V-ABCDE is a pyramid; ABODE is its base; AVB, BVC,... are its lateral faces, whose sum is its lateral surface ; A V, B V, . . . are its lateral edges ; AB, BC, . . . its basal edges ; V is. its vertex ; VO, its altitude. 274 GEOMETRY— BOOK VI. 9. A regular pyramid is a pyramid whose base is a regular polygon, and whose vertex is in the perpendicular to the base at its center. 10. The axis of a regular pyramid is the straight line joining its vertex with the center of its base. 11. The slant height of a regular pyramid is the alti- tude of any lateral face. 12. A pyramid is triangular, quadrangular, pentangular, . . . according as its base is a triangle, quadrilateral, pentagon, . . . 13. A truncated pyramid is the portion of a pyramid included between its base and a section cutting all its lateral edges. 14. A frustum of a pyramid is a truncated pyramid in which the cutting section is parallel to its base. 15. The base of the pyramid is called the lower base of the frustum, and the parallel section, its upper base. 16. The altitude of a frustum is the perpendicular from either base to the plane of the other. 17. The lateral faces of a frustum of a regular pyramid are the trapezoids included between its bases; the lateral surface is the sum of the lateral faces. 18. The slant height of a frustum of a regular pyramid is the altitude of any lateral face. 401. Proposition XY.— Theorem. Any section of a pyramid parallel to its base divides the edges and altitude proportionally, and is similar to the base. PYRAMIDS. 275 The section fghij of the pyramid Y-ABCDE, parallel to the base, divides the edges, VA, VB, . . and the altitude VO proportionally. For, conceive a plane through V par- allel to the base ; then the edges and altitude will be intersected by three par- allel planes. Therefore, Yf: YA :: Yg : VB . . . :: To : VO. The section fghij and the base ABCDE are similar polygons. For, fg, AB, are parallel; also, gh, BC...; hence, the angles, jfg, EAB, are equal; also, fgh, ABC... The triangles, Vfg, VAB, are similar; (?) also, Vgh, VBC. . . fg Yg gh Yh _ hi "'• AB " VB ~ ~BC~ YC ~~ CD .-. fg : AB :: gh : BC : : hi : CD : : . . . Therefore, fghij, ABCDE, are similar. (?) 402. Corollaries. 1. Ant/ section of a pyramid parallel to its base is to the base as Hie square of the perpendicular from the vertex to tlie plane of tlie section is to the square of Vie altitude of the pyramid. For, fghij : ABCDE : : ~fg 2 : AB 2 : : ff -VA 2 ::Yo 2 : Yif. 2. If two pyramids have equal altitudes, sections parallel to the bases and whose planes are at equal distances from Hie vertices are proportioml to the bases. For, let 6, b', denote the bases of two pyramids ; a, their 276 GEOMETRY— BOOK VI. common altitude ; s, s', sections parallel to the bases and whose planes are at the common distance d from the vertices. Then, s : b : : d 2 : a 2 , and s' : V : : d 2 : a 2 . s : b :: s' : V, .-. 8 : s' : : b : V. 3. If two pyramids have equal altitudes and equivalent bases, sections parallel to the bases and whose planes are at equal dis- tances from the vertices are equivalent. For, since s : s' :: b : b', if b = V, s = s'. 403. Proposition XYL— Theorem. The lateral area of a regular pyramid is equal to one-half the product of the perimeter of its base by its slant height. Let I denote the lateral surface of the regular pyramid V-ABCDE; e, a basal edge ; p, the perimeter of the base; t, one of the equal (?) lateral faces; n, the number of lateral faces; h, the slant height. Then, t = ie X h; -•. nt = \ne X h. But, I = nt, and p = ne • .-. l = ipXh. Remark. — Let b denote the area of the base, and s the entire surface. Then, s = ij>X^ + &- 404. Corollary. The lateral area of a frustum of a regular pyramid is equal to one-half the product of the sum of the perimeters of its bases by its slant height. PYRAMIDS. 277 For, let I denote the lateral surface of the frustum ; e', an edge of the upper base ; p', its perimeter ; t, one of the equal (?) lateral faces; h, the slant height; e, n, p, the same as above. Then, t = \(e -f e') X h; .-. nt = A(ne -f ne') X h. But, I = »ii, and p = ne, p' = ne' ; • •• l = i(p + ?0x*. Eemark. — Let b, b', denote the areas of the bases, and s the entire surface. Then, s = J {p + p') X h + b + V. 405. Proposition XVII.— Theorem. Two triangular pyramids having equivalent bases and equal altitudes are equivalent. Let P, P', denote the volumes of two triangular pyra- mids, V-ABC, V'-A'B'C, having equivalent bases and equal altitudes; then, P=P'. If the pyramids are not equivalent, let P > P'. Corresponding sections of the two pyramids, parallel to the bases, dividing the altitudes into the same number of equal parts, are equivalent. (?) 278 GEOMETRY.— BOOK VI. On the lower base and the several sections of the pyra- mid V-ABC, as lower bases, construct prisms whose lateral edges are parallel to the lateral edge VA, and whose alti- tudes are the equal divisions of the altitude of the pyramid. On the sections of the pyramid V'-A'B'C, as upper bases, construct prisms whose lateral edges are parallel to the lat- eral edge VA', and whose altitudes are the equal divisions of the altitude of the pyramid. Let S denote the siim of the volumes of the first set of prisms, S' that of the second set. Now, P S' ; .-. P — P' (b-b')+ab'l But, b : U :: a'*: (a'— a) 2 (402,1). Vb : Vb' :: a' : a'— a. (?) a' Vb — a Vb = a' Vb'. a'Cv/6— y'b') = aVb. PYRAMIDS. 281 .-. a'(] b— v'T/) (v'6 +1 6')= oV'6 (Vb + \/b'). (2) a'(6 — 6') = a (6 -f \/W). Substituting (2) in (1), F = £a (6 -f 6'+ VW)- 408. Proposition XIX— Theorem. The volume of any truncated triangular prism is equal to one-third of the product of its right section by the sum of its lateral edges. 1-F Let the truncated triangular prism ABC-D be divided by the planes, AEC, DEC, into three pyramids, E-ABC, E-ACD, E-CDF. The pyramid E-ABC has ABC for its base and E for its vertex. The pyramid E-ACD is equivalent to the pyramid B-ACD, having the same base, ACD, and an equal alti- tude. (?) But B-ACD is the same as D-ABC; hence, E-ACD is equivalent to D-ABC, which has ABC for its Iimsp and D for its vertex. The pyramid E-CDF is equivalent to the pyramid B-ACF, having an equivalent base (?) and an equal altitude. (?) But B-ACF is the same as F-ABC; hence, E-CDF is equivalent to F-ABC, which has ABC for its base and F for its vertex. 282 GEOMETRY.— BOOK VI. Hence, the truncated prism ABC-D is equivalent to three pyramids, E-ABC, D-ABC, F-ABC, whose common base is ABC and whose vertices are E, D, F. If the truncated prism is right, as in the second figure, its lateral edges are respectively the altitudes of the three pyramids. . • . ABC-D = IABC X BE+ %ABC XAD + $ABC X CF. .-. ABC-D = IABCX(BE+ AD + CF). The right section GHI of the truncated prism ABC-D, in the third figure, divides ABC-D into two right truncated prisms, OHI-A, GHI-D, having a common base GHI. . ■ . ABC-D = %GHI X (BH + AG + CI) + iGHIX(HE+GD+IF). .-. ABC-D = $GHI X (BE + AD + CF). 409. Proposition XX.— Theorem. Two tetrahedrons having a triJiedral angle of one equal to a trihedral angle of the other are proportional to the products of tlie three edges of these trihedral angles. The tetrahedrons, D-ABC, G-AEF, having coincident tri- hedral angles at A, have ABC, AEF, for their bases, and DO, GP, perpendiculars from the vertices, D, G, to the plane of the bases, for their altitudes. D-ABC _ ABC X DO _ ABC DO •'' G-AEF ~ AEFX GP ~ AEF X GP' W PYRAMIDS. 283 „. ABC ABXAC DO AD m D-A BC ^J7i - JCx ^Z) 410. Exercises. 1. The lateral surface of a pyramid is greater than the base. 2. The planes embracing the three edges of any trihedral angle of a tetrahedron and the medial lines of the opposite face intersect in a straight line. 3. The four lines joining the vertices of a tetrahedron with the intersections of the medial lines of the opposite faces, meet in a point which divides each of these lines in the ratio of 1 to 4. 4. The straight lines joining the middle points of the opposite edges of any tetrahedron meet in a point which bisects each of these lines. 5. The plane bisecting a dihedral angle of a tetrahedron divides the opposite edge into segments proportional to the adjacent faces. 6. A plane embracing any edge and the middle point of the opposite edge of a tetrahedron divides it into two equivalent solids. 7. The volume of a truncated triangular prism is equal to the product of its lower base by the perpendicular to the lower base from the intersection of the medial hues of the upper base. 284 GEOMETRY— BOOK VI. HE. SIMILAR POLYHEDRONS. 411. Definitions. 1. Similar polyhedrons, being identical in form, are polyhedrons having the same number of feces, respectively- similar and similarly placed, and their corresponding poly- hedral angles equal. 2. Homologous faces, lines, or angles of similar poly- hedrons are faces, lines, or angles similarly placed. 412. Proposition XXI.— Theorem. The ratio of similitude of any two Iwmologous faces of two similar polyliedrons is equal to the ratio of similitude of any other two homologous faces. For, the ratio of similitude of any two homologous faces is the ratio of any two homologous sides of these faces; and, since these sides are homologous sides of adjacent faces, the ratio of similitude of any two homologous faces is equal to that of any two adjacent homologous faces; and so on for all the faces. The ratio of similitude of any two homologous faces of two similar polyhedrons is called the ratio of similitude of the polyhedrons. 413. Corollaries. 1. The ratio of similitude of two similar polyhedrons is equal to the ratio of any two homologous edges. (?) 2. Any two homologous faces of two similar polyliedrons are proportional to the squares of any two homologous edges. (?) 3. The surfaces of two similar polyliedrons are proportional to tlie squares of any two homologous edges. (?) SIMILAR POLYHEDRONS. 285 414. Proposition XXII— Theorem. The portion of a tetrahedron cut off by a section parallel to any face is a tetrahedron similar to the given tetrahedron. Let ABCD be a tetrahedron ; EFG, a section parallel to BCD. Then, AEFG is a tetrahedron similar to ABCD. EFG divides the edges, AB, AC, AD, proportionally ; (?) hence, the faces, AEF, ABC, are similar; also, AFG, ACD, and AGE, ADB; (?) and EFG, BCD, are also similar. (?) The corresponding trihedral angles are equal. (?) fore, AEFG is a tetrahedron similar to ABCD. There- 415. Proposition XXIII —Theorem. Two tetrahedrons having a dihedral angle of one equal to a dihedral angle of Vie otlier, and the faces including tliese angles respective!)/ similar and similarly placed, are similar. Let the tetrahedrons, ABCD, EFGH, have the dihedral angles, AB, EF, equal, and the faces, ABC, ABD, respectively similar to EFG, EFH; then, these tetrahedrons are similar - . The trihedral angles, A, E, can be made to coincide in all their parts, (?) and hence are equal. Therefore the angles, CAD. GEH. are. equal. .D .F 286 GEOMETRY— BOOK VI. From the given similar faces we have the proportions, AC : EG :: AB : EF. ) [ .-, AC : EG :: AD : EH. AD : EH : : AB : EF. ) Therefore the faces, CAD, GEH, are similar. (?) In like manner, it can be proved that the trihedral angles, B, F, are equal, and that the faces, BCD, FGH, are similar. The trihedral angles, C, G, are equal, since their facial angles are respectively equal (?) and similarly placed; the trihedral angles, D, H, are equal. (?) Therefore the tetra- hedrons are similar. 416. Corollaries. 1. Two similar polyhedrons can be decomposed into the same number of tetrahedrons, respectively similar and similarly placed. Select two homologous vertices ; divide the homologous faces not adjacent to these vertices into the same number of triangles, respectively similar and similarly placed ; draw lines from the selected vertices to the vertices of these tri- angles. The planes of these lines will divide the polyhedrons as stated. (?) 2. Two polyhedrons composed of the same number of tetrahe- drons, respectively similar and similarly placed, are similar. (?) 3. The ratio of any two homologous lines of two similar polyhedrons is equal to the ratio of any two homologous edges, and therefore equal to the ratio of similitude of tlie polyhe- drons. (?) 4. In tlie corollaries of 413, lines may be substituted for edges. SIMILAR POLYHEDRONS. 417. Proposition XXIV. — Theorem. 287 Similar polyhedrons are proportional to Hie cubes of any two homologous lines. 1. Let T, t, denote the volumes of two similar tet- rahedrons ; E, E', E", and e, e', c" , respectively, homol- ogous edges of two homol- ogous trihedral angles ; L, I, any two homologous lines. Then, T ExE'xE" E E' E" E t~eXe'Xe"~e X e' X e"'' bUt » e T E E E E 3 ■ -. — = — X— X— = ~5-» or T : t : t e c e e 3 But, E : e ;: L : I, .-. E s : e 3 : 2. Let T, T',. . . t, t', . . . respectively, denote the similar tetrahedrons into which any two similar polyhedrons, P, p, . can be divided : L, L', . . . I, I', . . . respectively, homologous lines of the similar tetrahedrons. E' e' ' E" ~ e" E 3 : e 3 . L 3 : I 3 . Then, T :t :: L 3 : I 3 . T':t':: L'* : I' 3 . But, ( L :t ::L' : V ... \.-. L 3 : I 3 : : L' 3 : I' 3 ... T : t : : T ; f : : L 3 : I 3 . T+ T' +... : t + f + ... : : L 3 : I 3 . But. r t+ r +. ..=p."| I t + f +. . •=p.t •. P:p:: L 3 : I 3 . 288 GEOMETRY.— BOOK VI. IV. REGULAR POLYHEDRONS. 418. Definition. A regular polyhedron is a polyhedron all of whose faces are equal regular polygons, and all of whose polyhedral angles are equal. 419. Constructions. 1. To construct a regular tetrahedron. Construct the equilateral triangle BCD; at its center erect the per- pendicular OA, on which take the point A, whose distance from each of the vertices, B, C, D, is equal to a side of the triangle, and draw the straight lines, AB, AC, AD. The solid ABCD, thus determined, is a regular tetra- hedron. (?) 