2v, E L E M E N T S OF GE3OMETRY AND TRIGONOMRETRY, FROM TIIHE WORKS OF A. M. LEGENDRE. REVISED AND ADAPTED TO THE COURSE OF MATHEMATICAL INSTRUCTION IN THE UNITED STATES, BY CHARLES DAVIES, LL. D., AUrTHOR OF ARITHMETIC, ALGEBRA, PRACTICAL MATHEMATICS FOI PRACTICALT MI, ELEMENTS OF DESCRIPTIVE AND OF ANALYTICAL GEOMETRY, ELEMENTS OF DIFFERENTIAL AND INTEGRAL CALCULUS, AND SHADES, SHADO~WS, AND PERSPECTIVE, NEW-YORK: PUBLISHED BY A. S. BARNES & CO., No. 51 JOHN-STREET. CINCINNATI: H. W. DERBY & CO. 1802. DAVIES' COURSE OF MATHEMATICS. ~3abies' ffirst Lessons in Rfritbmetic-For Beginners. ]Jlabier' ritbtmetic-Designed for the use of Academies and Schools. We! to B1abies' aritmretic. 3labies' Unibersiftt Rrittmetic-EEmbracing the Science of Numbers and their numerous Applications. Btet! to Vabiefs' unibersit? aritDmetic. abtes' 3Eltmeantaqtr algetra-Being an introduction to the Science, and forming a connecting link between ARITHMETIC and ALGEBRA. Be! to abims' tllementar tlgebteva. 30abties' EClements of (comtetrl AND rtiOnnOlTetrt, with APPLICATIONS IN MENTSURATION.-This work embraces the elementary principles of Geometry and Trigonometry. The reasoning is plain and concise, but at the same time strictly rigorous. tabtes' jractical jVatbenmatics for Vracticarl nat-Embracing the Principles of Drawing, Architecture, Mensuration, and Logarithms, with Applications to the Mechanic Arts. sabie' 3Bourvon's ITgetbra-Including STURM's THEOREM-Being an abridgment of the Work of M. BOURDON, with the addition of practical examples. Il[abies' egentrr'a (ltonrettr: AND FrigotnomrtrN-From the works of A. M. Legendre, with the addition of a Treatise on MENSURATION OF PLANES AND SOLIDS, and a Table of LOGARITHMS and LOGAR1THMIIC SINES. 33abies' %turbeting-With a description and plates of the THEODOLITE, COMPASS, PLANE-TABLE, and LEVEL; also, Maps of the TOPOGRoaPtICAL SIGNS adopted by the Engineer Department-an explanation of the mletllod- of surveying the Public Lands, Geodesic and MariJi.im,, St:rveying, and an Elementary Treatise on NAVIGATION. Baebics' Delcriptibe 9cometr —With its application to SPHIIEICAL PROJECTIONS. bfabise' %I)a a.s, %latfoi)rs, AND iLintatr VespeCctibe. alabies' Mnalrticatl romettr,)-Embracing the EQUATIONS OF THE POINT AND STRAIGHT LINE-Of the CONIC SECTIONS —of the LINE AND PLANE IN SPACE; also, the discussion of the GENERAL EQUATION of the second degree, and of SUvFACES of the second order. 3labiet' Vtiffeerttial AND *ntetral Ctalutlus. Pabies' Ao.ic anrl U2tilit of Jtatbematics. E~NTERED according to Act of Congress, in the year one thousand eight hundred and fifty-one, by CHARLES DAVIES, in the Clerk's Office of the District Court of the United States for the Southern District of New York. J. P. JONES & CO., STEREOTYPERS. P R E F A C E. IN the preparation of the present edition of the Geometry of A. M. LEGENDRE, the original has been consulted as a model and guide, but not implicitly followed as a standard. The language employed, and the arrangement of the arguments in many of the demonstrations, will be found to differ essentially from the original, and also from the English translation by DR. BREWSTER. In the original work, as well as in the translation, the propositions are not enunciated in general terms, but with reference to, and by the aid of, the particular diagrams used for the demonstrations. It is believed that this departure from the method of Euclid has been.generally regretted. The propositions of Geometry are general truths, and as such, should be stated in general terms, and without reference to particular figures. The method of enunciating them by the aid of particular diagrams seems to have been adopted to avoid the difficulty which beginners experience in comprehending abstract propositions. But in avoiding this difficulty, and thus lessening, at first, the intellectual labor, the faculty of abstraction, which it is one of the primary objects of the study of Geometry to strengthen, remains, to a certain extent, unimproved. iv PREFACE. The methods of demonstration, in several of the Books, have been entirely changed. By regarding the circle as the limit of the inscribed and circumscribed polygons, the demonstrations in Book V. have been much simplified; and the same principle is made the basis of several important demonstrations in Book VIII. The subjects of Plane and Spherical Trigonometry have been treated in a manner quite different from that employed in the original work. In Plane Trigonometry, especially, important changes have been made. The separation of the part which relates' to the computations of the sides and angles of triangles from that which is purely analytical, will, it is hoped, be found to be a decided improvement. The application of Trigonometry to the measurement of Heights and Distances, embracing the use of the Table of Logarithms, and of Logarithmic Sines; and the application of Geometry to the mensuration of planes and solids, are useful exercises for the Student. Practical examples cannot fail to point out the generality and utility of abstract science. FISHKILL LANDING, July, 1851. CON TENTS. PAGE. ITRODUCTION, ----— 9 —---- -- --------------------- 9 BOOK I. Definitions,. 13 Propositions, ------------------------------------------------- 21 BOOK II. Ratios and Proportions, 47 BOOK III. The Circle, and the Measurement of Angles, - 57 Problems relating to the First and Third Books, 76 BOOK IV. Proportions of Figures-Measurement of Areas, 87 Problems relating to the Fourth Book,........... — 122 BOOK V. Regular Polygons-Measurement of the Circle, 135 BOOK VI. Planes and Polyedral Angles. 156 BOOK VII. Polyedrons, - 174 BOOK VIII. The Three Round Bodies, ----------------------------—. 202 BOOK IX. Spherical Geometry,....'227 vi CONTENTS. A PP ENDIX. PAGE. Note A, -............................. 245 The Regular Polyedrons, --------------------------------------- 247 Application of Algebra to the Solution of Geometrical Problems, 249 PLANE TRIGONOMETRY. Logarithms Defined, ----------------- -------- - - - - 255 Logarithms, Use of, -....... 256 General Principles, -------------------------------------- 256 Table of Logarithms, -------------- ------ - - - 257 To Find from the Table the Logarithm of a Number, ------- 258 To Find from the Table the Number corresponding to a Given Logarithm, ---------------------------------- 260 Multiplication by Logarithms, ----------------............ 261 Division by Logarithms, --- --- --------------- ---- 262 Arithmetical Complement, —----------------------- ---------- 263 To find the Powers and Roots of Numbers, by Logarithms, 265 Geometrical Constructions, ---------------------- 266 Description of Instruments, ---------- ----—, 266 Dividers, --------—.. —---------------- 266 Ruler and Triangle, ----------------------------------------- 266 Problems, -. —-------- -—. —-------------------------- 267 Scale of Equal Parts,. —-------------- ------—.; 268 Diagonal Scale of Equal Parts, -------------- 268 Semicircular Protractor, ---------------------------- 270 To Lay off an Angle with a Protractor,. —--------------------- 270 Parts of a Plane Triangle, 271 Plane Trigonometry, Defined, -- ------------------------ 271 Division of the Circumference,-.. —---------------—...... 271 Measures of Angles, ------------------------------ 271 Complement of an Are, -------------------------------------- 271 Definitions of Trigonometrical Lines, --------------------------- 272 Table of Natural Sines, ------------------------- ---- 273 Table of Logarithmic Sines, -------------------------------- 274 To Find from the Table, the Logarithmic Sine, &c., of an Are or Angle, 274 To Find the Degrees, &c., Answering to a Given Logarithmic Sine, &c., 276 Theorems,. —----------------.. —------------ 277 Solution of Triangles,.-...... —-- 281 Solution of Right-Angled Triangles, -287 Application to Heights and Distances,............................. 288 CONTENTS. vii ANALYTICAL PLANE TRIGONOMETRY'. PAGE. Circular Functions, - -- -- ----------------------- 297 Analytical Plane Trigonometry, Defined, ----------- ------------- 297 Quadrants of the Circumference, ------------- ---------—, 298 Versed-Sine, ---------------------- -- - - - 298 Relations of Circular Functions, -------------- -------------- 299 Table I., of Formulas, ------------------------------ 301 Algebraic Signs of the Functions, -------------- ------------------ 301 Table II., of Formulas,..................... —------- 306 General Formulas, - -307 HIomogeneity of Terms, -313 Formulas for Triangles, ------------------------------------ - 315 Construction of Trigonometrical Tables —, -- ----- 317 SPHERICAL TRIGONOMETRY. Spherical Triangle, Defined,_ 321 Spherical Trigonometry, Defined, ------------------------------ 321 First Principles, 321 Napier's Analogies, ------------------------ - 329 Napier's Circular Parts, -.......... 329 Theorems, -330 Solution of Right-Angled Spherical Triangles, by Logarithms, ------ 333 Of Quadrantal Triangles, ---------------------------- 335 Solution of Oblique-Angled Triangles, by Logarithms, ------------- 338 MENSURATION OF SURFACES. Area, or Contents of a Surface,. —---------------------...... 347 Unit of Measure for Surfaces, ------------------------------ 347 Area of a Square, Rectangle, or Parallelogram, 347 Area of a Triangle,..-........................... 348 Area of a Trapezoid,. —--------------------------------------- 350 Area of a Quadrilateral, - 351 Area of an Irremular Polygon, ---------------------------------- 351 Area of a Long and Irregular Figure bounded on One Side by a Right Line,.-................................ 351 Area of a Regular Polygon, ------------------------------------ 353 To Find the Circumference or Diameter of a Circle,. 354 To find the Length of an Arc, - 355 Area of a Circle, - 356 Area of a Sector of a Circle, 356 Area of a Segment of a Circle, 356 Area of a Circular Ring,-... -357 viii CONTENTS. MENSURATION OF SOLIDS. PAGE. Mensuration of Solids, divided into Two Parts, 358 Unit of Length, -.......... —---- 358 Unit of Solidity, -------------------------- -- 358 Table of Solid Measures,.-. 358 Surface of a Right Prism,- ------------------------------- 358 Surface of a Right Pyramid, --------------- -------------------- 359 Convex Surface of the Frustum of a Right Pyramid, -------------- 359 Solidity of a Prism, —------ ------------- ----------- 359 Solidity of a Pyramid, --------------------------- ----------- 360 Solidity of the Frustum of a Pyramid,-,. —------------ ---- - 360 The Wedge, --------------------------- 361 Rectangular Prismoid,. —-------------------------- 361 Solidity of the Wedge,. —------------------------- 361 Solidity of a Rectangular Prismoid, —--- 362 Surface of a Cylinder, ----- ------------------ ------------- 363 Convex Surface of a Cone, ------------------------------ 364 Surface of a Frustum of a Cone, 364 Solidity of a Cylinder,- --------- - 364 Solidity of a Cone, --- -- --------------------------- 365 Solidity of a Frustum of a Cone, ----- 365 Surface of a Spherical Zone, --------- ------------ ---------- 365 Solidity of a Sphere -—, —-------------------—, 366 Solidity of a Spherical Segment, -----------------—. —-------- 366 Surface of a Spherical Triangle, ---------------------- 366 Surface of a Spherical Polygon, - ---------------------- 367 Of the Regular Polyedrons, - -— 367 Theorem,. 367 Method of Finding the Angle included between two Adjacept Faces of a Regular Polyedron, ------------------------------------- 368 Table of Regular Polyedrons whose Edges are 1, —---------------- 369 Solidity of a Regular Polyedron, —-------- ------ 369 ELEMENTS OF G E O M E T R Y. INTRODUCTION. 1. SPACE extends indefinitely in every direction and contains all bodies. 2. EXTENSION is a limited portion of space, and has three dimensions, length, breadth, and thickness. 3. A SOLID, or BODY, is a limited portion of space. supposed to be occupied by matter. The difference between the terms, extension and solid, is simply this: the former denotes a limited portion of space, viewed in the abstract, while the latter denotes such a portion occupied by matter. The term, Solid, is generally used in Geometry, in preferenc.e to Extension, because the mind apprehends readily the forms and relations of tangible objects, while it often experiences much difficulty in dealing with the abstract notions derived from themr. It is, however, important to observe, that the geometrical properties of solids have no connection whatever with matter, and that thile demonstrations which, establish and make known those properties, are based on the attributes of extension only. 10 GEOMETRY. 4. A Solid being a limited portion of space, is necessarily divided from the indefinite space which surrounds it: that which so divides it, is called a Surface. Now, since that which bounds a solid is no part of the solid itself, it follows, that a surface has but two dimensions, length and breadth. 5. If we consider a limited portion of a surface, that which separates such portion from the other parts of the surface, is called a Line. This mark of division forms no part of the surfaces which it separates: hence, a line has length only, without breadth or thickness. 6. If we regard a limited portion of a line, that which separates such portion from the part, at either extremity, is called a Point. But this mark of division forms no part of the line itself: hence, a point has neither length, breadth, nor thickness, but place or position only. 7. Although we use the term solid to denote a given portion of space, the term surface to denote the boundary of a solid, the term line to denote the boundary of a surface, and the term point to designate the limit of a line, still, we may employ either of these terms, in an abstract sense, without any reference to the others. Thus, we may contemplate a river, as a solid, without considering its boundaries; may look upon the surface and perceive that it has length and breadth without refering to its depth; or, we may regard the distance across without taking into account either its depth or length. So likewise, we may consider a point without any reference to the line which it limits. In the definitions and reasonings of Geometry these terms are always used in an abstract sense; they are mere signs to the mind of the conceptions for which they stand. 8. ANGLE is a term which designates the portion of a surface included by two lines meeting at a common point; INTRODUCTION. 11 and it also denotes a portion of space included by two or more planes. 9. MAGNITUDE is a general term employed to denote those quantities which arise from considering the dimensions of extension, and is equally applicable to lines, angles, surfaces, and solids. Geometry is conversant with four kinds of magnitude. 1. Lines; which have length without breadth or thickness. 2. Angles; bounded by straight lines, by curves, and by planes. 3. Surfaces; which have length and breadth without thickness: and 4. Solids; which have length, breadth, and thickness. 10. FIGURE is a term applied to a geometrical magnitude and expresses the idea of shape or form. It is that which is enclosed by one or more boundaries. Thus, "A triangle is a plane Jigure bounded by three straight lines." 11. A PROPERTY of a figure is a mark or attribute common to all figures of the same class. 12. The portions of extension which constitute the geometrical magnitudes, are indicated to the mind by certain marks called lirnes. Thus, we say, the straight line A B, is the shortest distance between the two points A and B. The mark AB, on the paper, is A B not the geometrical line AB, but only the sign or representative of it-the geometrical line itself, having merely a mental existence. We also say, that the triangle C A CB is bounded by the three straight lines AB, A, CB. Now, the triangle A CB, is but the sign, to the mind, of a portion of a plane. That which the eye sees is not the geometrical A B conception on which the mind acts and reasons: but is, as it were, the word or sign which stands for and expresses the abstract idea. 12 GEOMETRY. These considerations have induced me to represent the geometrical magnitudes by the fewest possible lines, and to reject altogether the method of shading the figures. It is the conception of extension, in the abstract, with which the mind should be made conversant, and too much pains cannot be taken to exclude the idea that we are dealing with material things. ELEMENTS OF GEOMETRY. BOOK I. DEFINITIONS. * 1. EXTENSION has three dimensions, length, breadth, and thickness. 2. GEOMETRY is the science which has for its object: 1st. The measurement of extension; and 2dly. To discover, by means of such measurement, the properties and relations of geometrical figures. 3. A POINT is that which has place, or position, but not magnitude. 4. A LINE is length, without breadth or thickness. 5. A STRAIGHT LINE is one which lies in the same direction between any two of its points. 6. A BROKEN LINE is one made up of straight lines, not lying in the same direction. 7. A CURVE LINE iS one which changes its direction at every point. The word line when used alone, will designate a straight line; and the word curve, a curve line. 8. A SURFACE is that which has length and breadth without thickness. * See Davies' Logic and Utility of Mathematics. ~ 1. 14 GEOMETRY. 9. A PLANE is a surface, such, that if any two of its points be joined by a straight line, such line will be wholly in the surface. 10. Every surface, which is not a plane surface, or composed of plane surfaces, is a curved surface. 11. A SOLID, or BODY is that which has length, breadth, and thickness: it therefore combines the three dimensions of extension. 12. An ANGLE is the portion of a plane included between two straight lines which meet at a common point. The two straight lines are called the sides of the angle, and the common point of intersection, the vertex. Thus, the part of the plane includ- C ed between AB and A C is called an angle: AB and AC are its sides, and A its vertex. An angle is sometimes designated A B simply by a letter placed at the vertex, as, the angle A; but generally, by three letters, as, the angle BAC or CAB,-the letter at the vertex being always placed in the middle. 13. When a straight line meets another straight line, so as to make the adjacent angles equal to each other, each angle is called a right angle; and the first line is said to be perpendicular to the second. 14. An ACUTE ANGLE is an angle less than a right angle. 15. An OBTUSE ANGLE is an angle greater than a right angle. BOOK I. 15 16. Two straight lines are said to be parallel, when being situated in the same plane, they'cannot meet, how far soever, either way, both of them be produced. 17. A PLANE FIGURE is a portion of a plane terminated on all sides by lines, either straight or curved. 18. A POLYGON, or rectilineal figure, is a portion of a plane terminated on all sides by straight lines. The sum of the bounding lines is called the perimeter of the polygon. 19. The polygon of three sides, the simplest of all, is called a triangle; that of four sides, a quadrilateral/; that of five, a pentagon; that of six, a hexagon; that of seven, a heptagon; that of eight, an octagon; that of nine, an nonagon; that of ten, a decagon; and that of twelve, a dodecagon. 20. An EQUILATERAL polygon is one which has all its sides equal; an equiangular polygon, is one which has all its angles equal. 21. Two polygons are mutually equilateral, when they have their sides equal each to each, and placed in the same order: that is to say, when following their bounding lines in the same direction, the first side of the one is equal to the first side of the other, the second to the second, the third to the third, and so on. 22. Two polygons are mutually equiangular, when every angle of the one is equal to a corresponding angle of the other, each to each. 23. Triangles are divided into classes with reference both to their sides and angles. 1. An equilateral triangle is one which has its three sides equal. 16 GEOMETRY. 2. An isosceles triangle is one which has only two of its sides equal. 3. A scalene triangle is one which has its three sides unequal. 4. An acute-angled triangle is one which has its three angles acute. 5. A right-angled triangle is one which has a right angle. The side opposite the right angle is called the hypothenuse, and the other two sides, the base and perpen- / dicular. 6. An obtuse-angled triangle is one which has an obtuse angle. 24. There are three kinds of QUADRILATERALS: 1. The trapezium, which has none of its sides parallel. 2. The trapezoid, which has only two of its sides parallel. 3. The parallelogram, which has its opposite sides parallel. BOOK I. 17 25. There are four kinds of PARALLELOGRAMS: 1. The rhomboid, which has no right angle. 2. The rhombus, or lozenge, which is an equilateral rhomboid. 3. The rectangle, which is an equiangular parallelogram, but not equilateral. 4. The square, which is both equilateral and equiangular. 26. A DIAGONAL of a figure is a line which joins the vertices of two angles not adjacent. DEFINITIONS OF TERMS. 1. An axiom is a self-evident truth. 2. A demonstration is a train of logical arguments brought to a conclusion. 3. A theorem is a truth which becomes evident by means of a demonstration. 4. A problemn is a question proposed, which requires a solution. 5. A kmma is a subsidiary truth, employed for the demonstration of a theorem, or the solution of a problem. 2 18 GEOMETRY. 6. The common name, proposition, is applied indifferently, to theorems, problems, and lemmas. 7. A corollary is an obvious consequence, deduced from one or several propositions. 8. A scholium is a remark on one or several preceding propositions, which tends to point out their connection, their use, their restriction, or their extension. 9. A hypothesis is a supposition, made either in the enunciation of a proposition, or in the course of a demonstration. 10. A postulate grants the solution of a self-evident problem. EXPLANATION OF SIGNS. 1. The sign = is the sign of equality; thus, the expression A = B, signifies that A is equal to B. 2. To signify that A is smaller than B, the expression A < B is used. 3. To signify that A is greater than B, the expression A > B is used; the smaller quantity being always at the vertex of the angle. 4. The sign + is called plus; it indicates addition: 5. The sign - is called minuls; it indicates subtraction: Thus, A +B, represents the sum of the quantities A and B; A-B represents their difference, or what remains after B is taken from A; and A-B+C, or A+C-B, signifies that A and C are to be added together, and that B is to be subtracted from their sum. 6. The sign X indicates multiplication: thus AX B represents the product of A and B. The expression AX(B + C-D) represents the product of A by the quantity B+C-D. If A+D) were to be multiplied by A-B + C, the product would be indicated thus; (A+D)X (A-B+ C), whatever is enclosed within the curved lines, being consid BOOK I. 19 ered as a single quantity. The same thing may also be indicated by a bar: thus, A+B+ CXD, denotes that the sum of A, B and C, is to be multiplied by D. 7. A figure placed before a line, or quantity, serves as a multiplier to that line or quantity; thus 3AB signifies that the line AB is taken three times; 12A signifies the half of the angle A. 8. The square of the line AB is designated by ATBa; its cube by AB. What is meant by the square and cube of a line, will be explained in its proper place. 9. The sign / indicates a root to be extracted; thus V/2 means the square-root of 2; /A XB means the square-root of the product of A and B. AXIOMS. 1. Things which are equal to the same thing, are equal to one another. 2. If equals be added to equals, the wholes will be equal. 3. If equals be taken from equals, the remainders will be equal. 4. If equals be added to unequals, the wholes will be unequal. 5. If equals be taken from unequals, the remainders will be unequal. 6. Things which are double of the same thing, are equal to each other. 7. Things which are halves of the same thing, are equal to each other. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. 20 GEOMETRY. 10. All right angles are equal to each other. 11. From one point to another only one straight line can be drawn. 12. A straight line is the shortest distance between two points. 13. Through the same point, only one straight line can be drawn which shall be parallel to a given line. 14. Magnitudes, which being applied the one to the other, coincide throughout their whole extent, are equal. POSTULATES. 1. Let it be granted, that a straight line may be drawn from one point to another point. 2. That a terminated straight line may be prolonged, in a straight line, to any length. 3. That if two straight lines are unequal, the length of the less may always be laid off on the greater. 4. That a given straight line may be bisected: that is, divided into two equal parts. 5. That a straight line may bisect a given angle. 6. That a perpendicular may be drawn to a given straight line, either from a point without the line, or at a point of a line. 7. That a straight line may be drawn, making with a given straight line, an angle equal to a given angle. BOOK I. 21 PROPOSITION I. THEOREM. If one straight line meet another straight line, the sum of the two adjacent angles will be equal to two right angles. Let the straight line DC meet the straight line AB at C; then will the angle A CD plus the angle DCB, be equal to two right angles. At the point C6 suppose CE to E be drawn perpendicular to AB: then, A CE + ECB = two right angles (D. 13).* But ECB is equal A to ECD + DCB (A. 9): hence, A A CE + ECD + D CB = two right angles. But A CE J- ECD = A CD (A. 9): therefore, A CD + D CB = two right angles. Cor. 1. If one of the angles A CD or D CB, is a right angle, the other will also be a right angle. Cor. 2. If a straight line DE D is perpendicular to another straight line AB; then, reciprocally, AB will A be perpendicular to DE. For, since DE is perpendicular to AB, the angle A CD will be a E right angle (D. 13). But since A C meets DE at.the point C, making one angle A CD a right angle, the adjacent angle A CE will also be a right angle (c. 1). Therefore, AB is perpendicular to DE (D. 13). Coar.n 3. The sum of the succes C sive angles BA C, CAD, DAE, EAF, E formed on the same side of the line BF, is equal to two right an- F gles; for, their sum is equal to A that of the two adjacent angles BAC and CAF. * In the references, A. stands for Axiom —D. for Definition-B. for Book-P. for Proposition-C. for Corollary-S. for Scholium, and Prob. for Problem. 22 GEOMETRY. PROPOSITION II. THEOREM. Two straight lines, which have two points common, coincide the one with the other, throughout their whole extent, and form one and the same straight line. Let A and B be the two common points of two straight lines. In the first place, the two lines will coincide between the points A and B; for, otherwise there would be two straight lines between A and A B D B, which is impossible (A. 11). Suppose, however, that in being prolonged, these lines begin to separate at some point, as C, the one becoming CD, the other, CE. At the point C, suppose CF to be drawn, making with AC, the right angle ACF. Now, since A CD is a straight line, the angle FCD will be a right angle (P. I., c. 1): and since ACE is a straight line, the angle FCE will also be a right angle. Hence, the angle FCD is equal to the angle FCE (A. 10): that is, a whole is equal to one of its parts, which is impossible (A. 8): therefore the two straight lines which have two points, A and B, in common, cannot separate at any point, when prolonged; hence, they form one and the same straight line.* PROPOSITION III. THEOREM. If when a straight line meets two other straight lines at a common point, the sum of the two adjacent angles which it makes with them, is equal to two right angles, the two straight lines which are met, form one and the same straight line. Let the straight line CD meet the two lines AC, CB, at their common point C, and let the sum of the two adjacent angles, DCA, DCB, be equal to two right angles: then * See Note A. It is earnestly recommended to every pupil to read and understand this Note. Also, see Logic and Utility of Mathematics, ~ 262. BOOK I. 23 will CB be the prolongation of A C; or, A C and CB will form one and the same straight line. For, if CB is not the prolonga- D tion of AC, let CE be that prolongation. Then the line A CE being straight, the sum of the angles A CD, A — B D CE, will be equal to two right C E angles (P. I). But by hypothesis, the sum of the angles A CD, D CB, is also equal to two right angles: therefore (A. 1), ACD+DCE must be equal to ACD+DCB. Taking away the angle ACD from each, there remains the angle DCiE equal to the angle DCB: that is, a whole equal to a part, which is impossible (A. 8): therefore, AC and CB form one and the same straight line. PROPOSITION IV. THEOREM. When two straight lines intersect each other, the opposite or vertical angles, which they form, are equal. Let AB and DE be two straight lines, intersecting each other at C; then will the angle ECB be equal -to the angle A CD, and the angle ACE to the angle D CB. For, since the straight line DE A E is met by the straight line A, the sum of the angles A CE, A CD, c is equal to two right angles (P. I.); and since the straight line AB is met by the straight line EC, the sum of the angles ACE, and ECB, is equal to two right angles: hence (A. 1), ACE+ACD is equal to ACE+ECB. Take away from both, the common angle ACE, there remains (A. 3) the angle A CD, equal to its opposite or vertical angle ECB. In a similar manner it may be proved that ACE is equal to DCB. Scholium. The four angles formed about a point by two straight lines, which intersect each other, are together equal 24 GEOMETRY. to four right angles. For, the sum of the two angles A CE, A CB, is equal to two right angles (P. I); and the sum of the other two, A CD, DCB, is also equal to two right angles: therefore, the sum of the four, is equal to four right angles. In general, if any number of straight B lines CA, CB, CD, &c., meet in a cormn- A mon point C, the sum of all the sue- D cessive angles, ACCB, BCD, DCE, ECF, FCA, will be equal to four right an- C gles. For, if four right angles were F formed about the point C, by two lines E perpendicular to each other, their sum would be equal to the successive angles ACB, BCID, )CE, C ECF, ECA. PROPOSITION V. THEOREM. If two triangles have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each, the two triangles will be equal. In the two triangles EDF and BA, let the side ED be equal to the side BA, the side DEF to the side AC, and the angle D to the angle A; then will the triangle EDF be equal to the triangle BA C. For, if these trian- D A gles be applied the one to the other, they will exactly coincide. Let the side ED be placed on the equal side BA; E FB C then, since the angle D is equal to the angle A, the side 1DF will take the direction AC. But DF is equal to A C; therefore the point F will fall on C, and the third side EF, will coincide with the third side BC (A. 11): consequently, the triangle EDF is equal to the triangle BAC (A. 14). Cor. When two triangles have these three things equal, viz., the side EED=BA, the side.DF=AC, and the anglq D=A, the remaining three are also respectively equal, viz., the side EF=BC, the angle 1=B, and the angle F=C. BOOK I. 25 PROPOSITION VI. THEOREM. If two triangles have two angles and the included side of the one, equal to two angles and the included side of the other, each to each, the two triangles will be equal. Let EDF and BA C be two triangles, having the angle E equal to the angle B, the angle F to the angle C, and the included side E'F to the included side BC; then will the triangle EDF be equal to the triangle BAC. For, let the side EF D A be placed on its equal BC the point E falling on B, and the point F on C. Then, since the angle \ E is equal to the angle E F B, the side ED will take the direction BA; and hence, the point D will be found somewhere in the line BA. In like manner, since the angle F is equal to the angle C, the line FD will take the direction CA, and the point D will be found somewhere in the line CA. Hence, the point D, falling at the same time in the two straight lines BA and CA, must fall at their intersection A: hence, the two triangles ED, BA C, coincide with each other, and consequently, are equal (A. 14). Cor. Whenever, in two triangles, these three things are equal, viz.: the angle E=B, the angle F=C,.and the included side EF equal to the included side BC, it may be inferred that the remaining three are also respectively equal, viz.: the angle 1)=A, the side ED=BA, and the side DF=AC. Scholium. Two triangles which being applied to each other, coincide in all their parts, are equal (A. 14). The like parts are those which coincide with each other; hence, they are also equal each to each. The converse of this proposition is also true; viz., if two triangles have all the parts of the one equal to the parts of the other, each to each, the triangles will be equal: for, when applied to each other, they will mutually coincide. 26 GEOMETRY. PROPOSITION VII. THEOREM. The sum of any two sides of a triangle, is greater than the third side. Let ABC be a triangle: then will the sum of two of its sides, as AB, BC, be greater than the third side AC. B For the straight line A C is the shortest distance between the points A and C / (A. 12); hence, AB +BC is greater than AC. A C A/ \I Cor. If from both members of the inequality ACD; then will the side BC be greater than EE. Make the angle CA G =D; take A G=DE, and draw CG. Then, the triangles GA C and EDF will be equal, since they have an equal angle in each, contained by equal sides (P. 5); consequently, CG is equal to EF (P. 5, c). There may be three cases in this proposition. 1st. When the point G falls without the triangle BAC. 2d. When it falls on the base BC; and 3d. When it falls within the triangle. Case I. In the triangles A G C and ABC, we have, CI+IC> C; and A AI+IB >AB; therefore AGC+BC> GC+AB. B/ F Taking away A E from the one side and its equal AB from the other, and there will remain BC 28 GEOMETRY. greater than GC. But we have found that GGC is equal to BEF; therefore, BC will be greater than EF. A D Case IZI If the point G A fall on the side BC, it is evident that G C, or its equal EF, will be shorter than BC (A. 8). B G C E F Case II. Lastly, if the point G A D fall within the triangle BA C, we shall have AC + GCEF. Cor. Conversely: if two sides BA, AC, of a triangle BA C, are equal to two sides ED, DF, of a triangle EDF, C F each to each, while the third side B C of the first is greater than the third side EF of the second, then the angle BAC of the first triangle will be greater than the angle EDF of the second. For, if not greater, the angle BAC must be equal to EDF or less than it. In the first case, the side BC would be equal to EF (P. 5, c), in the second, BC would be less than EF; but either of these results contradicts the hypothesis: therefore, BA C is greater than ED)F. PROPOSITION X. THEOREM. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are equal. Let EDF and BA C be two triangles, having the side ED=BA, the side EF=BC, and the side DF=AC; then will the angle D=-A, the angle E=B, and the angle F= C, and consequently the triangle.EDF will be eqlual to the triangle BA C. BOO0K I. 29 For, since the sides D A ED, DF, are equal to BA, AC, each to each, if the angle D were greater than A, it would follow, by the last proposition, E FB C that the side EF would be greater than BC; and if the angle D were less than A, the side EF would be less than BC. But EF is equal to BC, by hypothesis; therefore, the angle D can neither be greater nor less than A; therefore it must be equal to it. In the same manner it may be shown that the angle E is equal to B, and the angle F to C: hence, the two triangles are equal (P. 6, s). Scholium. It may be observed, that when two triangles are equal to each other, the equal angles lie opposite the equal sides, and consequently, the equal sides opposite the equal angles: thus, the equal angles D and A, lie opposite the equal sides EF and BC. PROPOSITION XI. THEOREM. In an isosceles triangle, the angles opposite the equal sides are equal. Let BAC be an isosceles triangle, having the side BA equal to the side AC; then will the angle C be equal to the angle B. For, join the vertex A, and the mid- A dle point D, of the base BC. Then, the triangles BAD, DAC, will have all the sides of the one equal to those of the other, each to each. For, BA is equal to B C A C, by hypothesis, AD is common, and BD is equal to DOC by construction: therefore, by the last proposition, the angle B is equal to the angle C. Cor. 1. An equilateral triangle is likewise equiangular, that is to say, has all its angles equal. Cor. 2. The equality of the triangles BAD, DAC, proves also that the angle BAD, is equal to DA, and BDA to 30 GEOMETRY. ADUC; hence, the latter two are right angles. Therefore, the line drawn from the vertex of an isosceles triangle to the middle point of the base, divides the angle at the vertex into two equal parts, and is perpendicular to the base. Scholium. In a triangle which is not isosceles, any side may be assumed indifferently as the base; and the vertex is, in that case, the vertex of the opposite angle. In an isosceles triangle, however, that side is generally assumed as the base, which is not equal to either of the other two. PROPOSITION XII. THEOREM. Conversely: If two angles of a triangle are equal, the sides opposite them are also equal, or, the triangle is isosceles. In the triangle BA C, let the angle B be equal to the angle ACB; then will the side A C be equal to the side AB. For, if these sides are not equal, sup- A pose AB to be the greater. Then, take D BD equal to A C, and draw CD. Now, in the two triangles BD C, BA C, we l2ave BD =A (, by construction; the angle B equal to the angle A CB, by hypothesis; B/ and the side BC common: therefore, the two triangles, BDC', BA C, have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each: hence they are equal (P. 5). But the part cannot be equal to the whole (A. 8); hence, there is no inequality between the sides BA and AC; therefore, the triangle BAC is isosceles. PROPOSITION XIII. THEOREM. The greater side of every triangle is opposite to the greater angle; and conversely, the greater angle is opposite to the greater side. First, In the triangle CAB, let the angle C be greater than the angle B; then will the side AB, opposite C, be greater than A C opposite B. BOOK I. 31 For, make the angle BCGD=B. Then, in the triangle CDB, we shall A have CDt=BD (P. 12). Now, the side ACA C. C B Secondly. Suppose the side AB>AC; then will the angle C, opposite to AB, be greater than the angle B, opposite to A GC. For, if the angle CA C, the angle C cannot be less than B, nor equal to it; therefore, the angle C must be greater than B. PROPOSITION XIV. THEOREM. Fromn a given _point, without a straight line, only one perpendicular can be drawn to that line. Let A be the point, and DE the given line. Let us suppose that we can draw A two perpendiculars, AB, A C. Prolong either of them, as AB, till BF is equal to AB, and draw FC. Then D the two triangles CAB, CBF, will be equal: for, the angles CBA and CBEF are right angles, the side CB is common, and the side AB equal to BE, by construction; therefore, the two triangles are equal, and -the angle ACB=BCF (P. 5, c). But the angle ACB is a right angle, by hypothesis; therefore, BCF must likewise be a right angle. Now, if the adjacent angles.BCA, BCF, are together equal to two right angles, A CF must be a straight line (P. 3). Whence, it follows, that between the same two points, A and F, two straight lines can be drawn, which is impossible (A. 11): therefore, only one 32 GEOMETRY. perpendicular can be drawn from the same point to the same straight line. Cor. At a given point C, in the E D line AB, it is equally impossible to erect more than one perpendicular to that line. For, if CD, CE, were both perpendicular to AB, the angles A- B B/CD, BCE, would both be right angles; hence, they would be equal (A. 10), and a part would be equal to the whole, which is impossible. PROPOSITION XV. THEOREM. If from a point without a straight line, a perpendicular be let fall on the line, and oblique lines be drawn to different points: Lst. The perpendicular will be shorter than any oblique line. 2d. Any two oblique lines which intersect the given line at points equally distant from the foot of the perpendicular, will be equal. 3d. Of two oblique lines which intersect the given line at points unequally distant from the perpendicular, the one which cuts off the greater distance will be the longer. Let A be the given point, D)E the given line, AB the perpendicular, and AD, AC, AE, the oblique lines. Prolong the perpendicular AB till BF is equal to AB, and draw FC, A FD. First. The triangle BCF, is equal D to the triangle CAB, for they have C the right angle CBF= CBA, the side CB common, and the side BF=BA; F hence, the third sides, CF and CA are equal (P. 5, c). But ABF, being a straight line, is shorter than A CF, which is a broken line (A. 12); therefore, AB, the half of ABF, is shorter than A C, the half of ACF; hence, the perpendicular is shorter than any oblique line. BOOK I. 33 Secondly. Let us suppose BC=BE; then the triangle CAB will be equal to the triangle BAEi; for BC=BE, the side AB is common, and the angle CBA=ABE;hence, the sides A C and AE are equal (P. 5, c): therefore, two oblique lines, which meet the given line at equal distances from the perpendicular, are equal. Thirdly. Since the point C is within the triangle FDA, the sum of the sides OD, Di)A, is greater than the sum of the lines FGC, CA (P. 8): therefore AD, the half of the broken line FDA, is greater than AC, the half of FCA: consequently, the oblique line which cuts off the greater distance, is the longer. Ctor. 1. The perpendicular measures the shortest distance of a point from a line. Cor. 2. From the same point to the same strailght line, only two equal straight lines can be drawn; for, if there couldcl be more, we should have at least two equal oblique lines on the same side of the perpendicular, which is impossible. PROPOSITION XVI. TIIEOREM. if at the middle point of a straight line, a perpendicular to this line be drawn: st. Any. point of the perpencdicular will be euaclly distant froom the extremities of the line: 2d. Any point, without the perpendicular, will be unequtally distant from the extremnziies. Let AB be the given straight line, C its middle point, and ECF the perpendicular. First. Let D be any point of the per- F pendicular, and draw DA and DB. Then, since A C= CB, the two oblique lines AD, DB, are equal (P. 15). So, likewise, are the two oblique lines, AE, EB, the two A B A XF, FB, and so on. Therefore, any point in the perpendicular is equally distant from the extremities A and B. E 3 34 GEOMETRY. Secondly. Let I be any point out of F the perpendicular. If IA and lB be drawn, one of these lines will cut the perpendicular in some point as D; from this point, drawing DB, we shall have A B DB=DA. For, the straight line IB is less than ID-+DB, and ID~+DB=ID+DA=JIA; E therefore, IB<~IA; consequently, any point out of the perpendicular, is unequally distant from the extremities A and B. Cor. Conversely: if a straight line have two points E and F, each of which is equally distant from the extremities A and B, it will be perpendicular to AB at the middle point C. PROPOSITION XVII. THEOREM. If two right-angled triangZes have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, each to each, the triangles are equal. Let BA C and EDF be two right-angled triangles, having the hypothenuse A C=DF, and the side BA=ED: then will the triangle BAC be equal to the triangle ED)F. If the sides B C and EF are equal, the tri- D angles are equal (P. 10).' Now, suppose these two sides to be unequal, and B G C E F BC to be the greater. On BC take BG=EE, and draw AG. Then, in the two triangles BAG, EDF, the angles B and E are equal, being right angles,'the side BA=ED by hypothesis, and the side, BG=EF by construction; consequently, A G=DFP (P. 5, c). But by hyppthesis ACC=D)F; and therefore, AC=AG (A. 1). But the oblique line'AC cannot be equal to AG, since BC is greater than BG (P. 15); consequently, BC and EF cannot be unequal, and hence, the triangles are equal (P. 10). BOOK I. 35 PROPOSITION XVIII. THEOREM. If two straight lines are perpendicular to a third line, they are parallel to each other. Let the two lines A6, B/D, be perpendicular to AB; then will they be parallel. For, if they could meet in a point 0, on either side of B D -AB, there would be two perpendiculars OA, OB, let fall. from the same point on the A C same straight line; which is impossible (P. 14). PROPOSITION XIX. THEOREM. If two straight lines meet a third line, making the sum of the interior angles on the same side equal to two right angles, the two- lines are parallel. Let the two lines KC, HD, meet the line BA, making the angles BA, ABD, together equal to two right angles: then the lines tKC, HD, will be parallel. From G, the middle point of BA, draw the straight line EG, j E 4B D perpendicular to KU: then, it will also be perpendicular to HD. For, the sum BA C+ABD is K i C equal to two right angles, by hypothesis; the sum ABD+ABE is likewise equal to two right angles (P. 1): taking away ABD from both, there will remain the angle BA C=-ABE. Again, the angles EGB, AGF, are equal (P. 4); therefore, the triangles EGUB and AGF, have each a side and two adjacent angles equal; therefore, they are themselves equal, and the angle GEB is equal to the angle GFAt (P. 6, c). But GEB is a right angle by construction; therefore, GFA is a right angle; hence, the two lines KC, 36 GEOMETRY. RLD, are perpendicular to the same straight line, and are therefore parallel (P. 18). Scholium. When two parallel E straight lines AB, CD, are met by a third line FE, the angles A G which are formed take particular names. Interior angles on the same C side, are those which lie within the parallels, and on the same side of the secant line; thus, HGB, GIlD, are interior angles on the same side; and so also are the angles HGA, GHC. Alternate angles lie within the parallels. and on different sides of the secant line, but not adjacent; A GHI, GHD, are alternate angles; and so also are the angles GHCG BGH. Alternate exterior angles lie without the parallels, and on different sides of the secant line, but not adjacent: EGCB., CHUF, are alternate exterior angles; so also are the angles A GfE PHID. Opposite exterior and interior angles lie on the same side of the secant line, the one without and the other within the parallels, but not adjacent: thus, EGB, GfHD, are opposite exterior and interior angles; and so also, are the angles AGE, GHC. Cor.. If two straight lines meet a third line, making tile alternate acngles equal, the straight lines are parallel. Let the straight line ET meet the two straight lines CTD, AB, making the alternate angles A G1, GFID, equal to each other: then will AB and CD be parallel. For, to each of the equal angles, add the angle HGB; we E shall then have AGHI + HGB = GHD + HGB. A -B But AGH+HGB is equal to two right angles (P. 1): hence, C --- D GIID+tIIGB is also equal to two right angles (A. 1): then F CD and. AB are parallel (P. 19.) BOOK I. 37 Cor. 2. If a straight line EE, meet two straight lines tD, AB, making the exterior angle EGB, equal to the interior and opposite angle GHD, the two lines will be parallel. For, to each add the angle HGB: we shall then have, EGB +HGB = GHD+HGB: ibt EGB +1 GB is equal to two right angles; hence, GHD+ffGB is equal to two right angles; therefore, CD, AB, are parallel (P. 19). PROPOSITION XX. THEOREM. f,a straight line meet two parallel straight lines, the sum of the interior angles. on the same side will be equal to cto right angles. Let the parallels AB, CD, be met by the secant line?Ei: then will HGB + GHD, or HGA+ GHC, be equal to two right angles. For, if IIGB +GIHD be E not equal to two right angles, let IGL be drawn, nmaking the sum HGL- + A G.ID equal to two right an- /L gles; then IL and CD will be parallel (P. 19); and hence, /H wve shall have two lines GB, GL, drawn through the same point G and parallel to CD, which is impossible (A. 13): hence, 11GB+ +GHD is equal ta two right angles. In the same manner it may be proved that HGA+ GHC is equal to two right angles. Cor. 1. If 11GB is a right angle, GHD will be a right angle also: therefore, every straight line perpendicular to one of two parallels, is perpendicular to the other. Cor. 2. If a straight line meet two parallel straight lines, the alternate angles will be equal. Let AB, CD, be two parallels, and FE the secant line. 38 GEOMETRY. The sum HUGB + GHD is equal to two right angles. But E the sum HGB +HGA is also A equal to two right angles (P. 1). Taking from each the angle HGB, and there remains A GH= GHD. C- D In the same manner we may prove that GHC=IIGB. F Cor. 3. If a straight line meet two parallel lines, the opposite exterior and interior angles will be equal. For, the sum HUGB +GHD is equal to two right angles. But the sum HGB +EGB is also equal to two right angles. Taking from each the angle HGUB, and there remains GHD=EGB. In the same manner we may prove that GHC=A GE. Scholium. We see that of the eight angles formed by a line cutting two parallel lines obliquely, the four acute angles are equal to each other, and so also are the four obtuse angles. PROPOSITION XXI. THEOREM. If two straight lines mneet a third line, making the sum of the interior angles on the same side less than two right angles, the two lines will meet if sufficiently produced. Let the two lines CD, IL, meet the line EF, making the sum of the interior angles UGL, GID, less than two right angles: then will IL and C1) meet if sufficiently produced. For, if they do not meet they are parallel (D. 13). But they are not parallel, for if they were, the sum of the interior angles LGI, GID, L would be equal to two right C D angles (P. 20), whereas it is /HE less by hypothesis: hence, the F lines IL, CD, will meet if sufficiently produced. Cor. It is evident that the two lines IL, CD, will meet on that side of iEF on which the sum of the two angles HGL, GHD, is less than two right angles. BOOK I. 39 PROPOSITION XXII. TItEOREM. Two straight lines which are parallel to a third line, are parallel to each other. Let CD and AB be parallel to the third line EF; then are they parallel to' each other. Draw PQR perpendicular to EP, E F and cutting AB, CD, in the points R P and Q. Since AB is parallel to C 0 D EF, PR will be perpendicular to AB A B (P. 20, c. 1); and since CD is parallel P to E1; PR will for a like reason be perpendicular.to CD. Hence, AB and CD are perpendicular to the same straight line; hence, they are parallel (P. 18). PROPOSITION XXIII. THEOREM. Two parallels are everywhere equally distant. Let CD and AB be two parallel straight lines. Through any two points of AB, as F and E, suppose FEi and EG to be drawn perpendicular to AB. These lines will also be perpendicular to CD (P. 20, c. 1); and we are now to show that they will be equal to each other. If GF be drawn, the CH G D angles GFE, FGH, considered in reference to the parallels AB, CD, will be alter- B nate angles, and therefore, A F equal to each other (P. 20, c. 2). Also, the straight lines FH, EG, being perpendicular to the same straight line AB, are parallel (P. 18); and the angles EGF, GFH, considered in reference to the parallels FH, EG, will be alternate angles, and therefore equal. Hence, the two triangles EFG, FG,; have a common side, and two adjacent angles in each equal; therefore, the triangles are equal (P. 6); consequently, FH, which measures the distance of the parallels AB and CD at the point F, is equal to EG, which measures the distance of the same parallels at the point E. 40 GEOMETRY PROPOSITION XXIV. T'iiEOREM. If two angles have their sides parallel and lying in the same direction, they wzll be equal. Let BA C and RDET be. the two angles, having AB parallel to ED, and A C to EF; then will they be equal. For, produce DE, if necessary, till D it meets A C in G. Then, since FEl E is parallel to G C, -the angle DEF is equal to DGC (P.'20, c. 3); and since A X C DC is parallel to AB, the angle DCGC It. F is equal to BA C; hence, the angle DEF is equal to BAC (A. 1). Scholium. The restriction of this proposition to the case where the side EF lies in the same direction with AC, and ED in the same direction with AB, is necessary, because if 5FE were prolonged towards H, the angle DEH would have its sides parallel to those of the angle BAC, but would not be equal to it. In that case, DE~H and BAG would be together equal to two right angles. For, DEHI +2DET is equal to two right angles (P. 1); but DEAF is equal to sBAC: hence, DEIH+BAC is equal to two right angles. PROPOSITION XXV. THEOREM. In every triangle the sum of the three angles is equal to two right angles. Let ABC be any triangle: then will the sum of the angles C+A+B be equal to two right angles. For, prolong the side CA towards B E D, and at the point A, suppose AFE to be drawn, parallel to BC. Then, since AE, CB, are parallel, and CAD C D cuts them, the exterior angle DAE A is equal to its interior opposite angle C (P. 20, c. 3). In like manner, since AE, CB, are parallel, and AB cuts them, BOOK I. 41 the alternate angles B and BAE, are equal; hence, the three angles of the triangle BA C are equal to the three angles CAB, BA-], EAD; but the sum of these three angles is equal to two right angles (P. 1); consequently, the sum of the three angles of the triangle, is equal to two right angles (A. 1). Cor. 1. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles. Cor. 2. If two angles of one triangle are respectively equal to two angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular. Cor. 3. In any triangle there can be but one right angle: for if there were, two, the third angle must be nothing. Still less, can a triangle have more than one obtuse angle. Cor. 4. In every right-angled triangle, the sum of the two acute angles is equal to one right angle. C(or. 5. Since every equilateral triangle is also equiangular (P. 11, c. 1), each of its angles will be equal to the third part of two right angles; so, that, if the right angle is expressed by unity, each angle of an equilateral triangle will be expressed by -. Cor. 6. In every triangle ABC, the exterior angle BAD is equal to the sum of the two interior opposite angles B and C. For, AE being parallel to B, the part BAE is equal to the angle B, and the other part D)AE is equal to the angle C. PROPOSITION XXVI. THEOREM. The sum of all the interior angles of a polygon, is equal to twice as many right angles, less four, as the figure has sides. Let ABCDE be any polygon: then will the sum of its interior angles 42 GEOMETRY. A+B + C+D+E be equal to twice *as many right angles, less four, as the figure has sides. From the vertex of any angle A, D draw diagonals AC, AD, to the vertices of the other angles. It is plain that the polygon will be divided E into as many triangles, less two, as it has sides; for, these triangles may be considered as having the point A A B for a common vertex, and for bases, the several sides of the polygon, excepting the two sides which form the angle A. It is evident, also, that the sum of all the angles in these triangles does not differ from the sum of all the angles in the polygon: hence, the sum of all the angles of the polygon is equal to two right angles, taken as many times as there are triangles in the figure; that is, as many times as there are sides, less two. But this product is equal to twice as many right angles as the figure has sides, less four right angles. Cor. 1. The sum of the interior angles in a quadrilateral, is equal to two right angles multiplied by 4-2, which amounts to four right angles: hence, if all the angles of a quadrilateral are equal, each of them will be a right angle. IHence, each of the angles of a rectangle, and of a square, is a right angle (D. 25). Cor. 2. The sum of the interior angles of a pentagon is equal to two right angles multiplied by 5-2, which amounts to six right angles: hence, when a pentagon is equiangular, each angle is equal to the fifth part of six right angles, or to - of one right angle. Cor. 3. The sum of the interior angles of a hexagon is equal to 2X (6-2,) or eight right angles; hence, in the equiangular hexagon, each angle is the sixth part of eight right angles, or A of one. Cor. 4. In any equiangular polygon, any interior angle is equal to twice as many right angles, less four, as the figure has sides, divided by the number of sides. BOOK I. 43 Scholium. When this proposition is applied to polygons which have re-entrant angles, each re-entrant angle must be regarded as greater than two right angles. But to avoid all ambigtity, we shall henceforth limit our reasoning to polygons with salient angles, which are.named cQnvex polygons. Every convex polygon is such, that a straight line, drawn at pleasure, cannot meet the sides of the polygon in more, than two points. PROPOSITION XXVII. THEOREM. If the sides of any polygon be prolontged, in the same direction, the sum of the exterior angles will be equal to four right angles. Let the sides of the polygon ABCIDFG, be prolonged, in the same direction; then will the sum of the exterior angles a b+c+d +f +g, be equal to four right angles. For, each interior angle, plus its exterior angle, as A+a, is equal to two right angles (P. 1). But there are as many exterior as interior C angles, and as many of each as there are sides of the polygon: A hence the sum of all the interior and exterior angles, is equal to twice as many right angles as the polygon' has sides. Again, the sum of all the interior angles is equal to twice as many right angles as the figure has sides, less four right angles (P. 26). Hence, the interior angles plus four right angles, is equal to twice as many right angles as the polygon has sides, and consequently, equal to the sum of the interior angles plus the exterior angles. Taking from each the sum of the interior angles, and there remain the exterior angles, equal to four right angles. 4B14 GEOMETRY. PROPOSITION XXVIII. THEOREM. In every parallelogram, the opposite sides and angles are equal. Let ABCD be a parallelogram: then will AB=-DC, AD=BC, the angle. A=C, and the angle ADC=ABC. For, draw the diagonal BD, dividing the parallelogram into the two trian- B C gles, ABD, DBC. Now, since AD,.BC, -ire parallel, the angle ADB-DBC (P. 20, c. 2); and since AB, C1D, are parallel, A D the angle ABDZ=BDC: and since the side DB is common, the two triangles are equal (P. 6); therefore, the side AB, opposite the angle ADB, is equal to the side DC. opposite the equal angle DBC (P. 10, s.), and the third sides AD, BC, are equal: hence, the opposite sides of a parallelogram are equal. Again, since the triangles are equal, the angle A is equal to the angle C (P. 10, s.) Also, the angle ADC composed of the two angles, ADB, BDC, is equal to ABC, composed of the two equal angles DBC, ABD (A. 2): hence, the opposite angles of a parallelogram are equal. Cor. 1. Two parallels AB, CD, included between two other parallels AD, BC, are equal; and the diagonal DB divides the parallelogram into two equal triangles. Cor. 2. Two parallelograms which have two sides and the included angle in the one equal to two sides and the included angle in the other, each to each, are equal. Let the parallelogram ABCD, have D C the sides AB, AD, and'the included angle BAD equal to the sides AB, AD, and the included angle BAD, in the last figure: then will they be equal. A B For, in each figure, draw the diagonal I)B. By the last corollary, the diagonal divides each parallelogram into two equal triangles: but the triangle BAD in one parallelogram, is equal to the triangle BAD in the other (P. 5): hence, the parallelograms are equal (A. 6). BOOK I. 45 PROPOSITION XXIX. THEOREM. If the opposite sides of a quadrilateral are equal, each to each, the equal sides are parallel, and the figure is a parallelogram. Let ABCD be a' quadrilateral, having its opposite sides respectively equal, viz.: AB=DC, and AD=-BC; then will these sides be parallel, and the figure a parallelogram. For, having drawn the diagonal BD D C the two triangles ABD, BDC, have all the sides of the one equal to the corresponding sides of the other; therefore they are equal, and the angle ADB, A B opposite the side AB, is equal to DBC, opposite CD (P. 10, s.); therefore the side AD is parallel to BC (r. 19, c. 1) For a like reason ARB is parallel to CD': therefore, the quadrilateral ABCD is a parallelogram. PROPOSITION XXX. TI-IEOREMI. If two opposite sides of a quadrilateral are equal and parallel, the olter sides are equal and parallel, and the figure is a parallelogram. Let ABCD be a quadrilateral, having the sides AB, CD, equal and parallel; then will the figure be a parallelogram. For, draw the diagonal DB, divid- C in g the quadrilateral into two triangles. Then, since AB is parallel to D C, the alternate angles ABD, 13D) C are equal (P. 20, c. 2); moreover, the A B side DB is common, and the side AB-DC; hence, the triangle ABD is equal to the triangle DBC (P. 5); therefore, the side AD is equal to 3C, the angle ADB=DBC, and consequently AD is parallel to BC (P. 19, c. 1); hence, the figure ABCD is a parallelogram. 46 GEOMETRY. PROPOSITION XXXI. THEOREM. The two diagonals of a parallelogram divide each other into equal parts, or. mutually bisect each other. Let ADCB be a parallelogram, ACg and.lDB its diagonals, intersecting at E; then will AE=EC, and DE= EB. Comparing the triangles AED, BEC, B C we find the side AD= CB (P. 28), the angle ADB= CBE, and the angle DAE=ECB (P. 20, c. 2); hence, these A D triangles are equal (P. 6); consequently, A E, the side opposite the angle ADE, is equal to EC, opposite CBE, and DE opposite DAE is equal to EB opposite ECB. Scholium. In the case of the rhombus, the sides AB, BC, being equal, the triangles AEB, EBC, have all the sides of the one equal to the corresponding sides of the other, and are therefore equal: whence, it follows, that the angles AEB, BEC, are equal, and therefore, the two diagonals of a rhombus bisect each other at right angles. BOOK II. OF RATIOS AND PROPORTIONS. DEFINITIONS. 1. PROPORTION is the relation which one magnitude bears to another magnitude of the same kind, with respect to its being greater or less.* 2. RATIO is the measure of the proportion which one magnitude bears to another; and is the quotient which arises from dividing the second by the first. Thus, if A and B represent magnitudes of the same kind, the ratio of A to B is expressed by A and B are called the terms of the ratio; the first is called the antecedent, and the second, the consequent. 3. The ratio of magnitudes may be expressed by numbers, either exactly or approximatively; and in the latter case, the approximation may be brought nearer to the true ratio than any assignable difference. Thus, of two magnitudes, one may be considered to be divided into some number of equal parts, each of the same kind as the whole, and regarding one of. these parts as a unit of measure, the magnitude may be expressed by the number of units it contains. If the other magnitude contain an exact number of these units, it also may * See Davies' Logic of Mathematics: Proportion, ~ 267. 48 GEOME TRY. be expressed by the number of its units, and the two magnitudes are then said to be commensurable. If the second magnitude do not contain the measuring unit an exact number of times, there may perhaps be a smaller unit which will be containbd an exact number of times in each of the magnitudes. But if there is no unit of an assignable value, which is contained an exact number of times in each of the magnitudes, the magnitudes are said to be incommensurable. It is plain, however, that if the unit of measure be repeated as many times as it is contained in the second magnitude, the result will differ from the second magnitude by a quantity less than the unit of measure, since the remainder is always less than the divisor. Now, since the unit of measure may be made as small as we please, it follows, that magnitudes may be represented by numibers to any degree of exactness, or they will differ from their numerical representatives by less than any assignable magnitude. 4. We will illustrate these, principles by finding the ratio between the straight lilies CD and AB, which we will suppose commensurable. From the greater line AB,' cut off a part equal A C to the less CD, as many times as possible; for example, twice, with the remainder BE. From the line (7D, cut off a part, CF, equal to the remainder BE, as many times as possible; - once, for example, with the remainder DE. From the first remainder B1E, cut off a part equal to the second, DF, as many times as possi —E ble; once, for example, with the remainder BG. From the second remainder D1, cut off a part equal to BG, the third remainder, as many times B as possible. Continue this process, till a remainder occurs, which is contained exactly, a certain number of times, in the preceding one. Then, this last remainder will be the common measure of the proposed lines. IRegarding this as unity, we shall BOOK II. 49 easily find the values of the preceding remainders; and at last, those of the two proposed lines, and hence, their ratio in num.bers. Suppose, for instance, we find GB to be contained exactly twice in FD; BG will be the common measure of the two proposed lines. Put BG=1; we shall then have, ED= 2; but EB contains FD once, plus GB; therefore, we have E)i':3: CD contains EB once, plus FD; therefore, we have CDn-5 and lastly, AB contains CD twice, plus EB; therefore, we have AB=13; hence, the ratio of the lines is that of 5 to 13. If the line CD were takenfor unity, the line AB would be i5; if AB were taken for unity, CD would be ~ix.5. What has been shown, in respect to the straight lines, C19 and AB, is equally true of arty two magnitudes, A and B.. For, we may conceive A to be divided into Ai number of units, each equal. to A,. then A=MXA': let B be divided into NV number of equal units, each equal to A'; then B —NJVXA'; X and N being integer numbers. Now the ratio of A to B, will be the same as the ratio of AIx A' to ArXA'; that is, the same as the ratio of the numerical quantities Mt and.7 since A' is a common unit. 6. If there be four mlagnitudes, A, B, C, and D, having such values that B D then A/ is said to have the same rcvlio to.B, that C has to) D; or, tile ratio of A to) B is said to be equal to the ratio of C to D. WVhen four quantlties have this relation' to each o(llble, they are said to be in proportion. To i.l(licate that the ratio of A to B is equal to the ratio of C to D, the quantities are usually written thus, A:B:: C: D, and read, A is to B as C is to D. The quantities which are compa-red together are called the terms of the proportion. rThe first and last terms are. called the two extremes, and the second and third terms, the twao means. 4 50 GEOME-TRY. 7. Of four proportional quantities, the last is said to be a fourth proportional to the other three, taken in order. The first and second terms, are called the first couplet of the proportion; and the third and fourth terms, the second covlet: the first and third terms are called the antecedents, and the second and fourth terms, the consequents. 8. Three quantities are in proportion, when the first has the same ratio to the second, that the second has to the third; and then the middle term is said to be a mean proportional between the other two. 9. Magnitudes are in-.proportion by alternation, or alternately, when antecedent is compared with antecedent, and consequent with consequent. 10. Magnitudes are in proportion by inversion, or innersely, when the consequents are taken as antecedents, and the antecedents as consequents. 11. Magnitudes are in proportion by composition, when the sum of the antecedent and consequent is compared either with antecedent or consequent. 12. Magnitudes are in proportion by division, when the difference of the antecedent and consequent is compared either with antecedent or consequent. 13. Equimultiples of two quantities are the products which arise from multiplying the quantities by the same number: thus, mX A, mX B, are eqcuimnultiples of A and B, the common multiplier being m. 14. Two varying quantities, A and B, are said to be reciprocally proportional, or inversely:proportional, when their values are so changed that one is increased as many times as the other is diminished. In such case, either of them is always equal to a constant quantity divided by the other, and their product is constant. R]OOK -II. 51 PROPOSITION I. THEOREM. When four magnitudes are in proportion, the producb of the two extremes is equal to the product of the two means. Let A, B, C, D, be any four magnitudes, and N; N, P, Q, their numerical representatives; then, if. M: N:: P: Q, we shall have iMX Q=NxP. For, since the magnitudes are in proportion, we have (D. 6), MN= Q: therefore, N=lX Q': whence, NXP=MX Q. Cor. If there are three proportional quantities, the product of the extremes will be'equal to the square of the mean (D. 8). For, if N=-P, we have ~x Q=N2 or P2. PROPOSITION II. THEOREM. If the product of two magnitudes be equal to the product of two other magnitudes, two of them may be made the extremes and the other two the means of a proportion. If we have MX Q=NXP; then will M: N:: P: Q. For, if P have not to Q, the ratio which M has to N, let P have to Q', (a number greater or less than Q,) the same ratio which M has to N: that is, let: N::P: P': Q then (P. 1), HX Q'=Nx P; NxP NxP hence, Q'= but, Q= - Consequently, Q'=Q, and the supposition that it is either greater or less,-is absurd; hence, the four magnitudes d;, N, P, Q, are proportional. 52 GEOMETRY. PROPOSITION III. THEOREM. If four. magnitudes are in proportion, they will be in proportion when taken alternately. Let M, N, P, Q, be four quantities in proportion; so that M: N:: P: Q: then will': P::- N: Q. For, since M: N:: P: Q: we have MiX Q9=NXP; therefore M and Q may be made the extremes, and. N and P the means of a proportion (P. 2); hence, M: P:: N: Q. PROPOSITION IV. THEOREM. If there be four proportional magnitudes, and four other proportional magnitudes, having the antecedents the same in both, the consequents will be proportional. Let A.: N:: P: Q, giving X Q=NXP, and M: R:: P: S, giving MIXS=RXP, thenwillN: Q:: R: S. For, multiplying the equations crosswise, we have MX QXRXxP=lxSXNXP; cancelling MX P in both numbers, we have, QX R = S X N: hence (p. 2), N: Q:: R:S. Cor. If there be two sets of proportionals, in which the ratio of an antecedent and consequent of the one is equal to the ratio of an antecedent and consequent of the other, the remaining terms will be proportional. For, if we had the two proportions, Ml: P:: N: Q and: S:: T V, we shall also have ~_ a and _I M P 1' Now, if P then Q V and we shall have N Q T: V. BOOK II. 53 PROPOSITION V. THEOREM. If four magnitudes are in proportion, they will be in proportion when taken inversely. If M:: P Q, then will N::: Q: P. For, from the given proportion, we have XQ Q =NTP, or, NxP=Mx Q. Now, N and P may be made the extremes, and H and Q the means of a proportion (P. 2): hence V: M:: Q: P. PROPOSITION VI. THEOREM. If four magnitudes are in proportion, they will be in proportion by composition or division. If we have M::: P Q we shall also have M\N: M:: PtQ: P. For, from the first propdrtion, we have Xx Q=NxP, or NxP=Hx Q. Add each of the members of the last equation to, and subtract it from MXP, and we shall have, HX PtAX P=MXPPHx Q; or (AL~N) x-P=(PQ) X3+ But 3iTztN and P, may be considered the two extremes, and P+ Q and M, the two means of a proportion (P. 2): hence, (~N): H:: (P Q): P. PROPOSITION VII. THEOREM. Equimultiples of any two magnitudes, have the same ratio as the magnitudes themselves. Let H and N be any two magnitudes, and m any number whatever; then will m X M and m X N be equal mul GEOMETRY. tiples of i11 and N: and mX2 M will be to m VN, in the ratio of M to NV For, Lx N=ANX 1 multiplying each member by m, and we have m X X N= m XNXM: then (P. 2), mmxf: mXN-:: 211: N. PROPOSITION VIII. THEOREM. Offfour proportional magnitudes, if there be taken any equimultiples of the two antecedents, and any equimultiples of the two consequents, such equimultiples will be proportional. Let M, N, P, Q, be four magnitudes in proportion; and let mn and n be any numbers whatever, then will mXM: nXN:: mXP nX nQ. For, since M: N1V:: P Q, we have MX Q=NX P; hence, m XMX n X Q=n XNTXm x P, by multiplying both members- of the equation by m X n. But m X M and n X Q, may be regarded as the two extremes, and n X N and m X P, as the means of a proportion; hence, mXa1: nXN:: mXP: nxQ. PROPOSITION IX. THEOREM. Of four proportional magnitudes, if the two consequents be either augmented or diminished by magnitudes which have the same ratio as the antecedents, the resulting magnitudes and the antecedents will be proportional. Let M: N:: P: Q, and let M P:: m n: then will M: P:: Nzm: Q+n. For, since M: N::'P: Q,.1fXQ=NxP. and sinke 1. P: m: n, 1IX n=PXm, therefore, MX Q~I2MX n=NXPIPx m, or 1Mx (Q+n) =P X (Nm): hence (p. 2),: P:: N~m: Q4-n. BOOK II. 55 PROPOSITION X. TIEOlEM..If any number of magnitudes are proportionacZs, any one antecedent will be to its consequent, as the sum of all tAe antecedents to the ksum of the consequents. Let: N:: P::: R: S, S&c. Then since, M: N:: P: Q, we have HXQ —— NxP, and, 3: N:: R: S, we have MIXS=Nxt, add to each - X N-z M X AY then, MXANT+1iX Q +fMX S=xN+NX P+NX P, or, A x (N+ Q+S) =Nx (Mf+P +R); therefore(P. 2),f: N:: "+P+R: N+Q+S. PROPOSITION XI. THEOREM. If two magnitudes be each increased or diminished by like parts of each, the resulting magnitudes will have the same ratio as the magnitudes themselves. Let Al and nT be any two magnitudes and and like parts of each. We have I X N-=2I X 1 Ix N MxN add to both, or subt. - ---—, and we have (A. 2), MX Nx-tM m x - or, M (N I ) z=N(AI tI), m N that is (P. 2), M: N':: 1 —: N — PROPOSITION XII. THEOREM. if four magnitudes are proportional, their squares or cubes will also be proportional. Let M: -A: P: Q Then will, XMX Q=INX P. GEOMETRY. By squaring both members, M2x Q 2=NX P2, and by cubing both members, JI3X Q3=N3Xp3; therefore, if: N2 p alid M3: N3:: p3: Q3 Cor. In a similar way it may be shown that like powers or roots of proportional magnitudes are proportionals. PROPOSITION XIII. THEOREIM. If there be two sets of proportional magnitudes, the products of the corresponding terms will be proportionals. Let f:N:: P: Q, and R: S:: T:: V, then will MXR: NXS:: PxT QxV. For, since XX Q=Nx P, and RX V=SxT, we shall have MX QX R X V=Nx P x S x T or, M x QX V =TNx SxP xT; therefore, MXR: NXS:: PXT: QXV. PROPOSITION XIV. THEOREM. If any number of magnitudes are continued proportionals; t7hen, the ratio of the first to the third will be duplicate of the common ratio; and the ratio of the first to the' fourth will be triplicate of the common ratio; and so on. For, let A be the first term, and m the common ratio: the proportional magnitudes will then be represented by A, ml XA, nm2X A, m3XA, m4XA, &c.: Now, the ratio of the first to any one of the following terms exactly corresponds with the enunciation. BOOK III. THE CIRCLE, AND THE MEASUREMENT OF ANGLES. DEFINITIONS. 1. The CIRCUMFERENCE OF A CIRCLE is a curve line, all the points of which are equally distant from a point within, called the centre. The circle is the portion of the plane terminated by the circumference. 2. Every straight line, drawn from the centre to the circumference, is called a radius, or, semidiameter. Every line which passes through the centre, and is terminated, on both sides, by the circumference, is called a diameter. From the definition of a circle, it follows, that all the radii are equal; that all the diameters are also equal, and each double the radius. 3. Any part of the circumference is called an arc. A straight line joining the extremities of an arc, is called a chord, or subtense of the arc.* 4. A SEGMENT is the part of a circle included between an are and its chord. 5. A SECTOR iS the part of the circle included between an arc, and the two radii drawn to the extremities of the arc. * In all cases, the same chord belongs to two arcs, and consequently, also to two segments: but the smaller one is always meant, unless the contrary is expressed. 58 GEOME-TRY. 6. A STRAIGHT LINE is said to be inscribed in a circle, when its extremities are in the circumference. An inscribed angle is one which has its vertex in the circumference, and is included by two chords of the circle. 7. An inscribed triangle is one Which has the vertices of its three angles in the circumference. And generally, a polygon is said to be inscribed in a circle, when the vertices of all the angles are in the circumference. The circumference of the circle is then said to circumscribe the polygon. 8. A SECANT is a line which meets the circumference in two points, and lies partly within, and partly without the circle. 9. A TANGENT is a line which has but one point in common with the circumference. The point where the tangent touches the circumference, is called the point of contact. 10. Two circumferences touch each other when they have but one point in common. Trhe common point is called the point of tangency. 11. A polygon is circumscribed about a circle, when all its sides are tangents to the circumference. In the same case, the circle is said to be inscribed in the polygon. POSTULATE. 12. Let it be granted that the circumference of a circle may be described froin any centre, and with any radius. BOOK III. 59 PROPOSITION I. THEOREM. Every diameter divides the circle and its circumference each into two equal parts. Let AEBF be a circle, and AB a diameter. Now, if the figure AEB be applied to AFB, E their common base AB retaining its position, the curve line AEB must fall exactly on the curve line AFB, A B otherwise there would, in the one'oor the other, be points unequally distant from the centre, which is con- F trary to the definition of a circle. Hence, the diameter divides the circle and its circumference, each into two equal parts. PROPOSITION II. THIEOREM. Every chord is less than a diameter. Let AD be any chord. Draw the radii CA, CD, to its extremities. We shall then have (B. I., P. 7)* ADDA-A C (B. I., P. 7, c.) 68 GEOMETRY. PROPOSITION XIII. THEOREM. If the distance between the centres of two circles is equal to the sum of their radii, the circumferences will touch each other externally. Let C and D be the centres of two circles at a distance from each other equal to CA+AD. The circles will evidently have lE the point A common, and they will have no other; because if they have two points common, the distance A between their centres must be less than the sum of their radii, which is contrary to the supposition. Cor. If the distance between the centres of two circles is greater than the sum of their radii, the two circumfer—.ces will be exterior the one to the other. PROPOSITION XIV. THEOREM. If the distance between the centres of two circles is equal to the difference of their radii, the two circumferences will touch each other internally.. ILet C and D be the centres of two circles at a distance from each other equal to AD- A. It is evident, as before, that the E two circumferences will have the point A common: they can have no other; because if they had, the A distance between the centres would ( D be greater than AD- CA (P. 12); which is contrary to the supposition. Cor. 1. Hence, if two circles touch each other, either externally or internally, their centres and the point of contact will be in the same straight line. Car. 2. If the distance between the centres of two BOOK III. 69 circles is less than the difference of their radii, one circle will be entirely within the other. Scholium 1. All circles which have their centres on,the right line AD, and which pass through the point A, are tangent to each other at the point A. For, they have only the point A common, and if through A, AE be drawn perpendicular to AD, it will be a common tangent to all the circles. Scholium. 2. Two circumferences must occupy with respect to each other, one of the five positions above indicated. Ist. They may intersect each other in two points: 2d. They may touch each other externally: 3d. They may be external, the one to the other: 4th. They may touch each other internally: 5th. The one may be entirely within the other. PROPOSITION XV. THEOREM. in the same circle, or in equal circles, equal angles at the centre, intercept equal arcs on the circumference. And conversely: If the arcs intercepted are equal, the angles contained by the radii are also equal.'Let C and C be the centres of equal circles, and the angle A Ga =DCE. first. Since the angles ACB, DCLE, are equal, one of them may be placed upon the other. Let the angle AB B be placed on DCE. Then A HB since' their sides are equal, the p6int A will evidently fall on 2D, and the point B on E. The are AB will also fall on the are DE; for, if the arcs did not exactly coincide, there would, in the one or the other, be- points unequally distant from the centre; which is impossible: hence, the are AB is equal to DE (A. 14). Second. If the arec AB=DE, the angle A CB is equal 70 GEOMETRY. to DCE. For, if these angles are not equal, suppose one of them, as A CB, to be the greater, and let A CI be taken equal to DCE. From what has just been shown, we shall'then have AI= DE; but, by hypothesis, AB is equal to DE; hence, Al must be equal to AB, or a part equal to the whole, which is absurd (A. 8); hence, the angle CB is equal to DCE. PROPOSITION XVI. THEOREM. In the same circle, or in equal circles, if two angZes at the centre have to each other the ratio of two whole numbers, the intercepted -arcs will have to each other the same ratio: or, we shall have the angle to the angle, as the corresponding arc to the corresponding are. Suppose, for example, that the angles A CB, D)CE are to each- other as 7 is to 4; or, which is the same thing, suppose that the angle M, which may serve as a common measure, is contained 7 times in the angle ACB, and 4 C C -~- \ A\ / i times in DCIE. The seven partial angles A Cm, m Cn, n Cp, &c., ifito which A CB is divided, are each equal to any of the four partial angles into which D CE is divided; and each of the partial arcs, Amn, nne, np, &c., is equal to each of the partial arcs Dx, xy, &c. (r. 15). Therefore, the whole are.AB will be to the whole are DE, as 7 is to 4, But the. same reasoning would evidently apply, if in place of 7 and 4 any numbers whatever were employed; hence, if the angles ACB, DCE, are to each other as two whole numbers, they will also be to each other as the arcs AB, DiE. Cor. Conversely: If the arcs AB, DE, are to each other as two whole numbers, the angles A CB, DCUE will be to BOOK III. 71 each other as the same whole numbers, and we shall have AB ~DE:: ACB: D CE. For, the partial arcs, Am, mn, &c., and Dx, xy, &c., being equal, the partial angles A Cm, m CU, &c., and DCx, xCy, &c., will also be equal, and the entire arcs will be to each other as the entire angles. PROPOSITION XVII. THEOREM. In the same circle, or in equal circles, any two angles at the centre are to each other as the interce2ted arcs. Let A CB and A CD be two angles at the centres of equal circles: then will A CB A CD:: AB: AD. For, if the angles are C C equal, the arcs will be equal (P. 15). If they are unequal, // let the less be placed on the greater. Then, if the A\ proposition is not true, the angle A CB will be to the angle A CD as the are AB is to an are greater or less than AD. Suppose such are to be greater, and let it be represented by A O; we shall thus have,'the angle ACB: angle ACD:: are AB: are AO. Next conceive the are AB to be divided into equal parts, each of which is less than DO; there will be at least one point of division between' D and 0; let I be that point; and draw C(. Then the arcs AB, Al, will be to each other as two whole numbers, and by. the preceding theorem, we shall have, angle ACB: angle A C':: are AB: are AI. Comparing the two proportions with each other, we see that the antecedents in each are the same: hence, the consequents are proportional (B. II., P..4); and thus we find, the angle ACD: angle AGI:: are AO: are AL But the are A 0 is greater than the are Al; -hence, if this proportion is true, the angle A CD mast be greater than the 72 GEOMETRY. angle A CI: on the contrary, however, it is less; hence, the angle A CB cannot be to the angle A CD as the arc AB is to an are greater than AD. By a process of reasoning entirely similar, it may be shown that the fourth term of the proportion cannot be ess than AD; hence, it is AD itself; therefore, we have angle ACB: angle ACD:: are AB: arc AD. Scholium 1. Since the angle at the centre of a circle, and the arc intercepted by its sides, have such a connection, that if the one be augmented or diminished, the other will be augmented or diminished in the same ratio, we are authorized to assume the one of these magnitudes as the measure of the other; and we shall henceforth assume the arc AB as the measure of the angle A CB. It is only necessary, in the comparison of angles with each other, that the arcs which serve to measure them, be described with equal radii. Scholium 2. An angle less than a right angle will be measured by an are less than a quarter of the circumference: a right angle, by a quarter of the circumference: and an obtuse angle by an are greater than a quarter, and less than half the circumference. Scholium 3. It appears most natural to measure a quantity by a quantity of the same species; and upon this principle it would be convenient to refer all angles to the right angle. This being made the unit of measure, an acute angle would be expressed by some number between o and 1; an obtuse angle by some number between 1 and 2. This mode of expressing angles would not, however, be the most convenient in practice. It has been found more simple to measure them by the arcs of a circle, on account of the facility with which arcs can be made to correspond to angles, and for various other reasons. At all events, if the measurement of angles by the arcs of a circle is in any degree indirect, it is still very easy to obtain the direct and absolute measure by this method; since, by comparing the fourth part of the circumference with the arc which serves as a measure of any angle, we find the ratio of a right angle to the given angle, which is the absolute measure. BOOK III. 73 MSholium 4. All that has been demonstrated in the last three propositions, concerning the comparison of angles with arcs, holds true equally, if applied to the comparison of sectors with arcs. For, sectors are not only equal when their angles are so, but are in all respects proportional to their angles; hence, two sectors A CB, A CD, taken in the same circle, or in equal circles, are to each other as the arcs AB, AD, the bases of those sectors. Hence, it is evident that the arcs of equal circles, which serve as a- measure of corresponding angles, are proportional to their sectors. PROPOSITION XVIII. THEOREM. Any inscribed angle is measured by half the arc included between its sides. Let BAD be an inscribed angle, and let us first suppose the centre of the circle to lie within the angle BAD. Draw the diameter ACE, and the radii CB, CD. The angle BCE, being exterior to A the triangle ABC, is equal to the sum of the two interior angles CAB, A.BC (B. I., P. 25, C. 6): but the triangle BA C being isosceles, the angle CAB is equal C to ABC; hence, the angle BCE is double of BA C.. Since BCE is at the centre, it is measured by the are BE (P. 17, s. 1); E hence, BA C will be measured by the half of BE. For a like reason, the angle CAD will be measured by the half of ED; hence, BA C+ CAD, or BAD will be measured by half of BE +ED, or half of BED. Secondly. Suppose the centre C to lie without the angle BAD. Then, draw- A ing the diameter ACE, the angle BAE will be measured by the half of BE; the angle DAE by the half of DE: C hence, their difference, BAD, will be measured by the half of BE minus the half of ED, or by the half of BD. D E Hence, every inscribed angle is measured by half the are included between its sides. 