2. To construct a regular hexahedron. Construct a square ABCD, and an equal square on each side perpendicu- lar to the first. The upper sides of these squares will be the sides of an equal square. The solid ABCDF, thus determined, is a regular hexa- hedron. 3. To construct a regular octaliedron. Construct a square ABCD ; through its center pass a perpendicular EF to its plane ; from two points, E, F, on this perpendicular — one above and the other below the plane AC, whose JCr- I REGULAR POLYHEDRONS. 289 distances from each of the vertices. A, B, C, D, are equal to a side of the square — draw the straight lines, EA, EB, . . . FA, FB,... The solid EABCDF, thus determined, is a regular- octa- hedron. (?) 4. 2b cotistniet a regular dodecahedron. Construct a regular pentagon ABCDE, on each side of which construct an equal pentagon so inclined that trihedral angles shall be formed at A, B, C, D, E. The convex surface thus formed is composed of six regular pentagons. In like manner, upon an equal pentagon A'B'C'D'E' construct an equal convex surface. Apply one of these surfaces to the other, with their con- vexities turned in opposite directions, so that every isolated facial angle of one shall, with two consecutive facial angles of the other, form a trihedral angle. The solid thus determined is a regular dodecahedron. (?) 5. To construct a regular icosakedron. Construct a regular pentagon ABCDE; at its center erect a perpendicular to its plane; from a point P of this S. Q.-35. 290 GEOMETRY.— BOOK VI. perpendicular, whose distance from each of the vertices, A, B, C, D, E, is equal to a side of the pentagon, draw the straight lines, PA, PB, PC, PD, PE, thus forming a reg- ular pyramid whose vertex is P. Taking A and B as vertices, construct two pyramids, ABPEFO, BAGHCP, each equal to the first, and having two faces common with it, and two faces common with each other. There will thus be formed a convex surface com- posed of ten equal equilateral triangles. In like manner, upon an equal pentagon construct an equal convex surface. Apply one of these surfaces to the other, with their con- vexities turned in opposite directions, so that every com- bination of two facial angles of the one shall, with a combination of three facial angles of the other, form a pentahedral angle. The solid thus determined is a regular icosahedron. (?) 420. Proposition XXY— Theorem. Only five regular convex polyhedrons are possible. The faces must be equal regular polygons, and the poly- hedral angles convex, each having at least three facial angles. REGULAR POLYHEDRONS. 291 1. If the faces are equal equilateral triangles, convex polyhedral angles can be formed by arranging them in groups of three, four, or five, as in the tetrahedron, the octa- hedron, and the icosaliedron. No other combination of equilateral triangles can form a convex polyhedral angle; for, since each angle of an equi- lateral triangle is two-thirds of a right angle, a group of six or more would make the sum of the facial angles equal to or greater than four right angles, and therefore could not form a convex polyhedral angle, since the sum of the facial angles of a convex polyhedral angle is less than four right angles. 2. If the faces are equal squares, convex polyhedral angles can be formed by arranging them in groups of three, as in the regular hexaliedron or cube. A combination of four or more squares can not form a convex polyhedral angle. (?) 3. If the faces are equal regular pentagons, convex poly- hedral angles can be formed by arranging them in groups of three, as in the regular dodecahedron. A combination of four or more regular pentagons can not form a convex polyhedral angle. (?) 4. Convex polyhedral angles can not be formed with regular polygons having more than five sides ; (?) hence, polygons having more than five sides can not form the surface of regular polyhedrons, and therefore only five regu- lar convex polyhedrons are possible. Scliolium. — The regular polyhedrons can be formed thus : Draw the following diagrams on card-board ; cut through 292 GEOMETRY.— BOOK VI. in the exterior lines and half through in the interior; the cards cut out can be folded into the regular forms. Tetrahedron. Hexahedron. Octahedron. Dodecahedron. Jcosahedron. 421. Proposition XXVI.— Theorem. 1. The number of edges of a regular polyhedron is equal to one-half the product of the number of sides of one face by the number of faces. (?) 2. The number of vertices of a regular polyhedron is equal to the product of the number of vertices of one face by tiie number of faces divided by tlie number of facial angles at a vertex of the polyhedron. (?) 422. Table of Regular Polyhedrons. NAMES. PACES. ANGLES. EDGES. VERTICES. Tetrahedron 4 Triangles Trihedral 6 4 Hexahedron 6 Squares Trihedral 12 8 Octahedron 8 Triangles Tetrahedral 12 6 Dodecahedron 12 Pentagons Trihedral '30 20 Icosahedron 20 Triangles Pentahedral 30 12 SYMMETR Y.—SUPPLEMEXTAR Y. 293 423. Miscellaneous Exercises. 1. Define each of the regular polyhedrons. 2. Required the positions of four equidistant points. 3. Given the edge of a regular tetrahedron, required its altitude. 4. Required the entire surface and volume of a regular hexagonal pyramid whose altitude is 24 inches, and whose base is inscribed in a circle whose diameter ' is 16 inches. 5. The altitude of a regular tetrahedron is equal to the sum of the four perpendiculars from any point within to the four faces. V. SYMMETRY. — SUPPLEMENTARY. 424. Symmetry with respect to a Plane. 1. The symmetrical of a finite straight line with respect to a plane is an equal straight line. (?) 2. What is the symmetrical of an indefinite straight line with respect to a plane to which it is oblique, parallel, or perpendicular ? 3. The symmetrical of a polygon with respect to a plane is an equal polygon. (?) 4. What is the symmetrical of an indefinite plane with respect to a plane to which it is oblique, parallel, or per- pendicular ? 5. The symmetrical of a dihedral angle with respect to a plane is an equal dihedral angle. (?) 294 GEOMETRY.— BOOK VI. 6. Two polyhedrons symmetrical with respect to a plane have their homologous faces equal and their homologous polyhedral angles symmetrical. (?) 7. Two polyhedrons symmetrical with respect to a plane can be decomposed into the same number of tetrahedrons respectively symmetrical. (?) 8. Any two symmetrical polyhedrons are equivalent. (?) 425. Symmetry with respect to a Center. 1. The symmetrical of a polygon with respect to a center is an equal polygon. (?) 2. Two polyhedrons symmetrical with respect to a center have their homologous faces equal, and their homologous polyhedral angles symmetrical. (?) 3. The symmetrical of a polyhedron with respect to a center is equal to its symmetrical with respect to a plane. (?) 4. Two polyhedrons symmetrical with respect to a center are equivalent. (?) 426. Symmetry of a Single Figure. 1. The intersection of two planes of symmetry of a solid is an axis of symmetry. (?) 2. The intersections of three planes of symmetry of a solid are three axes of symmetry, and the common intersection of these axes is the center of symmetry. (?) BOOK VII. I. THE CYLINDER. 427. Definitions and Illustrations. 1. A cylindrical surface is a surface generated by a moving straight line which continually touches a given curve and has all of its positions parallel to a given straight line not in the plane of the curve. Thus, the surface ABCD, generated by the moving line AD continually touch- ing the curve ABC, and always parallel to a given straight line L, is a cylin- drical surface. 2. The moving line is called the generatrix; the curve which directs the motion of the generatrix is called the directrix; the generatrix in any position is called an element of the surface. 3. The generatrix , may be indefinite in extent, and the directrix a closed curve, as in the first diagram, or an open (295) 296 GEOMETRY— BOOK VII. curve, as in the second.. The only cylindrical surfaces con- sidered in this treatise are those whose directrices are closed 4. A cylinder is a solid bounded by a cylindrical surface and two parallel planes. 5. The lateral surface of a cylinder is its cylindrical surface. 6. The bases of a cylinder are the plane portions of its surface. 7. The axis of a cylinder is the straight line joining the centers of its bases. 8. The altitude of a cylinder is the perpendicular distance from one base to the plane of the other. 9. A section of a cylinder is a plane figure whose boundary is the intersection of its plane with the surface of the cylinder. 10. A right section of a cylinder is a section perpendic- ular to the elements. 11. A circular cylinder is a cylinder whose base is a circle. 12. The radius of a circular cylinder is the radius of the • 13. A right cylinder is a cylinder whose elements are perpendicular to its bases. Any element of a right cylinder is equal to its altitude. 14. An oblique cylinder is a cylinder whose elements are oblique to its bases. Any element of an oblique cylin- der is greater than its altitude. THE CYLINDER. 297 15. A cylinder of revolution is a cylinder generated by the revolution of a rectangle about one side as an axis. The axis of revolution is called the axis of the cylinder ; the side opposite generates the lateral surface ; the sides adjacent to the axis generate the bases ; the radius of cither base is the radius of the cylinder. 16. Similar cylinders of revolution are cylinders generated by similar rectangles re- volving about' homologous sides. 17. Conjugate cylinders are cylinders generated by the successive revolutions of a rectangle about two adjacent sides. 18. A tangent line to a cylinder is a line having only one point in common with the surface. 19. A tangent plane to a cylinder is a plane embracing an element of the cylinder without cutting its surface. 20. The element embraced by a tangent plane is called the element of contact. 21. A prism is inscribed in a cylinder when its lateral edges are elements of the cylinder, and its bases inscribed in the bases of the cylinder. 22. A prism is circumscribed about a cylinder when its lateral faces are tangent to the cylinder, and its bases cir- cumseribed about the bases of the cylinder. 428. Proposition I. — Theorem. Every section of a cylinder embracing an element embraces another element and is a parallelogram. 298 GEOMETRY— BOOK VII. Let ABCD be a section embracing AD, an element of the cylinder EF. Then it embraces an- other element; for, the element through B is in the cylindrical surface, and, since it is parallel to AD, it is also in the plane of the section, (?) and hence, is the intersection BG of the cylindrical surface and the section AG. Now, AD, BC, are parallel; also, AB, DC; (?) therefore, the section is a parallelogram. If the cylinder is right, the section AC is a rectangle. 429. Proposition II. — Theorem. The bases of a cylinder are equal. Any sections, AC, AG, embracing AD, an element of the cylinder EF, are parallelograms; (?) therefore, AB, DC, are equal ; also, AE, DG. Now, BC, EG, are equal and par- allel ; (?) .-. BE = CG; (?) hence, the triangles, ABE, DCG, are equal. (?) Let one base be applied to the other so that the equal triangles, ABE, DCG, coincide — -J3 coinciding with C, and E with G. Now, B, C, are in the same element; so also are E, G; therefore, any point in the perimeter of either base will coincide with the point in the same element in the other base. Hence, the bases coincide and are therefore equal. 430. Corollaries. 1. Any two parallel sections cutting all the elements of a cylindrical s.urface are equal. THE CYLINDER. 299 For, those sections are the bases of the cylinder included between them. '2. Any section of a cylinder parallel to Hie bate is equal to the bate. (?) 3. Any plane embracing an element and the center of either base of a cylinder, embraces the center of the other base, tlie axis, and the centers of all sections parallel to the base. (?) 4. . The axis of a cylinder pastes through the centers of all- sections parallel to the bases. (?) 431. Proposition III.— Theorem. The lateral area of a cylinder is equal to the product of the perimeter of a right section by an clement. Let / denote the lateral area of the cylinder AG; p, the perimeter of a right section IK; e, an element of the cylin- drical surface ; /', the lateral area of the inscribed prism BH; p', the perimeter of a right section of the prism ; e, an edge which is equal to an element. Then, I' =-- p' X e (379). Now, let the number of faces of the inscribed prism be indefinitely increased, the new edges continually bisecting the arcs in the right section; then, p' increases and approaches p as its limit, I' increases and approaches I as its limit, and e is constant. But, whatever be the number of foces, I'^p'X /=?Xf (301). 300 GEOMETRY.— BOOK VII. 432. Corollaries. 1. Tlw lateral area of a right cylinder is equal to the product of the perimeter of its base by its altitude. (?) 2. If s denote the entire area, b the area of either base, then, s = p X e + lb. 3. If I denote the lateral area, s the entire area, r the radius, a the altitude of a cylinder of revolution, then, I = 2nra, (?) s = 2-r (a + r). (?) 4. The lateral areas or the entire areas of similar cylinders of revolution are proportional to ilie squares of their radii or to the squares of their altitudes. Let I, V, denote the lateral areas ; s, s', the entire areas ; r, r 1 , the radii ; a, a', the altitudes, of two similar cylinders of revolution ; then, l___r"- 2-r (a -\- r) r a -|- r s' 2-r' (a' + /) r' ~ a'+ r' -r f r a a + r m But, - = —=— — ;• (?) r a a 4- r (?) (?) 433. Proposition IT. — Theorem. The volume of a cylinder is equal to the product of its base by its altitude. THE CYLINDER. 301 Let C denote the volume of the cylinder AG; b, its base; o, its altitude ; P, the volume of the inscribed prism BH; b', its base; a, its altitude, which is equal to that of the cylinder ; then, P = b' Xa (397). Now, let the number of faces of the inscribed prism be indefinitely increased, the new edges continually bisecting the arcs in the bases ; then, b' increases and approaches b as its limit, P increases and approaches C as its limit, and a is constant. But, whatever be the number of faces, P=U >< a; C=bXa (301). then, iM. Corollaries. 1. If the base of a cylinder is a circle whose radius is r, 2. The volumes of similar cylinders of revolution are propor- tional to the cubes of their radii or to €ie cubes of tlieir altitudes. Let 0, C", denote the volumes; r, r', the radii; a, a', the altitudes, of two similar cylinders of revolution ; then, C C ~~ sr'V ~~ r'2 X a'' But, r' a' [ ' C_ c 435. Exercises. 1. How can a cylindrical surface be generated by taking the generatrix for the directrix and the directrix for the generatrix ? 302 GEOMETRY.— BOOK VII. 2. What is the ratio of the lateral areas, the entire areas, and the volumes of the two conjugate cylinders generated by a rectangle whose base is b and altitude a? 3. The plane embracing an element of a circular cylinder and the tangent line to the base at the intersection of its circumference with the element, is tangent to the cylinder. 4. The intersection of two tangent planes to a cylinder is parallel to the elements. II. THE CONE. 436. Definitions and Illustrations. 1. A conical surface is a surface generated by a moving straight line continually touching a given curve and passing through a fixed point not in the plane of the curve. Thus, the surface generated by the moving line AA! , continually touching the curve A BCD, and passing through the fixed point V, is a conical surface. 2. The moving line is called the generatrix; the curve which directs its motion, the directrix; and any position of the generatrix, an element. 3. A conical surface generated by an indefinite line consists of two por- tions called nappes, — one the lower nappe, the other the upper nappe. 4. A cone is a solid bounded by a conical surface and a plane. THE CONE. 303 5. The lateral surface of a cone is its conical surface. 6. The base of a cone is the plane portion of its surface. 7. The vertex of a cone is the fixed point through which all the elements pass. 8. The altitude of a cone is the perpendicular distance from its vertex to the plane of its base. 9. The axis of a cone whose base has a center is the line joining that center with the vertex. 10. A section of a cone is a plane figure whose boundary is the intersection of its plane with the surface of the cone. 11. A right section of a cone is a section perpendicular to the axis. 12. A circular cone is a cone whose base is a circle. 13. A right cone is a cone whose axis is perpendicular to its base. The axis of a right cone is equal to its altitude. 14. An oblique cone is a cone whose axis is oblique to its base. The axis of an oblique cone is greater than its altitude. 15. A cone of revolution is a cone generated by the revolution of a right triangle about one of its perpendicular sides as an axis. 16. The side about which the triangle re- A volves is called the axis of the cone ; the // j ,\ / ; '< \ other perpendicular side generates the base ; j / j \ \ the hypotenuse generates the conical surface; /.-r— I— L„^ any position of the hypotenuse is an element ; ( / ' ) any element is culled the slant height 17. Similar cones of revolution are cones generated bv 304 GEOMETRY.— BOOK VII. the revolution of similar right triangles about homologous perpendicular sides. 18. Conjugate cones are cones generated by the successive revolutions of a right triangle about its perpendicular sides. 19. A truncated cone is the portion of a cone included between the base and a section cutting all the elements. 20. A frustum of a cone is a truncated cone in which the cutting section is parallel to the base. 21. The base of the cone is called the lower base of the frustum, and the parallel section the upper base. 22. The altitude of a frustum is the perpendicular distance from one base to the plane of the other. 23. The lateral surface of a frustum is the portion of the lateral surface of the cone included between the bases of the frustum. 24. The slant height of a frustum of a cone of revolution is the portion of any element of the cone included between the bases. 25. A tangent line to a cone is a line having only one point in common with the surface. 26. A tangent plane to a cone is a plane embracing an element of the cone without cutting the surface. The element embraced by a tangent plane is called the element of contact. 27. 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. 28. A pyramid is circumscribed about a cone when its lateral faces are tangent to the cone and its base is circum- scribed about the base of the cone. THE CONE. 305 437. Proposition V. — Theorem. Every section of a cone through its vertex is a triangle, Let VBD be a section of the cone V-ABC through the vertex V; then, VBD is a tri- angle. For, the straight lines joining V, B, and V, D, are elements of the surface ; (?) they also lie in the plane of the section ; (?) hence, they are the intersections of the con- ical surface with the plane of the section. BD is also a straight line ; (?) therefore the section VBD is a triangle. 438. Proposition VI— Theorem. Every section of a circular cone parallel to the base is a circle. The section EFG of the circular cone V-ABC, parallel to the base, is a circle. For, let be the center of the base, and I the point in which the axis VO pierces the parallel section. Planes embracing the axis VO and any elements, as VA, VB, intersect the base in the radii, OA, OB, and the parallel section in IE, IF, which are respectively parallel to OA, OB. (?) CIE-.OA: : VI : VO. I IF : OB :: VI : VO. But, OA = OB; .-. (?) IE: A:: IF: OB. IE — IF. Hence, any two straight lines drawn from I to the perimeter of the section are equal ; therefore the section is a circle whose center is /. S. G.-'JO. 306 GEOMETRY— BOOK VII. 439. Proposition Til.— Theorem. The lateral area of a cone of revolution is equal to one-half Hie product of the circumference of its base by its slant height. Let I denote the lateral area of the cone of revolution V-EFG; c, the cir- r cumference of its base ; h, the slant height, which is equal to an element of / the cone ; I', the lateral area of the reg- / ular circumscribed pyramid V-ABCD; ^(Tj p, the perimeter of its base; h, its Kf °^JJ^i3 slant height, which is equal to the slant a. f height of the cone. Then, I' = \p X h. Now, let the number of faces of the circumscribed pyramid be indefinitely increased, the new elements of contact continu- ally bisecting the arcs in the base; then, p diminishes and approaches c as its limit, V diminishes and approaches I as its limit, and h is constant. But, whatever be the number of faces of the pyramid, l'=$pXh; .-. i = i«X*. (?) 440. Corollaries. 1. If I denote the lateral area, s the entire area, r the radius, h the slant height of a cone of revolution, then, I = -rh, s — -xr (Ji -f- r). (?) 2. The lateral areas or the entire areas of similar cones of revolution are proportional to the squares of tlieir radii or to tlw squares of their slant heights. (?) (432, 4). 3. The lateral area of a frustum of a cone of revolution is equal to one-half tlie product of the sum of tlie circumferences of its bases by its slant height. THE CONK 307 Let I denote the lateral area of the frustum ; c, c', respect- ively, the circumferences of the lower and upper bases ; r, r', the radii of the bases ; h', h, h' — h, the slant heights of the cone, the frustum, and the part above the frustum. Then, I --^ -rh' — ->■' (h' — h) = - (rh' — r'h' + r'h). But, r : r' :: h' : h' — h ; (?) . • . rh' — rh = r'h'. rh' — r'h' =rh; . ■ . I = - {rh + r'h). £ = -(r + r')/t = i(e + c')h. (?) 4. The lateral area of a frustum of a cone of revolution is equal to the product of the circumference of a section equidistant from its bases by its slant height. Let r 1 ' denote the radius of this section ; c", its circum- ference ; ?•, r', the radii of the bases. Then, r" = i (r + r') ; (?) . ■ 2r" = r + ?•'. .-. l = 2-r"h = c"h. 441. Proposition VIII.— Theorem. Any section of a cone parallel to the base is to the base as the square of the altitude of tlie part above the section is to the square of the altitude of tlie cone. Let b denote the base of the cone J\ V-AC; a, its altitude; V , the section EG //A of the cone, parallel to the base ; a', the ^"CLiA altitude of the cone above the section; / -—-*5\ b,, the base of the inscribed pyramid; A k'i J^c ft.,, the section EFGH of the pyramid; ij^-^ «, a', the altitudes of the pyramid and the part above the section. Then, b 2 : b x : : a 2 ; a- (402, 1). 308 GEOMETRY.— BOOK VII. Now, let the number of faces of the inscribed pyramid be indefinitely increased, the new edges continually bisecting the arcs in the base of the cone; then, b x , b 2 , increase and approach their respective limits, b, b'. But, whatever be the number of faces of the pyramid, 6, : 6, :: a' 2 : a 2 ; .-. U : b : : a' 2 : a 2 . (?) 442. Proposition IX.— Theorem. The volume of a cone is equal to one-third of the product of its base by its altitude. Let C denote the volume of the cone V-AC; b, its base; a, its altitude; P, the volume of the inscribed pyramid V-ABCD; V , its base ; a, its altitude, which is equal to the altitude of the cone. Then, P = W xa. Now, let the number of faces of the inscribed pyramid be indefinitely increased, the new edges continually bisecting the arcs in the base of the cone ; then, V, P, increase and approach their respective limits, b, C. But, whatever be the number of faces of the pyramid, P = iV X a; .-. G = J6 X a. 443. Corollaries. 1. Any two cones are proportional to the products of their bases by their altitudes; (?) if their bases are equivalent, they are pro- portional to their altitudes ; (?) if their altitudes are equal, they are proportional to their bases. (?) 2. If C, r, a, respectively denote the volume, radius, and altitude of a circular cone, C=^r*a. (?) THE CONE. 309 3. Similar cones of revolution are proportional to the cubes of their altitudes or to the cubes of Vieir radii. (?) 4. Tlie volume of a frustum of any cone is equal to one-third of the product of its altitude by the sum of its lower base, its upper base, and a mean proportional between its bases. Let F denote the volume of the frustum ; b, V , its bases ; a', a, a' — a, the altitudes of the cone, the frustum, and the part above the frustum. Then, as in 407, 4, let the student develop the formula, F=$a(b + V + l '66'). 5. If F denote the volume of a frustum of a cone of revolution, a its altitude, r, r', the radii of its bases, 6 = -r\ b' = -r'~, i '66' = -it'. 444. Exercises. 1. What is the ratio of the lateral areas, the entire areas, and the volumes of the two conjugate cones generated by a triangle whose base is 6 and altitude a? 2. The plane embracing an element of a circular cone and the tangent line to the base at the intersection of its cir- cumference with the element is tangent to the cone. 3. A section of any cone parallel to the base is similar to the base. 4. How many barrels of water will that cistern contain the altitude of which is 8 ft., the diameter at the bottom 4 ft., and at the top 6 ft.? Am. 37. S bbl. 5. The volume of a frustum of a cone of revolution is 7050 cu. in. ; its altitude, 12 in. ; the diameter of the lower 310 GEOMETRY.— BOOK VII. base twice that of the upper base. What are the diameters of the^ bases? Ans. 35.8096 in., 17.9048 in. 6. If a frustum of a cone of revolution of which the alti- tude is 10 ft., and the diameters of the bases 6 ft., 4 ft., respectively, be divided into two equal parts by a plane parallel to its bases, what would be the altitude of each part? Ans. 4.0375 ft., 5.9625 ft. III. THE SPHERE. I. Sections and Tangents. 445. Definitions. 1. A sphere is a solid bounded by a surface all of whose points are equally distant from a point within, called the center. A sphere may be generated by the revolution of a semicircle about its diameter as an axis. 2. A radius of a sphere is the distance from the center to any point of the surface. All the radii of a sphere are equal. (?) 3. A diameter of a sphere is any straight line through the center whose extremities are in the surface. All the diameters of a sphere are equal, since each is twice the radius. 4. A plane section of a sphere is a plane figure whose boundary is the intersection of its plane with the surface of the sphere. 5. A line or plane is tangent to a sphere when it has only one point in common with the surface of the sphere. THE SPHERE. 311 6. Two spheres are tangent to each oilier when their sur- faces have only one point in common. 7. A polyhedron is circumscribed about a spliere when all of its faces are tangent to the sphere. In this case, the sphere is inscribed in the polyhedron. 8. A polyhedron is inscribed in a spliere when all its ver- tices are in the surface of the sphere. In this case, the sphere is circumscribed about the polyhedron. 9. A cylindrical or conical surface is circumscribed about a sphere when all its elements are tangent to the sphere. 10. A cylinder or a cone is circumscribed about a spliere when its bases and cylindrical surface, or the base and con- ical surface, are tangent to the sphere. In this case, the sphere is inscribed in the cylinder or cone. 446. Proposition X.— Theorem. Every plane section of a sphere is a circle. The plane section EFG of a sphere whose center is is a circle. For, from the center draw 01 perpendicular to the section, and the radii, OE, OF, OG, . . . to different points of the bound- ary of the section. The radii, OE, OF, OG, . . . are equal ; (?) therefore, IE, IF, IG, . . . are equal ; (?) hence, the section EFG is a circle whose center is J, the foot of the perpendicular 01. 312 GEOMETRY.— BOOK VII. 447. Corollai'ies and Definitions. 1. The line joining the centers of a spliere and a circle of the sphere is perpendicular to the circle. (?) 