74 GEOMETRY. D Ctor. 1., All the angles BA C, BDC, A BEC, inscribed in the same segment are equal; because they are each C measured by half of the same are B C BOC. 0 Cor. 2. Every angle BAD, inscrib- A ed in a semicircle is a right angle; because it is measured by half the D_ - semicircumference B3OD, that is, by D the fourth pa/rt of the whole circumference (P. 17, s. 2). Cor. 3. Every angle BAG, inscrib- A ed in a segment greater than a semicircle, is an acute angle; for it is measured by half the arc BO(, less than a semicircumference (P. 17, s. 2). B And every angle BOC, inscribed in a segment less than a semicircle, is 0 an obtuse angle; for it is measured by half the arc BA C, greater than a semicircumference. Cor. 4. The opposite angles A and B C, of an inscribed quadrilateral ABCD, are together equal to two right angles: for, the angle BAD is measured by half the arc BCD, the angle BCD is 1 measured by half the are BAD; hence, the two angles BAD, BCD, taken together, are measured by half the circumference; hence, their sum is equal to two right angles (P. 17, s. 2). PROPOSITION XIX. THEOREM. The angle formed by two chords, which intersect each other, is measured by half the sum of the arcs included between its sides. Let AB, CD, be two chords intersecting each other at E: then will the angle' AEC, or DEB, be measured by half of AC+~DB. BOOK III. 75 Draw AF parallel to DC: -the A C arc DF will be equal to AC (P. 10), and the angle FAB equal to the angle DEB (B. I., P. 20, C. 3). But the / angle FAB is measured by half the | E arc SFDB (P. 18); therefore, DEB is. measured by half of FDB; that F is, by half of DB-+Di, or half of D-B DB +A C. To prove the same for the angle DEA, or its equal BEC. Draw the chord AC. Then, the angle D CA will be measured by half the arc D.FA; and the angle BAC by half the arc CB (P. 18). But the outward angle AEiD, of the triangle EAC, is equal to the sum of the angles A and C (B. I., P. 25, c. 6); hence, this angle is measured by one-half of BC plus one-half of AFD; that is, by half the sum of the intercepted arcs. By drawing a chord BG, similar reasoning would apply to the angle AEC or DEB. PROPOSITION XX. THEOREM. The angle formed by two secants, is measured by half the difference of the arcs included between its sides. Let AB, AC: be two secants: then will the angle BA C be measured by half the difference of the arcs BEC and DF. Draw DE parallel to AC: the arc EC will be equal to DF (P. 10), and the angle BDE equal to D, F the angle BA C (B. I., P. 20, c. 3). But BiDE is measured by half the arc BE (P. 18); hence, BA C is also measured by half the arc BE (P. 17); B that is, by half the difference of RBE'C and EC, and consequently, by half the difference of BELC and DF. 76 GEOMETRY. PROPOSITION XXI. THEOREM. Either of the angles formed by a tangent and a chord, is measured by half the arc included between its sides. Let BE. be a tangent, and AC a chord. From A, the uoint. of contact, draw D the diameter [4,. The angle BAD is a right angle (P. 9), and is measured by half the semicircumference AMD (p. 17, M C s. 2); the angle DA C is measured by the half of D/C; hence, BAD+DA C, or BA C" is measured by the half of AMD plus B A E the half of DC;, or by half the whole are AID C. It may be shown, by taking the difference of the angles DAE, DAC, that the angle CAE is measured by half the are AC, included between its sides. PROBLEMS RELATING TO THE FIRST AND. THIRD BOOKS PROBLEM I. To bisect a given straight line. Let AB be the given straight line. From the points A and B as cen- D tres, with a radius greater than the half of AB, describe two arcs cutting each other in D; the point D will be equally distant from A and B. Find, in like manner, above or beneath the line AB, qA C B a second point E, equally distant from the points A and B; through the two - E points D and E, draw the line DE, and the point C, where this line meets AB, will be equally distant from A and B. BOOK III. 77 For, the two points D and E, being each equally distant from the extremities A and B, must both lie in the perpendicular raised at the middle point of AB (B. I., P. 16, C). But only one straight line can be drawn through two given points (A. 11); hence, the line DE must itself be that perpendicular, which divides AB into two equal parts. PROBLEM II. At a given point, in a given.straight line, to erect a peipendicular to that line. Let BC be the given line, and A the given point. Take the points B and C at equal distances from A; then from the points B and C as centres, with a radius greater than BA, describe two arcs intersecting eath other at D; draw AD and B A it will be the perpendicular required. For, the point D, being equally distant from B and C, must be in the perpendicular raised at the middle of BC (B. I., P. 16); and since two points determine a line, AD is that perpendicular. Scholium. The same construction serves for making a right angle BAD, at a given point A, on a given straight line B C. PROBLEM III. From a given point, without a straight line, to let fall a perpendicular on that line. Let A be the point, and BD the given straight line. From the point A as a centre, and with a radius sufficiently great, des- A cribe an are cutting the line BD in two points B and D; then mark a _ C point E, equally distant from the B points B and D, and draw AE: it will be the perpendicular required. For, the two points A and ]E are each equally distant from the points B and D; hence, the line AE is a perpendicular passing through the middle of BD (r. I., P. 16, c). 78 GEOMETRY. PROBLEM IV. At a point in a given line, to make an angle equal to a given angle. Let A be the given point, AB the given line, and IKL, the given angle. From the vertex K, as a L centre, with any radius, describe the are IL, terminating K B in the two sides of the angle. From the point A as a centre, with a distance AB, equal to Kl;, describe the indefinite arc BO; then take a radius equal to the chord L],; with which, from the point B as a centre, describe an are cutting the indefinite arc BO, in D; draw AD; and the angle BAD will be equal to the given angle E. For, the two arcs BD, L-T, have equal radii, and equal chords; hence, they are equal (P. 4); therefore, the angles BAD, IKL, measured by them, are also equal (P. 15). PROBLEM V. To bisect a given arc, or a given angle. iFirst. Let it be required to divide the arc AEB into two equal parts. From the points A and B, as centres, with equal radii, describe two arcs cutting each other in D; through the point D and the centre a, draw CD: it will bisect the are AB in the point E. For, the two points C and D are C each equally distant from the extremities A and B of the chord AB; hence, the line CD bisects the chord at right angles (B. I., P. 16, c); and'consequently, it bisects the arc AEB in the point D E (P. 6). Secondly. Let it be required to divide the angle A CB into two equal parts. We begin by describing, from the vertex C, as a centre, the arc A-EB; which is then bisect BOOK III. 79 ed as above. It is plain that the line CD will divide the angle A CB into two equal parts (p. 17, s. 1). S'cholium. By- the same construction, each of the halves AE, EB, may be divided into two equal parts; and thus, by successive subdivisions, a given angle, or a given arc, may be divided into four eqtial parts, into eight, into sixteen, and so on. PROBLEM VI. Through a given point, to draw a parallel to a given straight line. ILet A be the given point, and BC the given line. From the point A as a centre, with a radius greater than the 13- shortest distance from A to BC, describe the indefinite arc EO; A D. from the point E as a centre, O with the same radius, describe the are AF; lay off ED = AR, and draw AD: this will be the parallel required. For, drawing AE, the angles AEF, EAD, are equal (P. 15); therefore, the lines AD, EF, are parallel (B. I., P. 19, c. 1). PROBLEM VII. Two angles of a triangle being given, to find the third. Let A and B be the given angles. Draw the indefinite line DE.; at any point as E, make the angle C /IT DEC equal to the angle A, and the angle CEH equal to the other angle B: the remaining angle HEF D E F will be the third angle required; because,these three angles are together equal to two right angles (B. I., P. 1, c. 3), and so are the three angles of a triangle (B. I., P. 25); consepently, HEF is equal to the third angle of the triangle. 80 GEOMETRY. PROBLEM VIII. Two sides of a triangle, and the angle which they contain, being given, to describe the triangle. Let the lines B and C be equal to the given sides, and A the given angle. Having drawn the indefinite line DF, make at the point D, the angle fDE equal to the given angle A; then take DG=B, DH= C, and draw GH: DGGH will be D - the triangle required (B. I., P. 5). PROBLEM IX. A side and two angles of a triangle being given, to describe the triangle. The two angles will either be G F both adjacent to the given side, \ or one will be adjacent, and the other opposite: in the latter case find the third angle (PROB. 7), D E and the two adjacent angles will be known. Then drafw the straight line DE, and make it equal to the given side: at the point D, make an angle EDF, equal to one.of the adjacent angles, and at -E, an angle DEG equal to the other; the two lines DF, C,, will intersect each other in If; and DEH- will be the triangle required (B. I., P. 6). PROBLEM X. The three sides of a triangle being given, to describe the triangle. Let A, B, and C, denote the three given sides. Draw DE, and make it equal to the side A; from the point ]D as a centre, with a radius equal \ E to the second side B, describe an are; from E as a centre, with a B radius equal to the third side C, C.describe another arc intersecting the former in F; draw DF, EF; and DEF will be the triangle required (B. I., P. 10). BOOK III:. 8 Scholitnm. If one of' the'- sides were greater than: the sum of the other two, the -arcs would not intersect each other, for no such triangle could, exist (B. I., P, 7): but the solution will always be, possible,. when the surm of any two of the lines, is greater than the third. PROBLEM..XI. TWo sides of a triangle, and the angle opposite one of thel,, being given, to:describe the tia ti2Fagle., Let A and B be the given sides, and C the given angle. There are two cases. Ffirst. When the angle C is a right angle, or when it is obtuse. Draw DF and make the angle FDE-= C; take DEJ'=A: from the point E as a centre, with a radius equal to the given - side B, describe an are cutting it J)E in F; draw FEF; then' DIE will be the triangle required.. D In this case, the side B must be greater than A;- for the angle C being a right an:gle, or an obtuse angle, is the greatest angle of the triangle.(B.., P. 25, c. 3), and the side opposite to it must, therefore, also be the greatest (B. I., P. 13). Secondly. If the angle C is A - - jacute, and B greater than A, the B'- --—' same construction will again ap- E ply, and DEF will be the trian-' gle required. Fi ) But if the angle C is acute, A - f and the side B less than A, then Bi- -- the arc described from the centre E., with the radius EE=B, will cut the side DlP in two points D - F and G, lying on the same side., of Di: hence, there will be two triangles DEF, iDEG, either of which will satisfy all t. he conditions of the problem. 82 GEOMETRY. Scholium. If the arc described with E as a centre, should be tangent to the line D G, the triangle would be right angled, and there would be but one solution. The problem will be impossible in all cases, when the side B is less than the perpendicular'let fall from E on the line DF. PROBLEM XII. The adjacent sides of a parallelogram and their included angle being given, to describe the parallelogram. Let A and B be the given sides, and C the given angle. Draw the line DEf, and lay off DE equal to A: at F the point D, make the angle EDF=C; take DF=B; describe two arcs, the one from D E H f.i as a centre, with a radius Al -'G= DE, the other from E C as a centre, with a radius EG -1)F; to the point G, where these arcs intersect each other,,iraw FG, EG; DEUGF will be the parallelogram required. For, the opposite sides are equal, by construction; hence, the figure is a parallelogram (B. I., P. 29); and it is tormed with the given sides and the given angle. Cor. If the given angle is a right angle, the figure will be a rectangle; if, in addition to this, the sides are equaL it will be a square. PROBLEM XIII. To find the centre of a given circle or arc. Take three points, A, B, -, anywhere in the circumtarence, or in the are; draw AB, BC, or suppose them to be drawn; bisect these two \ 0O lines by the perpendiculars B)E, FG (PROB. 1): the point 0, where these perpendicu- A hlas meet, will be the centre D soulght (P. 6, s). BOOK III. 83 &S'holium. The same construction serves for making a circumference pass through three given points A, B, C; and also for describing a circumference, which shall circumscribe a given triangle ABC. PROBLEM[ XIV. Through a given point, to draw a tangent to a given circle. Let A be the given point, and C the centre of the given circle. A D If the given point A lies in the circumference, draw the radius CA, and erect AD perpendicular to it: AD will C be the tangent required (P. 9). If the point A lies without the circle, join A and the centre, by the straight line CA: bisect CA in 0; from 0 as a centre, with the radius.' OC, describe a circumference intersecting the given circumference in B; i \ draw AB: this will be the tangent required. \ For, drwing CB, the angle CBA being inscribed in a semicircle is a right angle (P. 18, c. 2); therefore, AB' is a perpendicular at the extremity of the radius CB; hence, it is a tangent (P. 9). Schoiunm 1. When the point A lies without the circle, there will be two equal tangents, AB, AD, passing through the point A: for, there will be two right-angled triangles, CBA, CDA, having the hypothenuse CA common, and the side CB.=CD; hence, there will be two equal tangents, AB, AD. The angles CAD, CAB, are also equal (B. 1., P. 17). Scholiurn 2. As there can be but one line bisecting the angle BAD, it follows, that the line which bisects the angle formed by two tangents, must pass through the centre of th1e circle. 84 GEOMETRY. PROBLEM XV. To inscribe a circl in a given triangle. Let ABC be the given triangle. Bisect the angles A and B, by the lines A O and BO, meeting in the point 0 (PRoB. D.5); from the point 0, let fall the perpendiculars OD, OE, OF (PROB. 3), on the three sides of A F the triangle: these perpendiculars will all be equal. For, by construction, we have the angle DA O=OAF; the right angle ADO=AFO; hence, the third angle AOD is equal to the third AOF (B. I., P. 25, C. 2). Moreover, the side A 0O is common to the two triangles A OD, A OF; and the angles adjacent to the equal side are equal: hence, the triangles themselves are equal (B. I., P. 6); and DO is equal to OF. In the same manner it may be shown that the two triangles BOD, BOE, are equal; therefore OD is equal to O0E; hence, the three perpendiculars OD, OE, OF,i are all equal. Now, if from the point 0 as a centre, with the radius OD, a circle be described, this circle will be inscribed in the triangle ABC (D. 11); for, the side AB, being perpendiecular to the radius at its extremity, is a tangent (P. 9); and the same thing is true of the sides BC, AC. Scholium. The three lines which bisect the three angles of a triangle meet in the same point. PROBLEM XVI. On a given straight line to describe a segment that 3hsall contaizn a given angle; that is to say, a segment such, that any angle inscribed in it shall be equal to a given angle. Let AB be the given straight line, and C the given angle. BOOK III. 85 (0 Produce AB towards D. At the point B, make the angle DBE= C; draw BO perpendicular to ZE, and at the middle point G, draw GO perpendicular to AB: from the point 0, where -these perpendiculars meet, as a centre, with the distance OB, describe a circumference: the required segment will be AJIB. For, since BEr is perpendicular to the radius OB at its extremity, it is a tangent (P. 9), and the angle ABEF is measured by half the arc AKB (P. 21). Also, the angle AJT1L, being an inscribed angle, is measured by half the tre AIMB (P. 18): hence, we have A11IB=AB —EBD- C: hence, any angle inscribed in the segment AM1B is equal to the given angle C'. Scholium. If the given angle were a right angle, the required segment would be a semicircle described on AB as a diameter. PROBLEM XVII. Twco agyles beeng given, to find their common measure, and by means of it, their ratio in numbers. Let A and B be the given angles. With equal radii describe the arcs CO), EF, to serve asA neasures for the angles. Afterwards, proceed in the comparison of the arcs CD, E, in the same manner as in the c(omparison of two straight lines (B. II., D. 4); since an arc may be cut off from an are of the same radius, as a straight 86 GEOMETRY. line from a straight line. WVe A shall thus arrive at the common measure of the arcs CD), EF, if they have one, and thereby at their ratio in num- D bers. This ratio will be the same as that of the given angles (P. 17); and if DO is the common measure of the arcs, the angle DAO 0 will be that of the angles. Scholizum. According to this method, the absolute value of an angle may be found by comparing the arc which measures it, to a quarter circumference. For example, if a quarter circumference is to the angle A as 3 to 1, then, the angle A will be 3 of one right angle, or -I- of four right angles. It may also happen, that the arcs compared have no common measure; in which case, the numerical ratios of the angles will only be found approximatively with more or less correctness, according as the operation is continued a greater or less number of times. BOOK IV. PROPORTIONS OF FIGURES- MEASUREMENT OF AREAS. DEFINITIONS. 1. SIMILAR FIGURES are those which are mutually equiangular (B. I., D. 22), and have their sides about the equal angles, taken in the same order, proportional. 2. In figures which are mutually equiangular, the angles which are equal, each to each, are called homologous angles: and the sides which are like situated, in respect to the equal angles, are called hovnologous sides. 3. AREA, denotes the superficial contents of a figure. The area of a figure is expressed numerically by the number of times which the figure contains some other figurQ regarded as a unit of measure. 4. EQUIVALENT FIGURES are those which have equal areas. The term equal, when applied to quantity in general, denotes an equality of measure; but when applied to geometrical figures it denotes an equality in every respect; and such figures when applied the one to the other, coincide in all their parts (A. 14). The term equivalent, denotes an equality in one respect only; viz.: an equality between the measures of figures. The sign s, denotes equivalency, and is read, is equivalent to. 5. Two sides of one figure are said to be reciprocally proportional to two sides of another, when one of the sides of the first is to one of the sides of the second, as the remaining side of the second is to the remaining side of the first. 88 G-EOM ETRY. 6. SIMILAR ARCS, SECTORS, or, SEGMENTS, are these, which in different circles, correspond to equal angles at tihe centre. Thus, if the angles A and 0 are A equal, the arc BEGt will be similar to 0 )GE,, the sector BAGC to the isector 1)OF, and the segment BCE, to the \D) E segment DEG. B C G F 7. The ALTITUDE of a triangle is the perpendicular let fall from the vertex of an angle on the -opposite side: this side is then called a base. i/ 8. The altitude of a parallelogram is the perpendicular distance between two opposite sides. These sides are called bases. 9. The altitude of a trapezoid is the perpendicular distance between its two / parallel sides. _ A PROPOSITION I. TIHEOREM. Parallelograms which have epqal bases and equal altitudes, are equvi'zlealt Since the two parallelograms have equal bases, those bases may be placed the one on the other. Therefore, let A.B be the common base of the two pairallelograms ABCD), A.BEF, which have the same altitude: then will they be equivalent. For, in the parallelogrm D CF E P F C E AB BCD, we have A B A AB =DC,..nd AD=BS(3I..28); A Baand in the parallelogram ABEE, we have, AB= EF, and' A=-B.E: hence, DC=-EF (A. 1). Now, if from the line DLE, we take away DC,'there will B.' O OK' V..' - Y;89 remain CE; and if from the same line we take away EF, there will remain):D)F; hence, CE= —F (A. 3); therefore, the triangles ADF.and BCE are mutually equilateral, and consequently, equal (B. I., P. 10). But if from the quadrilateral ABED, we takel away the triangle ADF, there will remain the parallelogram ABEE'; and if from the same quadrilateral, we take away the equal triangle BCE, there will remain the parallelogram ABCD). Hlence, any two parallelograms, which have equal bases and equal altitudes, are equivalent. Scholium. Since the rectangle and square:are. parallelograms (B. I., D. 25), it' follows that either is'q:tni-valent to *any parallelogram having an equal base and an equal altitude. And generally, whatever property is proved as belonging to a parallelogram, belongs equally to every variety of parallelogram. PROPOSITION II. THEOREM. If a trianglte agcd a parallelogramn have equal bases and equal altitudes, the triangle will be half the parallelogram. Place the base of the triangle on that of the parallelogram, so that the triangle A CB, and the parallelogram ABFD shall have the common.base AB: then will the triangle be half the parallelogram. For, since the triangle and the D F_ E. C parallelogram have equal altitudes, the vertex C, of the triangle, will be in the upper base of the paral-. lelogram (B. I., P. 23). Through A, A B draw AE parallel to BC, forming the parallelogram A BCE Now, the parallelograms ABFD, ABCE are equivalent, having the same base and altitude (P. 1). But the triangle ABC' is half the parallelogram BE (B. I., P. 2S, c.1): therefore, it is half the equivalent parallelogram B19 (A. 7). Cor. All triangles which have equal bases and equal altitudes are equivalent, being halves of equivalent paralleklograms. 90 GEOMETRY. PROPOSITION III. THEOREM. Two rectangles having equal altitudes are to eachl othter as their bases. Let ABCD, AEFD, be two rectangles having the conmon altitude AD: they are to each other as their bases AB, A-E. First. Suppose that the bases D F C are commensurable, and are to each other, for example, as the I numbers -7 and 4. If AB be diA E B vided into 7 equal parts, AE A E B will contain 4 of those parts. At each point of division erect a perpendicular to the base; seven partial rectangles will thus be formed, all equal to each other, because each has the same base and altitude (r. 1, s). The rectangle ABCD will contain seven partial rectangles, while ARED will contain four: hence, the rectangle ABCD: AEFD:: 7: 4, or as AB: AE. The same reasoning may be applied to any other ratio equally with that of 7 to 4: hence, whatever be the rhtio, we have, when its terms are commensurable, ABCD: AEFD:: AB: AE. Second. Suppose that the bases AB, D F K C AE, are incommensurable: we shall still have ABCD ALEFDD AB AB AE. _ A EIOB For, if the rectangles are not to each other in the ratio of AB to AE, they are to each other in a ratio greater or less: that is, the fourth term must be greater or less than AE. Suppose it to be greater, and that we have ABCD: AEFD:: AB:A O. Divide the line AB into equal parts, each less than EO. There will be at least one point I of division between LE and 0: from this point draw 1_K perpendicular to A]; BOOK IV. 91 forming the new rectangle AIK: then, since the bases AB, A., are commensurable, we have, ABUCD: A IKD:: AB: AL But by hypothesis we have ABCD: AEFD:: AB: A O. In these two proportions the antecedents are equal; hence, the consequents are proportional (B. II., P. 4), that is, AIKD: AEFD:: A: A O. But A0 is greater than Al; which requires that the rectangle AEFD be greater than AIKD: on the contrary, however, it is less (A. 8); hence, the proportion is not true; therefore ABCD cannot be to AET)D, as AB is to a line greater than AE. In the same manner, it may be shown that the fourth term of the proportion cannot be less than AE; therefore, being neither greater nor less, it is equal to AE. Hence, any two rectangles having equal altitudes, are to each other as their bases. PROPOSITION IV. THEOREM. Any two rectangles are to each other as the products of their bases and altitudes. Let ABCD, AEGF, be two rectangles; then will the rectangle, ABCD: AEGF:: ABxAD: -AEx AF. Having placed the two rect- H D C angles, so that the angles at A are opposite, produce the sides GE, CD, till they meet in H. A B Then, the two rectangles ABCD, AETD, having the same altitude G F AD, are to each other as their bases AB, AE: in like manner the two rectangles AELID, AEGF, having the same altitude AE, are to each other as their bases AD, AF: thus we have, ABCD: AFIID:: AB: AE AEHD: AEGF:: AD: AF. 92 GEOMETRY. Multiplying the corresponding terms of these proportions together (B. IL., P. 13), and omitting the term AE1JI), since it is common to both the-antecedent and consequent (B. II., P. 7), we have ABCD: AEGF:: ABXAD: AEx AF. Sclholiztm 1. If we take a line of a given length, as one inch, one foot, one yard, &c., and regard it as the linear unit of measure, and find how many times this unit is contained in the base of any rectangle, and also, how many times it is contained in the altitude: then, the product of these two ratios may be assumed as the reeasure of the rectangle. For example, if the base A of the rectangle A contains _ ten units and its altitude three, the rectangle will be repre- I sented by the number 10X3 =30; a number which is entirely abstract, so long as we regard the numbers 10 and 3 as ratios. But if we assume the square constructed on the linear unit, as the unit of surface, then, the product will give the number of superficial units in the surface; because, for one unit in height, there are as many superficial units as there are linear units in the base; for two units in height, twice as many; for three units in height, three times as many, &c. In this case, the measurement which before was merely relative, becomes absolute: the number 30, for example, by which the rectangle was measured, now represents 30 superficial units, or 30 of those equal squares described on the unit of linear measure: this is called the Area of the rectangle. AScholiam 2. In geometry, the product of two lines frequently means the same thing as their rectangle, and this expression has passed into arithmetic, where it serves to designate the product of two unequal numbers. The term square is employed to desigunate the product of a number multiplied by itself. BOOK IV. 93 The squares of the numbers 1, 2, 3, &c., are 1, 4, 9, &c. So likewise, the geometrical square constructed on a double line is evidently four times as great as the square on a single one; on ia triple line it is nine times as great, PROPOSITION V. THEOREM. The area of a parallelogram is equal to the product of its base and altitude. Let ABCD be any parallelogram, and BE its altitude: then will its area be equal to ABXBE. Draw A/v, and complete the rectangle ABEF. The parallelogram ABCD is equiv- F D E C alent to the rectangle ABEF (P. 1, s.); but this rectangle is measured by A B X F BE (P. 4, s. 1); therefore, AB x BE is A 1 equal to the area of the parallelogram ABC'D. CJor. Parallelograms of equal bases are to each other as their altitudes; and parallelograms of equal altitudes are to each other as their bases. For, let C and -D denote the altitudes of two parallelograms, and B the base of each: then, B X C: BXD::C: D (B. II., P. 7) If A and B are the bases, and C the altitude of each, we shall have, AxC': BXC:: A: B: and parallelograms, generally, are to each other. as the products of their bases and altitudes. PROPOSITION VI. THEOREM. The area of a triangle is eqtal to half the product of its batse and acltitude. Let BAC be a triangle, and AtD perpendicuplar to the base: then will its area be equal to one- half of'(/xAD). 94 GEOMETRY. For, draw CE parallel to BA, and A AE parallel to.BC, completing the parallelogram BE. Then, the triangle ABC is half the parallelogram ABCE, which B D) has the same base BC, and the same altitude AD (P. 2); but the area of the parallelogram is equal to BC x AD (P. 5); hence, that of the triangle must be ~BCXAAD, or BCX-AD. Cor. Two triangles of equal altitudes are to each other as their bases, and two triangles of equal bases are to each other as their altitudes. And triangles generally, are to each other, as the products of their bases and altitudes. PROPOSITION VII. THEOREM. The area of a trapezoid is equal to the product of its altitude, by half the sum of its parallel bases. Let ABCD be a trapezoid, EF its altitude, AB and CD its parallel bases: then will its area be equal to EF x. I(AB + CD). Through 1, the middle point of the D E C K side BG, draw KL parallel to the op- / posite side AD; and produce DC till -'I it meets KL. In the triangles IBL, ICi, we have A F L B the side IB=IC, by construction; the angle LIB= CHAT (B. I., P. 4); and since CKI and BL are parallel, the angle iBL=1CK (B. I., P. 20, C. 2); hence, the triangles are equal (B. I., P. 6); therefore, the trapezoid ABCD is equivalent to the parallelogram ALED, and consequently, is measured by ~FXAL (P. 5). But we have AL=DK; and since the triangles IBL and KCI are equal, the side BL= CK: hence AB+ CD= AL+DK'=2AL; hence, AL is the half sum of the bases AB, CD; hence, the area of the trapezoid ABCD, is equal to the altitude EF multiplied by the half sum of the bases AB, CD, a result which is expressed thus: AB + CD ABCD=EFx.2' BOOK IV. 95 Scholium. If through i; the middle point of BC, the line IH be drawn parallel to the' base AB, it will bisect AD at H/ For, since the figure ALI[H is a parallelogram, as also, HIEKD, their opposite sides are parallel, and we have AH-=IL, and DH=IK; but since the triangles LBi, IKC, are equal, we have IL=IK; therefore, AIt==HD. But since the line HI=AL, it is also equal to AB + CD 2 hence, the area of the trapezoid may also be expressed by EFx/III; consequently, the area of a trapezoid is equal to its altitude multiplied by the linle which connects the middle points of its inclined sides. PROPOSITION VIII. TIIEOREM. ThYe square described on' the sum of two lines is equivalent to the sum of the squares. described on the lines, together' witFh twice the rectangle contained by the lines. Let AB, BC, be any two lines, and A C their sum; then will AC2 or (AB+BC) > AB + BC + 2AB X BC. On A C describe the square A CDR; take AF=AB; draw FG parallel to A C, and BH parallel to AE. The square ACDE is made up of E TI D four parts; the first ABIF is the square described on AB, since we made AF= I AB: the second IGDH is the square G described on IG, or BC; for, since we have A C=AE and AB=AF, the difference, AC-AB must be equal to the A B C difference AE -AF, which gives BC = EF; but IQ is equal to BC, and DG to EF, because of the parallels; therefore, IGCDH is equal to a square described on BC. Now, if these two squares be taken away from the large square, there will remain the two rectangles BCGI; FIHJE, each of which is measured by ABXBC: hence, the square on the sum of two lines is equivalent to 96 GEOMETRY. the sum of the squares on the lines, together with twice the rectangle'contained by the lines. Cor. If the line AC were divided into two equal parts, the two rectangles Fl, BG, would become squares, and the square described on the whole line would be equivalent to four times the square described on half the line. Scholium liThis property is equivalent to the property demonstrated in algebra,' in obtaining the square of a binomial; which is expressed thus: (a+b)2=a2+2ab+b-. PROPOSITION IX. THEOREM. The square described on the difference of two lines, is equivaernt to the sum of the squares described on the lines, diminished by twice the rectangle contained by the lines. Let AB, BC, be two lines, and A their difference; then, ACf, or (AB-BC)2 +BC-2AB x BU. On AB describe the square ABIF; L F G I take A'E=AC; through C draw C parallel to B1i and through E draw K E Ei1 parallel to ABi and prolong it to lj making EIC=CB, and- then complete the square KEFL. Since KD=AB, and BC=KL, the two rectangles C, KG-, are each measured by A BXBC: the whole figure ABILKEA, is equivalent to AB2+BC; take from each the two rectangles C1, IKG, and there will remain the square ACDE, equivalent to AB 2+BC2 diminished by twice the rectangle of ABXBC. Scholium. This proposition is equivalent to the algebraical formula,: a-b)f2=a2- 2ab + b.2 BOOK IV. 97 PROPOSITION X. THEOREM. The rectangle contained by the sum and the difference of two lines, is equivalent to the difference of their squares. Let AB, BC, be two lines; then will (AB+BC) x (AB-BC j)o- AB2- -BC. Upon AB and A C, describe the F G I squares ABI~, ACDE; prolong AB I till BIf is equal to BC; and com- D plete the rectangle AKILE, and prolong CD to G. The base AK of the rectangle AL is the sum of the two lines AB BC; and its altitude AE is their difference; therefore, the rectangle AKLE is equivalent to (AB + BC)x (AB-BC). Again, DHIG is equal to a square described on CB; and since BH is equal to ED, and BK to EE, the rectangle B.L is equal to the rectangle 1'G: hence, the rectangle AKLE is equivalent to ABHIE plus EDGF, which is precisely the difference between the two squares AI and DI described on the lines AB, CB: hence, we have (A.1.), (AB +BC) X (AB-BC) A-B-BC. Scholium. This proposition is equivalent to the algebraical formula, (a+b)X (a-b)=a2-b2. PROPOSITION XI. THEOREM. The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the two other sides. Let BCA be a right-angled triangle, right-angled at A: then will the square described on the hypothenuse BC be equivalent to the sum of the squares described on the other two sides, BA, A C.'7 98 GEOMETRY. Having described a square K on each of the three sides, L let fall from A, on the hypothenuse, the perpendicular A AD, and prolong it to E; / and draw the diagonals AE, UNHB ID The angle ABF is made up of the angle ABC, together with the right angle CBF; the angle CBHis made " E F E G up of the same angle ABC, together with the right-angle ABN; hence, the angle A BF is equal to INtBC (A. 2). But we have AB =BI, being sides of the same square; and.BF=BC, for the same reason: therefore, the triangles ABE, IIBC, have two sides and the included angle in each equal; therefore, they are themselves equal (B. I., P. 5). But the triangle ABF is half the rectangle BE, because they have the same base BE, and the same altitude BD (r. 2). The triangle HBC is in like manner half the square Ai[: for, the angles BA C, BAL, being both right angles, A C and AL form one and the same straight line parallel to NB (B. I. P. 83); hence, the triangle and square have equal altitudes (B. i.,P. 23); they also have the common base BH; consequently, the triangle is half the square (P. 2). The triangle ABF has already been proved equal to the triangle NFBC; hence, the rectangle BDEF, which is double the triangle ABF, must be equivalent to the square A1, which is double the equal triangle HBC. In the same manner it may be proved, that the rectangle EGCD is equivalent to the square Al But the two rectangles _FEDB, EG CD, taken together, make up the square EFGCB: therefore, the square FGCB, described on the hypothenuse, is equivalent to the sum of the squares BALB-, GCIKA, described on the two other sides; that is, — 2 2-2 — BCr. enAB +AC.i Cor. 1. Hence, the square of one of the sides of a right BOOK IV. 99 angled triangle is equivalent to the square of the hypothenuse diminished by the square of the other side; thus, -2 — 2 -2 AB- -BC -A Cf. Cor. 2. If from the vertex of the right angle, a perpendicular be let fall on the hypothenuse, the parts of the hypothenuse are called segments: we shall then have, The square of the hypothenzuse to the square of either side about the right angle, as the hypothenuse to the segment adjacent to that side. For, by reason of the common altitude BF, the square BGC is to the rectangle BE, as BC to BYD (P. 3): but the square BL is equivalent to the rectangle BE: hence a12 p -2 BC2: BA.:: BC BD. We may show, in like manner, that BC: AC-:: BC DC. Cor. 3. The squares of the two sides containing the right angle, are to each other as the adjacent segments of the hypothenuse. For, the rectangles BDE1, DCGE, having the same altitude, are to each other as their bases BD, CD (P. 3). But these rectangles are equivalent to the squares A1, Al; therefore, we have -2 2 AB: A:: BD: DC. Cor. 4. The square described on the diagonal of a square is equivalent to double the square described on a side. Let ABCD be a square described on H D G AB, and EFGII a square described on the diagonal A C. The triangle ABC being right-angled and isosceles, we shall have E B F A-C3 AB +BC- 2AB. E B F It is plain, that of the eight equal right-angled triangles which compose the square EG, four will lie without the square ABCD, and four within it: hence, the square on the dcliagonal is equivalent to double the square on the side. 100 GEOMETRY. Cor. 5. By the last corollary, we have AC-: AB2:: 2: 1; hence, by extracting the square root (B. II., P. 12, c.), AC AB V'2 1: that is, the diagonal of a square is to the side as the square root of two to one: consequently, the diagonal and side of a square are incommensurable. PROPOSITION XII. THEOREM. In any triangle, the square of a side opposite an acute angle is equivalent to the squares of the base and the other side, diminished by twice the rectangle contained by the base and the distance from the acute angle to the foot of the perpendicular let fall from the opposite angle on the base, or on the base produced. Let ABC be a triangle, C one of the acute angles, and AD perpendicular to the base BC; then will AB 2 —BC 2+AC -2BCx CD. First. When the perpendicular falls within the triangle ABC, we have BD= BC- CD, and consequently, BD2_ -=2+ CD -2 BCX CD (P. 9). Adding AD2 to each, and observing that the right-angled triangles ABD, ADC, B D C -B2 -2 2 — 2 give AD-+B D AB-, and AD~+ C~ -OlC, we have AB2 BC2+A c —2BCX CD. Secondly. When the perpendicular AD A falls without the triangle ABC, we have BD= CD-BC; and consequently, -BD - CD+ BC —2 CBDX BC (P. 9). D B Adding AJD to both, we find, as before, D B C AB2 BC2- +AC -2BCx CD. BOOK IV. 101 PROPOSITION XIII. THEOREM. In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equivalent to the squares of the