2. If r, r', p, respectively denote the radius of a sphere, the radius of a circle of the sphere, and the perpendicular from the center of the sphere to the circle, then, r' = yr 2 — p 2 . 3. All circles of a sphere equally distant from the center are equal. (?) 4. As p increases to r, r 1 diminishes to ; and as p di- minishes to 0, r' increases to r; hence, 5. A great circle of a sphere is a circle passing through the center. 6. A small circle of a sphere is a circle not passing through the center. 7. All great circles of a sphere are equal. (?) 8. Any two great circles of a sphere bisect each other. (?) 9. Any great circle of a spliere bisects the sphere. (?) 10. An arc of a great circle can be drawn through any two points on the surface of a sphere ; (?) and only one, if the points are not at the extremities of a diameter ; (?) but an indefinite number, if the points are at the extremities of a diameter. (?) 11. An arc of a circle can he drawn through any three points on the surface of a sphere. (?) 12. An axis of a circle of a sphere is the -diameter of the sphere perpendicular to the circle. 13. The poles of a circle of a sphere are the extremities of its axis. THE SPHERE. 313 14. .1 pole of a circle of a sphere is equally distant from all the points in, tlie circumference of the circle. (?) lo. The poles of a great circle are equally distant from its circumference. (?) 16. The poles of a small circle are unequally distant from its circumference. (?) 17. All arcs of great circles drawn from a pole of a circle to points in tlie circumference of tlie circle are equal. (?) 18. The distance on the surface of a sphere, from one point of tlie surface to another, is usually understood to be {he arc of a great circle having the points for its extremities. 19. The polar distance of a circle of a sphere is the arc of a great circle drawn from any point of the circumference of the circle to its nearest pole. 20. The polar distance of a great circle is a quadrant of a great circle. (?) 21. The point on Hie surface of a spliere which is at a quadrant's distawe from tivo points, not at tlie extremities of a diameter, in an arc of a great circle, is the pole of the. arc. (?) 22. By means of these properties, arcs of great or small circles can be drawn with great facility on a spherical black- board, thus : 23. To draw an arc of a great circle through two given points on the surface of a sphere, describe about each of the points, as a pole, with an arc of a great circle equal to a quadrant, the arc of a great circle; then, the great circle described about the point of intersection of these arcs, as a pole, with an arc of a great circle equal to a quadrant, will pass through the two points. s. G.— 27. 314 GEOMETRY— BOOK VII. 448. Proposition XL — Theorem. A plane perpendicular to a radius at its extremity is tangent to the sphere at that point. Let be the center of a sphere, and MN a plane perpendicular to the radius OA at its .extremity A ; then, MN is tangent to the sphere ^ \___/^__l__ /___ J_^ at A. For, to any other point B, in the plane MN, draw the straight line OB. Since OA is perpendicular to MN, OB is oblique ; (?) . • . OB > OA; (?) hence, B is without the sphere ; (?) therefore, the plane MN has only one point in common with the sphere; (?) hence, MN is tangent to the sphere at A. 449. Corollaries. 1. A plane tangent to a sphere is perpendicular to the radius drawn to the point of contact. (?) 2. A straight line tangent to a circle of a sphere lies in the plane tangent to the sphere at the point of contact. (?) 3. Any straight line in a tangent plane through the point of contact is tangent to the sphere at that point. (?) 4. The plane of any two straight lines tangent to a spliere at the same point is tangent to the sphere at that point. (?) 450. Exercises. 1. The intersection of the surfaces of two spheres is the circumference of a circle, and the straight line joining the centers of the spheres is perpendicular to the circle at its center. THE SPHERE. 315 2. Given a material sphere, to find its radius. 3. Pass a plane tangent to a given sphere and embracing a given straight line without the sphere. 4. Through any four points not in the same plane one spherical surface can pass, and only one. 5. The four perpendiculars to the planes of the faces of a tetrahedron, at their centers, meet at the same point. 6. The six planes perpendicular to the edges of a tetra- hedron, at their middle points, meet in the same point. 7. A sphere can be inscribed in any tetrahedron. 8. The six planes bisecting the dihedral angles of a tetra- hedron meet in the same point. 9. The shortest distance on the surface of a sphere from one point of the surface to another is the arc of a great circle, not greater than a semi-circumference, whose extremi- ties are these points. II. Spherical Axgles axd Triangles. 451. Definitions and Remarks. 1. The angle of two curves having a common point is the angle of their tangents at that point. 2. A spherical angle is the angle of two arcs of great circles of a sphere. 3. A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. 4. The sides of a spherical polygon are the bounding arcs ; the angles are the angles included by consecutive sides ; the vertices are the intersections of the sides. 316 GEOMETRY.— BOOK VII. 5. A diagonal of a spherical polygon is an arc of a great circle dividing the polygon and terminating in two vertices not consecutive. 6. The planes of the sides of a spherical polygon form by their intersections a polyhedral angle whose vertex is the center of the sphere, and whose facial angles are measured by the sides of the polygon. 7. A spherical pyramid is a portion of a sphere bounded by a spherical polygon and the planes of the sides of the polygon. The spherical polygon is the base of the pyramid, and the center of the sphere is its vertex. A spherical pyramid is designated according to its base, as triangular, quadrangular, . . . 8. A convex spherical polygon is a spherical polygon whose corresponding polyhedral angle is convex. The only spherical polygons here considered are those in which no side exceeds a semi-circumference. 9. A spherical triangle is a spherical polygon of three sides. 10. A spherical triangle, like a plane triangle, is right or oblique, scalene, isosceles, or equilateral. 11. Two spherical triangles are symmetrical, if their suc- cessive sides and angles taken in a reverse order are equal. 12. Two spherical pyramids are symmetrical, if their bases are symmetrical triangles. (?) 13. The polar of a spherical triangle is a spherical tri- angle the poles of whose sides are respectively the vertices of the given triangle. THE SPHERE. 317 452. Proposition XII —Theorem. A spherical angle measures the dilwdral angle included by the planes of its sidcx. By definition, the spherical angle BPC is the angle TPT' of the tangents to its sides, BP, CP, at their point of intersection P. Since the tangent to an arc lies in the plane of that arc, and is perpendicular to the diameter drawn to the point of contact, the sides, TP, T'P, of the angle TPT', re- spectively, lie in the planes of the sides of the spherical angle BPC, and are both perpendicular, at the same point, to the diameter PP', which is the edge of the dihedral angle included by the planes of these sides, since they are arcs of great circles. Hence, the angle TPT' measures the dihedral angle included by the planes of the sides, BP, CP. 453. Corollaries. 1. The angles of a splierical polygon respectively measure the diJicdral angles of the corresponding splierical pyramid. (?) 2. .'1 spherical angle is measured by the arc of a great circle described about its vertex as a pole, and intercepted by its sides produced, if nec-essanj. For, if DE is the arc of a great circle described about the vertex P as a pole, the arcs, PD, PE, are quadrants; hence, DO, EO, are perpendicular to PP' ; (?) therefore, DOE measures the dihedral angle BPOC, and hence, the spherical angle BPC. Therefore, the arc DE, which meas- ures the angle DOE, measures the spherical angle BPC. 318 GEOMETRY.— BOOK VII. 3. The angle included by two ares of small circles is the same as the spherical angle having ilie same vertex, and whose sides Jiave the same tangents at the vertex. (?) 4. Two circumferences of great circles of a sphere intersect at right angles if either passes through the pole of the other, and conversely. (?) 5. The arc of a great circle described about a point on a given arc of a great circle as a pole, passes through all the points at a quadrant's distance from the pole, and is perpen- dicular to the given arc. (?) i I 454. Proposition XIII. — Theorem. If one triangle is the polar of another, the second is the polar of the first. The triangle A'B'C, c" r -'~" "~--4- whose sides are arcs of \ great circles described \ about the vertices, A, B, C, as poles, is, by defini- tion, the polar of the tri- angle ABC; then, ABC is the polar of A'B'C. \ /' For, since B is the I" pole of the arc A'C, and C the pole of the arc A'B', A' is at a quadrant's distance from each of the points B, C, and is, therefore, the pole of the arc BC. Also, B' is the pole of the arc AC, and C the pole of the arc AB; hence, ABC is the polar of A'B'C. Scholium. — The arcs of great circles described about A, B, C, as poles, will, if sufficiently produced, form three tri- angles exterior to the polar. THE SPHERE. 319 The polar triangles are distinguished by having their homologous vertices, A, A', on the same side of BC and of B'C ; B, B', on the same side of AC and of A'C ; C, C, on the same side of AB and of A'B'. 155. Proposition XIV. — Theorem. Each angle in either of two polar triangles is the supplement of the side opposite in Uie other, and each side is Hie supplement of the angle opposite in the oilier. Let ABC, A'B'C, be polar tri- angles in which a, b, c, and a', b', c', respectively, denote the sides opposite the angles A, B, C, and A', B', C". Since B" is the pole of the arc AC, and C the pole of the arc. AB, En = 90°, mC = 90° ; B'n + mC'= 180°. But, B'C'+ inn = B'm + mn -\- nC + mn, or, a' -f mn = B'n + mC'= 180° ; mn — ISO — a'. But mn measures the angle A ; (J? - ) A = 180° — a' ; Also, B = 180° — V ; C = 180° — e : ,4'= 180° — a; #=180° — b; C'=180° — c; a'= 180°— A. b' = 180° — J5. e' = 180° — C. a = 180° — A', b = 180° — F. c = 180° — C. In consequence of these relations, polar triangles are also called supplemental triangles. 320 GEOMETRY— BOOK VII. 456. Corollary. Tlie trihedral angles formed by tlie radii drawn from the center of the sphere to the vertices of the angles of two polar triangles are supplemental. (?) (358, 24), (451, 6), (453, 1). 457. Proposition XV. — Theorem. Two triangles on the same sphere or on equal spheres, having two sides and the included angle of one respectively equal to two sides and the included angle of the other, are either equal or symmetrical. 1. Let the two sides, AB, AG, and the in- cluded angle A, of the triangle ABC, be re- spectively equal to the two sides, DE, DF, and the included angle D, of the triangle DEF, and arranged in the same order. Then, these triangles are equal, for they can be applied so as to coincide. (?) (55). 2. If the equal parts are arranged in a reverse order, as in the triangles ABC, D'E'F', let DEF be the symmet- rical triangle of D'E'F' ; then, ABC and DEF are equal, since they can be made to coincide ; hence, ABC and D'E'F' are symmetrical. 458. Proposition XVI. — Theorem. Two triangles on the same spliere or on equal spheres, having a side and two adjacent angles of one respectively equal to a side and two adjacent angles of the other, are either equal or symmetrical. (?) (57). THE SPHERE. 321 459. Proposition XVII. — Theorem. Two mutually equilateral triangles on the same sphere or on equal splieres are mutually equiangular, and either equal or symmetrical. For, the facial angles of the corresponding trihedral angles at the center of the sphere are respectively equal, since they are measured by equal sides of the triangles ; hence, the dihedral angles are respectively equal (873) ; consequently, the angles of the triangles are respectively equal ; therefore, the triangles are equal or symmetrical, according as their equal sides are arranged in the same or in a reverse order. 4(J0. Proposition XVIII— Theorem. Two mutually equiangular triangles on the same sphere or on equal splteres are mutually equilateral, and either equal or symmetrical. Let T, T', be mutually equiangular spherical tri- angles; P, P', their re- spective polars. Since T, T', are mut- ually equiangular, P, P', are mutually equilateral ; (?) therefore, P, P', are mutually equiangular ; (?) hence, T, T', are mutually equilateral ; (?) consequently, T, T', are either equal or symmetrical. (?) 461. Corollaries. 1. Two mutually equiangular triangles on unequal spheres are not mutually equilateral, but their sides are respectively propor- tional to the radii of tlie spheres. (?) 322 GEOMETRY.— BOOK VII. 2. Either of two mutually equiangular triangles on unequal spheres is similar to the other or to the symmetrical of the other, according as the equal angles are arranged in the same or in a reverse order. (?) 462. Proposition XIX. — Theorem. The angles opposite the equal sides of an isosceles spherical triangle are equal. In the isosceles spherical triangle ABC, ji let AB = AC; then, B = C. For, the arc AD of a great circle drawn / from the vertex A to the middle point D I of the base BC, divides the triangle ABC IJ into two triangles, ABD, ACD, which are mutually equilateral, and hence, mutually equiangular; (?) hence, B and C are equal, since they are opposite the common side AD. 463. Corollaries. 1. The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the middle point of the base bisects tlie vertical angle, is perpendicular to the base, and divides tlie triangle into two symmetrical -triangles. (?) 2. Symmetrical isosceles spherical triangles are equal. (?) 464. Proposition XX. — Theorem. If two angles of a spherical triangle are equal, the sides oppo- site these angles are equal, or the triangle is isosceles. In the spherical triangle ABC, let B = C; then, AC = AB. THE SPHERE. 323 For, let A'B'C be the polar triangle of ABC, and denote the sides of the triangles respectively opposite the angles. A, B, C, and A', B', t", by a. b, c, and a', V , c' ; then, A b'= 180° — B; \ c' = 180° — 0; J .-. 6'=c'; 6 = 180° — -B' ; ) 180° ■ ( ';3 but, B=C; but, 2> v = C" ; 6 = c, or ;1C = AB. 165. Proposition XXI. — Theorem. Any side of a spherical triangle is less titan the sum of Hie other sides, and greater tfian tlieir difference. Let ABC be a spherical triangle _a on the sphere whose center is O. ,. ,\ Since any facial angle of the cor- / „ \ responding trihedral angle is less than ° \ ~f -- i c the sum of the other two, the side of ■- — — -<"<' the triangle which measures it is less than the sum of the sides which measure the other facial angles. r a < 6 + e ; . • . 6 > a — c, e ^> a — 6. 6 < a + e ; .". a > 6 — e, e>6 — a. , e < a + 6 ; .'. a > o — 6, 6 > c — -a. 466. Corollary. Any side of a spherical polygon is less than tiie sum of the oilier sides. (?) 324 GEOMETRY— BOOK VII. 467. Proposition XXII. — Theorem. In a spherical triangle, the greater side is opposite the greater angle, and the greater angle is opposite the greater side. 1. In the triangle ABC, let B> C; then, AC > AB. For, draw the arc BD of a great circle, making the angle CBD=BCD; then, DC = DB, (?) AD+DB>AB; .-. AD + DOAB; .-. AC> AB. 2. Let AC > ,4-B; then, B > C. For, if £ = C, AC = AB; if B < C, AC < AB. But these results are contrary to the hypothesis; hence, B is neither equal to C nor less than C; .-. B > C. 468. Proposition XXI1L— Theorem. The sum of the sides of a splierical triangle can have any value between the limits 0° and 360°. For, these sides respectively measure the facial angles of the corresponding trihedral angle at the center of the sphere ; but the sum of these facial angles can have any value be- tween the limits 0° and 360° (371) ; hence, the sum of the sides can have any value between these limits. 469. Corollaries and Definitions. 1. The sum of the angles of a splierical triangle can liave any value between the limits two right angles and six right angles. THE SPHERE. 325 For, employing the usual notation for a spherical triangle and its polar, ,1 = 180° — a', B = 180° — b', C= 180° — c'. .-. A + B + C = 540° — (a' + V + c'). But a' -\- b' -j- c' can have any value between 360° and 0° ; .". A -\- B -\- C can have any value between 180° and 540°. 2. One side of a splierical triangle can be a quadrant, two can be quadrants, or tliree can be quadrants. For, one of the facial angles of the corresponding trihe- dral angle can be a right angle, two can be right angles, or three can be right angles. 3. A spherical triangle is qvadrantal, bi-quadrantal, or tri- quadrantal, according as one of its sides is a quadrant, two are quadrants, or three are quadrants. 4. A spherical triangle can have one inght angle, two right angles, or three, right angles. For, one side of its polar triangle can be a quadrant, two can be quadrants, or three can be quadrants. 5. A spherical triangle is rectangular, bi-rectangular, or tri- rectangular, according as it has one right angle, two right angles, or three right angles. 6. A bi-quadrantal triangle is bi-rectangular. For, the vertex between the quadrants is the pole of the opposite side ; hence, the planes of the quadrants embrace the axis of the great circle of which the third side is an arc, and the planes of the quadrants are perpendicular to the plane of the third side; hence, the triangle is bi- rectangular. 326 GEOMETRY— BOOK VII. 7. A bi-rectangular triangle is bi-quadrantal. For, its polar triangle is bi-quadrantal, (?) and hence, bi- rectangular ; therefore, the given triangle is bi-quadrantal. 8. A tri-quadrantal triangle is tri-rectangular. For, each angle is measured by its opposite side, which is a quadrant. 9. A tri-rectangular triangle is tri-quadrantal. For, its polar triangle is tri-quadrantal, (?) and hence, tri-rectangular ; hence, the given triangle is tri-quadrantal. 10. A tri-rectangular triangle and its polar are coincident. For, each vertex is the pole of the opposite side. 11. A trisrectangular triangle is one-eighth of the surface of the sphere. For, the three great circles, each of which is perpendicular to the plane of the other two, divide the surface of the sphere into eight tri-rectangular triangles, which are equal, since they can be made to coincide. 12. One of tlie sides of a spherical triangle can be greater than a quadrant, two can be greater than quadrants, or three can be greater than quadrants. (?) 13. A spherical triangle can have one obtuse angle, two obtuse angles, or three obtuse angles. (?) 470. Proposition XXIV— Problem. To describe the arc of a circle through three given points on Vie surface of a spliere. Let A, B, C, be the three given points 1 . About A, B, THE SPHERE. 327 as poles, with quadrants, describe arcs intersecting in D; about B, C, as poles, with quad- rants, describe arcs intersecting in E. The arc of a great circle de- scribed about D as a pole will pass through A, B. (?) The arc of a great circle described about E as a pole will pass through B, C. The arc DH of a great circle described about the pole F on BA produced, at a quadrant's distance from H, the middle point of AB, is perpendicular to AB (453, 5). The arc EI of a great circle described about the pole G on CB produced, at a quadrant's distance from J, the middle point of BC, is perpendicular to BC. Any point of DH is equally distant from A, B ; (?) any point of EI is equally distant from B, C ; therefore, P, the intersection of DH, EI, is equally distant from A, B, C\ hence, the circle described about P as a pole, with the arc of a great circle equal to the distance from P to one of the points, A, B, C, will pass through A, B, C. The circle will, in general, be a small circle. 471. Proposition XXV.— Theorem. Symmetrical upheriral triangles are equivalent. Let ABC, A'B'C, be two , symmetrical spherical triangles having ABBA'S', AC=A'C, BC = B'C, A=A', B=B', C=C; then, ABC, A'B'C, are equivalent. / IV- 328 GEOMETRY— BOOK TIL For, let P, P', be the poles of the small circles whose circumferences pass through A, B, C, and A', B', C, respect- ively. These small circles are equal, x(~ — | ~^ p p ^ since they are circumscribed about plane triangles whose sides are respectively equal, being chords of equal arcs. Draw arcs of' great circles from the poles, P, P', to the vertices, A, B, C, and A', B', C, respectively. The triangles, PAB, P'A'B', are symmetrical and isos- celes; so also are PBC, P'B'C, and PAC, P'A'C. .-. PAB = P'A'B', PBC = P'B'C, PAG = P'A'C; .• . PAB + PBC — PAC = P'A'B' + P'B'C — P'A'C ; .-. ABC = A'B'C. If the poles are within the triangles, the three equations must be added. 472. Corollary. Two symmetrical spherical pyramids are equivalent. 473. Exercises. 1. Construct a right spherical triangle on a spherical black-board. 2. Construct a tri-rectangular triangle. 3. Construct a spherical triangle having two sides respect- ively 90°, 45°, and their included angle 60°, and construct its polar. 4. Construct a spherical triangle whose sides are 30°, 60°, 90°, respectively, and describe a circumference through its vertices. THE SPHERE. 329 5. The sum of two arcs of great circles drawn from any point within a triangle to the extremities of any side is less than the sum of the other sides. 6. Two spherical triangles whose vertices are respectively the opposite extremities of three diameters are symmetrical. 7. The angle of two intersecting curves on the surface of a sphere is equal to the dihedral angle of the planes em- bracing the tangents to the two curves at their point of in- tersection and the center of the sphere. 8. If two spherical triangles on the same sphere or on equal spheres have two sides of the one respectively equal • to two sides of the other, and the included angles unequal, the third sides are unequal and the greater third side be- longs to the triangle having the greater included angle, and conversely. HI. Measurements of the Sphere. 474. Definitions. 1. A lune is a portion of the surface of a sphere bounded bv two semi-circumferences of great circles. 2. The angle of a lune is the angle included by the semi- circumferences which form its boundary. 3. A spherical ungula, or wedge, is a portion of a sphere bounded by a lune and two great semicircles. 4. The base of an ungula is the bounding lune. fi. The angle of an ungula is the dihedral angle of its bounding semicircles, and is equal to the angle of its bound- ing lune. (?) 6. The edge of an ungula is the edge of its angle. S. G.— 2S. 330 GEOMETRY.— BOOK VII. 7. The spherical excess of a spherical triangle is the sum of its angles minus two right angles. 8. The spherical excess of a spherical polygon is the sum of its angles, minus two right angles taken as many times less two as the polygon has sides. 9. A zone is a portion of the surface of a sphere inter- cepted by the circumferences of two parallel circles of ' the ■ sphere. 10. The bases of a zone are the circumferences of the intercepting circles. If the plane of one base becomes tangent to the sphere, that base becomes a point, and the zone is said to have only • one base. If the planes of both bases become tangent to the sphere, the bases become points at the extremities of a diameter, and the zone becomes the surface of the sphere. 11. The altitude of a zone is the perpendicular distance from one base to the plane of the other. 12. A spherical segment is a portion of a sphere bounded by a zone and the two parallel circles whose circumferences are the bases of the zone. 13. The bases of a spherical segment are the bounding circles. A spherical segment has only one base when its bounding zone has only one base, and becomes the sphere when its bounding zone becomes the surface of the sphere. 14. The altitude of a spherical segment is the perpendic- ular distance from one base to the plane of the other. 15. A spherical sector is a portion of a sphere generated by a circular sector of the semicircle which generates the sphere. 16. The bounding surfaces of a spherical sector are, in THE SPHERE. 331 general, three curved surfaces, — two conical surfaces gen- erated by the radii of the generating circular sector, and the zone generated by the arc of the circular sector. 17. The base of a spherical sector is the bounding zone. 18. When will one of the conical surfaces of a spherical sector become a line? when a plane? when will one be convex and the other concave? when will both be concave? when will both be lines, and what will the sector be? 475. Proposition XXVI.— Theorem. A I une is to the surface of the spliere as tlie angle of the lune is to four right angles. Let L denote the lune ABA'CA, whose angle is A ; S, the surface of the sphere ; BCD, a great circle whose pole is A ; and R, a right angle. The arc BC measures the angle A of the lune ; the circumference BCD measures four right angles. 1. If the circumference BCD and the arc BC are com- mensurable, suppose the circumference divided into m equal parts, and that n of these parts are contained in the are BC. Planes through A A' and the respective points of division divide the surface of the sphere into m equal lunes; (?) the lune L contain' n of these lunes. L : S : : » : m ; ) [ .-. L : S :: A • 41?. A : 4R : : h : m ; ) 2. If the circumference BCD and the arc BC are incom- mensurable, the proposition can be proved by the method employed in 150, 2. 332 GEOMETRY.— BOOK VII. 476. Corollaries. 1. Two limes on the same sphere or on equal splwres are proportional to their angles. (?) 2. A lune is equivalent to the product of a tri-rectangular triangle by twice the angle of the lune, the right angle being taken as tlie unit of angles. For, let L denote a lune ; A, its angle ; T, a tri-rectangular triangle ; S, the surface of the sphere. Then, S=8T; .-. L : 8T :: A : 4; .-. L = T X 2A. 3. An ungula is equivalent to the product of a tri-reetangular spherical pyramid by twice the angle of the ungula, the right angle being tlie unit. (?) 477. Proposition XXVII— Theorem. If two semi-eircumfereiices of great circles intersect on tlie surface of a hemisphere, the sum of tlie opposite triangles thus formed is equivalent to a lune whose angle is equal to that included by the semi-circumferences. Let the semi-circumferences, BAD, CAE, intersect at A on the surface of a hemisphere ; then will the sum of the opposite triangles, BAG, DAE, be equivalent to a lune whose angle is BAG. For, the semi-circumferences pro- duced around the sphere intersect on the opposite hemisphere at A'. All great circles bisect each other; hence, each of the arcs, AD, A'B, is the supplement of AB ; . • . AD = A'B. THE SPHERE. 333 Likewise, AE — A'C, DE=BC; hence, the triangles, ADE, A'BC, arc symmetrical, and therefore equivalent; hence, ABC + ADE = ABC + A'BC = ABA'CA, a lune whose angle is Uvl C. 478. Corollary. The sum of two splierical pyramids, Hie sum of whose bases is equivalent to a lune, ts equivalent to ait, ungula whose base is Hie lune. (?) 479. Proposition XXVIII.— Theorem. The area of a spherical triangle is equal to its spherical excess multiplied by the area of the tri-rectangular triangle. Let ABC be a spherical triangle. Complete the circumference ABDE; produce AC, BC, to meet this circum- ference in D, E. ABC + BCD = lune A = Tx2A, ABC + ACE = lune B=T\2B, ABC + DCE = lune C =Tx 2C. 2ABC + ABC + BCD + ACE-j-DCE=Tx 2(A+B + C). But, ABC + BCD + ACE + DCE=4T. (?1 .-. 2ABC+4T =Tx2(A+B + C~). ABC=T(A+S + C—2). Tims, if .4^100°, £ = 120°, ('=140°, a right angle being the unit, 140° ABC \ 90° 100° a 120 c 90° + 90° '_>] = 2T. 334 GEOMETRY.