base and the other side, augmented by twice the rectangle contained by the base and the distance from the obtuse angle to the foot of the perpendicular let fall from the opposite angle on t4h base produced. Let A CB be a triangle, C the obtuse angle, and AD perpendicular to BC produced; then will -2 2 +W-2 AB AC -BC +2BC CD. A For, we have, BD=BC+ CD; and by squaring (P. 8), BD -cBC~D + +Dh + B C X CD. D B Adding AD'2 to both, and reducing as in the last theorem, and we have A-B =2BC ~+A-C +2BCX CD. Scholium. The right-angled triangle is the only one in which the sum of the squares described on two sides is equivalent to the square described on the third; for, if the angle contained by the two sides is acute, the sum of their squares is greater than the square of the opposite side; if obtuse, it is less. PROPOSITION XIV. THEOREM. In any triangle, the sumn of the squares described on two sides is equivalent to twice the square of half the third side, plus twice the square of the line drawn from the middle point of that side to the vertex of the opposite angle. Let ABC be any triangle, and AE a line drawn to the middle of the base BC; then will -A2 —+2 - 2 C+ 2AE. AB +AC ~-~-2BE +2AE. 102 GEOMETRY. For, on BC let fall the perpendic- A ular AD. Then, AC cAE +EC -2ECX ED. (P. 12). And, B ED C AB -AE +EB +2EB X ED (p. 13). Hence, by adding and observing that EB and EC are equal, we have AB2 +AC 2 —- 2EB +2AE2. Cor. 1. In any quadrilateral, the sum of.the squares of the four sides is equivalent to the sum of the squares of the two diagonals, plus four times the square of the line joining the middle points of the diagonals. Let ABCD be a quadrilateral, AC, BD, the diagonals, and EF a line join- D ing the middle points E and F. From the theorem, we have E 22,CD2~ CSB 2BF2 +2CF2, 1 A3 -2-2 2 -2 AD +AB =-,2BF +2AF' and from the same theorem, by multiplying by 2, 2CFK+2iFhc AF 4AE +4EP: hence, by addition, CDI + CB +A 2+AB,2 4BF +4AE +4E': whence (P. 8, c.), CD + CB +A-D +AB =BD +A C2+4 EC 2. Cor. 2. In the case of the parallelogram the points E and F will coincide, and the sum of the squares described on the sides will be equivalent to the sum of the squares described on the diagonals. BOOK IV. 103 PROPOSITION XV. THEOREM. If in any triangle, a line be drawn parallel to the base, it will divide the two other sides proportionally. Let ABC be a triangle, and DE a straight line drawn parallel to the base BC; then will AD: DB:: AE: EC. Draw the lines BE and CD. Then, A the triangles ADE, BDE, having a common vertex, E, have the same altitude, and are to each other as their D bases (P. 6, c.); hence we have ADE: BDE:: AD: DB. The triangles ADE, DEC, having a B C common vertex ), also have the same altitude, and are to each other as their bases; hence, ADE: DEC:: AE: EC. But the triangles BDE, DEC, are equivalent, having the same base DE, and their vertices B and C in a line parallel to the base: and therefore, we have (B. II., P. 4, c.) AD: DB:: AE: EC. Cor. 1. Hence, by composition, we have (B. II., P. 6), AD+DB: AD:: AE+EC: AE, or AB: AD:: AC: AE; and also, AB: BD:: AC: CE. Cor. 2. If any number of parallels A d EP1,'G1, BD, be drawn between two straight lines AB, CD, those straight lines will be cut proportionally, and we shall have A E: CF: EG: FH: GB: HD. For, let 0 be the point where AB o and CD meet. In the triangle O1F,/ the line AC being drawn parallel to the base EF, we shall have A C OE: AE:: OF: CF. E \F In the triangle O GH, we shall like- G/ wise have B OD O: EG:: OF F: 1\ 104 GEOMETRY. And, by reason of the common antecedents OE, OFF (B. II., P. 4), we have 0 AE: CF:EG FH. It may be proved in the same manner, A C that E F EG: FH:: GB: HD, G H and so on; hence, the lines AB, CD, are cut proportionally by the parallels A C, EF, Gi, &c. PROPOSITION XVI. THEOREM. Conversely: If two sides of a triangle are cut proportionlly by a straight line, this straight line will be parallel to the third side. In the triangle BA C, let the line DE be drawn, cutting the sides BA and CA proportionally in the points D and E; that is, so that BD DA:: CE EA: then will DE be parallel to BC. Having drawn the lines BE and A DQC we have (P. 6, c.), BDE DAE DAE, BD DA DEC: DAE:: CE EA: D but, by hypothesis, BD DA:: CE EA: hence (B. II:, P. 4, c.), B C BDE: DAE:: DEC: DAE, and since BDE and DEC have the same ratio to DAE, they have the same area, and hence are equivalent (D. 4). They also have a common base BC; hence, they have the same altitude (P. 6, c.); and consequently, their vertices D and E lie in a parallel to the base BC (B. I., P. 23). BOOK IV. 105 PROPOSITION XVII. THEOREM. The line which bisects the vertical angle of a triangle, divides the base into two segments, which are proportional to the adjacent sides. In the triangle A CB, let AD be drawn, bisecting the angle CAB; then will BD: CD: AB A C. Through the point C draw E CE parallel to AD, and prolong it till it meets BA produced in if. In the triangle BCE, the line A AD is parallel to the base CE; hence, we have the proportion (P. 15), BD:D C:: BA: AE. C D B But the triangle ACE is isosceles: for, since AD, CE, are parallel, we have the angle ACE —ZDAC, and the angle AEC=BAD (B. I., P. 20, c. 2, 3); but, by hypothesis, DA C =DAB; hence, the angle ACE=AEC, and consequently, AE=AC (B. I., P. 12). In place of AE in the above proportion, substitute A C, and we shall have, BD DC:: AB A C. Cor. If the line AD bisects the exterior angle CAE of the triangle BA G we shall have, BD: CD: AB A C. For, through C draw CF parallel to AD. Then, CAD=A CE, - and, EA4D=AFC; hence, (A. 1), A CF=AFC; consequently, AF is equal to A C. B C D But, since FC is parallel to the base AD, BD: D C: AB: AF; hence, BD: DCC: AB: AC. 106 GEOMETRY. PROPOSITION XVIII. THEOREM. Equiangular triangles have their homologous sides proportional, and are similar. Let BCA and CED be two equi- F angular triangles, having the angle / BAC= CDE, ABC=DCE, and ACB A/ " =DEC; then, the homologous sides will be proportional, viz.: B C E BC CE:: BA: CD:: AC: DE. Place the homologous sides BCn CE in the same straight line; and prolong the sides BA, ED, till they meet in F. Since BCE is a straight line, and the angle BCA equal to CED, it follows that A C is parallel to DE (B. I., P. 19, c. 2). In like manner, since the angle ABC is equal to /DCE, the line AB is parallel to D)C. Hence, the figure A CDF is a parallelogram, and has its opposite sides equal (B. I., P. 28). In the triangle BEE, the line A C is parallel to the base FE; hence, we have (P. 15,) BC: CE:: BA: AF; or putting CD in the place of its equal AF, BC: CE:: BA: CD. In the same triangle BEF, CD is parallel to BF; and hence, BC: CE:: FD: DE; or putting AC in the place of its equal FD, BC: CE:: AC: DIE. And finally, since both these proportions contain the same ratio BC to C E, we have (B. II., P. 4, C.), BA: CD:: AC: D-E. Thus, the equiangular triangles* CAB, CED, have their homologous sides proportional. But two figures are similar when they have their angles equal, each to each, and their BOOK IV. 107 homologous sides proportional (D. 1, 2); consequently, the equiangular triangles BA C, C, are two similar figures. Cor. Two triangles which have two angles of the one equal to two angles of the other, are similar; for, the third angles are then equal, and the two triangles are equiangular (B. I., P. 25, c. 2.) Scholium. Observe, that in similar triangles, the homologous sides are opposite to the equal angles; thus, the angle BCA being equal to CED, the side AB is homologous to DC; in like manner AC and DE are homologous, because opposite to the equal angles ABC, DCE. PROPOSITION XIX. THEOREM. Conversely: Triangles, which have their sides prop2ortional, are equiangular and similar. If, in the two triangles BA C, EDF, we have, BC: EF:: BA: ED:: A C: C D F; then will the triangles BA C, EDF,, have their angles equal, namely, A=D, B=E, C=F. At the point EY, make the angle A D FEG=B, and at F, the angle EFG =C; the third angle G will then E F be equal to the third angle A (B. I., P. 25, c. 2). Therefore, by the B < last theorem, we shall have B C G BC: EF::AB: EG: but, by hypothesis, we have BC: EF:: AB: DE; hence, EG=DE. By the same theorem, we shall also have BC: El:: AC: FG: and by hypothesis, we have BC: El:: AC: DF; hence, FG=DF. Hence, the triangles EGF, FED, having 108 GEOMETRY. their three sides equal, each to each, are themselves equal (B. I., A D P. 10). But, by construction, the triangles EGF and ABC are equi- E angular: hence, DEF and ABC are also equiangular and similar B C (A. 1). Scholium 1. By the last two propositions, it appears that triangles which are equiangular are similar: and conversely: if triangles have their sides proportional, they are equiangular, and consequently, similar. The case is different with regard to figures of more than three sides: even in quadrilaterals, the proportion between the sides may be altered without changing the allgles, or the angles may be changed without altering the proportion between the sides. Thus, in quadrilaterals, equality between the corresponding angles does not insure proportionality among the sides: and reciprocally: proportionality among the sides does not insure equality among the corresponding angles. It is evident, for example, that if in the quadrilateral ABCD, we draw D EF parallel to BC, the angles of the C quadrilateral AEFD, are made equal to those of ABCtD; though the proportion between their sides is different; and in A E B like manner, without changing the four sides AB, BC, CD, AD, we can change the angles by making the point B approach to D, or recede from it. Scholium 2. The two preceding propositions, are in strictness but one, and these, together with that relating to the square of the hypothenuse, are the most important and fertile in results of any in geometry. They are almost sufficient of themselves for every application to subsequent reasoning, and for solving every problem. The reason is, that all figures may be divided into triangles, and any triangle into two right-angled triangles. Thus, the properties of triangles include, by implication, those of all figures. BOOK IV. 109 PROPOSITION XX. THEOREM. Two triangles, which have an angle of the one equal to an angle of the other, and the sides containing those angles proportional, are similar. Let ABC, DEF, be two triangles, having the angle A equal to D; then, if AB: DE:: A C: DF, the two triangles will be similar. Make A G=DE, and draw GH A parallel to BC. The angle A Gi will be equal to the angle ABC (B. I., p. D 20, c. 3); and the triangles A G GH, ABC, will be equiangular: hence, we -- shall have, B C E F AB: AG AC: AH. But, by hypothesis, we have, AB: DE:: AC: DF; and by construction, AG=DE: hence AH=DF. Therefore, the two triangles A Glf, DEF, have two sides and the included angle of the one equal to the sides and the included angle of the other: hence, they are equal (B. I., P.5); but the triangle A G is similar to ABC: therefore, DEF is also similar to ABC. PROPOSITION XXI. THEOREM. Two triangles, which have their sides parallel, or perpendicular to each other, are similar. Let BA C, EDF, be two triangles, having their sides respectively parallel to each other; then will they be similar. First. If the side BA is parallel to G ED, and BC to EF, the angle ABC is equal to DEF (B. I., P. 24): if CA A is parallel to FD, the angle BCA \ D is equal to EFTD, and also, BA C to EDF; hence, the triangles CBA, I IE F FED, are equiangular; consequently B C they are similar (P. 18). 110 GEOMETRY. Secondly. If the side DE is per- A pendicular to BA, and the side FD to CA, the two angles I and HE of the quadrilateral DHAI are right E II angles; and since all the four angles /\ are together equal to four right angles (B. I., P. 26, c. 1), the remain- B G C ing two IA, IDH, are together equal to two right angles. But the two angles ED:, ID1, are also equal to two right angles (B. I., P. 1): hence, the angle EDF is equal to IA;, or BA C (A. 3). In like manner, if the third side EF is perpendicular to the third side BC, it may be shown that the angle DFE is equal to C and DEF to B: hence, the triangles ABC, DEF, which have the sides of the one perpendicular to the corresponding sides of the other, are equiangular and similar (p. 18). Scholium. In the case of the sides being parallel, the homologous sides are the parallel ones: in the case of their being perpendicular, the homologous sides are the perpendicular ones. Thus, in the latter case, DE is homologous with BA, DF with A C, and EF with BC. The case of the perpendicular sides may present a relative position of the two triangles different from that exhibited in the diagram. But we can always conceive a triangle FED to be constructed within the triangle ABC, and such that its sides shall be parallel to those of the triangle compared with BBAC; and then the demonstration given in the text will apply. PROPOSITION XXII. THEOREM. In any triangle, if a line be drawn parallel to the base, all lines drawn from the vertex will divide the base and the parallel into proportional _parts. Let BA C be a triangle, DE parallel to the base B, and the other lines drawn as in the figure; then will DI: BE:: IR: FGK:: KL: GI. BOOK IV. 111 For, since DI is parallel to BF, A the triangles IDA and FBA are equiangular; and we have DI: BF:: AI: AF; D and, since IK is parallel to FG, /I K we have, in like manner, B F G H C AI: AF:: IK: FG; hence (B. II., P. 4, c.), DI: BE:: IK: FG. In the same manner, we may prove that 1K: FG:: KL: GH; and so with the other segments: hence, the line DE is divided at the points 1, i; L, in the same proportion, as the base BC is divided, at the points F, G, H. Cor. Therefore, if BC were divided into equal parts at the points F, G, H; the parallel DE would be divided also into equal parts at the points 1 E L. PROPOSITION XXIII. THEOREM. In a right-angled triangle, if a perpendicular is drawn from the vertex of the right angle to the hypothenuse. 1st. The triangles on each side of the perpendicular are similar to the whole triangle, and to each other: 2d. Either side about the right angle is a mean proportional between the hypothenuse and the adjacent segment: 3d. The perpendicular is a mean proportional between the segments of the hypothenuse. Let BAC be a right-angled triangle, and AD perpendicular to the hypothenuse BC. First. The triangles BAD and A BA C have the common angle B, the right angle BDA=BA C, and therefore, the third angle BAD of the one, equal to the third angle B C, of the other (B. I., P. 25, c. 2): hence, these two triangles are similar (P. 18). In the same 112 GEOMETRY. manner it may be shown that the triangles DA C and BA C are similar; hence, the three triangles are all equiangular and similar. Secondly. The triangles BAD, /: BA G being similar, their homologous sides are proportional. But BD in the small triangle, and BA in the large one, are homologous sides, because they lie opposite the equal angles. BAD, BCA (P. 18, s.); the hypothenuse BA of the small triangle is homologous with the hypothenuse BC of the large triangle: hence, the proportion, BD: BA:: BA BC. By the same reasoning we have DC: AC:: AC: BC; hence, each of the sides AB, AC, is a mean proportional between the hypothenuse and the adjacent segment. Thirdly. Since the triangles DBA, DAC, are similar, we have, by comparing their homologous sides, BD AD: AD: DC; hence, the perpendicular AD is a mean proportional between the segments BD, DC, of the hypothenuse. Scholium. Since BD: AB:: AB: BC, 2 we have (B. II., P. 1, C.), AB2-BDXBC. -2 For the same reason, AC -C DCxBC; therefore, AB +A — BDX BD C+DCX BCc =(BD+DC)X BC- BC X BC —- BC 2; that is, the square described on the hypothenuse BC is equivalent to the sum of the squares described on the two sides BA, A C. Thus, we again arrive at this property of the right-angled triangle, and by a path very different from that which formerly conducted us to it: and thus it appears that, strictly speaking, this property is a consequence of the more general property, that the sidles of equiangular triangles are proportional. Thus, the fundamental propositions of geometry are reduced, as it were, to this single one, that equiangular triangles have their homologous sides proportional. BOOK IV. 113 It happens frequently, as in this instance, that by,deducing consequences from one or more propositions, we are led back to some proposition already proved. In fact, the chief characteristic of geometrical theorems, and one indubitable proof of their certainty is, that, however we combine them together, provided only our reasoning be correct, the results we obtain always agree with each other. The case would be different, if any proposition were false or only approximately true: it would frequently happen that on combining the propositions together, the error would increase and become perceptible. Examples in which the conclusions do not agree with each other, are to be seen in all the demonstrations, in which the reductio ad absurdum is employed. In such demonstrations, if the hypothesis is untrue, a train of accurate reasoning leads to a manifest absurdity: that is, to a conclusion in contradiction to a principle previously established: and from this we conclude that the hypothesis is false. Cor. If from the point A, in the circumference of a circle, two chords BA, AC, be drawn to. the extremities of a diameter BC, the triangle B) C BA C will be right-angled at A (B. ilI., P. 18, C. 2); hence, first, the perpendicular AD is a mean proportional between the two segments BD, DC, of the diameter, hence, AD)2 BDX B D. Furthermore, by the proposition, the chord BA is a mean proportional between the diameter BC, and the adjacent segment BD, that is, BA B =BC xBD, and A-C2 -BC X CD. PROPOSITION XXIV. THEOREM. Two triangles having an angle in each equal, are to each other as the rectangles of the adjacent sides. Let ABC, ADE, be two triangles having the equal angles A, placed, the one. on the other; then the triangle ABC: ADE:: ABXAC: ADXAE. 8 114 GEOMETRY. Draw BE. Then, the triangles ABE, A ADE, having the common vertex E, are to each other as their bases (P. 6, c.) D that is, BAE ~ DAE:: BA: DA. B C In like manner, since B is a common vertex, the triangle BAC: BAE:: AC: AE. Multiply together the corresponding terms of these proportions, omitting the common term ABE; and we have (B. II., P. 13), BAC: DAE: BAxAC: ADXAE. Cor. If the two triangles are equiva- A lent, we have, BAXAC=DAxAE: hence (B. II., P. 2), - BA: DA: AE: A C: consequently, JDC and BE are parallel B E (P. 16). PROPOSITION XXV. THEOREM. Similar triangles are to each other as the squares described on their homologous sides. Let ABC, DEFB, be two similar triangles, having the angle A equal to D, and the angle B=E: then will the triangle BA C be to the triangle -EDF, as a square described on any side of BA to a square described on the homologous side of EDF. First, by reason of the equal angles A and D, we have (P. 24), BAC: DEF:: BA X A: ~D-EX DF. D Also, because the triangles are similar (P. 18), B CE F AB: DE:: AC: DF, BOOK IV. 115 And multiplying the terms of this proportion by the corresponding terms of the identical proportion, AC: DF:: AC: DF, there will result -2 -2 ABXAC: DEXDF:: AC2: D Consequently (B. II., P. 4, C.), -2 -2 ADC: DEF:: AC: DF. Therefore, the similar triangles BA, EDF,' are to each other as the squares described on their homologous sides A C, DF, or as the squares described on any other two homologous sides. PROPOSITION XXVI. THEOREM. Two similar polygons may be divided into the same number of triangles, similar each to each, and similarly placed. Let AEDCB, FKIEHG, be two similar polygons. From any angle A, in the polygon AED CB, draw diagonals, AD, B G AC. From the homol- D ogous angle F. in the other polygon,,draw E K the diagonals F1 F1, to the other angles. The polygons being similar, the homologous angles, ABC, FGGII are equal, and the sides AB, BC, proportional to FG, G11, that is, AB: FG:: BC GH (D. 1). Wherefore, the triangles ABC, FGI, have an angle in each equal, and the adjacent sides proportional: hence, they are similar (P. 20); consequently, the angle BCA is equal to GHE. Taking away these equal angles from the equal angles BCD, GH_, and there remains A CD)=FLI. But since the triangles ABC, FGII are similar, we have AC: PXH:: BC: GH; and since the polygons are similar, 116 GEOMETRY. BC G CD: HI; hence, AC: FH:: CD HI. The angle A CD, we already know, is equal to PHI; hence, the triangles A CD, FFHf, are similar (r. 20). In the same manner, it may be shown that all the remaining triangles are similar, whatever be the number of sides in the polygons proposed: therefore, two similar polygons may be divided into the same number of triangles, similar, and similarly placed. AScholium. The converse of the proposition is equally true: -If two polygons are composed of the same number of triangles similar and similarly situated, the. two polygons are similar. For, the similarity of the respective triangles will give the angles, ABC=FGI, BCA=-GHF, AD=FPHI: hence, BCD=GHI, likewise, CDE=HIK, &c. Moreover, we have, AB: FG:: BC: GH:: CD: HI:: DE: 1KT, &c.; hence, the two polygons have their angles equal and their sides proportional; consequently, they are similar. PROPOSITION XXVII. THEOREM. The perimeters of similar polygons are to each other/ as their homologous sides: and the polygons are to each other as the squares described on these sides. Let AED CUB and FKIHG, be two similar polygons: then will per. AEDCB: per. FKIHG: ~AE: FK. First. Since the figures are similar, we have G AB: FG:: BC A DFI Gl CD: HI, &c., hence, the sum of the E antecedents AB +BC+ BOOK IV. 117 CD, &c., which makes up the perimenter of the first polygon, is to the sum of the consequents FG+GSH +H]; &c., which makes up the perimeter of the second polygon, as any one antecedent is to its consequent (B. II., P. 10); that is, as AB to FG, or as any other two homologous sides. Secondly. Since the triangles ABC, FG,; are similar, we have (P. 25), ABC: F:: AC FHP1; and from the similar triangles ACD, FHI, A CDa FHI:: A C H; therefore, by reason of the common ratio, A-C to F-2 we have (B. II., P. 4, c.) ABC: GH:: A CD: FHI. By the same reasoning, we should find A CD: FHI:: ADE: FIK; and so on, if there were more triangles. And from this series of equal ratios, we conclude that the sum of the antecedents ABC+ACD+ADE, which make up the poly. gon AEDCB, is to the sum of the consequents FGH+ ITHI~+FIK, which make up the polygon FKIHG, as one antecedent ABC, is to its consequent FGH (B.II., P. 10), or as AB2 is to FG2 (P. 25); hence, similar polygons are to each other, as the squares described on their homologous sides. Cor. If three similar figures are described on the three sides of a right-angled triangle, the figure on the hypothenuse is equivalent to the sum of the other two. Let A, B, C, denote three similar figures described on the hypothenuse and sides of a right-angled triangle, and a, b, c, the corresponding squares; then, A: B::: a: b: c; and, A B+C:: a: b+c (B.II.,P. 9): but, a is equivalent to b + c (P. 11); hence, A is equivalent B+C. 118 GEOMETRY. PROPOSITION XXVIII. THEOREM. If two chords intersect -each other in a circle, the segments are reciprocally proportional. Let the chords AB and CD intersect at O: then will, AO DO:: OC OB. Draw AO and BD. In the triangles CH B A GO, DOB, the angles at 0 are equal, \ being vertical angles (B. I., P. 4): the angle /1 \ A is equal to the angle D, because both are / inscribed in the same segment (B. III., P. 18, A D c. 1); for the same reason the angle C=B; the triangles are therefore similar (P. 18), and the homologous sides give the proportion AO DO:: CO ~ OB. Cor. Therefore, A Ox OB DOx CO: hence, the rectangle of the two segments of one chord is equivalent to the rectangle of the two segments of the other. PROPOSITION XXIX. THEOREM. -If from a point without a circle, two secants be drawn terminating in the concave arc, the whole secants will be reciprocally proportional to their external segments. Let the secants OB, 0 C, be drawn from the point 0: then will OB: OC OD: OA. For, drawing AC BD, the triangles o AOC, BOD have the angle 0 common; likewise the angle B-G C(B. III., P. 18, C. 1); these triangles are therefore similar (P. 18), A and their homologous sides give the proportion, OB: OC:: OD OA. B Cor. Hence, the rectangle - OBx OA o OC x OD. BOOK IV. 119 Scholium. This proposition, it may be observed, bears a close analogy to the. preceding, and differs from it only as the two chords AB, CD, instead of intersecting each other within, cut each other without the circle. The following proposition may'be regarded as a particular case of the proposition just demonstrated. PROPOSITION XXX. THEOREM. if from a point without a circle, a tangent and a secant be drawn, the tangent will be a mean proportional between the secant and its external segment. From the point 0, let the tangent OA, and the secant OG be drawn, then will OC OA:: OA: OD, or, OA-=- OCx OD. For, drawing AD and A C, the trian- 0 gles DAiO, CAO, have the angle 0 common; also, the -angle OAD, formed by a tangent and a chord, is measured by half the arc AD (B. III., P. 21); and the an- A. ---------— C gle C has the same measure (B. III., P. 18); hence, the angle OAD=C (A. 1): therefore, the two triangles are similar, and we have the proportion OC: OA:: OA OD. which gives OA OC0X OD. PROPOSITION XXXI. THEOREM. If either angle of a triangle is bisected by a line terminating in the opposite side, the rectangle of the sides'about the bisected angle, is equivalent to the square of the bisecting line, together with the rectangle contained by the segments of the third side. In the triangle BA C, let AD bisect the angle A; then will AB x A C A-+BDX DC'. 120 GEOMIETRY. Describe a circle through the three A points A,B, C (B. III., PROB. 13, S.); prolong AD till it meets the circumference in E, B'BR D C and draw CE. The triangle BAD is similar to the triangle EA C; for, by hypothesis, the E angle BAD=ERA C; also, the angle B=-E since:they are both measured by half the arc A C (B. III., P. 18); hence, these triangles are similar, and the homologous sides give the proportion BA: AE.: AD: AC; hence, BAxAC oAE xAD; but AE=AD+DE, and multiplying each of these equals by AD, we have AE x AD -AD+AD x DRE; now (P. 28, C.), ADxD E:.BDX DC; hence, finally, BAXA Ca AD +BDx D C. PROPOSITION XXXII. THEOREM. In any triangle, the rectangle contained by two sides is equiva. lent to the rectangle contained by the diameter of the circumscribed'circle, and the per pendicular let fall on the third side. In the triangle BA C let AD be drawn perpendicular to BC; and let PEC be the diameter of the circumscribed circle: then will ABX A C= ADx CRE For, drawing APE, the triangles DBA, CAE, are right-angled, the one E at D, the other at A: also, the angle B=E (B. III., P. 18, C. 1); these triangles are therefore similar, and we BDCC have AB: CE:: AD:AC; and hence, AB X A Cc CE x AD. Cor. If these equal quantities be multiplied by BC, there will result AB A CxBC= CEADx BC; BOOK IV. 121 now, ADXBC is double the area of the triangle (P. 6); therefore, the product of the three sides of a triangle is equal to its area multiplied by twice the diameter of the circumscribed circle. The product of three lines is sometimes represented by a solid, for a reason that will be seen hereafter. Its value is easily conceived, by supposing the lines to be reduced to numbers, and then multiplying these numbers together. Scholium. It may also be demonstrated, that the area of a triangle is equal to its perimeter multiplied by half the radius of the inscribed circle. For, the triangles A OB, B BOC, AOC, which have a E common vertex at 0, have for their common altitude the radius of th6 inscribed circle; hence, the sum of these triangles will be equal to the sum of the bases AB, BC, A C, multiplied by half the radius OD; hence, the area of the triangle ABC is equal to its perimneter multiplied by half the radius of the inscribed circle. PROPOSITION XXXIII. THEOREM. In every quadrilateral inscribed in a circle, the rectangle of the two diagonals is equivalent to the sum of the rectangles of the opposite sides. Let ABCD be a quadrilateral inscribed in a circle, and AC BD, its diagonals: then we shall have ACxBD —c-ABx CD+ADx BC. Take the are CO=AD, and draw B BO, meeting the diagonal A C in 1. The angle ABD= CBI, since the one has for its measure half of the are AD (B. III., P. 18), and the other, half A C of CO, equal to AD; the angle ADB =BC], because they are subtended by D 122 GEOMETRY. the same arc; hence, the triangle B ABD is similar to the triangle IBC, and we have the proportion AD CI:: BD: B; A and consequently, ADxBC CI X BD. Again, the triangle ABI is similar to the triangle BD C; for the arc AD being equal to CO, if OD be added to each of them, we shall have the arc A0O=DC; hence, the angle ABI is equal to DBC; also, the angle BAI to _BDC, because they stand on the same arc; hence, the triangles AB], DBC, are similar, and the homologous sides give the proportion AB: BD:: AI: CD; hence, AB X CD c AI X BD. Adding the two results obtained, and observing that AIx BD)+ CIx BD=(AI + CI) X BD=A C BD, we shall have AD x BC+ABx CD-A-AC xBD. PROBLEMS RELATING TO THE FOURTH BOOK. PROBLEM I. To divide a given straight line into any number of equal parts, or into parts proportional to given lines. First. Let it be proposed to divide the A line AB into five equal parts. Through the extremity A, draw the indefinite straight C line A G: take AC of any magnitude, and D K apply it five times upon A C; join the last E point of division G, and the extremity B F 1of the given line, by the straight line GB; Gthen through C draw CI parallel to GB: / - - BOOK IV. 123 AI will be the fifth part of the line AB; and by applying AI five times upon AB, the line AB will be divided into five equal parts. For, since CI is parallel to GB, the sides A G, AB, are cut proportionally in C and I (P. 15). But AC is the fifth part of AG, hence, Al is the fifth part.of AB. Secondly. Let it be I F B proposed to divide the A line AB into parts pro- - portional to the given P lines P, Q, R. Through Q'' C A, draw the indefinite R! line AG; make'AC=P, E G CD= Q, DE=R; join the extremities E and B; and through the points C and D, draw C1; D1F parallel to EB; the line AB will be divided into parts Al, IF, FB, proportional to the given lines P, Q, W. For, by reason of the parallels C;, DF, EB, the parts AI, IF, EB, are proportional to the parts AC, CD, DE (r. 15, c. 2); and by construction, these are equal to the given lines P., Q, R. PROBLEM II. To find a fourth proportional to three given lines, A, B, C. Draw the two indefi- D nite lines DE, DF, forming any angle with each X Ai l other. Upon DE take Bi —---- DA=A, and DB=B; C - X upon DF take DC=-C 7 A\ draw AC2; and through E F the point B, draw BX parallel to A C; and DX will be the fourth proportional required. For, since BX is parallel to AC, we have the proportion (r, 15, c. 1), DA: DB:: DC: DX; now, the first three terms of this proportion are equal to the three given lines: consequently,) DX is the fourth proportional required. 124 GEOMETRY. Cor. A third proportional to two given lines, A, B, may be found in the same manner, for it will be the same as a fourth proportional to the three lines, A, B, B. PROBLEM III. To find a mean proportional between thro given lines A and B, Upon the indefinite line DF, G take DE=A, and EF=B; and upon the whole line DF, as a diameter, describe the semicircum- D E F ference DGF; at the point E, B —---- erect, upon the diameter, the per- Al H pendicular EGU meeting the semicircumference in G; EG will be the mean proportional required. For, the perpendicular EG,. let fall from a point in the circumferentce upon the diameter, is a mean proportional between the two segments of the diameter DE, EF (P. 23, c.); and these segments are equal to the given lines A and B, PROBLEM IV. To divide a given line into two such parts, that the greater part shall be a mean proportional between the whole line and the other part. Let AB be the given line. At the extremity B, erect the E perpendicular BC, equal to the c half of AB; from the point C, D as a centre, with the radius CB, describe a semicircle; draw A C' cutting the circumference in D; A F B and take AF=AD: then F will be the point of division, and we shall have, AB: AF:AF F A FFB. For, AB being perpendicular to the radius at its extremity, is a tangent (B. III., P. 9); and if AC be prolonged BOOK IV. 125 till it again meets the circumference, in E, we shall have (P. 30), AE: AB:: AB A: D; hence, by division, AE-AB AB:: AB-AD ~ AD. But, since the radius is the half of AB, the diameter DE is. equal to AB, and consequently, AE-AB=AD=AF; also, because AF=AD, we have AB-AD=FB: hence, AF: AB:FB AD, or AF; whence, by inversion, AB AF AF FB. Scholium. This sort of division of the line AB, viz., so that the whole line shall be to the greater part as the greater part is to the less, is called division in extreme and mean ratio. It may further be observed, that the secant AE is divided in extreme and mean ratio at the point D; for, since AB= DE, we have, AE: DE:: DE: AD. PROBLEM V. Through a given point, in a given angle, to draw a line so that the segments comprehended between the point and the two sides of the angle, shall be equal. Let BCD be the given angle, and A the given point. Through the point A, draw AE parallel to CD, make BE= CE, and C through the points B and A, draw BAD; this will be the line required. For, AE being parallel to iCD, we D have, BE: EC:: BA1: AD; but BE=EC; therefore, BA=AD. 126 GEOMETRY. PROBLEM VI. To describe a square that shall be equivalent to a given parallelogram, or to a given triangle. First. Let ABCD be X Y the given parallelogram, D C AB its base, and DE its altitude: between AB and DE find a mean A E B proportional XY; then will the square described upon XY be equivalent to the parallelogram ABCD. For, by construction, AB:.XY:: XY: DE; therefore, XY2 AB X DE; but ABXDE is the measure of the parallelogram (P. 5), and XY2 that of the square; consequently, they are equivalent. Secondly. Let BA C A be the given triangle, BC its base, AD its al- I ty titude: find a mean proportional between BC and the half of AD, and B D C let XY be that mean; the square described upon XY will be equivalent to the triangle ABC. For, since BC: XY:: XY:'AD, it follows, that XYV BCx AD; hence, the square described upon XY is equivalent to the triangle BA C. PROBLEM VII. Upon a given line, to describe a rectangle that shall be equivalent to a given rectangle. Let AD be the line, and ABFC the given rectangle. BOOK IV. 127 Find a fourth pro- X E portional to the three C F lines, AD, AB, A C, and let AX be that fourth proportional; a rectan- A B A D gle constructed with the lines AD and AX will be equivalent to the rectangle ABFC. For, since AD: AB:: AC: AX, it follows, that ADx AX - AB X A C; hence, the rectangle ADEX is equivalent to the rectangle ABFC. PROBLEM VIII. To find two lines whose ratio shall be the same as the ratio of two rectangles contained by given lines. Let AXB, CX D, be the rectangles contained by the given lines A, B, C, and D. Find X, a fourth proportional to the three lines, B, O D; then will the two Bl lines A and X have the same ratio to each other as the rectangles AXB and DCx D. X For since, B: C:: SD: X, it follows that Cx D c B XX; hence, AxB: CxD:: AXB: BXX:: A:X. Cor. Hence, to obtain the ratio of the squares described upon the given lines A and C find a third proportional X, to the lines A and C, so that A: C:: C: X; you will then have AXX C2, or A2XX AxC2; hence, A2: C:: A X. 128 GEOMETRY..PROBLEM IX. To find a triangle that shall be equivalent to a given polygon. Let AEDCB be the given polygon. First. Draw the diagonal CE C cutting off the triangle CDE; through the point D, draw D)F D parallel to CE, and meeting AE / / prolonged; draw CF: the polygon AEDUB. is equivalent to the polygon AFCB, which has one side less than the given polygon. For the triangles CDE, CFE, have the base CE common, they have also equal altitudes, since their vertices D and F, are situated in a line DF parallel to the base: these triangles are therefore equivalent (P. 2, c.) Add to each of them the figure AECB, and there will result the polygon AEDCB, equivalent to the polygon AFCB. The angle B may in like manner be cut off, by substituting for the triangle ABC, the equivalent triangle A GC, and thus the pentagon AEDCB will be changed into an equivalent triangle GCF. The same process may be applied to every other figure; for, by successively diminishing the number of its sides, one being retrenched at each step of the process, the equivalent triangle will at last be found. Scholium. We have already seen that every triangle may be changed into an equivalent square (PROB. 6); and thus a square may always be found equivalent to a given rectilineal figure, which operation is called squaring the rectilineal figure, or finding'the quadrature of it. The problem of the quadrature of the circle consists in finding a square equivalent to a circle whose diameter is given. BOOK IV. 129 PROBLEM X. To find the side of a square which shall be equivalent to tlhe sum or the difference of two given squares. Let A and B be the sides of the given squares. First. If it is required to find F a square equivalent to the sum G of these squares, draw the two Al indefinite lines, ED,'EE, at right / I angles to each other; take ED)=A, and EG=B; and draw DG: H E D this will be the required side of the square. For the triangle DEG being right-angled, the square described upon the hypothenuse DG, is equivalent to the sumr of the squares upon ED and EG (P. 11). Secondly. If it is required to find a square equivalent to the difference of the given squares, form, as before, the right angle FEHT; take GE equal to the shorter of the sides A and B; from the point G as a centre, with a radius GH, equal to the other side, describe an are cutting Eili in H: the square described upon EH will be equivalent to the difference of the squares described upon the lines A and B. For, the triangle GEtII is right-angled, the hypothenuse GH=A, and the side GE=B; hence, the square described upon EH, is equivalent to the difference of the squares A and B (P. 11, c. 1). Scholium. A square may thus be found, equivalent to the sum of any number of squares; for a construction similar to that which reduces two of them to one, will reduce three of them to two, and' these two to one, and so of others. It would be the same, if any of the squares were to be subtracted from the sum of the others. 9 130 GEOMETRY. PROBLEM XI. To find a square which shall be to a given square as a given line to a given line. Let AC be the given square, and M and N the given lines. Upon the indefi- D C M — nite line EG, take EF..=M, and FG=N; upon EG as a diameter describe a semicircum- K ference, and at the point F erect the perpendicular FH. From the point L, draw the chords HG, HE, which produce indefinitely: upon the first, take HK equal to the side AB of the given square, and through the point K draw KI parallel to EG; HI will be the side of the required square. For, by reason of the parallels K;, GSE, we have HI: HK:: HE: HG; hence, 2' 2 E IE2 -G' but in the right-angled triangle GHE, the square of HE is to the square of HG as the segment EF is to the segment FG (P. 11, c. 3), or as M is to NC; hence, -Ad ~2 2 M N. But HK=AB; therefore, the square'described upon HI is to the square described upon AB as M is to N. PROBLEM XII. Upon a given line, to describe a polygon similar to a given polygon. Let FG be the given line, and AEDCB the given polygon. BOOK IV. 131 In the given poly- C gon, draw the diago- B G nals AC, AD; at the point F make the angle DF GFH=BA C, and at the AF point G, the angle FGHl E =ABC; the lines F1, GH will intersect each other in 1; and the triangle FGH will be similar to ABC (P. 18). In the same manner upon FH, homologous to AC; describe the triangle FIN/ similar to ADC; and upon Fi, homologous to AD, describe the triangle FIIK similar to ADE. The polygon FGHIHIK will be similar to ABCDE, as required. For, these two polygons are composed of the same number of similar triangles, similarly placed (P. 26,,s.) PROBLEM XIII. Two similar figures being given, to describe a similar figure which shall be equivalent to their sum or difference. Let A and B be homologous sides of the given figures. Find a square equivalent to the sum or difference of the squares described upon A and B; let X be the side of that square; then will X be that side in the figure required, X which is homologous to the sides A and B in the given figures. Let the figure itself, then, be constructed on the side X, as in the last problem. This figure will be equivalent to the sum or difference of the figures described on A and B (P. 27, c.) PROBLEM XIV. To describe a figure similar to a given figure, and bearing to it the given ratio of M to N: Let A be a side of the given figure, X the homologous side of the required figure. 132 GEOMETRY. Find the value of X, such, that its square' shall be to the square of A, as M to N (PROB. 11). Then upon X describe a fignre similar to the given figure (PRoB.12): this will be the figure required. A X PROBLEM XV. To construct a figure similar to the figure P, and equivalent to the figure Q. Find A;- the side of a square equivalent to the figure P, and N the side of a P square equivalent to the figure Q (PROB. 9, s.) Let X be a A B fourth proportional to the three given lines, M,; N, AB; upon the side X, homologous to AB, describe a figure similar to the figure P; it will also be equivalent to the figure Q. For, calling Y the figure described upon the side X, we have, P: Y:: A12 X2 but by construction, AB: X:: M N; or, AB: X M' N hence, P: Y:: ML: N2. But, by construction also, M'2- P, and N2o Q therefore, P Y:: P Q; consequently, Y- Q; hence, the figure IY is similar to the figure P, and equivalent to the figure Q. PROBLEM XVI. To construct a rectangle equivalent to a given square, and having the sum, of its adjacent sides equal to a given line. Let C be the square, and the line AB equal to the sum of the sides of the required rectangle. BOOK IV. 133 Upon AB as a diameter, describe a semicir- D cumference; at A, draw AD perpendicular to AB, C and make it equal to the side of the square C; A FB then draw the line DE parallel to the diameter AB; from the point E, where the parallel cuts the circumference, draw EF perpendicular to the diameter; AF and FB will be the sides of the required rectangle. For,- their sum is equal to AB; and their rectangle AF X FB is equivalent to the square of EF, or to the square of AD; hence, this rectangle is equivalent to the given square C. &holium. The problem is impossible, if the distance AD exceeds the radius; that is, the side of the square C must not exceed half the line AB. PROBLEM XVII. To construct a rectangle that shall be equivalent to a given square, and the difference of whose adjacent sides shall be equal to a given line. Let C denote the given square, and A B the difference of the sides of the rectangle. Upon the given line AB, as a D diameter, describe a circumference. At the extremity of the diameter, draw the tangent AD, and make it equal to the side of the square C; through the point D and the cen- A B tre 0 draw the secant DOE, intersecting the circumference in E and F; then will DE and DF be the F adjacent sides of the required rectangle. For, the difference of these sides is equal to the diameter EF or AB; and the rectangle DE, DF is equal to AD2 (P. 30); hence, the rectangle DFx DE, is equivalent to the given square C. 134 GEOMETRY. PROBLEM XVIII. To find the common measure, between the side and diagonal of a square. Let ABCG be any square, and AC its diagonal. We first apply CB upon CA. E For this purpose let the semicircumference DBE be described, from the centre C, with the radius CB, C and produce A C to E. It is evident that CB is contained once in AC, with the remainder AD. The result of the first'operation is, therefore, a A F, quotient 1, with the remainder AD. This remainder must now be compared with BC, or its equal AB. Since the angle ABC is a right angle, AB is a tangent, and since AAE is a secant drawn from the same point, we have (P. 30), AD AB: AB:: AB: AE. Hence, in the second operation, where AD is compared with AB, the equal ratio of AB to AE may be taken instead: but AB, or its equal CD, is contained twice in AE, with the remainder AD; the result of the second operation is therefore a quotient 2 with the remainder AD, and this must be again compared with AB. Thus, the third operation consists in comparing again AD with AB, and may be reduced in the same manner to the comparison of AB or its equal CD with AE; from which there will again be obtained a quotient 2, and the remainder AD. Hence, it is evident that the process will never terminate, and consequently that no remainder is contained in its divisor an exact number of times; therefore, there is no common measure between the side and the diagonal of a square. This property has already been shown, since (P. 11, c. 5), AB: A 1: /2, but it acquires a greater degree of clearness by the geometrical investigation. BOOK V. REGULAR POLYGONS-MEASUREMENT OF THE CIRCLE. DEFINITION. A REGULAR POLYGON is one which is both equilateral and equiangular. A regular polygon may have any number of sides. The equilateral triangle is one of three sides; the square, is one of four. PROPOSITION I. THEOREM. Regular polygons of the same number of sides are similar figures. Let ABCDEF, abcedf, be two regular polygons. Then, either angle, E D as A, of the polygon /e ~ABCDEl, is equal to F C twice as many right angles less four, as the fig- a b ure has sides, divided by A B the number of sides; and the same is true of either. angle of the other polygon (B. I., P. 26, c. 4); hence (A. 1), the angles of the polygons are equal. Again, since the polygons are regular, the sides AB, BC, CD, &c., are equal, and so likewise the sides ab, be, cd (D.), &c.; hence AB: ab BC: be: CD: cd, &c.; therefore, the two polygons have their angles equal, and their sides taken in the same order proportional; consequently, they are similar (B. IV., D. 1). Cor. 1. The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their surfaces are to each other as the squares of those sides (B. IV., P. 27). 136 GEOMETRY. Cor. 2. The angle of a regular polygon, like the angle of an equiangular polygon, is determined by the number of its sides (B. I., P. 26, c. 4). PROPOSITION II. THEOREM.. 4 regular polygon may be circumscribed by the circumference of a circle, and a circle may be inscribed within it. Let IIGFE, &c., be any regular polygon. Through the three points A, B C, B describe the circumference of a circle: ACC the centre O will lie in the line OP, drawn perpendicular to B C at the middle point P (B. III., P. 6, s.) Then I draw OB and 0. If the quadrilateral OPCRD be G E placed upon the quadrilateral OPBA, F they will coincide; for, the side OP is common; the angle OPC= OPB, each being a right angle; hence, the side PC will apply to its -equal PB, and the point C will fall on B: besides, the polygon being regular, the angle PCD=PBA (D.); hence, CD will take the direction BA; and since CD=BA, the point D will fall on A, and the two quadrilaterals will coincide. Hence, OD is equal to A 0; and consequently, the circumference which passes through the three points A, B, C, will also pass through the point ). In the same manner it may be shown, that the circumference which passes through the three points B, G, D, will also pass thrqugh the point E; and so of all the pther vertices; hence, the circumference which passes through the points A, B, C, passes also through the vertices of all the angles of the polygon, consequently, the circum-'frence of the circle circumscribes the polygon (B. III., D. 7). Again,'in reference to this circle, all the sides AB, BC, CD, &c., of the polygon, are equal -chords; they are therefore equally distant from the centre (B. III., P. 8): hence, if from the point 0 as a centre, with the distance O0P, a circumference be described, it will touch the side B/, and all the other sides of the polygon, each in its middle point, and the circle will be inscribed in the polygon (B. III., D. 11). BOOK V. 137 Scholium. The point 0, the common centre of the inscribed and circumscribed circles, may also be regarded as the centre of the polygon; and the angle A OB is called the angle at the centre, being formed by two lines drawn from the centre to the extremities of the same side AB. The perpendicular OP, is called the apothem of the polygon. Cor. 1. Since all -the chords AB, BC, CD, &c., are equal, all the angles at the centre are likewise equal (B. III, P. 4); and therefore, the value of any angle will be found by dividing four right angles by the number of sides of the polygon. Cor. 2. To inscribe a regular E D polygon of any number of sides in a given circle, we have only to divide the circumference into C as many equal parts as the polygon has sides; for, when the arcs are equal, the chords AB, BC, CD, A B &c., are also equal (B. III., P. 4); hence, likewise the triangles AOB, BOC, COD, must be equali because their sides are equal each to each (B. I., P. 10); therefore, by addition, all the angles ABC, BCD, CDE, &c., are equal (A. 2); hence, the figure ABCDE,, is a regular polygon. PROPOSITION III. PROBLEM. To inscribe a square in a,given circle. Draw two diameters A C, BD, intersecting each other at right angles; join their extremities A, B, C, D, the figure ABCD will be a square. For, the angles A OB, BOC, &c., B being equal, the chords AB, BC, &c., are also equal (B. III., P. 4): and the angles ABC, BCD, &c., A C being inscribed in semicircles, are right angles (B. III., P. 18, c. 2).. Scholiunm. Since the triangle BCO is right-angled and isosceles, we have (B. IV., P. 11, C. 5), BC: BO:: V2: 1, hence, the side of the inscribed square is to the radius, as the square root of two, to unity. 138 GEOMETRY. PROPOSITION IV. THEOREM. If a regular hexagon be inscribed in a circle, its side will be equal to the radius. Let ABCDEE,, be a regular hexagon, inscribed in a circle: then will its side AB be equal to the radius OA. For, the angle A OB is equal to one-sixth of four right angles, E (P. 2, c. 1), or one-third of two right angles: hence, the sum of the remaining angles -OAB, H c OBA, is equal to two-thirds of two right angles (B. I., P. 25). But the triangle A OB is isos- A B celes, hence, the angles at the base are equal (B. I., P. 11): therefore each is one-third of two right angles.: hence, the triangle A OB is equiangular: hence, AB =AO (B. I., P. 12). PROPOSITION V. PROBLEM. To inscribe in a given circle, a regular hexagon. Let 0 be the centre, and OB -the radius of the given circle. Beginning at any point, as B B, apply the radius BO, six times as a chord to the circumference, and we shall form the A C regular hexagon BCDEFA (P. 4).- Hence, to inscribe a regular hexagon in a given circle, the radius must be applied six times as a chord, to the circumference; which will bring us round to the point of begin- E ning. Cor. 1. If the vertices of the alternate angles be joined BOOK V. 139 by the lines A C, E, EA, there will be inscribed in the circle an equilateral triangle A CE since each of its angles will be measured by one-sixth of four right angles, or onethird of two (B. I., P. 25, C. 5). Cor. 2. If we draw the radii OA, OC, the figure OCBA will be a rhombus: for, we have OC0= CB=BA= OC. Hence, the sum of the squares of the diagonals is equivalent to the sum of the squares of the sides (B. IV., P. 14, c. 2): that is, AC B2 OB2 4AB 2~ 40B; and by taking away OB2, we have, A C 3 O0B2; hence, -A2 -2 A-C: OB -- 3' 1; or, AC: OB: /V3: I: hence, the side of the inscribed equilateral triangle is to, the radius, as the square root of three, to unity. PROPOSITION VI. PROBLEM. In a given circle to inscribe a regular decagon. Let 0 be the centre, and OA the radius of the given circle. Divide the radius OA in G extreme and mean ratio at the point M (B. IV., PROB. 4): Take OM, the greater segment, and lay it off from A to B; the chord AB will \ be the side of the regular decagon, and by apply- M ing it ten times to the circurnference, the decagon will A\C be inscribed in the circle. B For, drawing MB, we have by construction, AO 0: O:: OM AfM; or, since AB= OM, A 0: AB:: AB AM. 140 GEOMETRY. But since the triangles ABO, AMB, have a -common angle G A, included between proportional sides, they are similar (B. IV., P. 20). Now the triangle IE BA 0O being isosceles, AMB must be isosceles also, and \ AB=BM; but AB = OM; hence, also MB=MO; hence, M the triangle BMIO is isosceles. Again, in the isosceles tri- AC angle BMO, the angle AMB B being exterior, is double the interior angle 0 (B. I., P. 25, c. 6): but the angle AMB=MAB; hence, the triangle OAB is such, that each of the angles OAB or OBA, at its base, is double the angle 0, at its vertex; hence, the three angles of the triangle are together equal to five times the angle 0, which consequently, is the fifth) part of two right angles, or the tenth part of four; hence, the arc AB is the. tenth part of the circumference, and the chord AB is the side of the regular decagon. Cor. 1. By joining the vertices of the alternate angles of the decagon, a regular pentagon ACEGI will be inscribed. Cor. 2. Any regular polygon being inscribed, if the arcs subtended by its sides be severally bisected, the chords of those semi-arcs will form a new regular polygon of double the number of sides.: thus it is plain, that the square will enable us to inscribe, successively, regular polygons of 8, 16, 32, &c., sides. And in like manner, by means of the hexagon, regular polygons of 12, 24, 48, &c., sides may be inscribed; and by means of the decagon, polygons of 20, 40, 80, &c., sides. Cor. 3. It is further evident, that any of the inscribed polygons will be less than the inscribed polygon of double the number of sides, since a part is less than the whole. BOOK V. 141 PROPOSITION VII. PROBLEM. A regular inscribed polygon being given, to. circumscribe a similar polygon about the same circle. Let 0 be the centre of the circle, and CDEFAB regular inscribed polygon. At T, the middle H T G point of the arc AB, draw a tangent 1 GH, and do the same at the N S middle point of each of the arcs BC, C6D, &c.; I C - these tangents will be C parallel to the chords AB, BC, CD, &c. (B. III., P P. 10, c.); and will, by their intersections, form K L the regular circumscribed polygon GH1K &c., similar to the one inscribed. For, since T is the middle point of the arc B1RA, and N the middle point of the equal arc BNC, it follows, that BT=BN; or that the vertex B of the inscribed polygon, is at the middle point of the arc NBT. Draw OH. The line OH will pass through the point B. For, the right-angled triangles OTI, NO1,1 having the common hypothenuse OH., and the side OT= ON, are equal (B. I., P. 17), and consequently the angle TOH=10OXV wherefore the line OH passes through the middle point B of the arc TN (B. III., P. 15). In the same manner it may be shown that OI passes through Cl; and similarly for the other vertices. But since GH is parallel to AB, and III to BC, the angle GHI =ABC (B. I., P. 24); in like manner, HIK=BCDand so for the other angles: hence, the angles of' the cir cumscribed polygon are equal to those of the inscribed. And further, by reason of these same parallels, we have GH: AB:: OH: OB, and HI: BC:: OH: OB; therefore, GH AB:: HI: BC. 142 GEOMETRY. But AB=BC, t T G therefore GH=Hi.L For a like reason, N HtI= Ix; &c.; hence, the sides of the circumscribed polygon are all I I equal; hence, this polygon is regular and similar to the inscribed polygon. K Q L Cor. 1. - Reciprocally: if the circumscribed polygon GHIK &c., be given, and the inscribed one ABC &c., be required, it will only be necessary to draw from the vertices of the angles G, H, 1 &c., of the given polygon, straight lines OG, OH; &c., meeting the circumference in the points A, B, C, &c.; then to join these points by the chords AB, BC, &c.; this will form the inscribed polygon. An easier solution of this problem would be, simply to join the points of contact T, N, P, &c., by chords T1V NP, &c., which likewise would form an inscribed polygon similar to the circumscribed one. Cor. 2. Hence, we may circumscribe about a circle any regular polygon, which can be inscribed within it, and conversely. Cor. 3. It is plain that NH+HT=HT+TG =HG, one of the equal sides of the polygon. Cor. 4. If through B, A, F, &c., the middle points of the arcs NBT, TAS, SFR, &c., we draw tangent lines, we shall thus form a new regular circumscribed polygon having double the number of sides: and this process may be repeated as often as we please. The new polygon will be regular, because it will be similar to a new inscribed polygon which may be formed (P. 6, c. 2) of double the number of sides of the first.' It is plain, that each new circumscribed polygon will be less than the one from which it was derived, since a part is less than the whole. BOOK V. 143 PROPOSITION VIII. THEOREM. The ares of a regular polygon is equal to its perimeter multiplied by half the radius of the inscribed circle. Let there be the regular polygon GHIK, and ON; OT, radii of the inscribed circle drawn through points of tangency: then will its area be equal to the perimeter multiplied by one-half of OT. For, the triangle G OH is II T G measured by GH X OT; the triangle OH1, by HIX ON: but ON= N OT; hence, the two triangles taken together are measured by I (GH+HI)x R 0T. And, by finding the measures of the other triangles, it will appear K that the sum of them all, or the whole polygon, is measured by the sum of the bases GH, H;, &c., or, the perimeter of the pol3ygon, multiplied by one-half of OT; that is, the area of the polygon is equal to its perimeter multiplied by half the radius of the inscribed circle. PROPOSITION IX. THEOREM. The perimeters of regular polygons, having the same number of sides, are to each other as the radii of the circumscribed circles; and also, as the radii of the inscribed circles; and their areas are to each other as the squares of those radii. Let AB be the side of one polygon, 0 the centre, and consequently OA the radius of the circumscribed circle, and OD, perpendicular to AB, the radius of the inscribed circle. Let ab, be a side of the other A D B polygon, o the centre, oa and od, the a d b radii of the circumscribed and the inscribed circles. Then, the perimeters of the two polygons are to each other as the sides AB and ab (aB.Iv., P. 27): but o the angles A and a are equal, being 0 144 G EOMETRY. each half of the angle of the polygon; so also are the angles B and A I) B b; hence, the triangles ABO, abo, a d b are similar, as are, likewise, the right-angled triangles ADO, ado; therefore, AB: ab:: AO ao:: DO: do; hence, the perimeters of the poly- 0 gons are to each other as the radii AO, ao, of the circumscribed circles, and also, as the radii DO, do, of the inscribed circles. The surfaces of these polygons are to each other as the squares of the homologous sides AB, ab (B. IV., P. 27); they are therefore likewise to each other as the squares of AO, ao, the radii of the circumscribed circles, or as the squares of OD, od, the radii of the inscribed circles. PROPOSITION X. THEOREM. Two regular polygons, of the same number of sides, can always be formed, the one circumscribed about a circle, the other inscribed in it, which shall dfffer from each other by less than any given surface. Let Q be the side of a square less than the given surface. Bisect AC, a fourth part of the circumference, and then bisect the half of this fourth, and proceed in this manner, always bisecting one of the, arcs formed by the last bisection, until an arc is found whose chord AB is less than Q. As this arc will be an exact part of the circumference, if we apply the chords AB, BC, CD, &c., each equal to AB, the last will terminate at A, and there will be formed a regular polygon ABCDE &c., inscribed in the circle. Next, describe about the circle a similar polygon abcde &c. (P. 7): the difference of these two polygons will be less than the square of Q. For, from the points a and b, draw the lines aO, bO, to the centre 0: they will pass through the points A and B (P. 7). Draw also OK to the point of contact K: it will BOOK V. 145 bisect AB in I, and be per- ~ pendicular to it (B. III., P. 6, S.) b d Prolong A 0O to E, and draw BE. K Let p represent the circumscribed polygon, and P the a 0 E inscribed polygon: then since the triangles a Ob, A OB, are Q like parts of p and P, we have (B. II., P. 11), aOb: AOB:: p: P: But the triangles being similar (B. IV., P. 25), aOb- A OB:: 2: OA2,or O-i2. Hence, p: P:: Oa 2 Again, since the triangles OaK,; EAB are similar, having their sides respectively parallel (B Iv., P. 21). Oa2. K 02: A.2: E2, hence P: P:: AE:. EB, or by division (B.II., P. 6), -2 -2 p: p -:: AE: A2_ —EB2, or AB. But p is less than the square described on the diameter AE (P. 7, c.4); therefore, p-P is less than the square described on AB: that is, less than the- given square on Q: hence, the difference between the circumscribed and inscribed polygons may, by increasing the number of sides, always be made less than any given surface. PROPOSITION XI. PROBLEM. The suiface of a regular inscribed polygon, and that of a similar circumscribed polygon, being given; to find the surfaces of the regular inscribed and circumscribed polygons having double the number of sides. Let AB be a side of the given inscribed polygon; EF, parallel to AB, a side of the circumscribed polygon, and C the ceitre of the circle. If the chord AM and the tangents AP, BQ, be drawn, AiM will be a side of an in10 146 GEOMETRY. scribed polygon, having twice the number of sides; and AP +Plf=2P JI or PQ, will be a side of the similar circumscribed polygon (P. 7, c. 3). Now, as the same construe- E p M q F tion will take place at each angle corresponding to A C /A D it will be sufficient to consider \ AC1 CH by itself; for the triangles connected with it are evidently to each other as the whole polygons of which they form part. Let P, then, be C the surface of the inscribed polygon whose side is AB, p, that of the similar circumscribed polygon; P' the surface of the polygon whose side is AM, p' that of the similar circumscribed polygon: P and p are given; we have to find P and p'. First. Now the triangles A CD, AClf, having the common vertex A, are to each other as their bases CD, CHu[ (B. IV., P. 6, c.); they are likewise to each other as the polygons P and P', of which they form part (B. II., P. 11): hence, P: P':: CD: C. Again, the triangles CAMf, CifE, having the common vertex Md; are to each other as their bases CA, CE; they are likewise to each other as the polygons P' and p of which they form part; hence, P': p:: CA: CE. But since AD and MIE are parallel, we have, CD: C:: CA:' CE; hence, P: P':: P': p; hence, the polygon P' is a mean proportional between the two given polygons P and p, and consequently, P'= /Px. Secondly. The altitude CM being common, the triangle CPrIl is to the triangle CPE, as Pl is to PE; but since CP bisects the angle ACE, we have (B. IV., P. 17), P17: PE:: GC: CE:: CD: CA:: P P'; hence, CPM: CPE:: P: P'; BOOK V. 147 and consequently, CPM: CPM +CPE, or CME:: P: P +P'; and hence, 2CUPM, or CIMPA: UME:: 2P: P+P'. But CMPA is to CME as the polygons p' and p, of which they form part: hence, p'::: 2P: P+P'. Now as P' has been already determined; this new proportion will serve to determine p', and give us, 2P xp and thus by means of the polygons P and _p it is easy to find the polygons P' and p', which shall have double the number of sides. PROPOSITION XII. PROBLEM. Yb find the approximate area of a circle whose radius is unity. Let the radius of the circle be 1; the side of the inscribed square will be -V/2 (P. 3, s.); that of the circumscribed square will be equal to the diameter 2; hence, the surface of the inscribed square will be two, and that of the circumscribed square will be 4. Hence, P=2, and p=4; by the last proposition we shall find the inscribed octagon P'= V/8=2.8284271, 16 circumscribed octagon 2'=+ 8 =83.3137085. The inscribed and the circumscribed octagons being thus determined, we shall easily, by means of them, determine the polygons having twice the number of sides. We have only in this case to put P=2.8284271, p=3.3137085; we shall find P'= /-PXp=3.0614674, 21 p,=3.1825979. These polygons of 16 sides will enable us to find the polygons of 32 sides; and the processes may be continued 148 GEOMETRY. until the difference between the inscribed and circumscribed polygons is less than any given surface- (P.10). Since the circle lies between the polygons, it will differ from either polygon by less than the polygons differ from each other: and hence, in so far as the figures which express the areas of the two polygons agree, they will be the true figures to express the area of the circle. We have subjoined the computation of these polygons, carried on till they agree as far as the seventh place of decimals. NUMBER OF SIDES. INSCRIBED POLYGONS. CIRCUMSCRIBED POLYGONS. 4... 2.0000000... 4.0000000 8.. 2.82842-71... 3.3137085 16.. 3..0614674... 3.1825979 32 3.1214451 3.1517249 64... 3.1365485.. 3.1441184 128.., 3.1403311....1422236 256... 3.1412772 3.1417504 612... 3.1415138... 3.1416321 1024... 3.1415729... 3.1416025 2048... 3.1415877.. 3.1415951 4096.. 3.1415914 3.1415933 8192... 3:1415923... 3.1415928 16384... 3.1415925... 3.1415927 32768... 3.1415926... 3.1415926 The approximate -area of the circle, we infer, therefore, is equal to 3.1415926. Some doubt may exist perhaps about the last decimal figure, owing to errors proceeding from the parts omitted; but the calculation has been carried on with an additional figure, that the final result here given might be absolutely correct even to the last decimal place. The number generally used, for computation, is 3.1416, a number very near the true area. Scholium 1. Since the inscribed polygon has the same number of sides as the circumscribed polygon, and since the two polygons are regular, they will be similar (P. 1): and, therefore, when their areas approach to an equality with the circle, their perimeters will approach to an equality with the circumference. BOOK V. 149 Scholium 2. That magnitude to which a varying magnitude approaches continually, and which it cannot pass, is called a limit. Having shown that the inscribed and circumscribed polygons may be made to differ from each other by less than any given surface (P. 10), and since each differs from the circle less than from the other polygon, it follows that the circle is the limit of all inscribed and circumscribed polygons, resulting from continued bisections, and that the circumference is the limit of their perimeters. Hence, no sensible error can arise in supposing that what is true of a polygon resulting from an indefinite number of bisections of the subtended arcs, is also true of its limit, the circle. Indeed, the circle is but a polygon of an infinite number of sides. PROPOSITION XIII. THEOREM. Tlhe circumferences of circles are to each other as their radii, and the -areas are to each other as the squares of their radii. Let us designate the circumference of the circle whose radius is CA by circ. CA; and its area, by area CA: it is then to be shown that,ctrc. CA: circ. OB:: CA: OB, and that area CA' area OB:: CA OB. E B F ci itn it o Inscribe within the circles two regular polygons of the same number of sides. Then, whatever be the number of sides, their perimeters will be to each other as the radii CA and OB (P. 9). Now, if the arcs subtending the sides 150 G-EOMETR~Y. MA N P E F P G T L of the polygons be continually bisected, the perimeter of each polygon-will continue to approach the circumference of the circumscribed circle, and when we pass to the limit (P. 12, s. 2), we shall have circ. CA circ. OB:: CA OB. Again, the ar'eas of the inscribed polygons are to each other as CA- to OB2 (P. 9). But when the number of sides of the polygons is increased, and we pass to the limit; we shall have -2 -2 area CA: area OB CA OB2. Cor. 1. It is plain that the limit of any portion of the perimeter of an inscribed regular polygon lying between the vertices of two angles, is the corresponding are of the circumscribed circle. Thus, the limit of the perimeter intercepted between G and F is the are GFE. Cor. 2. If we multiply the antecedent and consequent of the second couplet of the first proportion by 2, and of the second by 4, we shall have circ. CA ~ circ. OB:: CA:2 0B; and area CA: area OB:: 4CA 40B; that is, the circumferences of circles are to each other as their diameters, and their areas' are to each other as the squares of their diameters. BOOK V. 151 PROPOSITION XIV. THEOREM. Similar arcs are to each other as their radii: and sbimilar sectors are to each other as the squares of their radii. Let AB, DE, be similar arcs, and ACEB, DOE, similar sectors: then will AB: DE:: CA: OD; and ACB: DOE:: CA: O-. For, since the arcs are sim- A B ilar, the angle C is equal to E the angle 0 (B. IV., D. 6). But we have (B. III., P. 17), O C angle C 4 right angles:: AB: circ. CA, and, angle 0: 4 right angles:: DE: circ. OD; hence (B. II., P. 4, c.), AB: DE:: circ. CA: circ. OD; but these circumferences are as the radii A C, DO (P. 13); hence, AB: DE:: CA: OD. For a like reason, the sectors A CB, DOE, are to each other as the whole circles: which again are as the squares of their radii (P. 13); therefore, sect. A CB sect. DOE:: CA: OD2. PROPOSITION XV. THEOREM. The area of a circle is equal to. the product of half -the radius by the circumference. Let A CDE be a circle whose cen- A C tre is 0 and radius OA: then will IF area OA=~OAXcirc. OA. For, inscribe in the circle any regu- 0 D lar polygon, and draw OF perpendicular to one of its sides. The area of 152 GEOMETRY. the polygon is equal to 4 OF, multiplied by the perimeter (P. 8). Now, F let the number of sides of the polygon be increased by continually bisecting the arcs which subtend the sides: 0 the limit of the perimeter is the circumference of the circle; the limit of the apothem is the radius OA, and the limit of the area of the polygon is the area of the circle (P. 12, s. 2). Passing to the limit, the expression for the area becomes area OA=- OA X circ. OA; consequently, the area of a circle is equal to the product of half the radius by the circumference. Cor. The area of a sector is equal to the arc of the sector multiplied by half the radius. For, we have (B. III., P. 17, s. 4), M A B sect. A CB: area CA:: AMB: circ. CA; or, sect. ACB: area CA:: AMiBX 4CA: circ. CAx ~CA. But, circ. CAX 4 CA is equal to the area CA; hence, AMB X 4 CA is equal to the area of the sector. D PROPOSITION XVI. THEOREM. The area of a circle is equal to the square of the radius multiplied by the ratio of the diameter to the circumference.'Let the circumference of the circle whose diameter is unity be denoted by -r: then, since the diameters of circles are to each other as their circumferences (P. 13, c. 2), r will denote the ratio of any diameter to its circumference. We shall then have 1:: 2CA: cire. CA: therefore, circ. CA =rX 2 CA. Multiplying both members by 4 CA, we have 4CA xcirc. CA=y x CA2, BOOK V. 153 or (P. 15) area CA=t X CA2, M that is, the area of a circle is equal A B to %r into the square of the radius. Scholium 1. To find the numerical value of %, let us suppose the C radius CzA=1; we shall then have area CGA=r. D But we have found the area of the circle whose radius is 1 to be 3.1415926 (P. 12): therefore, we have qr=3.1415926. In common calculations, we take ~=3.1416. Scholium 2. The problem of the quadrature of the circle, as it is called, consists in finding a square equivalent in surface to a circle, the radius of which is known. Now it has just been proved, that a circle is equivalent to the rectangle contained by its circumference and half its radius (P. 15); and this rectangle may be changed into an equivalent square, by finding a mean proportional between its length and its breadth (B. IV., PR&B. 3). To square the circle, therefore, is to find the circumference when the radius is given; and for effecting this, it is endugh to know the ratio of the diameter to the circumference. Hitherto the ratio in question has never.been determined except approximatively; but the approximation has been carried so far, that a knowledge of the exact ratio would afford no real advantage whatever beyond that of thel approximate ratio. Accordingly, this problem, which engaged geometers so deeply, when their methods of approximation were less perfect, is now degraded to the rank of those idle questions, with which no one possessing the slightest tincture of geometrical science, will occupy any portion of his time. Archirnedes showed that the ratio of the diameter to the circumference is included between 310- and 31-~; hence, 31 or 22 affords at once a pretty accurate approximation to the number above designated by ~; and the simplicity of this first approximation has brought it into very general 154 GEOMETRY. use. Metius, for the same number, found the much more accurate value "3. At last, the value of r, developed to a certain order of decimals, was found by other calculators to be' 3.1415926535897932 &c.: and some have had patience enough to continue these decimals to the hundred and twenty-seventh, or even to the hundred and fortieth place. Such an approximation is practically equivalent to perfect accuracy: the root of an imperfect power is in no case more accurately known. PROPOSITION XVII. THEOREM. If the circumferences of two circles intersect each other, the arc of the common chord in the less circle will be longer than the corresponding arc of the greater.* Let A and B be the centres of two circles, AC, BC, their radii, C and D) the - points in which their circumferences intersect and CD their common chord: then will the arc DEC described with the radius BC, be longer than the arc DFC described with the greater radius A C. Join the centres A and B, C and prolong AB to -E. Then, since A B bisects *the chord K CD at right angles (B. III., P. 11); it also bisects the arcs at the points F and E (B. III., P. F 6). Draw CE and DE which A B will be equal to each other (B. III., P. 4); also CF and DF. Bisect the arcs C-E, ED, and also the arcs CF, ED, and draw chords subtending the new arcs: there will thus be inscribed in the two segments:DEC, JFC, portions of two polygons, having the same number of sides.in each. Now, since the point F is within the triangle DEC, * The arc considered in this demonstration -is the one which is less than a semicircle. BOOK V. 155 EC plus ED is greater than CT plus FD (B. I., P. 8): hence, the half, CE is greater than the half, CF. If now, with C as a centre, and CE as a radius, we describe an arc EH, the chord CE being greater than CF, the arc CElf will be greater than the arc CF (B. III., P. 5). If we suppose the arc CYKE to move with the chord CE then, when the chord CE becomes the chord CE, the arc CKE will pass through the points C and J, and will have with CFH, the common chord CH. If, now, we bisect the arc which is equal to CKJE, and also the are CFH, we know from what has already been shown, that the chord of half the outer arc will be greater than the chord of half the inner arc CF11 much more will it be greater than the chord of CL, which is half the arc CF; that is, the chord of the arc CE; one-half of CE, will always be greater than the chord of the arc CL, onehalf of CF. Hence, the perimeter of that portion of the polygon inscribed in the segment CED, will be greater than the perimeter of the corresponding polygon inscribed in the segment CED.' If, then, we continue the bisections indefinitely, the limit of the outer perimeter will be the arc CED, and of the inner, the arc CFD: hence, the arc CED is greater than the are CFD. Cor. If equal chords be taken in unequal circles, the arc of the chord in the greatest circle will be the shortest; for, the circles may always be placed as in the figure. BOOK VI. PLANES AND POLYEDRAL ANGLES. DEFINITIONS. I. A straight line is perpendicular to a plane, when it is perpendicular to every straight line of the plane, which passes through its foot: conversely, the plane is perpendicular to the line. The point at which the perpendicular meets the plane, is called the foot of the perpendie, ular. 2. A line is parallel to a plane, when it cannot meet that plane, to what distance soever both be produced. Conversely, the plane is parallel to the line. 3. Two planes are parallel to each other,-when they cannot meet, to what distance soever both be produced. 4. The indefinite space included between two planes which intersect each other, is called a diedral angle: the planes are called the faces of the angle, and their line of common intersection, the edge of the angle. A diedral angle is measured by the angle contained between two lines, one drawn in each face, and both perpendicular to the common intersection at the same point. This angle: may be acute, obtuse, or a right angle. If it is a right angle,.the two faces are perpendicular to each other. 5. A POLYEDRAL angle is the indefinite space included by several planes meeting at a common point. Each plane is called a face: the line in which any two faces intersect, is called an edge: and the common point of meeting of all the planes, is called the vertex of the.polyedral angle. BOOK VI. 157 Thus, the polyedral angle whose ver- S tex is S, is bounded by the four faces, ASB, BSC, CSIQ, DSA. Three planes, at least, are necessary to form a polyedral D L:C angle. A polyedral angle bounded by three planes, is called a triedral angle. B POSTULATES. 1. Let it be granted, that from a given point of a plane, a line may be drawn perpendicular to that plane. 2. Let it be granted, that from a given point without a plane, a perpendicular may be let fall on the plane. PROPOSITION I. THEOREM. A straight line cannot be partly in a plane, and partly out of it. For, by the definition of a plane (B. I., D. 9), when a straight line has two points common with it, the line lies wholly in the plane. Scholitnm. To discover whether a surface is plane, apply a straiglht line in different ways to that surface, and ascertain if it coincides with the surface throughout its whole extent. PROPOSITION II. THEOREM. Two straight lines which intersect each other, lie in the same plane, and determine its position. Let AB, A C be two straight lines which intersect each other in A; a plane may be conceived in which the straight line AB is found; if this plane be turned B round AB, until it pass through the point C, then the line A,, which has two of its points A and 6, in this plane, lies wholly in it; hence, the position of the plane is determined by the single condition of containing the two straight lines AB, A C. 158 GEOMETRY. Cor. 1. Any three points A, B, C, not in a straight line, determine A the position of a plane. Hence, a triangle BA C, determines the posi- B tion of a plane. Cor. 2. Hence, also, two paral- E lels AB, CD, determine the posi- A / tion of a plane; for, drawing the secant EF, the plane of the two straight lines AE, EF, is that of the parallels AB, CD. But the lines AE, EF, determine this plane; therefore, so do the parallels, AB, CD). PROPOSITION III. THEOREM. If two planes cut one another, their common section will be a straight line. Let the two planes AB, CD, cut C one another, and let E and F be two points, of their common section. Draw the straight line EE. This line lies wholly in the B plane AB, and also, wholly in the E plane CD (B. I., D. 9): therefore, D it is in both places at once. But since a straight line and a point out of it cannot lie in two planes at the same time (P. II., c. 1), EF contains all the points common to both planes, and consequently, is their common intersection. PROPOSITION IV. THEOREM. If a straight line be perpendicular to two straight lines at their point of intersection, it will be perpendicular to the plane of those lines. Let MN be the plane of the two lines BB, CI6 and let AP be perpendicular to each of them at their point of intersection P; then will AP be perpendicular to every line of the plane passing through P, and consequently to the plane itself (D. 1). BOOK VI. 159 For, through P draw in the A plane MN, any straight line as PQ. Through any point of this line, as Q, draw BQC, so that BQ shall M be equal to QC (B. IV., PROB. 5); B draw AB, A Q, AC. The base BC being divided into P two equal parts at the point Q, the C triangle BPC gives (B. Iv., P. 14). N P —2+2 P -,2 2 iC2. The triangle BAC in like manner gives,.x-2 -`2 2. AC 2+AB2 c2AQ2 +2QC2. Taking the first of these equals from the second, and observing that the triangles APC, APB, being right-angled at P, give a - Pv2 2 - 2 2 -2 -PC APi, and AB_-PB - AP, we shall have, A 2+AA 2 2A -Q-2PQ2. Therefore, by taking the halves of both, we have AP2 — AQ-PQ, or AQ2APK+$PQ; hence, the triangle APQ is right-angled at P; hence, AP is perpendicular to PQ. Sclholium. Thus, it is evident, not only that a straight line may be perpendicular to all the straight lines which pass through its foot, in a plane, but that it always must be so, whenever it is perpendicular to two straight lines drawn in the plane: hence, a line and plane may fulfil the conditions of the first definition. Cor. 1. The perpendicular AP is shorter than any oblique line A Q; therefore, it measures the shortest distance from the point A to the plane 2IN[ Cor. 2. At a given point P, on a plane, it is impossible to erect more than one perpendicular to the plane; for, if there could be two perpendiculars at the same point P, draw through these two perpendiculars a plane, whose section with the plane MN is PQ; then these two perpen 160 GEOMETRY. diculars would be both perpendicular to the line PQ, at the same point, which is impossible (B. I., P. 14, c.) It' is also impossible to let fall from a given point, out of a plane, two perpendiculars to that plane; for, if AP, A Q, be two such perpendiculars, the triangle APQ will have two right angles APQ, A QP, which is impossible (B. I., P. 25, C. 3). PROPOSITION V. THEOREM. If from a point without a plane, a perpendicular be drawn to the plane, and oblique lines be drawn to different points: 1st. Th/e oblique lines which pierce the plane at points equally distant from thie foot of the perpendicular, are equal: 2d. Of two oblique lines which pierce the plane at unequal distances, the'one passing through the remote point is the longer. Let AP be perpendicular to the plane MN; AB, A, AD, oblique lines intercepting the equal distances PB, PC, PD, and AE a line intercepting the' larger distance PE: then will AB=A C=AD; and A-E will be greater than AD. For, the angles APB, APR, APD, A being right angles, and the distances PB, PC, PD, equal to each other, N the triangles APB, APC, APD, have in each an equal angle con- / tained by two equal sides: there- B E fore they are equal (B. I., P. 5); DRC hence, the hypothenuses, or the oblique lines AB, A C, AD, are etqual to each other. Again, since the distance PE is greater than PD, or its equal PB, the oblique lifie AE is greater than AB, or its equal AD (B. 1, P. 15). Cor. All the equal oblique lines, AB, AC, AD, &c., terminate in the circumference BC'D, described from P, the foot of the perpendicular, as a centre; therefore, a point A being given out of a plane, the point P at which the per BOOK VI. 161 pendicular let fall from A would meet that plane, may be found by marking upon that plane three points, B, C, D, equally distant from the point A, and then finding the centre of the circle which passes through these points; this centre will be P, the point sought, Scholium. The angle ABP is called the inclination of the oblique line AB to the plane MN; which inclination is evidently equal with respect to all such lines AB, AC, AD, as make equal angles with the perpendicular; for, all the triangles ABP, ACP, ADP, &c., are equal to each other. PROPOSITION VI. THEOREM. If from the foot of a perpendicular a line be drawn at right angles to any line of a plane, and the point of intersection be joined with any point of the perpendicular, this last line will be perpendicular to- the line of the plane. Let AP be perpendicular to the plane NMlf, and PD perpendicular to BC; join D) with any point of the perpendicular, as A; then will AD also be perpendicular to BC. Take DB=DC, and draw PB, A PC, AB, A C. Now, since DB is. equal to D C, the oblique line PB M is equal to PC (B. 1, P. 5): and l since PB is equal to PC, the p D oblique line AB is equal to A C (P. 5); therefore, the line AD N has two of its points A and 1) equally distant from tlhe extremi- E ties B and C; therefore, AD is a perpendicular to BC, at its middle point D (B. I., P. 16, c.) Cor. It is evident, likewise, that BC is perpendicular to the plane of the triangle APD, since it is perpendicular to the two straight lines AD, PD of that plane (P. 4). Scholium 1. The two lines AE, BC, afford an instance of two lines which are not parallel, and yet do not meet, because they are not situated in the same plane. The short11 162 GEOMETRY. est distance between these lines is A the straight line PD, which is at once perpendicular to the line AP M and to the line BC. The distance PD is the shortest distance between D them: because, if we join any other two points, such as A and B, we shall have AB>AD, AD>PD; N therefore, still more, AB > PD. E?' Scholium 2. The two lines AE, CB, though not situated in the same plane, are conceived as forming a right angle with each other; because ALE and the line drawn through any one of its points parallel to BC, would make with each other h right angle.' In the same manner, AB, PD, which represent any two straight lines not situated in the same plane, are supposed to form with each other the same angle, as would be formed by AB and a straight line drawn through any point of AB, parallel to PD. PROPOSITION VII. THEOREM. If one of -two parallel liles be perpendicular to a plane, the other will also be perpendicular to the same plane. Let ED, AP, be two parallel lines; if AP is perpendicular to the plane NM; then will ED be also perpendicular to it. For, through the parallels A E AP, DE, pass a plane; its intersection with the plane MN will M X be PD; in the plane MN draw BC perpendicular to PD, and then draw AD. D Now, BC is perpendicular to the plane APRDE (P. 6, c.) there- N fore, the angle BDE is a right angle; but the angle EDP is also a right angle, since AP is perpendicular to PD, and DE parallel to AP (B. I., P. 20, c. 1); therefore, the line DE is perpendicular to the two straight lines DP, DB; consequently it is perpendicular to their plane MN (P. 4). BOOK VI. 163 Cbr. 1. Conversely: if the straight lines AP, DE, are perpendicular to the same plAne MN,; they will be parallel. For, if they be not parallel, draw, through the point D, a line parallel to AP, this parallel will be perpendicular to the plane A2IN; therefore, through the same point D more than one perpendicular will be erected to the same plane, which is impossible (P. 4, c. 2). Cor. 2. Two lines A and B; parallel to a third C, are parallel to each other; for, conceive a plane perpendicular to the line C; the lines A and B, being parallel to C, are perpendicular to this plane; therefore, by the preceding corollary, they are parallel to each other. The three parallels are supposed not to be in the same plane; otherwise the proposition would be already known (B. I., P. 22). PROPOSITION VIII. THEOREM. if a straight line is parallel to a line of a plane, it is parallel to the plane. Let the straight line AB be parallel to the line CD of the plane NM; then will it be parallel to the plane NM. For, if the line AB, which lies A B in the plane ABDC, could meet the plane 1HN it could only be in some _ point of the line CD, the common intersection of the two planes; but C D the line AB cannot meet CD) since \ they are parallel (B. I., D. 16): hence, it will not meet the plane MN; therefore, it is parallel to that plane (D. 2). PROPOSITION IX. THEOREM. Two planes which are perpendicular to the same straight line are parallel to each other. Let the planes MY1, PQ, be perpendicular to the line AB, then will they be parallel. 