— BOOK VII. 480. Corollaries. 1. The volume of a triangular spherical pyramid is equal to the splierical excess of its base multiplied by the area of the tri- rectangular spherical pyramid. (?) 2. Two triangular splierical pyramids, on the same sphere or on equal spheres, are proportional to their bases. (?) 481. Proposition XXIX— Theorem. The area of a spherical polygon is equal to its spherical excess multiplied by the area of the tri-rectangular triangle. Let P denote the area of a spherical polygon ; S, the sum of its angles ; n, the number of its sides ; t, t', t", . . . the areas of the triangles formed by drawing diagonals from any vertex ; s, s', s", . . . respectively, the sums of the angles of these triangles; T, the area of the tri-rectangular triangle. Then, t = (s — 2)T, If = (s' — 2)T, if' = (a" — 2)T, . . . t + t' -\-t" +... =[s+a' + s"+... — 2(n — 2)]T. But, t + t' +t" + ... =--P, s +s' +«"+...= S. P = [S— 2(n — 2)]3T. 482. Corollaries. " 1. .Aim/ too spherical pyramids, on the same sphere or on equal spheres, are proportional to their bases. (?) 2. Any splierical pyramid is to the sphere as its base is to the surface of the sphere. (?) 483. Proposition XXX.— Lemma. The area of the surface generated by the revolution of a straight line about an axis in the same plane, is equal to the THE SPHERE. 335 product of the projection of Hie line on the axis by the circum- ference 10/iose radhts is Hie perpendicular to the line erected at its middle point and terminated by the axis. Let the straight line AB revolve about the axis IT in the same plane, and let /! PQ be its projection on the axis ; CO, the %rr perpendicular to AB at its middle point 0, B /_ L^. terminating in the axis. Then, area AB = PQ X 2-.OC. The surface generated by AB is the lateral surface of a frustum of a cone of revolution ; hence, drawing CR per- pendicular and AD parallel to XI', (1) area AB = ABx2-.CR (440, 4). The triangles, ABD, COB, are similar; (?) AD : AB :: CR : OC. But, AD = PQ, CR : OC :: 2- .CR : 2-.0C*. PQ : AB :: 2- .CR : 2-. OC. PQX2-.0C = AB X2--CR. . ■ . (2) area AB = PQx'2~- OC. If either extremity of AB is in the axis XI' AB gen- erates the lateral surface of a cone of revolution, and for- mula (2) still holds true. (?) If AB is parallel to the axis XI", it generates the lateral surface of a cylinder of revolution, and formula (2) still holds true. (?) If AB is perpendicular to the axis XI', it generates a circular ring ; PQ becomes zero ; CO becomes infinite ; formula (2) gives an indeterminate result, and therefore fails; but formula (1) holds true. (?) 336 GEOMETRY.— BOOK VII. 484. Proposition XXXI— Theorem. The area of tlie surface 'of a sphere is equal to the product of its diameter by the circumference of a great circle. Let the semicircle ABD, and any reg- ular inscribed semi-polygon, as ABCD, revolve together about the diameter AD. The semi-circumference will generate the surface of a sphere, and the semi- perimeter, a surface equal to the sum of the surfaces generated by its sides. OE, OF, OG, drawn from the center to the middle points of the chords, AB, BC, CD, are perpendicular to the chords and equal to each other. (?) .-. area.AB = APx2-.OE; (?) also, area BC = PQ X 2* .OE; (?) and, area CD = QD X 2*.OE; (?) . • . area ABCD = (AP + PQ + QD) X 2* . OE. . ■ . area ABCD = AD X 2- . OE. Now, if the number of sides of the regular inscribed semi- polygon be indefinitely increased, the surface generated by the semi-perimeter will approach the surface of the sphere as its limit, and OE will approach OA. .-. area of surface of sphere = AD X 2th. A (301). / If s denote the surface, d the diameter, c the circum- ference, , s = d X c. 485. Corollaries. 1. The surface of a sphere is equivalent to four great circles. For, if r denote the radius of the sphere, s = dXc = 2rX 2kv = 4w 2 . THE SPHERE. 337 2. The surfaces of two spheres are proportional to Hie squares of their radii. For, s : s' :: 4-r 2 : 4~r' 2 : : r 2 : r' 2 . 3. The surface of a sphere is equivalent to a circle wlwse radius is equal to the diameter of Hie sphere. For, s = 4-r 2 = -(2r) 2 = -d 2 . 4. The surfaces of two spheres are proportional to the squares of their diameters. For, 8 : s' : : -d 2 : -d' 2 : : d 2 : d' 2 . 486. Proposition XXXII— Theorem. The area of a zone is equal to the product of its altitude by the circumference of a great circle. If the semicircle DBE, the arc AC, ~ together with the equal chords, AB, BC, ZT" revolve about the diameter DE as an af-—^ axis, the semi-circumference will gen- \\ erate the surface of a sphere, the arc c ^5$ AC a zone, and the chords, AB, BC, a surface whose area is expressed by the formula, area ABC = PQ X 2*. OF. (?) If the number of equal arcs into which AC is divided be increased indefinitely, the surface generated by the chords of these arcs will approach the zone as its limit, the per- pendicular OF will approach the radius of the sphere .as its limit, and PQ will remain constant. .-. area zone AC = PQ X 2-. OZ). (?) Denoting the area of the zone by z, its altitude by a, and the radius of the sphere by r, the above formula becomes z = 2*ra- s. G— 29. 338 GEOMETRY— BOOK VII. 487. Corollaries. 1. Zones on the same sphere or on equal spheres are propor- tional to their altitudes. (?) 2. A zone is to the surface of ihe sphere as the attitude of the zone is to the diameter of the sphere. (?) 3. A zone of one base is equivalent to the circle whose radius is the chord of tlie generating arc. For, zone DA = DP X 2* . OD ; zone DA = -.DP X DE = -.DA 2 (240, 3, 2d). 4. If a becomes 1r, z becomes s, the surface of the sphere. Then, z = 2-ra becomes s = 4~r 2 . 488. Proposition XXXIII.— Lemma. The volume generated by the revolution of a plane triangle about an exterior axis, in the plane of the triangle and through its vertex, is equal to the product of the area generated by the base by one-third of tlie altitude of the triangle. 1. When one side of the triangle lies in the axis : E A Let the side AB of the triangle ABC lie in the axis XY. Draw AD perpendicular to BC, and CE perpendicular to XY. The volume generated by ABC is the sum or the differ- ence of the volumes of the cones generated by the right THE SPHERE. 339 triangles, BEC, A EC, according as CE is within or without the triangle. But, vol. BEC=i-.EC 2 X BE, (?) and, vol. AEC = i-.EC 2 X AE\ (?) . • . vol. BEC ± vol. AEC = §-.EC 2 X {BE ± AE). Taking the plus signs for the first figure, and observing that BE + -AE = AB, and taking the minus signs for the second figure, and observing that BE — AE = J.U, we have vol. ABC = is. EC" X AB = |-.£CX«X AB. But EC X AB = BC X AD, since each is twice the area of the triangle ; .-. vol. ABC = ^-.EC X BCX AD. But -. EC X BC = area of surface generated by BC (439) ; .• . vol. ABC = area BC X J AD. 2. When only the vertex is in the axis, and the base not parallel to the axis : In this case, the volume gen- erated by ABC is the difference of the volumes generated by the triangles, ACF, ABF. But, vol. ACF = area CF X i AD, and vol. ABF= area BF X I AD. . • . vol. ACF— vol. ARF=(area CF — area BF) X JAD. vol. ABC = area £C X * AD. 340 GEOMETRY— BOOK VII. 3. When only the vertex is in the axis, and the base parallel to the axis: In this case, the volume generated by the triangle ABC is the sum or the difference of the volumes generated by the right triangles, ADB, ADC, according as the perpen- dicular AD is within or without the triangle. Drawing BE, CF, perpendicular to the axis, the volume generated by the right triangle ADB is the volume of the cylinder generated by the rectangle ADBE, minus the vol- ume of the cone generated by AEB. But, vol. ADBE = -.AD 2 X BD, (?) and vol. AEB ■■= £-.AD 2 X BD. (?) .-. vol. ADBE — vol. AEB --= f-.AD 2 X BD. I*. AD 2 XBD = 2- .ADxBDx §AD = area BD X \AD. vol. ADB — area BD X\AD. Likewise, vol. ADC — area CD X IAD. . • . vol. ADB ± vol. ADC = area (BD ± CD) X \AD. Taking the plus signs for the first figure, and observing that area (BD-{-CD) = area BC, and taking the minus signs for the second figure, and observing that area (BD ^- CD) = area BC, and that the first member is equal to vol. ABC, we have vol. ABC = area BC X $AD. THE SPHERE. 341 489. Corollary. In each of the three preceding cases, if the triangle ABC is isosceles, AD is the perpendicular to BC erected at its middle point and terminated by the axis. Denoting this perpendicular- by p, and the projection of BC on the axis A' I" by a, we have area BC = a X 2^p = 2-pa. (483). .• . vol. ABC — 2-pa X ip — ?-p 2 a. 490. Proposition XXXIV.— Theorem. The volume of a sphere is equal to die product of the area of its surface by one-third of its radius. Let the semicircle ABD, and any reg- ular inscribed semi-polygon, as A BCD, with radii drawn to the vertices, revolve together about the diameter AD as an axis. The semicircle will generate a sphere, and the semi-polygon, a volume equal to the sum of the volumes generated by the triangles, AOB, BOC, COD. OE, OF, 00, drawn from the center to the middle points of the chords, AB, BC, CD, are perpendicular to the chords and equal to each other. (?) vol. AOB =areaJJ5 X §OE, (?) also, vol. BOC = area BC X iOE, (?) and vol. COD = area CD X $OE, (?) vol. ABCD = area ABCD X i OE. Whatever be the number of sides of the regular semi- polygon, the volume generated is equal to the product of 342 GEOMETRY.— BOOK VII. the area of the surface generated by the semi-perimeter by one-third of the apothem. Now, if the number of sides be increased indefinitely, the volume generated will approach the volume of the sphere as its limit, the surface generated will approach the surface of the sphere, and the apothem will approach the radius. .•. vol. of sphere = area of surface X $ radius (303). Denoting the volume of the sphere by v, the area of its surface by s, the radius by r, we have v = s X ir. 491. Corollaries. 1. (1) v = far* ; (?) (2) v = J*d». (?) 2. The volumes of two spheres are proportional to the cubes of their radii or to the cubes of their diameters. (?) 492. Proposition XXXV— Theorem. The volume of a spherical sector is equal to the product of the area of the zone which forms its base by one-third of the radius of the sphere. If the semicircle DBE, the arc AG, and the equal triangles, AOB, BOC, revolve about the diameter DE as an axis, the triangles will generate a vol- ume expressed by the formula, vol. OABC= area ABC X &0F. (?) Whatever be the number of equal triangles, the volume generated will be equal to the area of the surface generated THE SPHERE. 343 by their bases, multiplied by one-third of their common altitude. Now, if the number of equal triangles be increased in- definitely, the volume generated will approach the spherical sector OAC as its limit, the surface generated by the bases will approach the zone generated by the arc AC, and the common altitude will approach the radius. .•. vol. spherical sector OAG = zone AC X $ radius (303). Denoting the volume of the spherical sector by v, the area of the zone which forms its base by z, the radius by r, we have v = z X $r. 493. Corollaries. 1. If a denotes the altitude of the zone, z = 2-ra; . •. v = §-r 2 •' = 0, the segment has only one base, and v = i«.-r s -|- 4~ a3 - 2. If -1 coincides with G, and .B with JH", then f = 0, »• = 0, a = rf, the diameter of the sphere, the segment becomes the sphere, and v = $-nd s . 498. Formulas for the Sphere. 1. r = id. 11. c = 1 ' -$. 2- r=^- 12. c — f 6v--. 3. r =4 X <|- 13. 14. $ = 4-r-- 8 = "d 2 - s ,'Gc 4. r .-= A ^/ — • 15. _ C 2 5. d = 2r. 16. s = t 36;rw 2 . 6. d = f -- 17. t> = |^- 3 . 18. „ = *«/«. 8. d^Vl- 19. c 3 9. c = 2-r. 10. c = -rf. 20. * <8 346 GEOMETRY.— BOOK VII. 499. Exercises. 1. If L denotes the area of a lune, A its angle, r the radius of the sphere, prove that L = A-r 2 . 2. If P denotes the area of a spherical polygon, e its spherical excess, r the radius of the sphere, prove that P = i- r 2 e . 3. The volume of any spherical pyramid is equal to the product of the area of its base by one-third of the radius of the sphere. 4. If v denotes the volume of a spherical pyramid, e the spherical excess of its base, r the radius of the sphere, prove that v = ^-r s e. 5. A sphere 6 in. in diameter is bored through the center with a 3-inch auger ; required the volume remaining. 6. Find the ratio of the surfaces of a sphere and the circumscribed cube, also the ratio of their volumes. 7. Find the ratio of the surfaces of a sphere and the circumscribed cylinder, also the ratio of their volumes. 8. The diameter of a hollow sphere of glass is 4 ft., the thickness of the shell is 1 in. If this sphere be melted and run into a solid sphere, what would be its diameter? 9. The volume of a cone is 1,000 cu. in., the radius of its base is 6 in. ; required the volume of the circumscribed sphere. 10. The surface of a sphere is s ; what is the surface of a sphere whose volume is n times as great? 11. The volume of a sphere is v; what is the volume of a sphere whose surface is n times as great? BOOK Till. MODERN GEOMETRY. I. TEANSVERSALS. 500. Definitions. 1. A transversal is a straight line intersecting a system of lines. Thus, TV, intersecting two sides of the triangle ABC and the third side produced, or the three sides produced, is a trcuisversal of the triangle. A transversal through a vertex is an angk-tmnsvcmil. 2. The segments of the sides are the distances of the points of intersection of the transversal and sides from the vertices. Thus, Ac, Be; Ab, Cb; Ba, Ca. (347) 348 GEOMETRY—BOOK VIII. 3. Adjacent segments are segments having a common extremity. Thus, Ac, Be; Ab, Cb; Ba, Ca; Ab, Ac; Ba, Be; Ca, Cb. 4. Non-adjacent segments are segments not having a common extremity. Thus, Ac, Ba, Cb; Ab, Be, Ca. 5. A complete quadrilateral is the figure formed by four straight lines intersecting in six points. Thus, ABCDEF is a complete quadrilateral; AC, BD, EF, are its diagonals, the external diago- nal EF being called the third diagonal. 