164 GEOME TRY. For, if they can meet any - M where, let 0 be one of their 0.... common points, and draw OA, AV OB. NTow, the line AB, which P" N is perpendicular to the plane IMN, is perpendicular to the B C straight line OA, drawn through its foot in that plane (D. 1); for the same reason AB is perpendicular to BO; therefore, there are two perpendiculars, OA and OB, let fall from the same point 0, upon the same straight line, which is impossible (B. I., -P. 14); therefore, the planes MN, PQ, cannot meet each other; consequently, they are parallel. PROPOSITION X. THEOREM. If a plane cut two parallel planes, the lines of intersection will be parallel. Let the parallel planes NM; QP, be intersected by the plane E/I; then will the lines of intersection EF, GH, be parallel. For, if the lines EF, G1; M lying in the sambe plane, were E F not parallel, they would meet each other when prolonged; and then the planes MN, PQ, in which p those lines lie, wofild also meet; and hence, the planes would not G H be parallel, which is contrary to Q the hypothesis. PROPOSITION XI. THEOREM. If two planes are parallel, a straight line which is perpendicular to one, is also perpendicular to the other. Let MIV; PQ, be two parallel planes, and let AB be perpendicular to the plane NHI; then will it also be perpendicular to QP. BOOK VI. 165 For, draw any line BC in the M plane PQ, and through the lines O AB and BC, pass a plane AB, A intersecting the plane YiN in p N AD; the intersection AD is parallel to BC (P. 10). But the line \ AB, being perpendicular to the plane MN;, is perpendicular to the straight line AD (D. 1); therefore, also, to its parallel BC (B. I., P. 20, c. 1); hence, the line AB being perpendicular to any line BC, drawn through its foot in the plane PQ, is perpendicular to that plane (D. 1). PROPOSITION XII. THEOREM. All parallels included between two parallel planes are equal. Let MN, PQ, be'two parallel planes, and HF, GE, two parallel lines: then will GE=HF. For, through the parallels GE, M E JHF, draw the plane EGHF, intersecting the parallel planes in EF and GL. The intersections EF, Gil, are parallel to each other (P. 10); and since GE, HF are p G parallel, the figure EGHF is a parallelogram; consequently, H Q EG=FH (B. I., P. 28). Cor. Hence, it follows, that two parallel planes are everywhere equidistant. For, suppose EG to be perpendicular to the plane PQ; then, the parallel El is also perpendicular to it (P. 7), and the two parallels are likewise perpendicular to the plane MN (P. 11); and being parallel, they are equal, as shown by the proposition. 166 GEOMETRY. PROPOSITION XIII. THEOREM. If two angles, not situated in the same plane, have their sides parallel and lying in the same direction, these angles will be equal and their planes will be parallel. Let the angles CAE and DBF, have the side A C parallel to BD, and lying in the same direction: also, AE parallel to BF, and lying in the same direction; then will the angles CABE and DBF be equal, and their planes parallel. For, take A C and BD equal to each other, and also AE=BF; / and draw CE, DF, AB, CD, EF./ H Since AC is equal and parallel to A G,E BD, the figure ABDC is a paral- N lelogram (B. I., P. 30); therefore, CD) is equal and parallel to AB. For P a similar reason, EF is equal and parallel to AB; hence, also, CD is equal and parallel to EF (P. B F 7, c. 2); hence, the figure DFEC Q is a parallelogram, and the side CE is equal and parallel to DF; therefore, the triangles CAE, DBF, have their corresponding sides equal; consequently, the angle CAE=IDBF. Again, the plane ACE is parallel to the plane BDF. For, if not, suppose a plane to be drawn through the point A, parallel to BDF. If this plane be different from ACE, it will meet the lines CD, EF, in points different from Ca and E, for instance in G and H; then, the three lines BA, DG, FH, will be equal (P. 12), and each equal to AB: but the lines AXh, CD, EF, are already known to be equal; hence, DC=DG, and HF=FE, which is absurd; hence, the plane ACE is parallel to BDF. Cor. If two parallel planes MXIV; PQ, are met by two other planes CAABD, EABF, the angles CAE, DBF, formed by the intersections of the parallel planes are equal; for, the intersection A C is parallel to B/D, and AE to BF (P. 10); therefore, the angle CABE=DBF. BOOK VI. 167 PROPOSITION XIV. THEOREM. If three straight lines, not situated in the same plane, are equal and parallel, the triangles formed by joining the extremities of these lines will be equal, and their planes parallel. Let AB, CD, EF, be three equal and parallel lines. Since AB is equal and paral- M lel to CD, the figure ABDC is a parallelogram; hence, the side AC is equal and parallel to BD A (B. I., P. 30).'For a like reason, A the sides AE, BF, are equal and N parallel, as also C'E, DF; hence, the two triangles A CE, BDE; have their sides equal, and are therefore D equal (B. I., P. 10); and as their sides, are parallel and lie in the same B F directions, their planes are parallel (P. 13). PROPOSITION XV. THEOREM. If two straight lines be cut by three parallel planes, they will be divided proportionally. Suppose the line AB to meet the parallel planes MNr PQ, RS, at the points A, E, B; and the line CD to meet the same planes at the points C, F, D: then will AE ~ElB:: CF ~ FD. Draw AD meeting the plane M PQ in G, and draw ACG, EG, GF, ~ BD. Since the parallel planes PQ, A RS, are cut by the third plane BAD, the intersections BD and EUG P are parallel (P. 10): and we have E AE:EB:: A GD. AG\ Q and the intersections A C, GF, R being parallel, B-f AG GD:: CF: ED; D hence (B. II., P.4, c.), AE: EB:: CF: ED. 168 GEOMETRY.'PROPOSITION XVI. THEOREM. If a line is perpendicular to a plane, every plane passed through the perpendicular, is also perpendicular to the plane. Let AP be perpendicular to the plane NMf; then will every plane passing through AP be perpendicular to NH. Let BF be any plane passing through AP, and BC its intersection with the plane ITV.: In the plane V1YN, draw DE perpendicular to BP: then the line AP, M ) F being perpendicular to the plane MN, B is perpendicular to each of the two \ straight lines B(,'DE. Now, since AP \ and DE are both perpendicular to the C common intersection BC, the angle which N they form will measure the angle between the planes (D. 4): but the angle APD, or APE, is a right angle: hence, the two planes are perpendicular to each other. Scholium. When three straight lines, such as AP, BP, -DP, are perpendicular to each other, any two may be regarded as determining a plane, and the three will determine three planes. Now, each line'is perpendicular to the plane of the other two, and the three planes are perpendicular to each other. PROPOSITION XVII. THEOREM. Conversely: If two planes are perpendicular to each other, a line drawn in one of them perpendicular to their common intersection, will be perpendicular to the other plane. Let the plane BA be perpendicular to VI~; then, if the line AP be perpendicular to the intersection BC, it will also be perpendicular to the plane NHi B For, in the plane MN, draw PD perpendicular to PB; then, because the / planes are perpendicular, the angle APD D is a right angle (D. 4); therefore, the line N BOOK VI. 169 AP is perpendicular to the two straight lines PB, PD, passing through its foot; therefore, it is perpendicular to their plane MN (P. 4). Cor. If the plane BA is perpendicular to the plane i~VN; and if at a point P of the commoni intersection we erect a perpendicular to the plane M1, that perpendicular will be in the plane BA. For, let us suppose it will not, then, in the plane BA draw AP perpendicular to PB, the common intersection, and this AP at the same time, is perpendicular to the plane MiV; by the theorem; therefore at the same point P there are two perpendiculars to the plane tfN; one out of the plane BA, and one in it, which is impossible (P. 4, c. 2). PROPOSITION XVIII. THEOREM. If two planes which cut each other are perpendicular to a third plane, their common intersection is also perpendicular to that plane. Let the planes BA, DA, be perpen- A dicular to NM; then will their intersection AP be perpendicular to NM; M For, at the point P, erect a perpen- B dicular to the plane MVN; that perpendicular must be at once in the plane p AB and in the plane AD (P, 17, c.); D therefore, it is their common intersection N AP. PROPOSITION XIX. THEOREM. The sum of either two of the plane angles which include a triedral angle, is less than the third. The proposition requires demonstration only when the plane angle, which is compared with the sum of the two others, is greater than either' of them. Therefore, suppose the triedral angle S to be formed by the three plane angles ASB, ASC. BSC, and that the angle ASB is the greatest; we are to show that ASBAABC (P. 19), and the same may be shown at each of the other vertices, C, D, XE, A: hence, the sum of the angles at the bases, in the triangles whose common vertex is S, is greater than the sum of the angles at the bases, in the set whose common vertex is 0: therefore, the sum of the vertical angles about S is less than the sum of the angles about 0: that is, less than four right angles. Scholium. This demonstration is founded on the supposition that the polyedral angle is convex, or that the plane of no one face produced can ever meet the polyedral angle; if it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude. PROPOSITION XXI. THEOREM. If two triedral angles are included by plane angles which are equal each to each, the planes of the angles are equally inclined to each other. Let S and T be the vertices of two triedral angles, and let the angle ASC=DTiF, the angle ASB=-DTE, and the angle BSC=ETF; then will the -inclination of the planes ASC, ASB,'be equal to that of the planes DTF,: DTE. For, having taken SB at S T pleasure, draw BO perpendicular to the plane ASC; from the point O, where the perpendicular meets the plane, draw OA, A OC, perpendicular to SA, SC; 0 draw AB, BC. Next take TE==SB; draw EP perpendicular to the plane DTF; from the point P draw PD, PF, perpendicular respectively to TD, TF; lastly, draw )DE and EF. '172 GEOMETRY. The triangle SAB, is right-' S T angled at A, and the triangle TDE at D (P. 6): and since the angle ASB=DTE, we haveC SBA= TED. Moreover, since A B D SB=TE, the triangle SAB is P equal to the' triangle TDE; therefore, SA=TD, and AB=DE. In like manner, it may be shown, that SC=TE and BC=EF. That proved, the quadrilateral ASCO is equal to the quadrilateral DTFP: for, place the angle ASC upon its equal DT.F; because SA=TD, and SC'= T, the point A will:fall on D, and the point C on,F; and, at the same time, A 0, which is perpendicular to SA, will fall on DP, which is perpendicular to TD, and, in like manner,, OC ofi PF; wherefore, the point 0 will fall on the point P, and hence, AO is equal to DP. But the triangles A OB, DPE, are right-angled at 0 and P; the hypothenuse AB=.DE, and the side A O=DP: hence, those triangles are equal (B. I., P. 17); and, consequently, the angle OAB=PDIE. But the angle OAB measures the inclination of the two faces ASB, ASC; and the angle PDE measures that of the two faces DTLE, DTF; hence, those two inclinations are equal to each other. Scholium 1. It must, however,- be observed, that the angle A of the right-angled triangle A OB is properly the inclination of the two planes ASB, ASC, only when the perpendicular BO falls on the same side of SA, with SC; for, if it fell on the other side, the angle of the two planes would be obtuse, and the obtuse angle together with the angle A of the triangle OAB would make two right angles. But in the same case, the angle of the two planes TDE, TDF, would also be obtuse, and the obtuse angle together with the angle D of the triangle DPE, would make two right angles; and the angle A being thus always equal to the angle D, it would follow that the inclination of the two planes ASB, ASC, must be equal to that of- the two planes TDE, TDF. Scholium 2. If two triedral angles are included by three BOOK VI. 173 plane angles, respectively equal to each other, and if, at the same time, the-equal or homologous angles are disposed in the same order, the two triedral angles will coincide when applied the one to the other, and consequently, are equal (A. 14). For, we have already seen that the quadrilateral SA 0C may be placed upon its equal TI9PF; thus, placing SA upon TD, SC falls upon TE, and the point 0'upon the point P. But because the triangles AOB, DPE, Are equal, OB, perpendicular to the plane ASC, is equal to PE, perpendicular to the plane TDF; besides, these perpendiculars lie in the same direction; therefore, the point B will fall upon the point E, the line SB upon TE, and the two angles will wholly coincide. Scholium 3. The equality of the triedral angles does not exist, unless the equal faces are arranged in the same manner. For, if they were arranged i an n inverse order, or, what is the same, if the perpendiculars OB, PE, instead of lying in the same direction with regard to the planes ASC, DTlF lay in opposite directions, then it would be impossible to make these triedral angles coincide the one with the other. The theorem would not, however, on this account, be less true, viz.: that the faces containing the equal angles must be equally inclined to each other; so that the two triedral angles would be equal in all their constituent parts, without, however, admitting of superposition. This sort of equality, which is not absolute, or such as admits of superposition, deserves to be distinguished by a particular name: we shall call it, equality by symmetry. Thus, those two triedral angles, which are formed by faces respectively equal to each other, but disposed in an inverse older, will be called trieclral angles egual by symmetry, or simply symmetrical angles. BOOK VII. POLYEDRONS. DEFINITIONS. 1. POLYEDRON is a name given to any solid bounded by polygons. The bounding polygons are called faces of the polyedron; and the straight line in which any two adjacent faces meet each other, is dalled an edge of the polyedron. 2. A PRISM is a polyedron in which two of the faces are equal polygons with their planes and homologous sides parallel, and all the other faces parallelograms. 3. The equal and parallel polygons are called bases of the prism-the one the lower, the other, the upper baseand the parallelograms taken together, make up the lateral or convex surface of the prism. 4. The ALTITUDE of a prism is the distance between its two bases, and is measured by a line drawn from a point in one base, perpendicular to the plane of the other. 5. A right prism is one whose edges, formed by the intersection of the lateral faces, are perpendicular to the planes of the bases. Each edge is then equal to the altitude of the prism. In every other case, the prism is oblique, and each edge is then greater than the altitude. BOOK VII. 175 6. A TRIANGULAR PRISM is one whose bases are triangles: a quadrangular prism is one whose bases are quadrilaterals: a pentangular prism is one whose bases are pentagons: a hexangular prism is one whose bases are hexagons, &c. 7. A PARALLELOPIPEDON is a prism whose bases are parallelogramns. 8. A RECTANGULAR PARALLELOPIPEDON is one whose faces are all rectangles. When the faces are squares, it is called --------- a cube, or regular hexaedron. 9. A PYRAMID is a solid bounded by a polygon, and by triangles meeting at a common point, called the vertex. The polygon is called the base of the pyramid, and the triangles, taken together, the convex, or lateral surface. The pyramid, like the prism, takes different names, according to the form of its base: thus, it may be triangular, quadrangular, pentangular, &c. 10. The ALTITUDE of a pyramid is the perpendicular let fall from the vertex on the plane of the base. 11. A RIGHT PYRAMID is one whose base is a regular polygon, and in which the perpendicular let fall from the vertex upon the base passes through the centre of the base. This perpendicular is then called the axis of the pyramid. 12. The SLANT HEIGHT of a right pyramid, is the perpendicular let fall from the vertex to either side of the polygon which forms the base. 13. If a pyramid is cut by a plane parallel to its base, forming a second base, the part lying between the bases, is called a truncated pyramid, or frustum of a py?ramid. 176 GEOMETRY. 14. The altitude of a frustum is the perpendicular distance between its bases: and the slant -height, is that portion of the slant height of the pyramid intercepted between the bases of the frustum. 15. The diagonal of a polyedron is a line joining the vertices of any two of its angles, not in the same face. 16. Similar polyedrons are those whose polyedral angles are equal, each to each, and which are bounded by the same number of similar faces. 17. Parts which are like placed, it similar polyedrons, whether faces, edges, or angles, are called homologous. 18. A regular polyedron is one whose faces are equal and regular polygons, and whose polyedral angles are equal. PROPOSITION I. THEOREM. The convex surface of a right prison is equal to the perimeter of either base multiplied by its altitude. Let ABCDE-KI be a right prism: then will its convex surface be equal to (A.B +B~+ CD+]DE+ EA)X AF. For, the convex surface is equal to the sum of all the rectangles AG, B-F,, CI;, ID)I EF, which compose it. Now, K.-\ the altitudes A B, UG, CII, &c., of the F rectangles, are equal to the altitude of the prism, and the area of each rect- D angle is equal to its base multiplied by E its altitude (13. IV., P. 5). Hence, the sum of these rectangles, or the convex surface of the prism, is equal to (AB ~+BC CD+ DE+EA)x AF; that is, to the perimeter of the base of the prism multiplied by the altitude. Cor. If two right prisms have the same altitude, their convex surfaces are to each other as the perimeters of their bases. BOOK VII. 177 PROPOSITION II. THEOREM. In every prism, the sections formed bly parallel planes, are equal polygons. Let the prism AH be intersected by the parallel planes NP, SV; then are the polygons NOPQR, STVXY; equal. For, the sides SJT NO, are parallel, being the intersections of two parallel I planes with a third plane ABGF; these F.' H same sides, ST, NO, are included be- Y O tween the parallels NS, OT, which are S - edges of the prism: hence, NO is TV equal to ST. For like reasons, the N sides OP, PQ, QR,' &c., of the section / p NOPQR, are equal to the sides TV,;. /. D VX, XY; &c., of the section STVXY, A each to each; and since the equal B sides are at the same time parallel, -it follows that the angles' NOP, OPQ, &c., of the first section, are equal to the angles STY; TVX; &c., of the second, each to each (B. VI., P. 13). Hence, the two sections NOPQR,,STVXY; are equal polygons. Cor. Every section of a prism, parallel to the bases, is equal to either base. PROPOSITION III. THEOREM. Tf a pyramid be cut by a plane parallel to its base: 1st. The edges and Wlie altitude will be divided proportionally: 2d. The section will be a polygon similar to the base. Let the pyramid SABC DE, of which SO is the altitude, be cut by the plane abcde; then will Sa: SA:: So: SO, acnd the same for the other edges;.and the polygon abcde, yvill be similar to the base ABC'DE. 12 178 GEOMETRY. First. Since the planes ABC, abc, are S parallel, their intersections AB, ab, by the third plane SAB, are also parallel (B. VI., P. 10); hence, the triangles SAB, Sab, are similar (B. IV., P. 21), and we have a c SA.: ~Sa SB Sb; /; for a like reason, we have A SB: Sb: SC: c; and so on. Hence, the edges SA, SB, SC, B &c., are cut proportionally in a, b, c, &c. The altitude SO is likewise cut in the same proportion, at the point o; for BO and bo are parallel, therefore, we have SO: So:: SB ~Sb. Secondly. Since ab is parallel to AB, be to BC, cd to CD, &c., the angle abe is equal to ABC, the angle bed to BCD, and so on (B. VI., P. 13). Also, by reason of the similar triangles SAB, Sab, we have AB: ab:: SB: Sb; and by reason of the similar triangles SBC, Sbc, we have SB: Sb:: BC: be; hence, AB: ab:: BC: be; we might likewise have BC: be:: CD: cd, and so on. Hence, the polygons ABCDE, abcde have their angles equal, each to each, and their sides, taken' in the same order, proportional; hence, they are similar (B. IV., D. 1). Cor. 1. Let S-ABCDE, S S.XYZ, be two pyramids, having a common vertex and their bases in the same plane; if these pyramids are cut by a plane parallel to the plane of their bases, the sections, abcde, xyz, will be to each other as the bases ABCDE, B XYZ. Y BOOK VII. 179 For, the polygons ABClDE, abode, being similar, their surfaces are as the squares of the homologous sides AB, ab; that is, B. IV., P. 27), ABCDE: abcde:: AB: ab2. but, AB: ab:: SA: Sa; hence, ABCDE: abcde:: 5A: Sa-. For the same reason, XYZ: xyz:: SX x. But since abc and xyz are in one plane, we have likewise (B. VI., P. 15), SA Sa:: SX: Sx; hence, ABUDE: abcde:: XYZ: xyz; therefore, the sections abcde, xyz, are to each other as the bases ABCDE, XYZ. Cor. 2. If the bases ABCDE, XYZ, are equivalent, any sections abcde, xyz, made at equal distances from the bases, are also equivalent. PROPOSITION IV. THEOREM. The convex surface of a right pyramid is equal to the perimeter of its base multiplied by half the slant height. Let S be the vertex, ABCDE the base, and XSF the slant height of a right pyramid; then will the convex surface be equal to SF x (AB +BC+ riD+DE). For, since the pyramid is right, the point 0, in which the axis meets the base, is the centre of the polygon e. - ABCDE (D. 11); hence, the lines OA,, OB, OC, &c., drawn to the vertices of the base, are equal. In the right-angled triangles SAO, E' SBO, the bases and perpendiculars are F g equal: hence, the hypothenuses are equal: and it may be proved in the same way, that all the edges of the right pyramid are 180 GEOMETRY. equal. The triangles, therefore, which form the convex surface of the prism S are all equal to each other. But the area of either of these triangles, as ESA, e is equal to its base EA, multiplied by i half the perpendicular Si, which is thie slant height of the pyramid: hence, the c area of all the triangles, or the convex E surface of the pyramid, is equal to the perimeter of the base multiplied by half A the slant height. Cor. The convex surface of the frustum of a right pyramid is equal to half the sum of the perimeters of its upper and lower bases multiplied by its slant height. For, since the section abcde is similar to the base (p. 3), and since the base ABCD.E is a regular polygon (D. 11), it follows that the sides ea, ab, be, cd, and de, are all equal to each other. Hence, the convex surface of the frustum ABCDE-d is composed of the equal trapezoids EAae, ABba, &c., and the perpendicular distance between the parallel sides of either of these trapezoids is equal to Ff, the slant height of the frustum. But the area of either of the trapezoids, as AEeca, is equal to ~(EA + ea)X Ff (B. Iv., P. 7): hence, the area of all of them, or the convex surface of the frustum, is equal to half the sum of the perimeters of the upper and lower bases multiplied by the slant height. PROPOSITION V. THEOREM. gf the three faces which i7tclude a triedral angle of a prism are equal to- the three faces which include a triedral angle of a second prism, each to each, and are like placed, the two prisms are equal. Let B and b be the vertices of two triedral angles included byi faces respectively equal to each other, and similarly placed; then will the prism ABCD-E-K be equal to the prism abcde-k. For, place the base abcde upon the equal base ABCDE; BOOK VII. 181 K k F I ia A BC b c then, since the triedral angles at b and B are equal, the parallelogram bh will coincide with BE, and the parallelogram bf with BF. But the two upper bases being equal to their corresponding lower bases, are equal to' each other, and consequently, will coincide: hence, hi will coincide with H1,; ik with IX1 kf with EF; and therefore, the lateral faces of the prisms will coincide: hence, the two prisms coinciding throughout, are equal (A. 14). Gor. Two right prisms, which have equal bases and equal altitudes, are equal. For, since the side AB is equal to ab, and the altitude BG to bg, the rectangle ABGF is equal to abgf; so also, the rectangle BGHC is equal to bghc; and thus the three planes, which include the'triedral angle B, are equal to the three which include the triedral angle b. Hence, the two prisms are equal. PROPOSITION VI. THEOREM. In every paralelo2pipedon, the opposite faces are equal and parallel. Let ABCD be a parallelopipedon, then will its opposite faces be equal and parallel. For, the bases ABCD, EFG;, are E:I equal parallelograms, and have their planes parallel (D. 7). It remains only F G to show, that the same is true of any A two opposite lateral faces, such as BCGF, ADHE. B C 182 GEOMETRY. Now, BC is equal and parallel to E I AD, because the base ABCD is a parallelogram; and since the lateral faces / are also parallelograms, BF is equal and At- D parallel to AE, and the same may be shown for the sides FG and E;, CG and B C DH; hence, the angle CBF is equal to the angle DAE, and the planes DAE, CBF, are parallel (B. VI., P. 13); and the parallelogram BUCGF, is equal to the parallelogram ADHE. In the same way, it may be shown that the opposite parallelograms ABFE, DCGt, are equal and parallel. Cor. 1. Since the parallelopipedon is a solid bounded by six faces, of which any two lying opposite to each other, are equal and parallel, it follows that any face and the one opposite to it, may be assumed as the bases of the parallelopipedon. Cor. 2. The diagonals of a parallelopipedon bisect each other. For, suppose two diagonals B;, ZD, to be drawn through opposite vertices. Draw also BD, FH. Then, since BF is equal and parallel to D1;, the figure BDHF is a parallelogram; hence, the diagonals BH, D)E mutually bisect each other at E (B. I., P. H 31). In the same manner, it may be F shown that the diagonal BE and any other diagonal bisect each other; hence, the four diagonals mutually bisect each other, in a common point. If the six B faces are equal to each other, this point may be regarded as the centre of the parallelopipedon. Scholium. If three straight lines AB, AE, AD, passing through the same point A, and making given angles with each other, are known, a parallelopipedon may be formed on these lines. For this purpose, conceive a plane to be passed through the extremity of each line, and parallel to the plane of the other two, that is, tlirough the point B pass a plane parallel to DAE, through D a plane parallel to BAE, and through E a plane parallel to BAD. The mutual intersections of these planes will form the parallelopipedon required. BOOK VII. 183 PROPOSITION VII. THEOREM. If a plane be passed through the opposite diagonal edges of a parallelopipedon, it will divide the solicd into two equivalent triangular prisms. Let the parallelopipedon ABCD-H be dividqd by the plane BDtHF, passing through the opposite edges BF, DI)I: then will the triangular prism ABD-1I, be equivalent to the triangular prism BCD-H. For, through the vertices B and F, H pass the planes Bccda, FghTe,~ at right angles to the edge BY, the former cut- E ting the three other edges of the par- e I G allelopipedon prolonged in the points / D c, d, a, the latter in the points g, h, e. Now, the sections Bcda, Fghe, are A d equal parallelograms. For, the cutting a; planes being perpendicular to the same / c straight line BF, are parallel (B. vI., P. B 9): hence, the sections are equal (P. 2); and they are parallelograms because Ba, cd, two opposite sides of the same section, are formed by the meeting of a plane aBcd, with two parallel planes ABF_, DCGIT (B. vI., P. 10). For a similar reason Be and ad are parallel; hence, the figures are equal parallelograms. For a like reason the figure aBFe is a parallelogram; so also, are BcyF, cghd, adhe, the other lateral faces of the solid aBcd-h; hence, that solid is a prism (D. 2), and that prism is right, since the edge BF is perpendicular to its bases. But the right prism aBcd-h is divided by the plane BH into two equal right prisms aBd-h, Bcd-h; for, the bases aBd, Bed, are equal, being halves of the same parallelogram, and since the prisms have the common altitude BEf, they are equal (P. 5, c.) It is nqw to be proved that the oblique triangular prism ABD-H is equivalent to the right triangular prism 5aBd-h. Since these prisms have a common part ABD-h, it will only be necessary to prove that the remaining parts, 184 GEOMETRY. namely, the solids aBd-D,. eFh-H, are H equivalent. Since ABFE, aBFe, are parallelograms, the sides AE, ae, are E each equal to BF; hence, they are equal to each other; and taking away the common part Ae, there remains Aa=Ee. In the same manner it may A / a-,-, be proved that Dd =Hh. C To bring about the superposition of the two solids, eFh-f, a.B&-D, let the base eFh be placed on the equal bate aBd-the point e falling on a, the point h on d: the edges eE, h/, will then coincide with aA, dD, since all the edges are perpendicular to the same plane aBcd. Hence, the \two solids will coincide exactly with each other; consequentlyr, the oblique prism ABD-tH is equivalent to the right prism aBd-h. In the same manner, it may be shown that the oblique prism BCUD-It is equivalent to the right prism Bcd-h. But the two right prisms have been proved equal: hence, the two triangular prisms ABD-;, BCD-BU, being equivalent to equal right prisms, are equivalent to each other. Cor. Every triangular prism ABD-H is half the parallelopipedon AG, having the same triedral angle A, and the same edges AB, AD, AE. PROPOSITION VIII. THEOREM. If two parallelopipedons have a common lower base, and their upper bases in the same plane and between the same parallels, they are equivalent. Let the parallelopipedons AG, AL, have the common base ABCD, and their upper bases EG, IL, in the same plane, and between the same parallels EA;, HL; then will they be equivalent. There may be three cases, according as 1E is greater than, equal to, or less than -EF; but the demonstration, for each case, is the same. We will show, in the first place, that the triangular prisms.AIE-l, BKF-G are equal. Since EF and 1K are BOOK VII. 185 each equal to AB (B. I., H GM L P. 28), they are equal to each other. Add FI -to K each, and we have EI=FK:' and since the angle AEF D -- is equal to BFK (B. I., P. 20, c. 3); the triangle AtEI is equal to the triangle BFK (B. I., P. 5). Again, since El is equal to FI, and EEH equal and parallel to FG, the parallelogram EMI is equal to the parallelogram FL (B. I.,P. 28, C. 2): also, the parallelogram AHT is equal to the parallelogram CF (P. 6): hence, the three planes which include the polyedral angle at E are respectively equal to the three which include the polyedral angle at F, and being like placed, the triangular prism AI-E-H is equal to the triangular prism BKF- G (P. 5). But, if the triangular prism AIE-H be taken away from the solid AL, there will remain the parallelopipedon ABCD-iM; and if the equal triangular prism BKFT-G be taken away from the same solid, there will remain the parallelopipedon ABCD-ET; hence, the two parallelopipedons ABCD-MI, ABCD-I, are equivalent. PROPOSITION IX. THEOREM. Two parallelopipedons, having their lower bases equal,= and equal altitudes, are equivalent. Let the parallelopipedons AG, AL, have the common base ABCD, and equal altitudes; then will their upper bases, EFGfE, IKLM, be in the same plane; and the two parallelopipedons will be equivalent. For, let the edges FE, GU, be prolonged, as also, KL and IM; till, by their intersections, they form the parallelogram NOPQ, in the-plane of the upper bases: this parallelogram will be equal to either of the bases 1L, EG. For, the upper bases IL, EUG, being each equal to the common base A C, are equal to each other. But OP which is equal to FG, is also equal to KL, and ON is 186 GEOMETRY. equal to ZK,; being be Q p E G tween the same parallels: hence, the paral- N E F lelogram NVP is equal to IL or EG (B. I., P. K 28, c. 2). Now, if a third par- / allelopipedon be conceived, having for its lower base the paral- / lelogram ABCD, and for B its upper base NOPQ, this third parallelopipedon will be equivalent to the parallelopipedon A G, since they have the same lower base, and their upper bases lie in the same plane and between the same parallels, QG, NF (P. 8). For a like reason, this third parallelopipedon will also be equivalent to the parallelopipedon AL; hence, the two parallelopipedons AG, AL, which have equal bases:and equal altitudes, are equivalent. PROPOSITION X. THEOREM. Any parallelopipedon may be changed into an equivalent rectangular parallelopipedon having an equal altitude and an equivalent base. Let ABCD-H be any parallelopipedon. From the vertices A, H G B, C, D, draw AJ, BK, CL, DM,1 perpendicular to F the plane of the lower M base, and equal to the altitude of AG: there will I // thus be formed the paral- lelopipedon AL equiva- / / lent to AG (P. 9), and having its.lateral faces D AIL, JBL, &c., rectangles. A' B Now, if the base ABCID is a rectangle, AL will be a rectangular parallelopipedon BOOK VII. 187 equivalent to AG, and consequently, the parallelopipedon required. But if ABCD is not a rectangle, draw M Q L P A 0 and BN perpendicular to D (, and OQ and NP perpendicular to the base; we shall then have, a rectangular parallelopipedon ABNO-Q: for, by construction, the bases ABNO, and IK'PQ, are rectangles; so also, are the lateral faces, o.Ci i the edges AI, OQ, &c., being perpendicu- D lar to the plane of the base; hence, the solid AP is a rectangular parallellopipedon. But the two parallelopipedons AP, AL, may be conceived as having the same base ABKI, and the same altitude AO: hence, the parallelopipedon AG, which was at first changed into an equivalent parallelopipedon AL, is now changed into an equivalent rectangular parallelopipedon AP, having the same altitude Al, and a base ABNO equivalent to the base ABCD. PROPOSITION XI. THEOREM. Two rectangular parallelopipedons, which have equal bases, are to each other as their altitudes. Let the parallelopipedons AG, AL, have the common base BD, then will they be to each other as their altitudes AE AA. First. Suppose the altitudes AE, Al, to E H be to each other as two whole numbers, as 15 is to 8, for example. Divide AE F G into 15 equal parts, whereof Al will con- 0 m tain 8; and through x, y, z, &c., the points I M pf division, pass planes parallel to the common base. These planes will divideK l the solid AG into 15 parallelopipedons, all T equal to each other, because they have A D equal bases and equal altitudes-equal B bases, since every section KLMIJ, parallel to the base ABCD, is equal to that base (P. 2), equal alti 188 GEOMETRY. tudes, because the altitudes are the equal divisions, Ax, xy, yz, &c. But of those 15 E I equal parallelopipedons, 8 are contained in AL; hence, the solid AG is to the F solid AL as 15 is to 8, or generally, as O2 the altitude AE is to the altitude AI. Second. If the ratio of AE to AI cannot be expressed exactly in numbers, Hi it may still be shown, that we shall A D have B C solidAG: solid AL:: AE: AI. For, if this proportion is not correct, suppose we have, sol. AG: sol. AL:: AE: AO greater than Al. DIivide AE into equal parts, such that each shall be less than OI; there will be at least one point of division m, between 0 and 1. Let P denote the parallelopipedon, whose base is ABCD, and altitude Am; since the altitudes AE, Am, are to each other as two whole numbers, we have sol. AG: P:: AE: An. But by hypothesis, we have sol. AG: ol. AL:: AE: AO; therefore (B. ii., P. 4), sol. AL: P::AO Am. But AO is greater than Am; hence, if the proportion is correct, the solid AL must be greater than P. On the contrary, however, it is less: therefore, A O cannot be greater than- Al. By the same mode of reasoning, it may be shown that the fourth term cannot be less than AI; therefore, it is equal to AI: hence, rectangular parallelopipedons having equal bases, are to each other as their altitudes. BOOK VII. 189 PROPOSITION XII. THEOREM. Two rectangular 2arallelop2ipedons, having equal altitudes, are to each other as- their bases. Let the parallelopipedons AG, AK, have the same altitude,4E; then will they be to each other as their bases For having placed the E AN For, having placed the I E II two solids by the side of each other, as- the \figure L represents, prolong the plane NFIL 0 till it meets the plane x DCGH in PQ; we thus - have a third parallelopipe- Z V don AQ, which may be compared with each of the I 1 parallelopipedons AG,: AK. M D _ — The two solids A G,'A QI having the same base ADHE N are to each other as their altitudes AB, A 0: in like B C manneyr, the two solids A Q, AK; having the same base A OLE, are to each other as their altitudes AD, AM. Hence, we have sol. AG: sol. AQ:: AB AO; also, sol. 4Q: sol. AK:: AD: AIJ. Multiplying together the corresponding terms of these proportions, and omitting, in the result, the common multiplier sol. A Q; we. shall have sol. AG so1. 1AK, ARX AD o AOx AA But ABXAD represents the area of the base ABCD; and AOXAM represents the area of the base AMINO; hence, two rectangular' parallelopipedons having equal altitudes, are to each other as their bases. 190 GEOMETRY. PROPOSITION XIII. THEOREM. Any two rectangular parallelopipedons are to each other as the products of their bases by their altitudes; that is, as the products of their three dimensions. Having placed the two I E solids AG, AZ, so that their faces have the cornmon angle BAE, produce the planes necessary for F i completing the third par- Y — l - allelopipedon A,; which Z z will have an equal altitude with the parallelopipedon AG. By the last proposi- MI_.-I-. D. tion, we have \ sol. AG sol. AK:: N P ABCD AMNO. But the two parallelopipe- B C dons AKI, AZ, having the same base NA, are to each other as their altitudes AE, AX; hence, we have, sol. AK: sol. AZ:: AE: AX. Multiplying together the corresponding terms of these proportions, and omitting in the result the common multiplier sol. AK; we shall have, sol. AG so. AZ:: ABCDI)XAE AMNOxAX. Instead of the bases ABCD and AMNO, put ABX AD and AOxAAM; and we shall have, sol. A G sol. AZ:: ABXAD xAE: AOxAMixAX: hence, any two rectangular parallelopipedons are to each other, as the products of their three dimensions. Scholium 1. The magnitude of a solid, its volume oI extent, is called its solidity; and this word is exclusively employed to designate the measure of a solid: thus, we say the solidity of a rectangular parallelopipedon is equal to the product of its base by its altitude, or to the product of its three dimensions. BOOK VII. 191 In order to comprehend the nature of this measurement, it is necessary to consider, that the number of linear units in one dimension of the base multiplied by the number of linear units in the other dimension of the base, will give the number of superficial units in the base of the parallelopipedon (B. IV., P.4, S.) For each unit in height, there are evidently, as many solid units as there are superficial units in the base. Therefore, the number of superficial units in the base multiplied by the number of linear units in the altitude, gives the number of solid units in the parallelopipedon. If then, we assume as the unit of measure, the cube whose edge is equal to the linear unit, the solidity will be expressed numerically, by the number of times which the solid contains its unit of measure. Scholium 2. As the three dimensions of the cube are equal, if the edge is 1, the solidity is l XX 1=1: if the edge is 2, the solidity is 2X2X2=8; if the edge is 3, the solidity is 3X3X3=27; and so on. Hence, if the edges of a series of cubes are to each other as the numbers 1, 2, 3, &c., the cubes themselves, or their solidities, are as the numbers 1, 8, 27, &c. Hence it is, that in arithmetic, the cube of a number is the name given to a product which results from three equal factors. If it were proposed to find a cube double of a given cube, we should have, unity to the cube-root of 2, as the edge of the given cube to the edge of the required cube. Now, by a geometrical construction, it is easy to find the square root of 2; but the cube-root of it cannot be found, by the operations of elementary geometry, which are limited to the employment of the straight line and circle. Owing to the difficulty of the solution, the problem of the duplication of the cube became celebrated among the ancient geometers, as well as that of the trisection of an angle, which is a problem nearly of the same species. The solutions of these problem.s have, however, long since been discovered; and though less simple than the constructions of elementary geometry, they are not, on that account, less rigorous or less satisfactory. 192 GEOMETRY. PROPOSITION XIV. THEOREM. The solidity of a paraZlelopipedon, and generally of any prism, is equal to the product of its base by its altitude. -First. Any parallelopipedon is equivalent to a rectangular parallelopipedon, having an, equal altitude and an equivalent base (P. 10). But, the solidity of a rectangular parallelopipedon is equal to its base multiplied by its height; hence, the solidity of any parallelopipedon is equal to the product of its base by its altitude. Second. Any triangular prism is half a parallelopipedon so constructed as to have an equal altitude and a double base (P. 7). But the solidity of the parallelopipedon is equal to its base multiplied by its altitude; hence, that of the triangular prism is also equal to the product of its base, which is half that of the parallelopipedon, multiplied into its altitude. Third. Any prism may be divided into as many triangular prisms of the same altitude, as there are triangles formed by drawing diagonals from a common vBrtex in the polygon which constitutes its base. But the solidity of each triangular prism is equal to its base multiplied by its altitude; and since the altitudes are equal, it follows that the sum of all the triangular prisms must be equal to the sum of all the triangles which constitute their bases, multiplied by the common altitude. Hence, the solidity of any polygonal prism, is equal to the product of its base by its altitude. Cor. Since any two prisms are to each other as the products of their bases and altitudes, if the altitudes be equal, they will be to each other as their bases simply; hence, two prisms of the same altitude are to each other as their bases. For a like reason, two prisms having equivalent bases are to each other as their altitudes. BOOK VII. 193 PROPOSITION XV. THEOREM. Two trziangtlar pyramids, Ihaving equivalent bases and, equal altitudes, are equivalent,.or equal in solidilty. Let S-ABC, S-abc, be two such pyramids; let their eqlLivalent bases ABC, abc, be situated in the sa'ne plane, and let AT be their common altitude: then will they be equivalent. T____ S / i/ X- 1/ d. B b For, if these pyramids are not equivalent, let S-abc be the smaller; and suppose Aa to be the altitude of a prism which, having ABC for its base, is equal to their difference. Divide the altitude AT into equal parts Ax, xy, yz, &c., each less than Aa, and let le denote one of those l)arts; through the points of division pass planes parallel to the planes of the bases; the correspondilng sections formed by these planes in the two pyramids are respectively equivalent, namely, DE~F to def GtlI to ghi, &c. (P. 3, c. 2). This being done, upon the triangles ABC, DEF, GHII &c., taken as bases, construct exterior prisms having for edges the parts AD, _DG, GIA, &c., of the edge SA; in like manner, on bases def; ghi, klm., &c., in the second pyramid, construct interior prisms, having for edges the correspond13 194 GEOMETRY. T S /P P~f yN G0 x1D A E ing parts of Sa. It is plain, that the sum of all the e'xterior prisms of the pyramid S-ABC is greater than this pyramid; and also, that the sum of all the interior prisms of the pyramid S-abc is less than this pyramidifference the difference, between tthe sprism of all the exterior pisms of one pyramid, and the sum of all the interior prisms of the other, is greater than the difference between the two pyramids themselves. Now, beginning with the bases, the second exterior prism FEi~D-C, is equivalent to the first interior prism efd-a, because they have the same altitude k, and their bases FD, efd are equivalent; for a like reason, the third exterior prism HIG-Ef, and the second interior prism ltig-d are equivalent; the fourth exterior and the third interior; and so on, to the last in each series. Hence, all the exterior prisms of the pyramid S-ABC, excepting the first prism BCA-D, have equivalent corresponding ones in the interior prisms of the pyramid S-abc: hence, the prism BCA-D, is the difference between the s-am of all the exterior prisms of the pyramid S-ABq, and the sum of the interior prisms of the pyramid S-abc. But the difference between these two sets of prisms has already been proved to be greater than that between the two pyramids; which latter difference We supposed to be equal to the prism BCA-ca: hence, the BOOK VII. 195 prism BCA-D, should be greater than the prism BCA-a. But in reality it is less; for they have the same base ABC, and the altitude Ax of the first is less than the altitude Aa of the second. Hence, the supposed inequality between the two pyramids cannot exist; therefore, the two pyramids S-ABC; S-abc, having equal altitudes and equiva lent bases, are themselves equivalent. PROPOSITION XVI. THEOREM. Every triangular prism may be divided into three equivalent triangular pyramids. Let ABC-DEF be a triangular prism; then may it be divided into three equivalent triangular pyramids. Cut off the pyramid F-ABC E D from the prism, by the plane FA C; there will remain the solid F-AC'DE, which may be' " F considered as a quadrangular pyramid, whose vertex is F, and whose base is the parallelogram ACDE. Draw the diagonal CE; and pass the plane FCE, which will cut the quadrangular pyramid into two triangular pyramids F-A CE, F- CDE. These two triangular pyramids have for their common altitude the perpendicular let fall from F, on the plane A CDE; they have equal bases; for the triangles A CE, CDE, are halves of the same parallelogram; hence, the two pyramids F-ACE, F- CDE, are equivalent (P. 15). But the pyramid F-CDE, and the pyramid F-ABC, have equal bases ABC, DEF; they have also the same altitude, namely, the distance between the parallel planes ABC,; DEF; hence, the two pyramids are equivalent. Now, the pyramid F- CDEL has already been proved equivalent to F-ACE; hence, the three pyramids F-AB', F-CDE, F-ACE, which compose the prism, are all equivalent. Cor. 1.' Every triangular pyramid is a third part of a 196 GEOMETRY. triangular prism, which has an equivalent base and an equal altitude. Cor. 2. The solidity of a triangular pyramid is equal to a third part of the product of its base by, its altitude. PROPOSITION XVII. THEOREM. The solidity of every pyramid is equal to a third part of the product of its base by its altitude. Let 8-ABCDE be a pyramid: then will its solidity be equal to one-third of the product of the base ABCUDE by the altitude SO. Pass, the planes SEB, SEC, through S the vertex S, and the diagonals EB, ECU; the polygonal pyramid S-ABUDE will then be divided into several triangular,e pyramids, all having the same altitude SO. But each of these pyramids is mea- bk sured by the product of its base ABE, F BCE, CDE, by a third part of its alti- A// / tude SO (P. 16, c. 2); hence, the sum of these triangular pyramids, or the polygonal pyramid S-ABC-DE is measured by the sum of the triangles ABE, BCE, CDE, or the polygon ABCDE multiplied by one-third of SO; hence, every pyramid is measured by a third part of the product of its base by its altitude. Cor. 1. Every pyramid is the third part of a prism which has the same base and the same altitude. Cor. 2. Two pyramids having the same altitude are to each other as their bases. Cor. 3. Two pyramids having equivalent bases are to each other as their altitudes. Cor. 4. Pyramids are to each other as the products of their bases by their altitudes. Scholium. The solidity of any polyedral body may be computed, by dividing the body into pyramids; and this BOOK VII. 197 division may be accomplished in various ways. One of the simplest is to pass all the planes of division through the vertex of the same polyedral angle; in that case, there will be formed as many pyramids as the polyedron has faces, less those faces which bound the polyedral angle whence the planes of division proceed. PROPOSITION XVIII. THEOREM. Tht solidity of the frustum of a pyramid is equal to that of three pyramids having for their common altitude the altitude of the frustum, and for bases the lower base of the frustum, the upper base, and a mean proportional between the two bases. Let ABCDE-e be the frustum of a pyramid: then will its solidity be equal to that of three pyramids having the common altitude of the frustum, and for bases the polygons ABCDE, abcde, and a mean proportional between them. Let T-FGH be a triangular pyramid having the same altitude, and an equivalent base with the pyramid S-ABCDE. These two pyramids are equivalent (P. 17, c. 3). Now, if we regard their bases as situated in the S T same plane; the plane of the section abed, will form a - in the triangular pyramid a / section ftgh, at the same F -- II distance above the common A C plane of the bases; and, B G therefore, the section fgh will be to the section abcde, as the base FGH is to the base ABCDE (P. 3, c. 1): and since the bases are equivalent, the sections will also be equivalent. Hence, the pyramids S-abede, TI-fgh will be equivalent (P. 17, C. 3). If these be taken from the entire pyramids S-ABCDE, T-FGIH, the frustums ABCDE-e, FGH-h which remain, will be equivalent: hence, if the proposition is true, in the single cage of the frustum of a triangular pyramid, it is true in every other. 198 GEOMETRY. Let FGH-h be the frustum of a triangular pyramid. Through the three points, F, g, H, pass the plane EgIH; it cuts off from the frustum the triangular pyramid g-FGH. This " pyramid has for its base the lower base FGH of the frustum; its alti- F tude is equal to that of the frustumna, because the vertex g lies in the K / plane of the upper base fgh. G This pyramid being cut off,' there remains the quadrangular pyramid g-fhlHti whose vertex is g, and base fhHF. Pass the plane gfH through the three points f g, H; it divides the quadrangular pyramid into two triangular pyramids g-fF4, g-fhf. The latter has for its base the upper base gfh of the frustum; and for its altitude, the altitude of the frustumn, because its vertex H lies in the lower base. Thus we already know two of the three pyramids which compose the frustum. It remains to examine the third pyramid g-FfH. Now, if gK be drawn parallel to fF, and if we conceive a new pyramid K-fFH, having iK for its vertex and fFH for its base, these two pyramids have the same base HfF; they also have the same altitude, because their vertices g and K lie in the line gK, parallel to Ff, and consequently, parallel' to the plane of the base: hence, these pyramids are equivalent (P. 17, c. 3). But the pyramid lKJ-fFH may be regarded as having FKH for its base, and its vertex at f: its altitude is then the same as that of the frustum. We are now to show that the base FKH is a mean proportional between the bases FGH and fgh. The triangles FHl, fgh, have the angle F=f; hence (B. Iv., P. 24), -PI1K: fgh:: FKXFH: fg X fh; but because of the parallels, FK=fg, FPK: fgh:: FH: fh. We have also, FH G: FEK:: FG: FK, or fg. But the similar triangles GIH, fgh, give BOOK VII. 199 FG: fg:: FH: fh; hence, FGL::: F PITK: fgh; that is, the base FHK is a, mean proportional between the two bases FGI, fgh. Hence, the solidity of the frustum of a triangular pyramid is equal to that of three pyramids whose common altitude is that of the frustum, and whose bases are the lower base of the frustum, the upper base, and a mean proportional between the two bases. PROPOSITION XIX. THEOREM. Similar triangular prisms are to each other as the cubes of their homologous edges. Let CBD-P, cbd-p, be two similar triangular prisms, and BC,. bc, two homologous edges: then will the prism CBD-P be to the prism cbd-p, as BC to b3. For, since the prisms are p similar, the homologous angles B and b are equal, A p and the faces which 1bound' -. a a them are similar (D. 16). Hence, if these triedral angles.. be applied, the one to the C c B c other, the angles cbd will coincide with CBD, the edge ba with BA, and the prism cbd-p will take the position Bcd-p. From A draw All perpendicular to the common base of the prisms: then will the plane BAH be perpendicular to the plane of the common base (B. VI., P. 16). Through a, in the plane BAH, draw aht perpendicular to BLH: then will ah also be perpendicular to the base BDC (B. VI., P. 17); and AH, ah will be the altitudes of the two prisms. Since the bases CBD, cbd, are similar, we have (B. IV., P. 25), base CBD: base cbd CB2 Now, because of the similar triangles ABE, aBh, and of the similar parallelograms A, ac, we have AH: ah:: AB: ab:: GB: cb; hence, multiplying together the corresponding terms, we have base CBDXA-L: base cbd X ah:: - B3: b3. 200 GEOMETR1Y. But the solidity of a prism is equal to the base multiplied by the altitude (P. 14); hence, ~ — 3 prism BCID-P: prism bcd-p:: BC3 be, or as the cubes of any other of their homologous edges. Cor. Whatever be the bases of similar prisms, the prisms are to each other as the cubes of their homologous edges. For, since the prisms are similar, their bases are similar polygons (D. 16); and these similar polygons may each be divided into the same number of similar triangles, similarly placed (B. IV., P. 26); therefore, each prism may be divided into the same number of triangular prisms, having their faces similar and like placed; hence, their polycdral angles are equal (B. VI., P. 21, s. 2); and consequently, the triangular prisms are similar (D. 16). But these triangular prisms are to each other as the cubes of their homologous edges, and being, like parts of the polygonal prisms, their'sums, that is, the polygonal prisms, are to each other as the cubes of their homologous edges. PROPOSITION XX. THEOREM. Two similar pyramids are to each other as the cubes of their homologous eclges. For, since the pyramids are similar, the homologous polyedral angles at the vertices are equal (D. 16). tIence, the polyedral angles at the vertices may be made to coincide, or the two pyramids may be so placed as to have the polyedral angle S common. In that position the bases ABCDE, S abcde, are parallel; for, the homologous faces being similar, the angle Sab is equal to SAB, and Sbc to SBC; hence, the plane ABC, is parallel to the plane abe (B. VI., / P. 13). This being proved, let SO be drawn from the vertex S, perpendicu- E D lar to the plane ABC, and let o, be the A C point where this perpendicular pierces the plane abc: from what has already been, B BOOK VII. 201 shown, we have (P. 3), SO: So:S: A: Sa:: AB: ab; and consequently, ~ SO: So:: AB: ab. But the bases ABCDE, abcde, being similar figures, we have (B. Iv., P. 27), ABCDE: abcde:: AB ab2; multiply the corresponding terms of these two proportions, there results,, ABCDEEx-SO: abcde x So: AB ab3. Now, ABCDE x LSO measures the solidity of the pyramid S-ABCDIE, and abcdeX So measures that of the pyramid S-abcde (P. 17); hence, two similar pyramids are to each other as the cubes of their homologous edges. GENERAL SCHOLIUMS. 1. The chief propositions of this Book relating to the solidity of polyedrons, may be expressed in algebraical terms, and so recapitulated in the briefest manner possible. 2. Let B represent the base of a prism; t its altitude: then, solidity of prism=B X.'! 3. Let B represent the base of a pyramid; II its altitude: then, solidity of pyramid=B X ~-H. 4. Let H represent the altitude of the fhustum qf a pyramid, having the parallel bases A and P; i/AXB is the mean proportional between those bases; then solidity of frustum=~-H(A +B+ WfA x B.) 5. In fine, let P and p represent the solidities of two similar prisms or pyramids; A and a, two homologous edges: tHen, P: p:: A3 a3. BOOK VIII. THE THREE ROUND BODIES. DEFINITIONS. 1. A CYLINDER is a solid which may be generated by the revolution of a rectangle ABCD, turning about the immovable side AB. In this movement, the sides A D, BC, p continuing always perpendicular to AB, A describe the equal circles DHP, CGQ, which are called the bases of the cylinder; the side CD, describing, at the same I K L time, the convex surface. The immovable line AB is called F the, axis of the cylinder. C Every section MNHKL, made in the cylinder, by a plane, at right angles to the axis, is a circle equal to either of the bases. For, whilst the rectangle ABCD turns about AB, the line KI, perpendicular to AB, describes a circle, equal to the base, and this circle is nothing else than the section made by a plane, perpendicular- to the axis at the point I. Every section QPIG, made by a plane passing through the axis, is a rectangle double the generating rectangle ABCD. 2. SIMILAR CYLINDERS are those whose axes are proportional to the radii of their bases: hence, they are generated by similar rectangles (B. IV., D. 1). BOOK VIII. 203 3. If, in the circle ABCDE, which N K forms the base of a cylinder, a polygon F ABCDE be inscribed, and a right prism, constructed on this base, and equal G _o in altitude to the cylinder; then, the prism is said to be inscribed in the cylinder, and the cylinder to be circumscribed about the prism. The edges AF, BG, CH,1 &c., of the prism, being perpendicular to the plane A D of the base, are contained in the convex surface of the cylinder; hence, the prism and the cylinder touch one another along these edges. 4. In like manner, if ABCD is a. QI polygon, circumscribed about the base F of a cylinder, a right prism constructed on this base, and equal in altitude to the cylinder, is said to be circumscribed about the cylinder, and the cylinder to be inscribed in the prism. Let M, N; &c., be the points of con- A O tact in the sides AB, BC, &c.; and. C through the points 1 N, &c., let X, M NY; &c., be drawn perpendicular to the plane of the base: these perpendiculars will then lie both in the surface of the cylinder, and in that of the circums6ribed prism; hence, they will be their lines of contact. 5. A CONE is a solid which may be generated by the revolution of a right-angled triangle SAB, turning about the immovable side SA. S In this movement, the side AB describes a circle BDC~F, called the base of the cone; the hypothenuse SB describes H the convex surface of the cone. The point S is called the vertex of the cone, SA the axis, or the altitude, and SB the slant height. C A B Every section HKLFl; made by a D 204 GEOMETRY. plane, at right angles to the axis, is a circle. Every section EDS, made S by a plane passing through the axis, is an isosceles triangle, double the generating triangle SAB. F di 6. If, from the cone S-C1DB, the cone S-FlIA be cut off by a plane i' parallel to the base, the remaining c solid CFHIB is called a truncated cone, or the frustum of a cone. The frustum may be generated by the revolution of the trapezoid ABHG, turning about the side AG. The immovable line A G is called the axis, or altitude of the frustum, the circles BDC, HFI~ are its bases, and BH its slant height. 7. SIMILAR. CONES are those whose axes are proportional to the radii of their bases: hence, they are generated by similar right-angled triangles (B. IV., D. 1). 8. If, in the circle ABCDE, which S forms the base of a cone, any polygon ABCDE is inscribed, and from the vertices A, B, q D, E, lines are drawn to S. the vertex of the cone, these lines may be regarded as the D edges of a pyramid whose base is A the polygon ABCDE and vertex S. The edges of this pyramid are in the B convex surface of the cone, and the pyramid is said to be inscribed in the cone. The cone is also said to be circumscribed about the pyramid. 9. The SPHERE is a solid terminated by a curved surface, all the points of which F G are equally distant from a point within, called the centre. The sphere may be generated by the revolution of a semicircle DALE, about its diameter DE: for, the surface described in this movement, E BOOK VIII. 205 by the semicircumference DAE, will have all its points equally distant from its centre C. 10. Whilst the semicircle DAE, revolving round its diameter DE, describes the sphere, any circular sector, as DCF, or fCA, describes a solid, called a spherical sector. 11. The radius of a sphere is a straight line drawn from the centre to any point of the surface; the diameter or axis is a line passing through the centre, and terminated, on both sides, by the surface. All the radii of a sphere are equal; all the diameters are equal, and each is double the radius. 12. It will be shown (P. 7,) that every section of a sphere, made by a plane, is a circle: this granted, a great circle is a section which passes through the centre; a small circle, is one which does not pass through the centre. 13. A plane is tangent to a sphere, when it has but one point in common with the surface. 14. A zone is the portion of the surface of the sphere included between two parallel circles, which form its bases. If the plane of one of these circles becomes tangent to the sphere, the zone will have only a single base. 15. A slhlerYical segment is the portion of the solid sphere, included between two parallel circles which form its bases. If the plane of one of these circles becomes tangent to the sphere, the segment will have only a single base. 16. The Zllitudec of a zone, or of a segment, is thle distance between the pl)anes of the two parallel circles, which form the bases of the zone or segment. 17. The Cylinder, the Cone, and the Sphere, are the dtree routiLd bodies treated of in the Elements of Geometry. 206 GEOMETRY. PROPOSITION I. THEOREM. The convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude. Let CA be the radius of the base of a cylinder, and H its altitude; denote the circumference whose radius is CA by circ. CA: then will the convex surface of the cylinder be equal to circ. CAXH. Inscribe in the base of the cylinder any regular polygon, F BDEFGA, and construct on this polygon a right prism having its altitude equal to A C 11, the altitude of the cylinder: this prism will be inscribed in the cylinder. The B D convex surface of the prism is equal to the perimeter of the polygon, multiplied by the altitude 1T (B. VII., P. 1). Let now the arcs which subtend the sides of the polygon be continually bisected, and the number. of sides of the polygon continually increased: the limit of the perimeter of the polygon is circ. CA (B. 5, P. 12, s. 2), and the limit of the convex surface of the prism is the convex surface of the cylinder. But the convex surface of the prism is always equal to the perimeter of its base multiplied by HI; hence, the convex surface of the cylinder is equal to tlze circumference of its base multiplied by its altitude. PROPOSITION II. THEOREM. The solidity of a cylinder is equal to the product of its base by its altitude. Let CA be the radius of the base of the cylinder, and H the altitude. Let the circle whose radius is CA be denoted by area CA: then will the solidity of the cylinder be equal to area C'Ax H. BOOK VIII. 207 For, inscribe in the base of the cylinder any regular G polygon BDEFGA, and construct on this polygon a right prism having its altitude equal A E to R, the altitude of the cylinder: this prism will be inscribed in the cvlinder. The solidity of this prism will be B equal to the area of the polygon multiplied by the altitude H (B. VII., P. 14). Let now the number of sides of the polygon be con tinually increased, by bisecting the arcs; the solidity of each new prism will still be equal to its base multiplied by its altitude: the limit of the polygon is the area CA, and the limit of the prisms, the circumscribed cylinder. But the solidity of each new prism is equal to the base multiplied by the altitude: therefore, the solidity of the cylinder is equal to the product of its base by its altitude. Cor. 1. Cylinders of equal altitudes are to each other as their bases; and cylinders of equal bases are to each other as their altitudes. Cor. 2. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases. For, the bases are as the squares of their radii (B. V., P. 13); and the cylinders being similar, the radii of their bases are to each other as their altitudes (D. 2); hence, the bases are as the squares of the altitudes; therefore, the bases multiplied by the altitudes, or the cylinders themselves, are as the cubes of the altitudes. Scholium. Let R denote the radius of a cylinder's base, and H the altitude; then we shall have, surface of base=nX R2, convex surface= 2i -X R X1 solidity= r X RH.I 208 GEOMETRY. PROPOSITION III. THEOREM. The convex smfcace of a cone is equal to the circumference of its base, mulntiplied by half the slant height. Let the circle ABCD be the base of a cone, S the vertex, SO the altitude, and SA the slant height: then will the convex surface be equal to circ. OAX -SA. For, inscribe in the base of the cone any regular polygon AB3CD, and on this polygon as a base conceive a right pyramid to be constructed, A D having S for its vertex: this 0 pyramid will be inscribed in tile cole. From S, draw SG perpendicular to one of the sides of the polygon. The convex surface of the inscribed pyramid is equal to the perimeter of the polygon,which forms its base, multiplied by half the slant height SG (B. VIr., P. 4). Let now the number of sides of the inscribed polygon be continuaily increased, by bisecting the arcs: the limit of the perimeters of the polygons is cire. OA; the limit of the slant heIigllt of the pyramids is the slant height of the cone, and the limit of their surfaces, is the convex surface of the circumscribed cone. But the convex surface of each new pyramnid is equal to the perimeter of the base multiplied by half the slant height (B. VII., P. 4); hence, the convex.. szfytce of the cone is equal to the circumference of its base.mZultijlied by half its slant height. Scholtlm. Let L denote the slant height, and R the radius of the base: then, convex surface=2,rX RX 2L=rX R X L. BOOK VIII. 209 PROPOSITION IV. THEOREM. The convex surface of the frustum of a cone is equal to its slant height, multiplied by half the sum of the circumferences of its bases. Let BIA-DE be a frustum of a cone: then will, convex surface-ADX (circ. OA+circ. CD.) For, inscribe in the bases of the frustum two regular poly- E gons of the same number of sides, and having their sides L - K I parallel, each to each. The lines joining the vertices of the corresponding angles may be regarded 0 d i as the edges of the frustum of a right pyramid inscribed in the frustum of the cone. The convex surface of the frustum of the pyramid is equal to half the sum of the perimeters of its bases multiplied by the slant height fh (B. vII., P. 4, c.) Let the number of sides of the inscribed polygons be continually increased by bisecting the arcs: the limits of the perimeters of the polygons are circ. OA and circ. CD; the limit of the slant height is the slant height of the frustum, and the limit of the convex surface, the convex surface of the frustum; hence, the convex sutface of the frustum of a cone is equal to its slant height multiplied by half the sum of the circumferences of its bases. Cor. Through 1, the middle point of AD, draw MlKL parallel to AB, also li, Dd, parallel to CO. Then, since Al, lD, are equal, Ai, id,.are also equal (B. Iv., P. 15, c. 2): hence, Kl is equal to ~(OA+CD). But since the circumferences of circles are to each other as their radii (B. v., P. 13), circ. Kl=1(circ. OA+circ. CD); therefore, the convex surface of the frustum of a cone is equal to its slant height multiplied by the circumference of a section at equal distances from the two bases. 14 210 GEOMETRY. Scholium 1. If from the middle point I and the two extremities A and D, of a line AD, lying wholly on one side of the line OC the perpendiculars DC, I~, and A 0, be drawn, and then B0 d the line AD be revolved around 0C, we shall have I surf. described by AD=-ADx (circ. OA+circ. CD)); that is, =ADX cir. 1(1. For, it is evident that the surface described by AD is that of the frustum of a cone, having OA and CD for the radii of its bases. Scholium 2. The measure found above applies equally to the case when the point D falls at C, and the surface becomes that of a cone; and to the case in which AD becomes parallel to OC, and the surface becomes that of a cylinder. In the first case, CD is nothing: in the second, it is equal to OA. PROPOSITION V. THEOREM. The solicity of a cone is equal to its base rnultiplied by a third of its altitude. Let SO be the altitude of a cone, OA the radius of its base, and let the area of the base be designated by area QOA; then will, solidity-area OA X ~-SO. Inscribe in the base of the cone any regular polygon ABDEF, and join the vertices A, B, C,,&c., S with the vertex S of the cone: then will there be inscribed in A the cone a right pyramid having the same vertex as the cone, and having for its base the polygon ABDEE The solidity of this pyramid will be equal to its base multiplied by one-third of its altitude (B. vII., P. 17). B O O K VIII. 211 Let the arcs be bisected and the number of sides of the polygon be continually increased: the limit of the polygons will be the area OA, and the limit of the pyramids will be the cone whose vertex is S: hence, the solidity of the cone is equal to its base multiplied by a third of it.> altitude. Cor. 1. A cone is the third of a cylinder having the same base and the same altitude; whence it follows, 1. That cones of equal altitudes are to each other as their bases; 2.- That cones of equal bases are to each other as their altitudes; 3. That similar cones are as the cubes of the diameters of their bases, or as the cubes of their altitudes. Cor. 2. The solidity of a cone is equivalent to the solidity of a pyramid having an equivalent base and the same altitude. Scholium. Let R be the radius of a cone's base, H its altitude; then, solidity=- X R 2x H. PROPOSITION VI. THEOREM. The solidity of the frustum of a cone is equivalent to the sum of the solidities of three cones whose common altitude is the altitude of the frustum, and whose bases are, the lower base of the frustum, the upper base of the frustum, and a mean prooortional between them. Let AEB- CD be the frustum of a cone, and OP its altitude; then will its solidity be equivalent to D P C 3,~x OPx(O6B'+PCF + OB XPC). For, inscribe in the lower and upper bases two regular polygons having the same number of sides, and having their sides parallel, B each to each. Join the vertices of the corresponding angles, and there E 212 GEOMETRY. will then be inscribed in the frustum of the cone, the frustum of a regular pyramid. The D solidity of the frustum of this pyramid will be equivalent to tree pyramids having the comrnlon altitude of the frustum, and for bases, the lower base of the A B frustum, the upper base of the frustum, and a mean proportional E between them (B. VII., P. 18). Let the number of sides of the inscribed polygons be continually increased, by bisecting the arcs: the limits of the polygons will be, area OB and area PC; and the limit of the frustums of the pyramids will be the frustum of the cone: the expression for the solidity will then become: of the first pyramid, OP x OBx x, of the second 31OPx P xr, of the third OP X OB X PC X:. hence, the solidity of the frustum of the cone is equivalent to 1rX OPX (0B2+PC2+ OBXPC.) PROPOSITION VII. THEOREM. Every section of a sphere, made by a plane, is a circle. Let AMlIB be any section made by a plane, in the sphere whose centre is C: then will it be a circle. For, from the point C, draw CO perpendicular to the plane D AMB; and different lines CM, A1 CM; to different points of the 1 0 curve AMB, which terminates the section. C The oblique lines CM, CM, CA, are equal, being radii of the sphere; hence, they pierce the plane AIB at equal distances from the perpendicular CO (B. VI., P.5, c.); therefore, all the lines O0M OM OB, are BOCK VIII. 213 equal; consequently, the section AMB is a circle, whose centre is O. Cor. 1. If the section pass through the centre of the sphere, its radius will be the radius of the sphere; hence, all great circles are equal. Cor. 2. Two great circles always bisect each other; for their common intersection, passing through the centre, is a diameter. Cor. 3. Every great circle divides the sphere and its surface into two equal parts: for, if the two parts were separated and afterwards placed on the common base, with their convexities turned the same way, the two surfaces would exactly coincide, no point of the one being nearer the centre than any point of the other. Cor. 4. The centre of a small circle, and that of the sphere, are in the same straight line, perpendicular to the plane of the small circle. Cor. 5. The radius of any small circle is less than the radius of the sphere; and the further its centre is removed from the centre of the sphere, the less is its radius: for, the greater CO is, the less is the chord AB, the diameter of the small circle AMB. Cor. 6. An arc of a great circle may always be made to pass through any two given points of the surface of the sphere: for, the two given points, and the centre of the sphere make three points, which determine the position of a plane. But if the two given points were at the extremities of a diameter, these two points and the centre would then lie in one straight line, and an infinite number of great circles might be made to pass through the two given points. Cor. 7. The distance between any two points on the surface of a sphere is less when measured on the arc of a great circle than when measured on the arc of a small circle. For, let A and B be any two points on the surface of a sphere, let ADB be the are of a great circle, and AMB 214 GEOMETRY. the arc of a small circle passing through them, and AB the common chord. Then, since the radius CA is greater than the radius OA, the arc ADB is less than the arc AMB (B. V., P. 17). PROPOSITION VIII. THEOREM. Every plane perpendclicular to a radius at its extremity is tangent to the sphere. Let FAG be a plane perpendicular to the radius OA, at its extremity A: then will it be tangent to the sphere. For, assuming any other point / M1 in this plane, draw OA, 02f: then the angle OAM is a right angle, and hence, the distance O /2f is greater than OA: therefore, the point M lies without the sphere; o hence, the plane FAG, can have no point but A common to it and the surface of the sphere; consequently, it is a tangent plane (D. 13). Scholium. In the same way it may be shown, that two spheres are tangent the one to the other, when the distance between their centres is equal to the sum or the difference of their radii; in which case, the centres and the point of contact lie in the same straight line. PROPOSITION IX. LEMMA. -If a regular semi-polygon be revolved about a line passing through the centre and the vertices of two opposite angles, the sufface described by ilts perimeter will be equal to the axis multiplied by the circumference of the inscribed circle. Let the regular semi-polygon ABCDEF, be revolved about the line AF as an axis: then will the surface described by its perimeter be equal to AF multiplied by the circumference of the inscribed circle. BOOK VIII. 215 For, from E and D, the extremities of one of the equal sides, let fall the E perpendiculars E/IJ DI, on the axis AF; H and from the centre 0, draw ON per- M pendicular to the side DE: ON will be DN I the radius of the inscribed circle (B. V., P. 2). Now, the surface described in the revolution, by any one side of the reg- C P ular polygon, as DE, has been shown to be equal to DE X circ. NJ1 (P. 4, s. 1). B Q But since the triangles EDK, ONM, are A similar (B. IV., P. 21), ED: El or III:: ON: N:: circ. ON: circ. NM; hence, EDX circ. NA= HIX circ. ON; and since the same may be shown for each of the other sides, it is plain that the surface described by the entire perimeter is equal to (FH+HI~+P+PQ+QA)Xcrc. O=VAF Xcirc. ON Cor. The surface described by any portion of the perimeter, as EDC, is equal to the distance between the two perpendiculars let fall from its extremities on the axis, multiplied by the circumference of the inscribed circle. For, the surface described by DE is equal to HtIxcirc. ON, and the surface described by DC is equal to IPXcirc. ON: hence, the surface described by EYD+DC is equal to (HI+IP)Xcirc. ON, or equal to IHPXcirc. ON3 PROPOSITION X. THEOREM. The surface of a sphere is equal to the product of its diameter by the circumference of a great circle. Let ABCDE be a semicircle. Inscribe in it a regular semi-polygon, and from the centre 0 draw OF perpendicular to one of the sides. Let the semicircle and the semi-polygon be revolved about the common axis AE: the semicircumference ABCDE will describe the surface of a sphere (D. 9); and the peri 216 GEOMETRY. meter of the semi-polygon will describe E a surface which has for its measure DF AE X circ. OF (P. 9), and this will be true whatever be the number of sides of C the semi-polygon. If now, the arcs be continually bisected, the limit of the perimeters of the semi- B polygons will be the semicircumference ABCDE; the limit of the area described A by the perimeter will be surface of the sphere, and the limit of the perpendicular OF will be the radius OE: hence, the surface of the sphere is equal to AExcirc. OE. Cor 1. Since the area of a great circle is equal to the product of its circumference by half the radius, or onefourth of the diameter (B. v., P. 15), it follows that the surface of a sphere is equal to four of its great circles: that is, equal to 4,tX OA, (B. v., P. 16). Cor. 2. The szuface of a zone is equal to its altitude multiplied by the circuamference of a great circle. For, the surface described by any por- A tion of the perimeter of the inscribed B'H polygon, as BC+ CD, is equal to EHX circ. OF (P. 9, c.): and. when we pass to C the limit, we have the surface of the zone equal to EHXcirc. OA. 0 Cor. 3. When the zone has but one DE base, as the zone described by the arc ABCD, its surface will still be equal to the altitude AE multiplied by the circumference of a great circle. Cor. 4. Two zones, taken in the same sphere or in equal spheres, are to each other as their altitudes; and any zone is to the surface of the sphere as the altitude of the zone is to the diameter of the sphere. BOOK VIII. 217 PROPOSITION XI. LEMMA. If a triangle and a rectangle, having the same base and the same altitude, turn together about the common base, the solicl generated by the triangle is a third of the cylinder generated by the rectangle.Let BA C be a triangle, BEEC a rectangle, having the common base BC, about which they are to be revolved. On the axis, let fall the perpendicular AD: then, the cone F A E generated by the triangle BAD is a third part of the cylinder generated by the rectangle BElAD (. / v., c. 1): also, the cone generated B D C by the triangle DA AC is a third part of the cylinder generated by the rectangle DAEC: hence, the sum of the two cones, or the solid generated by BA C, is a third part of the cylinders generated by the two rectangles, or a third part of the cylinder generated by the rectangle BEE C. If the perpendicular AD falls F E A without the triangle; the solid generated by CBA is, in that case, the difference of the two cones generated by BAD and B C D CAD; but at the same time, the cylinder generated by B.EC, is the difference of the two cylinders generated by BFAD and CEAD. Hence, the solid, generated by the revolution of the triangle, is still a third part of the cylinder generated by the revolution of the rectangle having the same base and the same altitude. Scholium. The circle of which AD is the radius, has for its measure r X A-L); hence, ~ X AD2 X BC measures the cylinder generated by BFEC, and -l rl A-D~XBC measures the solid generated by the triangle BAC. 218 GEOMETRY. PROPOSITION XII. LEMMA. If a triangle be revolved about any line dcraftn thirozigh its vertex in the same plane, the solicd generated will have for its measure, the area of the triangle multiplied by two-thirds of the circumferelnce traced by the middle poinzt of the base. Let CAB be a triangle, I the middle point of the base, and CD the line about which it is to be revolved: then will the solid generated be measured by area CAB X 3 circ. IK. Prolong the base AB till it p meets the axis CD in D; from I the points A and B, draw Aj / 0, BN, perpendicular to the axis, and draw CP perpendicular to C K N D DA produced. The scholium to the last proposition gives the following measures: solid generated by CAD= 3~rX AiT 2X CD, solid generated by CBD — ~X B3i X CD: hence, the difference of these solids, which is the solid generated by the triangle CAB, has for its measure x(AKi 2-BN2)X CD. To this expression another form may be given. From i; the middle point of AB, draw 1K perpendicular to CD; and through B, draw BO parallel to CD. We shall then have (B. IV.. P. 7, s.), A]i+BN=2H11, and AJI-BN=A O; hence, (AIf~+BN)X (AJI —BN)=AMi 2 —BN2=2IKxA O hence, the measure of the solid is also equal to r2 x IKX A OX CD. But CP being perpendicular to AB produced, the triangles AOB and CPD are similar; hence, AO CP: AB: CD. and, A Ox CD= CP xAB. BOOK VIII. 219 But CPTXAB is double the area of the triangle CAB; therefore, A Ox CD=2 CAB: hence, the solid generated by the thriangle CAB is measured by 4 rX CABxIM=U CABX -~1 IK; and since 2 X IK-circ. IK; we have, solid= CAB X -- circ. 17. Cor. If the triangle is A isosceles, the perpendicular \I CP will pass through 1/, the middle point of the base; and we shall have CAB =AB x- C. M K Substituting this value of CO - D CAB in the measure of the solid before found, viz.: solid= CAB X3 Xr X /I, gives, solid= R ~ X AB x IT x CI. But the triangles A OB, CKfi; are similar (B. Iv., P. 21); hence, AB BO or LiA T:: CI: I; which gives, AB x IKI= JN x Cf. Substituting for AB XIIE, we have, solid= W lX XMNV: that is, the solid generated by the revolution of an isosceles triangle about any line draw'ii through its vertex, is measured by two-thirds of r into the square of the perpendicular let fall on the base, into the distance between the two perpendiculars let fall from the extremities of the base on the axis. Scholium. The demonstration appeals to involve the supposition that AB prolonged will meet the axis: but the results are equally true if AB is parallel to the axis. Thus, the cylinder generated by p A B MNBA is measured by t X AMl X MN: the cone generated by CAM is measured by,rxAAM X C,; and the cone generated by CBN is measured C M N by l3 XAM2 x CiV. 220 GEOMETRY. Add the first two solids, and from the sum p A B subtract the third: we shall then have solid by CAB=-XAi X 2 A (MN+ 3 - I CV) // =r X A-ff2X (-MN + C M N C - I CIv + Z 11N); and since ~n11V+Cl- 1 =CN, we have solid by CAB-=, XAM2X M1N. But AB= CP and Af1V=AB; hence, solid by CAB=AB x CUP X X CP= CAB X circ. CP. But the circumference traced by P is equal to the circumference traced by the middle point of the base: hence, the result agrees with tlhe general enunciation. PROPOSITION XIII. LEMMA. Jf a regualcar sermi-polygon be revolved about a line passing through its centre and the vertices of two opposite angles, the solid generated,will be measutred by tiwo-thirds the area of the inscribed circle multtip2lied by the axis. Let GDBF be a regular semi-polygon and 01 the radius of the inscribed circle: then, if this semi-polygon be revolved about GF, the solid generated will have for its measure, 2 area OIX GF. For, since the polygon is regular, A M the triangles, OFA, OAB, OBC, &c., are isosceles and equal; then, all the per- BgT pendiculars let fall from 0 on their bases, will be equal to 01; the radius C 0 of the inscribed circle. Now, we have the following measures for the solids generated by these DQ triangles (P. 12, c.): viz., OFA is measured by r X OI X:F, P1G OAB" " " -aeX O01XAIP OBC " " " 2X01 ONO &c.; BOOK VIII. 221 hence, the entire solid generated by the semi-polygon is measured by 3X 01 (_FAI +rINrfN+N+ O Q+ QR +RG): that is, by 7r X o0IX GE. But, ~ X OI =area 0O (B. v., P. 16): hence, solidity = r X area OI X GF. PROPOSITION XIV. THEOREM. The solidity of a sphere is equal to its surface mrnutilied by a third of its radius. Let 0 be the centre of a sphere and OA its radius: then its solidity is equal to its surface into one-third of OA. For, inscribe in the semi-circle A ABC-DE a regular semi-polygon, havy- BF ing any number of sides, and let 01 be the radius of the circle inscribed in the polygon. If the semicircle and semi-polygon be revolved about EA, the semicircle will generate a sphere, and the semipolygon a solid which has for its measure w2 6-IX EA (P. 13); and this is true whatever be the number of sides of the semi-polygon. But if the number of sides of the polygon be continually increased, the limit of the solids generated by the polygons will be the sphere; and when we pass to the limit the expression for the solidity will become Zr X O-A 2X EA, or by substituting 2 0A for EA, it becomes $- X OA X OA, which is also equal to 4a X OA x 3 OA. But 4,X OAi is equal to the surface of the sphere (P. x., c. 1): hence, the solidity of a sphere is equal to its surface multiplied by a third of its radius. Scholium 1. The solidity of every spherical sector is equal to the zone which forms its base, multiplied by a third of the radius. 