501. Proposition I.— Theorem. A transversal of a triangle divides the sides into segments such that the product of three non-adjacent segments is equal to the product of the other three segments. Let TV be a transversal of the triangle ABC. Draw CD parallel to AB; then, from similar triangles, Ba : Ca : : Be : CD, Cb : Ab : : CD : Ac; BaXCb : Ab X Ca : : Be : Ac; AcXBaX Cb=Ab X BcXCa. TRANSVERSALS. 349 502. Corollary. Three points in the sides of a triangle, one heing in a side produced, or all iliree being in the sides pfrodueed, dividing tlie sides so tliat the 'product of tliree non-adjacent segments is equal to the product of tlie other tliree segments, are in the same straight line. Let a, 6, e, be points such that (1) Ac X Ba X Cb = Ab X Be X Ca. Let the straight line ab cut the third side in c'; then, (2) Ac'X BaX Cb = AbX Be' X Ca (501). (l)--(2) = (3) Ac : Ac':: Be : Be'. This is true only when c, c', coincide. (Axg., 317, 3; 116, 1.) But abc' is a straight line; .•. abc is a straight line. 503. Proposition II.— Theorem. Tlie three angle-tramversah of a triangle through a common point divide the sides so tliat Hie product of Uiree non-adjacent segments is equal to tlie product of the other three segments. Let Aa, Bb, Ce, through P be angle-transversals of ABC. The transversals, Bb, Ce, of the triangles, AaC, ABa, give APXaB xCb = Ab X aP X CB; Ac X BCX aP= APX Be X aC; .-. Ac X Ba X Cb = Ab X Be X Ca. 350 GEOMETRY.— BOOK VIII. 504. Corollaries. 1. Three angle-transversals of a triangle, dividing the sides so tliat the product of three non-adjacent segments is equal to the product of the other three segments, pass through a common point (?) 2. If one of three angle-transversals of a triangle, through the same point, bisects one side, the line joining Hie points of division of the other sides is parallel to that side. (?) 3. If three angle-transversals of a triangle pass through the same point, and the line joining the points of division of two sides is parallel to the third side, this side is bisected by the other point. (?) 505. Exercises. 1. The three medial lines of a triangle pass through the same point (51, 17), (504, 1). 2. The three bisectors of the angle of a triangle pass through the same point (221), (504, 1). 3. The three altitudes of a triangle pass through the same point (227, 1), (504, 1). 4. The three angle-transversals of a triangle to the points of tangency of the inscribed circle pass through the same point (176, 1), (504, 1). 5. The transversal through the intersection of the non- parallel sides of a trapezoid and the intersection of the diagonals bisects the parallel sides (503), (217). 6. The middle points of the three diagonals of a complete quadrilateral are in the same straight line (501), (502). HARMONIC PROPORTION. 351 II. HARMONIC PROPORTION. 506. Definitions. 1. Three quantities are in harmonic propoHion if the differ- ence of the first and second is to the difference of the second and third as the first is to the third ; and the second is the harmonic mean between the first and third. 2. A pencil is a system of lines diverging from a point. 3. A ray is one of the lines of a pencil. 4. The vertex of a pencil is the point from which the rays diverge. 5. A harmonic pencil is a pencil of four rays dividing a transversal harmonically (250, 3, 4, 5, 6). 6. Conjugate rays are the alternate rays of a harmonic pencil. 7. Two circles intersect each other orthogonally if the tan- gents at the common point are at right angles. 507. Proposition in.— Theorem. The distance between tlw points of division of a line divided harmonically is tlie harmonic mean between the distances from the extremities of 2/ie line to tJie point of division not behveen them. Let C, D, divide AB j- d 5 3 harmonically. Then, AC : BC :: AD : BD. That is, AD— CD : CD — BD : : AD : BD. .•. CD is the harmonic mean between AD and BD. 352 GEOMETRY.— BOOK VIII. 508. Corollary. A line divided harmonically is the harmonic mean between the distances from the extremity wot between, the points of division to the points of division (250, 4), (507). 509. Proposition IV. — Theorem. Every transversal of a harmonic pencil is divided harmonic- ally at the points of intersection. If the pencil V-ACBD divides AB harmonically, it will divide any other transversal A'B' harmonic- ally. Through B, B', draw MN, M'N', parallel to AV. .-. AC : BC : : AV : BM, AV : BN : : AD : BD. By hypothesis, we have AG : BC :: AD : BD; •. BM = BN; .■ Similar triangles give the proportions, A'C : B'C :: A'V : B'M', AD' : B'D' : : A'V Since B'M' = B'N', A'C : B'C : : A'D' Hence, A'B' is divided harmonically at C, U. 510. Exercises. 1. If the straight line AB is divided internally at G and externally at D, so that CD is the harmonic mean between AD and BD, then AC X BD = BC X AD (506, 1). AV : BM :: AV B'M' = B'N' (237). BN; B'N'. B'H. ANHARMONIC RATIO. 353 2. If the straight line AD is divided by the points B, C, so that AG X BD = BC X AD, then AB is divided har- monically at the points C, D, and CD at the points A, B. 3. Any diagonal of a complete quadrilateral is divided harmonically by the other two. 4. If two circles cut each other orthogonally, any diameter of either intersecting the circumference of the other is divided harmonically by that circumference, and conversely (248). in. ANHARMONIC RATIO. 511. Definitions. 1. The anharmonic ratio of four points in a straight line is the ratio of the rectangle of the distances between the first and third and the second and fourth to the rect- angle of the distances between the first and fourth and the second and third. Thus, AC X BD : AD X BC is the anharmonic ratio of the - 1 no d points A, B, C, D, which is briefly expressed by writing the letters in order in brackets, thus, [ABCD-] = AC X BD + AD X BC. 2. The anharmonic ratio of a pencil of four rays is the anharmonic ratio of the four points of intersection of these rays by any transversal. 3. The anharmonic ratio of four points in the circum- ference of a circle is the anharmonic ratio of the pencil formed by joining these points with any point in the cir- cumference. S. G.— 30. 354 GEOMETRY.— BOOK VIII. 4. The anharmonic ratio of four tangents is the an- harmonic ratio of the points of intersection of the four tangents by any* fifth tangent. 512. Proposition V.— Theorem. The anliarmonic ratio of four points is not changed by inter- changing two of the letters designating the points, provided the other two are . also interchanged. \_ABCD~] = AC X BD : AD X BG. [BADC~] = BD X AG : BC X AD. [CDAB] = CA X DB : CB X DA. \DCBA~] = DB X CA : DA X CB. .-. IABCD] = [BADC] = \CDAB~] = [DGBA]. 513. Scholiums. 1. Since four letters taken all in a set have twenty-four permutations, four points have six different anharmonic ratios, each having four expressions. AC AD 2. [ABCD] =AC X BD : AD X BC=2Q : ^- AC AD 3. If four points, A, B, C, D, are harmonic, -^- = -^-j- , and -jj~ : 777; = 1. But four points taken at random in BC bD a straight line are not, in general, harmonic, -=^ is not , t AD , AC AT) . , equal to ^y; , and -=-^ : ^yr is anharmonic. 514. Proposition VI. — Theorem. Any anharmonic ratio of the points of intersection of a pencil of four rays by a variable transversal is constant. ANHARMONIG RATIO. 355 Let ABCD, A'B'G'D', be any two positions of a variable transversa] of the pencil V-LMNP. Draw Bd, B'd', parallel to AV; then, by similar tri- angles, AV _AC BC (1) (3) (1) -j- (2) = (5) AD Be A'V A'C Be' Bd Be (2) 4H C ' Bd ~ BD' B'C" (4) A'V A'V Bd' ~ BD' ' AC X BD _ AD X BC - t^BCDJ- BV Vc' A'C'X B'H [A'BC'D'l (3) - (4) = (6) Wc , A , Ux BC , {ABCD'] = [A'B'C'D'l „ , Bd B'd' ,„-„. But - ^ = w< 237); 515. Corollaries. 1. If Hie cuigles of two pencils are respectively equal, ilieir anharmonic ratios are equal. 2. The anliarmonic ratio of four fixed points in tlie circum- ference of a circle is constant (511, 3), (154), (515, 1). 356 QEOMETRY.—BOOK VIII. 516. Proposition VII.— Theorem. The anharmonic ratio of four fixed tangents to a circle is constant. Let A, B, C, D, be the points of tangency of the fixed tangents ; L, M, N, P, the intersections of these tangents by a fifth tangent. Then, by 176, 3, LVM is measured by \AT—\BT=\AB, (?) MVN is measured by \BT + \ TC = \BC, (?) NVP is measured by \DT— \CT = IDC. (?) Hence, the angles of the pencil V-LMNP are constant; therefore, the anharmonic ratio [LMNP] is constant (515, 1). 517. Corollary. The anharmonic ratio of four tangents to a circle is equal to the anharmonic ratio of the four points of tangency. (?) 518. Proposition VIII. — Theorem. If two pencils have the same anharmonic ratio and a common homologous ray, the intersections of the other homologous rays are in the same straight line. Let the pencils V-ABCD, V'-ABCD, have the same enharmonic ratio and a com- mon homologous ray, VA. Let the transversal through B, C, two points of inter- section of homologous rays, intersect the ray VA in A, the ray VD in D', the ray V'D ANHARMOMC RATIO. 357 in D". Then, [ABCD 1 ] = [>1BCZ)"] ; (?) hence, D, D', are coincident; (?) and, since D' is in VD, D" in VD, both X>', D", must be at D, the intersection of VD, VD. Hence, the points B, C, D, are in the same straight line. 519. Corollary. If one anliarmonie ratio of a pencil is equal to one anliar- monie ratio of anotiier pencil, tlie other anliarmonie ratws of ilie first are respectively equal to tliose of tlie second. (?) 520. Proposition IX. — Theorem. If two straight lines have one anliarmonie ratio of four points of Hie one equal to one anliarmonie ratio of four points of tlie other, and. two homologous points coincident, the straight lines through the other three pairs of homologous points meet in a common point. Let [ABCD] = [AE'CU] ; A, the coincident points; V, the intersection of B'B and C'C. Draw VA, YD', and denote the point in which VD' cuts AB by D". Then, [ABCD"] = [AB'C'D'] (514). [ABCD"] = [ABCD]. Hence, D", D, coincide; (?) .". the straight line DD passes through V. 521. Corollary. If one anliarmonie ratio of a system of four points is equal to one a>ikarmonic ratio of another system, the other anharmonic 358 GEOMETRY.— BOOK VIII. ratios of the first system arc. respectively equal to those of the second. (?) 522. Proposition X.— Theorem. The intersections of the three pairs of opposite sides of an inscribed hexagon lis in the same straight line. The pencils formed by drawing rays from B, F, to A, C, D, E, have the same anharmonic ratio (515, 2). The first is cut by the transversal LPDE, the second by NCDQ; .-. [LPDE] = [NCDQ]; -■ . LN, PC, EQ, meet in the same point M (520); .-. L, M, N, the in- tersections of the three pairs of opposite sides of the inscribed hex- agon, lie in the same straight line. 523. Corollaries. 1. The intersection of one side of an inscribed pentagon with the tangent at the opposite vertex, and the intersection of the other non-consecutive sides, are tliree points in the same straight line. For, as the vertex D of the last figure approaches C, the side CD approaches the tangent at C; but the theorem for the hexagon holds for all positions of D up to C, when the hexagon becomes a pentagon and CD a tangent. 2. If tangents be drawn at two consecutive vertices of an in- scribed quadrilateral, the point of intersection of each with the AXHAJRMOXIC RA HO. 359 side through tlie point of tangency of the other, and Hie inter- section of tlie other two sides, are three points in the same- straight line. This becomes evident by supposing D in the hexagon to move to 0, and E to F. 3. The intersections of the tangents draicn at the opposite vertices of an inscribed quadrilateral, and tlie intersections of the pairs of opposite sides, are four points in Vie same straight line. For, if in the inscribed hexagon F move to A, and D to C, the intersection of the tangents at the opposite vertices, A, C\ will be on. the straight line through the intersections of the opposite sides of the quadrilateral ABCE; but the same quadrilateral can be obtained from another hexagon so as to form tangents at the other opposite vertices, B, E; but the intersection of these tangents is on the line through the intersections of the opposite sides ; hence, the four points of intersection are in the same straight line. 4. The intersections of the sides of an inscribed triangle- with the tangents at tlie opposite vertices lie in the same straight line. This becomes evident by supposing F in the hexagon to move to A, C to B, E to D. Remake. — Theorem X is due to Pascal, and Theorem XI to Brian- chon. These theorems with their corollaries afford a beautiful exempli- fication of the method of anharmonic ratio, and will well repay tlie student for the labor which their mastery will require. 524-. Proposition XI.— Theorem. The tiiree diagonals joining the opposite vertices of a circum- scribed hexagon pass through the same point. 360 GEOMETRY.— BOOK VIII. Regard AB, BC, CD, EF, as fixed tangents, cut by the tangent LN in L, N, D, E, and by the tangent FM in A, P, M, F; then, (516), \_LNDE-] = \_APMF\ Hence, the pencils, B-LNDE, C-APMF, have equal an- harmonic ratios and a common homologous ray PN; there- fore the intersections, A, D, 0, of BL, CA ; BD, CM; BE, CF, lie in the same straight line (518) ; hence, the diago- nals, AD, BE, CF, pass through the same point. 525. Corollaries. 1. The line joining a vertex of a circumscribed pentagon and the point of tangency of the opposite side, and the diagonals drawn from the extremities of this side to the extremities of the adjacent sides, meet in the same point. This becomes evident by moving C to the circumference. 2. The lines joining the points of tangency of each of two ANHABMONIC RATIO. 361 adjacent sides of a circumscribed quadrilateral with the vertex at tlie extremity of the oilier, ami the diagonal joining the other ver- tices, meet in Hie same point. This becomes evident by moving C and E to the cir- cumference. 3. The diagonals of a circumscribed quadrilateral and tlie lines joining tlie points of ta>igency of the opposite sides meet at the same point. For, if C and F move to the circumference, the line joining the points of tangency of two opposite sides passes through the intersection of the diagonals of the quadrilateral ; but the same quadrilateral can be obtained from another hexagon by moving two vertices to the points of tangency of the other opposite sides, the diagonal joining these ver- tices becoming the line joining these points of tangency ; but this line passes through the intersection of the diagonals of the quadrilateral ; hence, the diagonals and the lines joining the points of tangency of the opposite sides meet in the same point. 4. The straight lines joining tlie points of tangency of a cir- cumscribed triangle rvith Hie opposite vertices pass through the same point. This becomes evident by moving the alternate vertices of the circumscribed hexagon to the circumference. 526. Exercises. 1. If the straight lines through the corresponding vertices of two triangles meet in the same point, the intersections of their corresponding sides lie in the same straight line (518). 2. If the intersections of the corresponding sides of two triangles lie in the same straight line, the straight lines S. G.-31. 362 GEOMETRY.— BOOK VIII. through their corresponding vertices meet in the same point (514). 3. If the three sides of a triangle pass through three fixed points in a straight line, and two vertices of the tri- angle move in two fixed straight lines, the third vertex moves in a straight line which passes, through the intersec- tion of the other two lines (52G, 2). 4. If the three vertices of a triangle move in three fixed straight lines which meet in a point, and two sides of the triangle pass through two fixed points, the third side passes through a fixed point which is in the straight line passing through the other two (526, 1). IV. POLE AND POLAR TO THE CIRCLE. 527. Definitions. 1. A point is the pole of a fixed line, or a line is the polar of a fixed point, with respect to a circle, if the line and point are so situated that the chord of a revolving secant through the point is divided harmonically at the fixed point and the intersection of the fixed line with the secant. 2. The polar point is the intersection of the polar with the diameter through the pole. Thus, P is the pole of the fixed line GC, and CC the POLE AND POL AM TO THE -CIRCLE. 363 polar of the fixed point P, if P and CC are so situated that the chord AB of a revolving secant through P is divided harmonically at'P and C (250, 3); and C" is the polar point, if PC passes through the center. 528. Proposition XII.— Theorem. The polar of a given point with respect to a circle is a straight line perpendicular to the diameter through the point at the harmonic conjugate of the point ivith respect to ilie extremities of that diameter. Let P be the given point; 0, the center of the circle; AB, the chord of any secant through P; C, the harmonic conjugate of P with respect to A and B. Draw CC per- pendicular to the diameter A'B' through P, and BC meeting the circumference in D; also, AC, BA', BB'. Since PCC is a right angle, C" is on the circumference whose diameter is PC (179, 3) ; and, since AB is divided harmonically at P and C, CP bisects the angle ACT) (260). Hence, the arcs, AA', A'D, are equal; (?) therefore, BA' bisects the angle PBC ; hence, BB', perpendicular to BA', bisects the angle exterior to PBC (34, 4). Therefore, PC is divided harmonically at A', B' (221), (223), (250, 3) ; hence, G" is the harmonic conjugate of P with respect to A', B' ; and, since P is fixed, C is fixed; hence, CC is the polar of P (527, 1). 364 GEOMETRY.— BOOK VIII. 529. Corollaries. 1. The radius of the circle is a mean proportional between the distances of the pole and its polar from the center. (?) 2. If the pole is without the circle, its polar is tlw chord joining the points of tangency of the tangents drawn from the pole. (?) 3. If the pole is on the circumference, its polar is the tangent at the pole. (?)■ 4. The pole and polar point are interchangeable. (?) 530. Proposition XIII.— Theorem. 1. The polar of every point of a straight line passes through the pole of that line. 2. The pole of every straight line passing through a given point is on the polar of that point. 1. Let CC be a straight line; P, its pole; and C, any point of CC OC through P is perpendicular to the polar CC (528), and C" is the polar point. Draw PP' perpendicular to OC ; .-. OC : OP :: OC : OP'; .-. OCxOP'=OP XQC But, OPxOC'=OA 2 (529,1); .-. OC xOP'=OA 2 . Therefore, PP' through P is the polar of C. RECIPROCAL POLARS. 365 2. Let P be a given point; CO', its polar; and PP', any straight lino through P. Draw OC perpendicular to PP'. Thru oc x op' =.- or. Therefore, C, on the polar of P, is the pole of PP'. 531. Corollaries. 1. The pole of a stmight line is tte intersection of Hie polars of any two of its points. (?) 2. The polar of any point is the straight line joining the poles of any two straight lilies Hirough that point. (?) 0. If the pole is within Hie circle, its polar is Hie locus of the intersection of Hie pair of tangents at the extremities of any chord through the pole. (?) 4. //' the vertices of one of two polygons are respectively Hie poles of the sides of the oHier, the vertices of the second are respectively the poles of Hie- sides of Hie first (?) 5. If through a fixed point in the plane of a circle any two secants be drawn, and their intersections iviHi the circumference be joined by chords, the locus of Hie intersections of these chords is the polar of the fixed point. (?) V. RECIPROCAL POLARS. 532. Definitions. 1. Reciprocal polars are two polygons so related that the vortices of either are respectively the poles of the sides of the other with respect to the same circle, which is called the auxiliary circle. 2. A reciprocal theorem is a theorem inferred from another by means of reciprocal polars. 366 GEOMETRY.— BOOK VIII: 533. Proposition XIV— Problem. Infer Brianckon's theorem (524) from Pascal's (522). Chords joining the consecutive points of tangency of the sides of the circumscribed hexagon form an inscribed hexa- gon, the vertices of which are respectively the poles of the sides of the circumscribed hexagon (529, 3). Hence, the vertices of the circumscribed hexagon are respectively the poles of the sides of the inscribed hexagon (531, 4) ; there- fore, the pole of each of the three diagonals joining opposite vertices of the circumscribed hexagon is on each of two opposite sides of the inscribed hexagon (530, 2), and con- sequently at their intersection. Hence, by Pascal's theorem, the poles of the three diagonals lie on the same straight line ; therefore, the three diagonals, which are the polars of these poles, pass through the pole of this straight line (530, 1). Let Pascal's theorem be inferred from Brianchon's. VI. EADICAL AXES. 534. Definitions. 1. The power of a point in the plane of a circle is the rectangle of the segments, external or internal, into which the point divides the chord passing through it. Thus, the power of P is PA X PB. The power is RADICAL AXES. 367 positive or negative, according as the point is without or within the circle, since, in the first case, the segments have like signs, and in the second case, unlike. 2. The radical axis of two circles is the locus of the point whose powers with respect to the circles are equal. 535. Proposition XV.— Theorem. The power of a given point with respect to a given circle is the constant equal to the square of tlie distance from the point to the coda; minus tlw square of the radius. In the diagrams (534), let p denote the power; d, the distance from the point to the center; and r, the radius. The power is equal to the rectangle of the segments of the chord through the center (247), (245) ; then, p = (d — r) (d + r) —d 2 — r 2 , 1st diagram ; p — — (r — d) ()• + d) = d°- — r", 2d diagram. 536. Corollaries. 1. The power of a point without tlie circle is equal to Hie square of tiie tangent drawn from that point (248). 2. The power of a point within the circle is equal to minus the square of half {he least clwrd through tlie point (136, 3), (246). 3. The power of a point on the circumference is equal to zero. (?) 4. The power of tlie center is equal to minus tlw square of tlie radius. (?) 537. Proposition XVI.— Theorem. The radical axis of two circles is the straight line perpendic- ular to tlie straight line joining their centers, dividing it so tiiat 368 GEOMETRY.— BOOK VIII. the difference of the squares of the segments is equal to the differ- ence of the squares of the radii. Let 0, (J, be two circles ; r, /, their radii ; P, any point of the radical axis ; d, d', the distances of P from the centers. Then, d 2 — r 2 = d' i — r' 2 (535); .-. d 2 — d' 2 =r 2 — r' 2 . Let PR be perpendicular to 00'. Then, OR 2 = PO 2 — PR 2 , OR? = P0 2 OR 2 — O'R 2 : PO ■ P0 2 = d*- — PR; d' 2 = r 2 — r n Hence, PR, perpendicular to 00' at R, is the radical axis, since it embraces any point P of that axis. 538. Corollaries. 1. There is an infinite number of pairs of circles having the same radical axis and their centers in the same straight line. (?) 2. The radical axis of hvo circles external to each other lies between them, tangent to neiOier. (?) 3. The radical axis of two intersecting circles is the straight line through their points of intersection. (?) 4. The radical axis of two circles tangent externally or inter- nally is the common tangent at their point of tangency. (?) 5. The radical axis of two circles, one of which lies wholly within the other, is exterior to both circles. (?) CENTERS OF SIMILITUDE. 369 6. The tangents to two circles drawn from any point of their radical axis are equal. (?) 7. The radical axis bisects any common tangent. (?) 8. The radical axes of a system of tiiree circles, taken two and two, meet in a common point called tlw radical center. (?) VII. CENTERS OF SIMILITUDE. 539. Definitions. 1. The external and internal centers of similitude of two circles are two points which divide the line joining their centers harmonically in the ratio of the two radii. 2. Homologous points are the alternate intersections of two circumferences with a transversal through the external center of similitude, or the mean and extreme intersections of two circumferences with the transversal through the in- ternal center of similitude. 3. Anti-homologous points are the extreme and mean intersections of two circumferences with a transversal through the external center of similitude, or the alternate intersec- tions of two circumferences with a transversal through the internal center of similitude. 4. Anti-homologous chords are chords joining anti-homol- oirous points of two transversals through the same center of similitude. 540. Proposition XYIL— Theorem. 1. The transversal through the e.rtremities of parallel radii hlhig in Hie same direction passes through tlte external center of similitude. 370 GEOMETRY.— BOOK VIII. 2. TJie transversal through the extremities of parallel radii lying in opposite directions passes through the internal center of simili- The similar triangles, CO A, CO 1 A', and C'OA, CO' A', give' CO : CO' : : OA : OA', CO : CO' :: OA : O'A'; .• C, C, are the external and internal centers of similitude. 541. Corollaries. 1. The radii drawn to the intersections of two circumferences with a transversal through either center of similitude are parallel two and two. (?) 2. The extremities of parallel radii are homologous points, and tangents at homologous points are parallel. (?) 3. The distances from a center of similitude to two homologous points are proportional to the radii of the circles. (?) 542. Proposition XYIIL— Theorem. The product of the distances from a center of similitude to two anti-homologous points is constant. o" CENTERS OF SIMILITUDE. 371 The transversals, CA, CO, intersect the circumferences in the homologous points, A, A', and B, B' ; also, M, J/', and X X'. . CA CB O/_0J/ r . A1 „ " ' CA' CB' ~ CM' ~ OM' l °* 1 ' " , - But, CA' X CB' = OJ/' X OX' (247). Multiplying the first and second of the above ratios by CA' X CB', and the third by 0.1/' X OX', we have CA X CB' = Ci' X 0£ = OJ/" , OX'. But OJ/X OX' is constant; .-.. CUX OB' and 0.1' X OB are constant. 543. Corollaries. 1. The two pairs of anti-homologous points of two transversals through the same center of similitude lie on the same circumfer- ence. (?) 2. Anti-homologous chords of two circles intersect on their radical axis. (?) 3. Tangents at two anti-homologous points in tivo circles in- tersect on tlieir radical axis. (?) 54-t. Proposition XIX.— Theorem. If three circles be taken two and two, then, 1. TJie straight lines joining the center of each with the in- ternal center of similitude of the oilier two meet- in a point. 2. The three external centers of similitude are in a straight line. 3. Tlie external center of similitude of any pair, and the internal centers of similitude of tlie otlier pairs, are in a straight line. 372 GEOMETRY.— BOOK VIII. Let 0, O, 0", be three circles whose radii are R, R', R" ; E, I, and E', I', and E", I", respectively, the external and internal centers of similitude of 0, O, and (7, 0", and 0", 0. 01 OI R R'' OF OT 01 X OT X O'l" R_ R"' pry I" R" ~ (539, 1). R X R' X R" OIX 0"I'X 01" " R'x R"X R 01 X OT X OT' = OI X OT X 01". OF, OI", 0"I, meet in a point (504, 1). 1. 2. 0E OE'~ R O E' # 0"E" R" R"' E"~ R (6dy ' 1) - R' ' O'E' 0E X OE' X 0"E" = O'E X O'E' x 0E" E, E', E", are in a straight line (502). This line is called the external axis of similitude. O'E" R" 01 3. E"~~ R ' O'l o"E" x oi x or * °L - * r539 n R' ' OT ~ R" k 6 ' h OE" x OI X OT. E", I, T, are in a straight line (502). Likewise, E, T, I", also, E', I", I, are in a straight line. These three lines arc called the internal axes of similitude.