222 GEOMETRY. For, the solid described by any A portion of the regular polygon, as the B isosceles triangle OAB, is measured by 2 0-2XAF (P. 12, c.); and when we pass to the limit which is the spherical sector, the expression for this measure C becomes 2 X A X AF, which is equal to 2srxAOxAFx~ AO. But 2,rXAO is the circumference of a great circle of the sphere (B. v., P. 16), which being multiplied by AF gives the surface E of the zone which forms the base of the sector (P. x., c. 2); and the proof is equally applicable to the spherical sector described by the circular sector BOC: hence, the solidity of the spherical sector is equal to the zone which fbrms its base, multiplied by a third of the radius. Scholium 2. Since the surface of a sphere whose radius is X, is expressed by 4srXR2 (P. x., c. 1), it follows that the surfaces of spheres are to each other as the squares of their radii; and since their solidities are as their surfaces multiplied by their radii, it follows that the solidities of spheres are to each other as the cubes of their radii, or as the cubes of their diameters. Scholium 3. Let R be the radius of a sphere; its surface will be expressed by 4 XR2, and its solidity by 4YrxR 2X3R, o or0 rXR 3. If the diameter be denoted by D, we shall have R1-2 D, and R3=D3: hence, the solidity of the sphere may be expressed by X D3=1 X X D3. PROPOSITION XV. THEOREM. The surface of a sphere is to the whole surface of the circurnscribed cylinder, including its bases, as 2 is to 3: and the solidities of these two bodies are to each other in the same ratio. Let JPNQ be a great circle of the sphere; ABCD the BOOK VIII. 223 circumscribed square; if the semicircle Pi1IQ and the half square D C PADQ are at the same time made B; make the angle BAD / =B; then we shall have AD=DB (P. 11); but ADA B, D C is greater than A C; hence, C putting DB in place of AD, D we shall have DB+DC>AC, or BC>AC. Secondly. If we suppose BC>AC, the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC=A C; if BA C were less than ABC, we should then, as has just been shown, find BC and P the position of the',-' boat from which the angles A APCz= 330 45', CPB= 220 30',, and A.PB = 56~ 15', have been measured. Q Subtract twice APC- 67~0 30' from 1800, and lay off at \ A and C two angles, CAO, \ / ACO, each equal to half the " / remainder = 56~ 15'. With p. the point 0, thus determined, as a centre, and OA or 0 C as a radius, describe the circumference of a circle: then, any angle inscribed in the segment APC, will be equal to 330 45'. Subtract, in like manner, twice CPB 45~, from 180~, and lay off half the remainder =- 67 30', at B and C, determining the centre Q of a second circle, upon the circumference of which the point P will be found. The required point P will be at the intersection of these two circumferences. If the point P fall on the circumference described through the three points A, B, and C, the two auxiliary circles will coincide, and the problem will be indeterminate. ANALYTICAL PLANE TRIGONOMETRY. 40. WE have seen (Art. 2) that Plane Trigonometry explains the methods of computing the unknown parts of a plane triangle, when a sufficient number of the six parts are: given. To aid us in these computations, certain lines were employed called, sines, cosines, tangents, cotangents, &c., and a certain connection and dependence were found to exist between each of these lines and the angle or are to which it belonged. All these lines exist and may be computed for every conceivable angle, and each will experience a change of value where the angle or are passes from one state of magnitude to another. Hence, they are called functions of the angle or are; a term which implies such a connection between two varying quantities, that the value of the one shall always change with that,of the other. 41. In Plane Trigonometry, the numerical values of these functions were alone considered (Art. 13), and the. angles from which they were deduced were all less than 180 degrees. Analytical Plane Trigonometry, explains all the processes for computing the unknown parts of rectilineal triangles, and also, the nature and properties of the circular functions, together with the methods of deducing all the formulas which express relations between them. 298 ANAL Y TICAL 42. Let C be the centre of a circle, B and DA, EB, two diameters at right angles to each other-dividing the circumference into four quadrants. Then, C AB is called the first quadrant; BD the second quadrant; DE the third E quadrant; and FEA the fourth quadrant. All angles having their vertices at C, and to which we attribute the plus sign, are reckoned from the line CA, and in the direction from right to left. The arcs which measure these angles are estimated from A in the direction to B, to D, to E, and to A; and so on. 43. The value of any one of the circular functions will undergo a change with the angle to which it belongs, and also, with the radius of the measuring arc. When all the' functions which enter into the same formula are derived from the same circle, the radius of that circle may be regarded as unity, and represented by 1. The circular functions will then be expressed in terms of 1: that is, in terms of the radius. Formulas will be given for finding their values when the radius is changed from unity to any number dienoted by R (Art. 87). 44. We have occasion to refer to but one circular function not already defined. It is called the versed sine. The versed sine of an angle or arc, is that part of the diameter intercepted between the point where the measuring arcs begin and the foot of the sine. It is designated, ver-sin. 45. The names which have been given of the circular functions (Art. 11) have no reference to the quadrants in wvhich the measuring arcs may terminate; and hence, are equally applicable to all angles. T First quadrant. PM = sin a, C.if = cos a, AT = tan a A CT = sec a, AM= ver-sin a. PLANE TRIGONOMETRY. 299 Second quadrant. PMf = sin a, CM = cos a, AT = tan a, IA CT = sec a, AiM = ver-sin a. T Third quadrant. T PM = sin a, CM = cos a, AT = tan a, A[ C1' = sec a, AH = ver-sin a. p Fouarth quadrant. PM = sin a, UM = cos a, \ AT = tana, C CT = see a, AM = ver-sin a. 46. We will now proceed to established some of the iinportant general relations between the T circular functions. p Regarding the radius CP of the circle as unity, and denoting it by 1 (Art. M - A 43); we have in the right-angled trian- / gle CPiL PM'~ + CM = R, = I, W ~ 2 2 that is, sin2a +cos a=,*. (1) 47. Since the triangle CPM and CTA are similar, we have, A T PH CA = C' that is, tan a n a,. (2) cos a * The symbols sin" a, cos' a, tan' a, &c., signify the squae of the sine, the euare of the cosin, &c. 300 ANALYTIC AL 48. Substituting in equation (2), 90 - a for a, we have, sin (90 - a) tan (90 - = cos (90- a)' cos a that is (Art. 12), cot a = sin a.(3) 49. Multiplying equations (2) and (3), member by member, we have, tan aX cot a=l..... (4) 50. From the two similar triangles CPA! and C1TA, we have, CT CP CA - CH' that is, seca = -—. (5) 51. Substituting for a, 90 - a, we have, 1 sec (90-a) = cos (90-a) 1 that is, cosec a = sin a -. (6) 52. In the right-angle CTA, we have, 2-" 2 C2,'o CA- +A-2; that is, sec2 a = 1 + tan a... (7) 53. From equation (3) we obtain, 2 Cosa 2 a = cot a: sina adding 1 to each member, we have, 2 1 + cos a: sin a 2 2 sia a- cos a that is, sin + cot a: sin a that is (Eq. ) cosec a 1 + a. (8) that is (Eq. 6), cosec a =1 + cot a... (8) PLANE TRIGONOMETRY. 301 54. We: have, AiL equal to the versed sine of the arc AP; hence, ver-sin a = -- cos a. - - (9) 55. These nine formulas being often referred to, we shall place them in a table. They are used so frequently, that they should be com mitted to memory. TABLE I. 1.... sin. a + cos a =R =1. sin a 2. tall a. = cos a. cos a 3. cot a-. a sinl a 2 4.. tan a X cot a -R2 =1. 5.. sec a cos a 6.. coseca sin a 7... sec a = + tan a. 8... cosec a = 1 + cot a. 9... ver-sin a - 1 - cos a. 56. We will now explain thle principles which determine the alyebrcaic signs of the trigonometrical functiors. There are but two. 1.st. All lines estimated from DA, upwacrds, are considered positive, or have the sign +: and all lines estimated from DA, in the opposite direction, that is, dcltnwcwrds, are considered negative, or'have the sign -. 2d. All lines estimated from EB along CA, that is, to the right, are considered positive, or have the sign 4+: and all lines estimated from EB along CD, that is, in the opposite direction, are considered negative, or have the sign -. 302 ANALYTICAL 57. Let us determine, from the above principles, the algebraic signs of the sines and cosines in the different quadrants. First quadrant. 58. In the first quadrant. Pl = sin a P alnd P in CH= cos a, D A are both positive, the former being above C M the line DA, and the latter being estimated from C to the right (Art. 56). E Second quadrant. 59. In the second quadrant, P111= sin a, and Pm = CM-= - cos a: the sine is positive, being Above the line A M IC DA, and the cosine negative being estimated to the left of BE. E Third quadrant. 60. In the third quadrant, PH' — sin a, B and Pmn = CM= - cos a: / the sine is negative, falling below the Dt-MA line DA, and the cosine is negative, being estimated to the left of the cen.tre C. E Fourth quadrant. 61. In the fourth quadrant, B PM =- sin a, and Pm= iVCM = cos a: the sine is negative, falling below the D \ A line DA, and the cosine is positive, fall- m ing on the right of EB. Hence, we conclude, that PLANE TRIGONOMETRY. 303 1st. The sine is positive in, the.first and second quadrants, and negative in the third and fourth: 2d. The cosine is positive in the first and fourth quadrants, and negative in the second and third: In other words, 1st. The sinee of an angle less than 180~ is positive; and the sine of an angle greater than 180~ and leas than 360~, is negative: 2d. The cosine of an angle less than 90~ is positive; the cosine of an angle greater than 90~, and less than 2700, is negative; and the cosine of an angle greater than 2700, and less than 360~, is positive. 62. The algebraic signs of the sine and cosine being determined, the signs of all the other trigonometrical functions may be at once established by means of the formulas of Table I. Thus, for example, sin a tan a= cos a Now, if the algebraic signs of sin a and cos a are alike, the tangent is positive; if they are unlike, it is negative. Hence, the tangent is positive in the first and third quadrants, and negative in the second and fourth. The same is also true of the cotangent: for, cos a sin a 63. Again, since sec a — cos a the sign of the secant is always the same as that of the cosine. And since, cosec a- a sin a the sign of the cosecant is always the same as that of the sine. 304- ANALYTICAL 64. The versed sine is constantly positive. For, it is always found by subtracting the cosine from radius, and the remainder is a positive quantity, since the cosine can never exceed radius. When the cosine is negative, the versed sine becomes greater than radius. 65. Let q denote a quadrant: then the following table will show the algebraic signs of the trigonometrical lines in the different quadrants. First q. Second q. Tflird q. Fourth q. sine + + - - cosine + - - + tangent + - + - cotangent + - + - 66. WVe have thus far supposed all angles to be estimated from the line CA from right to left, that is in the direction from A to B, to D), &c., and B have also regarded such angles as positive. It is sometimes convenient to give different signs to the angles them- D selves. If we suppose the angles to be estimated from CA, in the direction from left to right, that is, in the direction from A to E, to D, wc., ve must treat the angles themselves as negative, and affect then with the sign -. For a negative angle less than 90~, the sine will be negative, and the cosine positive: for one greater than 90~ and less than 180~, the sine and cosine will both be negative. The algebraic sign of the sine always changes, when we change the sign of the arc, but the sign of the cosine remains the same. IIence, calling x the arc, we have in general, sin (-x)=-sin x, cos (- x)= cos xI, tan (- x)= - tan x, cot (-xj =-cot x. 67. We shall now examine the changes which take PLANE TRIGONOME TRY. 305 place in the values of the trigonometrical lines, as the angle increases from 0 to 360~, and shall begin with the sine and cosine. When the arc is zero, the sine is 0, and the cosine equal to R = 1. At 90~ the sine becomes equal to R = 1, and the cosine becomes 0. At 180~, the sine becojnes 0, and the cosine equal to - R = - 1. At 2700, the sine becomes equal to- R = - 1, and the cosine equal to 0. At 3600~, the sine becomes equal to 0, and the cosine to R= 1. Hence, First quadrant. As the arc increases from 0 to 90~: The sine increases from 0 to 1: The cosine decreases from 1 to 0. Second quadrant. As the are increases from 90~ to 180~: The sine decreases from. 1 to 0: The cosine increases, numerically, from 0 to - 1. Third quadrant. As the arc increases from 180~ to 270~: The sine increases, numerically, from 0 to - 1: The cosine decreases, numerically, from - 1 to 0. Fourth quadrant. As the arc increases from 270~ to 360~: The sine decreases, numerically, from - 1 to 0: The cosine increases from 0 to R = 1. 68. By substituting the values found above for sine and cosine, in the expressions for tangent, cotangent, secant, and cosecant, in Table I., and recollecting that the quotient of 0 divided by a finite quantity is 0 (Bourdon, Art. 109); and that the quotient of a finite quantity divided by 0, is infinite (Bourdon, Art. 110); we have the following table. 20 306 ANALYTICAL TABLE II. sin 0 = 0, sin (180~ + a) = - sin a, cos 0 = 1, cos (1800~ + a) = - cos a, tan 0 = 0, tan(180~ + a) = tal a, cot 0 =. cot (180~ + a) = cot a. sin (900 - a) = cos a, sin (270~ - a) = - cos a, cos (90~ - a) = sin a, cos (270~ - a) - - sin a, tan (900 - a) = cot a, tan(2700 - a) = cot a, cot (900 - a) = tan a. cot (2700 - a) = tan a. sin 90~ = 1, sin 270~ = - cos 900~ = 0, cos 270~ = 0, tan 90~ = 0O tan 270~ = - C cot 900 = 0. cot 270~ = 0. sin (900 + a) = cos a, sin (270~ + a) -- cos a, cos (90 + a = - sin a, cos(270 +a) = sin a, tan (90~ + a) = - cot a, tan(270 + a) = - cot a, cot (900 + a) = - tan a. cot (270 +a) = - tan a. sin (180 -- a)= sin a, sin (360 - a) = -sin a, cos (1800 - a) = - cos a, cos (360 - a) = cos a, tan (180 - a) = - tan a, tan (3600 - a) = - tan a, cot (180~- a) = - cot a, cot (3600 - a) = - cot a. sin 1800 = 0, sin 360~ = 0, cos 180~ — 1, cos 360 = 1, tan 1800 = O, tan 360~ = 0, cot 180~ = -. cot 360~ = o0. 69. The examinations thls far, have been limited to angles and arcs wltich do not exceed 360~. It is easily shown, however, that the.addition of 360~ to any arc as x, will make no difference in its trigonometrical functions; for, such addition would terminate the arc at precisely the same point of the circumference. Hence, if C represent an entire circumference, or 360~, and n any whole number, we shall have, sin (Cu +x) = sinx; or, sin (n X C + x) = sinx. The same is also true of the other functions. PLANE TRIGONOMETRY. 307 70. It will further appear, that whatever be the value of an arc denoted by x, the sine may be expressed by that of an arc less than 180~. For, in the first place, we may subtract 360~ from the arc x, as often as they are contained in it: then denoting the remainder by y, we have, sin x =sin y. Then, if y is greater than 1800, make y - 180 = z, and we shall have, sin y =- sin z. Thus, all' the cases are reduced to that in which the arc whose functions we take, is less than 1800; and since we also know that, sin (90 + x) = sin (90 - x), they are ultimately reducible to the case of arcs between 0 and 90~. GENERAL FORMULAS. 71. To find the formula for the sine of the diference of two angles or arcs. Let A CB be a triangle. From the vertex C let fall the perpendicular CD, on the base AB, produced. Denote the outward angle CBD by A B D a, and the angle CAB by b. Then, AB = AD - DB. But (Art. 25), AD = AC cos b, and BD = BC cos CBD. Hence, AB = AC cos b - BC cos a. Dividing both members by AB, we have AC BC 1 =AB cos b - -B cos a. But, since sin a = sin UBA, we have (Art. 21), AC sin a a BC sin b and AB- sin G' AB ='sin:LC' 808 ANALYTICAL sin a sin b hence, 1= cos b - cos a, sn CG sin C or, sin C = sin a cos b- sin b cos a. But the angle C is equal to the difference between the angles a and b (Geom. B. I., P. 25, C. 6): hence, sin (a- b) = sin a cos b - cos a sin b;.. (a) that is, The sine of the difference of any two arcs or angles i equal to the sine of the first into the cosine of the second, minus the cosine of the first into the sine of the second. It is plain that the formula is equally true in whichever quadrant the vertex of the angle C be placed: hence, the formula is true for all values, of the arcs a and b. 72. To find the formula for the sine of the sum of two anges or arcs. By formula (a) sin (a - b) =sin a cos b - cos a sin b, substitute for b, 1800~- b, and we have sin [a - (180~- b)] = sin a cos (180~ — b) - cos a sin (1800-b) = -sin a cos b - cos a sin b. But, sin [(a - (180~- b)] = sin [(a + b) - 1800)] - sin [180~- (a -t b)] =- sin (a + b) (Art. 68). Making the substitutions and changing the signs, we have, sin (a + b) = sin a cos b + cos a sin b.. (b) 73. To find the formula for the cosine of the sum of two angles or arcs. By formula (b) we have, sin (a + b) = sin a cos b + cos a sin b, substitute for a, 900 + a, and we have, sin [(90~ + a) + b] = sin (90~ + a) cos b + cos (90~ + a) sin b. PLANE TRIGONOMETRY. 309 But, sin [90~ + (a + b)] = cos (a + b) (Table II.): sin (90~ + a)= cos a, and, cos (90~ + a) - sin a; making the substitutions, we have, cos (a + b)= cos a cos b - sin a sin b.. (c) 74. To find the formula for the cosine of the do'erence between two angles or arcs. By formula (b) we have, sin (a + b) = sin a cos b + cos a sin 6. For a substitute 90~ - a, and we have, sin [90~ - (a - b)] = sin (90~ - a) cos b + cos (90~ - a) sin b. But, sin [90~ - (a - b)] = cos (a - b) (Table II.), sin (90~ - a) = cos a, cos (90~ - a) = sin a; making the substitutions, we have, cos ( - b)=cos a cos b + sin a sin b.. (d) 75. To find the formula for the tangent of the sum of two arcs. By Table I., tan (a + b) s= ( + ) cos (a+ b) sin a cos b + cos a sin b cosa cos b - sin a sin b' by (b) and(c), dividing both numerator and denominator by cos a cos b, sin a cos b cos a sin b =cos a cos b +cos a'cos b sin a sin b cos a cos b. tan a + tan b. ( tan (a + b) = 1 - tan a tan b 310 ANALYTICAL 76. To find the tangent of the difference of two angles. By Table I., tan (a - b)= sin (a- b) cos (a- b) sin a cos b - cos a sin b cos a cos b + sin a sin b by(a)and(d). Dividing both numerator and denominator by cos a cos b, and reducing, we have, tan a - tan b tan (a-b) =... (g) tan (a- = tan a tan b 77. The student will find no difficulty in deducing the following formulas. cot a cot b - 1 cot (a +b) cot acotb (h) cot a cot b+ cot b - cot a' 78. To find tle sine of twice an arc, in functions of the are. By formula (b) sin (a + b) = sin a cos b + cos a sin b. Make a = b, and the formula becomes, sin 2a = 2 sin a cos a. (k) If we substitute for a, -, we have, sin a = 2 sin a cos a... () 79. To find the cosine of twice an arc in functions of the arc. By formula (c) cos (a + b)= cos a cos b - sin a sin b. Make a = b, and we have, cos 2a = Cos2 a-sin2 a.... () By Table I., sin2 a = 1- cos2 a; hence, by substitution, cos 2a=2cos2a-1.... (11) Again, since cos2 a = 1 sin2 a, we also have, cos 2a = 1- 2 sin2 a... (12) PLANE TRIGONOMETRY. 811 80. To determine the tangent of twice or thrice a given angle in functions of the angle itsef By formula (f) tan a + tan b tan (a + b) = 1 — tan a ta b Make b - a, and we have, ta~ 2- = 2 tan a (n) 1 - tan" a Making b - 2a, we have, tan a + tan 2a tan 3a = 1- tan a tan 2a' substituting the value of tan 2a, and reducing, we have, 3 tan a- tan3 a tan 3a = 2 - (m ) -1- 3 tan2 a The student will readily find cot a- tan a cot 2a = 2 (n) 81. To find the sine of half an arc in terms of the functions of the arc. By formula (I 2) cos 2a = 1- 2 sin2 a. For a, substitute ~a, and we have, cos a = 1 - 2 sin 2 la; hence, 2 sine2 -a = 1 — cos a, sin a =/1-cosa (o) 2 82. To find the cosine of half a given angle in terms of the functions of the angle. By formula (1 1) cos 2a = 2 cos2 a - 1. For a, substitute 2a, and we have, cos a = 2 cos2 a-1; hence, cos a (p) 2OSaa b 2 812 ANALYTICAL 83. To find the tangent of half a given angle, in functions of the angle. Divide formula (o) by (p), and we have, 1 + cos a Multiplying both terms of the second member by V1/ -cos a, 1 1- cos a tan -la =... (qi) 2 sin a Multiplying both terms by the denominator 1 + cos a, sin a tan -a = 1 + cos. (q 2) 2 1 + cos aGENERAL FORMULAS. 84. The formulas of Articles 71, 72, 73, 74, furnish a great number of consequences; among which it will be enough to mention those of most frequent use. By adding and subtracting we obtain the four which follow, sin (a+b) + sin (a - b) = 2 sin a tos b,. () sin (a + b) - sin (a -b) = 2 sin b cos a,. (s) cos (a + b) + cos (a - b) = 2 cos a cos b,. (t) cos (a - b) - cos (a + b) = 2 sin a sin b,. (u) and which serve to change a product of several sines or cosines into linear sines or cosines, that is, into sines and cosines multiplied only by constant quantities. 85. If in these formulas we put a + b = i, a - b =q, which gives a = 2 + q, we shall find sin p + sin q = 2 sin (p + ) cos (p-q),.. (v) sinp - sin q = 2 sin (p - q) cos (p + q) (x) cos p + cos q = 2 cos ~(p + q) cos( - q), (). (y) cos q - cos p = 2 sin (p + q) sin (p-q),. (z) PLANE TRIGONOMETRY. 313 If we make q = 0, we shall have, sin p = 2 sin 2p cos ~p,.... (x 1) 1 - cos p = 2cos2 2,.... (y1) 1-cos p=2 sin2 -,.(z1) 86. From formulas (v), (x), (y), (z), and (k 1), we obtain; sin p + sin q sin ~ (p + q) cos (p - q)_ tang (p + q) sin p - sin q cos ( q) sin (p- ) tang 2 (p- q)' sin p + sin q sin I (p + q) -= tang 21 (P + N). cos p + cos q cos ) (p+q) + ) sin p + sin q _ cos I (p q) cot~ (p - ). cos q- cosp sin 2( - q) - sin p - sin q sin 2 (p - q) - tang 2 (P - ) cos p + cos q cos ~ (p - ~q) 2 sin p - sin q cos (p q)cot 2 =) - cot ~ (2 + q). cos q - cos p sin ~ (P + q) cosp+cos o p + q) os cos (p + ) (p - q) cot (p + q) cos q - cosp sin sin l(p + q) sin 1 (p - q) tang (p - q) sin p + sin q _ 2sin- q) os ( q)_ C - q) sin (p + q) 2sin (p +q) cos (p + q) cos (p + q)' sin p - sin q 2sin (p -q) cos ~ (p + q) sin (p - q) sin (p + q) 2sin I (p + q) cos -1 (p + q) sin -2 (p + )' These formulas are the expressions of so many theorems. The first shows that, the sum of the sines of two arcs is to the difference of those sines, as the tangent of half the sum of the arcs is to the tangent of half their difference. HOMOGENEITY OF TERMS. 87. An expression is said to be homogeneous, when each of its terms contains the same number of literal factors. Thus, sin2 a + GOS2 a&= R2 (1) is homogeneous, since each term contains two literal factors. 314 ANALYTICAL If we suppose R = 1, we have, sin2a+coss a=-...... (2) T'his equation merely expresses- the numerical relation between the values of sin2 a, Cos2 a, and unity. If we pass from the radius 1 to any other radius, as R, it becomes necessary to replace these abstract numbers by their corresponding.literal factors. For this, we must observe, that the radius of a circle bears the same ratio to any one of the functions of an arc, (the sine for example,) as the radius of any other circle, to the corresponding function of a similar arc in that circle. For example,, 1: sin a:: R: sin a; sin a sin a hence, in which the sin a, in the first consequent, is calculated to the radius 1, and in the second, to the radius R. If, now, we substitute this value of sin a to radius 1, in equation.(2), we have, sin a sin a cos a cos a - X + X =1; R R R or, sin2 a + cos2 a = R2, an expression which is homogeneous: and any expression may be made homogeneous in the same manner; or, it may be made so, by simply multi2plyiig each term by such a power of R as shall give the same number of linear factors in all the terms. 88. Since the sine of an are divided by its radius is equal to the sine of another are containing an equal number of degrees divided by its radius, we may, if we please, define the sine of an are to be the ratio of the radius to the perpendicular let fall from one extremity of the are on a diameter passing through the other extremity. Or, if in a right-angled triangle, we let A = right angle; B -'angle at base; C = vertical angle; a = hypothenuse; c = base; b = perpendicular, PLANE TRIGONOMETRY. 315 we may deduce all the functions of the angle without any reference to the circle. For, let us call, by definition, b C sin B= —, cos B =-, a a b c tan B --—, cot B=-, c a a sec B = —, cosec B -. c b Each of these expressions, regarded as a ratio, is a mere abstract number. If we make the hypothenuse a = 1, the abstract numbers will then represent parts of a rightangled triangle, or the corresponding lines of a circle whose radius is unity. Formulas for Triangles. 89. Let A CB be any triangle, and C designate the sides by the letters a, b, c; then (Art. 21), sin A a sin A a sin B b =_ A B in.B b sin C c sin C c that is, the sines of the angles are to each other as thei~ oppo0 site sides. 90. We also have (Art. 22), a+b a-b tan ~(A+B) tan ~(A-B) that is,'the sum of any two sides is their difference, as the tangent of hadf the sum of the opposite angles to the tangent of half their difference. 91. In case of a right-angled triangle, in which the right angle is B, we have (Art. 24), 1: tan A:: c: a; hence, a= c tan A,.(2) And again (Art. 25), 1: cosA:: b: c; hence, c =b cosA,. (3) 316 ANALYTICAL 92. There is but one additional case, that in which the three sides are given to find an angle. Let A CB be any triangle, and CD C a perpendicular upon the base. Then, whether the perpendicular falls without or within the triangle, we shall have (B. IV., P. 12), B B 2 = A2 +AB2 - 2AB X AD. But, AD = AC cos A; and representing the sides by letters, and substituting for AD, its value, we have, b +2 C2 a2 cos A- - 2be If we now substitute for cos A, its value from formula (Art. 81), we shall have, b2 +- c2 a2 2sin2 ~-A = -1 -b + c2be 2bc - (b2 + c2 - a2) 2be a2 - b - c2 + 2bc a2- (b- c)2 2bc 2bc (a + b - c) (a + c- b) 2bc sin 2-A = (a + b-c) (a + c-b) 4be If now, we make 2 (a + b + c) = s, we have a + b + c = 2s, and a+b-c = 2s-2c; also, a+c-b =2s-2b: hence, sin) (s-c), bc 93. If we add 1 to each member of the equation above, in which we have the value of cos A, we shall have, 1 cosA 2bc + b2 + c2-a2 _ (b + c)2- a2 2bc 2bc PLANE TRIGONOMETRY. 317 _(b + c + a) (b + c-a) and 2be;and, 1 + cos A 2s (s- a) be Substituting for 1 + cos A, its value (Art. 82), and reducing, we have, cos A = A /-. bc 94. If, now, we recollect that the tangent is equal to the sine divided by the cosine (Art. 47), we have, tan A= >/(s - b) (s - c) 2 s (s - a) and observing that the same formula applies equally to either of the other angles we have, tan l B = V/ (s- a) (s - c tan 2 as (s - b) ta~n =/ ( - (sa) (s - b) s (s - C) CONSTRUCTION OF TRIGONOMETRICAL TABLES. 95. If the radius of a circle is taken equal to 1, and the lengths of the lines representing the sines, cosines, tangents, cotangents, &c., for every minute of the quadrant be calculated, and written in a table, this would be a table of natural sines, cosines, &c. 96. If such a table were known, it would be easy to calculate a table of sines, &c., to any other radius; since, in different circles, the sines, cosines, &c., of arcs containing the same number of degrees, are to each other as their radii (Art. 87). 97. Let us glance for a moment at some of the methods of calculating a table of natural sines. When the radius of a circle is 1, the semi-circumfer 318 ANALYTICAL ence is known to be 3.14159265358979. This being divid ed.successively, by 180 and 60, or at once by 10800, gives.0002908882086657, for the aye of 1 minute. Of so small an arc, the sine, chord, and are, differ almost imperceptibly from each other; so that the first ten of the preceding figures, that is,.0002908882 may be regarded as expressing the sine of 1'; and, in fact, the sine given in the tables, which run to seven places of figures is.0002909. By Art. 46, we have, cos = 2/(1-sin2). This gives, in the present case, cos 1'=.9999999577. Then we have (Art. 84), 2 cos 1' X sin 1' - sin 0' = sin 2' =.0005817'764, 2 cos 1' X sin 2' - sin 1' = sin 3' =.0008726646, 2 cos 1' X sin 3' - sin 2' = sin 4' =.0011635526, 2 cos 1' X sin 4/ - sin 3' = sin 5' =.0014544407, 2 cos 1' X sin 5' - sin 4' = sin 6' =.0017453284, &c., &c., &c. Thus may the work be continued to any extent, the whole difficulty consisting in the multiplication of each successive result by the quantity 2 cos 1' = 1.9999999154. Or, having found the sines of 1' and 2', we may determine new formulas applicable to further computation. If we multiply together formulas (a) and (b) (Art. 71-72), and substitute for cos2 a, I — sin2 a, and for cos~ b, 1 - sin2 b, we shall obtain, after reducing, sin (a + b) sin (a - b) = sin2 a - sin2 b; and hence, sin,(a + b) sin (a - b) = (sin a + sin b) (sin a - sin b); or, sin (a b): sin a- sin b:: sin a + sin b: sin (a +I- b). Applying this proportion, we have, sin 1': sin 2' —sin 1':: sin 2' + sin 1': sin 3', sin 2': sin 3'-sin 1':: sin 3' + sin 1l': sin 4', sin 3': sin 4!- sin l':: sin 4' + sin': sin 5', sin 4': sin 5'-sin 1':: sin 5''+ sin l' sin 6',. &e., &c., &c. PLANE TRIGONOMETRY. 319 In like manner, the computer might proceed for the sines of degrees,-&c., thus: sin ~: sin 2~ — sin ~:: sin 2~ + sin 1: sin 3~, sin 2: sin 3 — sinl:: sin 3~ + sin l: sin 4~, sin 3~: sin 4 - sinl ~:: sin 41 sinlV: sin 5~, &c., &c., &c. Having found the sines and cosines, the tangents, cotangents, secants, and cosecants, may be computed from them (Table I). 98. There are yet other methods of computation and verification, which it may be well to notice. Let AP be an arc of 60~: then the chord AP is equal to the radius CA (B. V., P. 4): and the triangle CPA is equilateral. Hence, Pil bisects CA, or C M cos 60'~ = R, or equal to one-half, when R 2 1. But cos, 60~ = sin 30~ (Art. 12): hence, sin 300 =2; and, cos 30~ = /1-sin2 30 = i 3 Then, by formulas of Articles 81, and 82, we can find the sine and cosine of 15~, 70 30', 3~ 45', &c. 99.-If the are AP were 45~, the right-angled triangle CPM would be isosceles, and we should have Cf = PM; that is, sin 45~ = cos 45~. Hence, sin2 a + cos2 a = 1, gives 2 sin2 45 = 1; or, sin 45~ = cos 45~- /= l. -sin 45~ Also, tan 450 - so 450 1 = cot 45~. Above 45~, the process of computation may be simplified by means of the formula for the tangent of the sum of two arcs (Art. 75). 1 +- tan b tan (45~1 h)= 1_ tau1. 320 PLANE TRIGONOMETRY. 100. If the trigonometrical lines themselves were used, it would be necessary, in the calculations, to perform the operations of multiplication and division. To avoid so tedious a method of calculation, we use the logarithms of the sines, cosines, &c.; so that the tables in common use show the values of the logarithms of the sines, cosines, tangents, cotangents, &c., for each degree and minute of the quadrant, calculated to a given radius. This radius is 10,000,000,000, and consequently, its logarithm is 10. The logarithms of the secants and cosecants are not entered in the tables, being easily found from the cosines and sines. The secant of any arc is equal to the square of radius divided by the cosine, and the cosecant to the square of radius divided by the sine (Table I): hence, the logarithm of the former is found by subtracting the loga-:rithm of the cosine from 20, and that of the latter, by subtracting the logarithm of the sine from 20. SPHERICAL TRIGONOMETRY. 1. A SPHERICAL TRIANGLE is a portion of the surface of a sphere included by the arcs of three great circles (B. IX., D. 1). Hence, every spherical triangle has six parts; three sides and three angles. 2. SPHERICAL TRIGONOMETRY explains the processes of determining, by calculation, the unknown sides and angles of a spherical triangle, when any three of the six parts are given. For these processes, certain formulas are employed which express relations between the six parts of the triangle. 3. Any two parts of a spherical triangle are said to be of the siame species when they are both less or both greater than 90~; and they are of different species, wlien one is less and the other greater than 90~. 4. Let ABC be a spherical trian- B gle, and P the centre of the sphere. N A The angles of the triangle are equal to the diedral angles included P between the planes which determine its sides: viz.: the angle A to the angle included by the planes PAB C and PA C; the angle B to the angle included by the planes PBC and PBA; the angle C to the angle included by the planes PCB and PCA (B. IX., D. 1). rl'he sides CB, CA, AB, of the spherical triangle, measure the angles CPB, CPA, APB, at the centre of the sphere. Denote these sides or angles, respectively, by a, b, and c. 5. Assume any point on PA, as M, and suppose PM= R = 1. Then, in the planes APB, A PC, draw MN and 21 322 SPHERICAL TRIGONQMETIRY. dO, both perpendicular to the com- B mon intersection PA: then, OMN N~/~ will measure the angle between these / c planes (B. VI., D. 4), and'hence, will P A a be equal to the angle A of the tri-o 6 angle. Join 0 and N by the straight line ON. C In the triangles NPO and NMO, we have (Plane Trig., Art. 92). PN 2+ 5 — Y6 ~N+ PO -- _____ cos P - cos a;cos cos A 2MOXM and by reducing to entire terms, 2PNX P O Xcos a= PN2-'- NO02; 2O X MVNXcos A= lN2 +MO'2 —N2O By subtracting the second equation from the first, we have, 2(PN XPOX cos a - MO x MN cos A) = -PN- fN - ~- 5 - 2P-~-; and by dividing both members by 2PN x PO, we have, O IMN PH PM cos a - - x P X cos zA = B X -' But (Plane Trig., Art. 88), gives HMO.MN PM PM =sin b, = sin c, = -cosc, = cos b; substituting these values and reducing to entire terms, we have, cos a - sin b sin c cos A = cos b cos c; and by transposing, cos a = cos b cos c + sin b sin c cos A. A similar equation may be deduced for the cosine of either of the other sides: hence, cos a = cos b cos c + sin b sinec cos A, cos b = cos a cos c + sin a sin c cos B, (1) cos c = cos a cos b + sin a sin b cos. J That is: The cosine of either side of a spherical triangle is equal to the product of the cosines of the two other sides plus the product of their sines into the cosine of their included angle. The three equations (1) contain all the six parts of the spherical triangle. If three of the six quantities which SPHERICAL TRIGONOMETRY. 323 enter into these equations be given or known, the remaining three can be determined. (Bourdon, Art. 103): hence, if three parts of a spherical triangle be known, the other three may be determined from them. These are the primary'formulas of Spherical Trigonometry. They require to be put under other forms to adapt them to logarithmic computation. 6. Let the angles of the spherical triangle, polar to ABC, be denoted respectively by A', B', C', and the sides by a', b', c'. Then (B. IX., P. 6), a' = 180~-A, b' = 180~ —B, c' = 180- C, A' = 180 - a, B' = 180 - b, C' = 1800 - c. Since equations (1) are equally applicable to the polar triangle, we have, cos a' = cos b' cos c' + sin b' sin c' cos A': substituting for a', b', c' and A', their values from the polar triangle, we have, - cos A = cos B cos C - sin B sin C cos a; and changing the signs of the terms, we obtain, cos A = sin B sin C cos a - cos B cos C. Similar equations may be deduced from the second and third of equations (1); hence, cos A =sin B sin C cos a- cos B cos C, 1 cos B = sin A sin C cos b - cos A cos C, (2) cos C = sin A sin B cos c - cos A cos B. That is: The cosine of either angle of a spherical triangle, is.equal to the product of the sines oj the two other angles into the cosine of their included side, minus the product of the cosines of those angles. 7. The first and second of equations (1) give, after transposing the terms, cos a - cos b cos c = sin b sin c cos A, cos b -'cos a cos c = sin a sin c cos B; by adding, we have, cos a + cos b - cos c (cos a + cos b) = sin c (sin b cos A + sin a cos B); 324 SPHERICAL TRIGONOMETRY. and by substracting the second from the first, cos a - cos b + cos c (cos a - cos b) =: sin c (sin b cos A - sin a cos B); these equations may be placed under the forms, (1 - cos c) (cos a + cos b) = sin c (sin b cos A + sin a cos B), (1 + cos c) (cos a — cos b) = sin c (sin b cos A - sin a cos B); multiplying these equations, member by member, we obtain, (1 - Cos2 c) (coS2 a - cos2 b) = sin2 c (sin2 b cose A- sin2 a cos2 B): substituting sin2 c for 1 - cos2 c, 1 - sin2 A for cos2 A, and. 1 - sin2 B for cos2 B, and dividing by sin2 c, we have, cos2 a - cos2 b- sin2 b - sin2 b sin2 A - sin2 a + sin2 a sin2 B: then, since cos2 a - os2 sin2 b - sin2 a, we have, sin2 b sin2 A = sin2 a sin2 B; and, by extracting the square root, sin b sin A = sin a sin B. By employing the first and third of equations (1) we shall find, sin c sin A = sin a sin C; and, by employing the second and third, sin b sin C = sin c sin B: hence, sin A sin a -- sin b or sin B sin A:: sin b: sin a, siml B sln b - sin A sin a sin C sin c; or sin C sinA:: sin c: sin a, (3) sin G sin c si — sin; or sin B: sin C:: sin b: sin c. sin B sin b That is: In every spherical triangle, the sines of the angles are to- each other as the sines of their opposite sides. S. Each of the formulas designated (1) involves the three sides of the- triangle together with one of the angles. These formulas are used to determine the angles when the three sides are known. It, is necessary, however, to put SPHERICAL TRIGONOMETRY. 325 them under another form to adapt them to logarithmic computation. Taking the first equation, we have, cos a- cos b cos a cosA= sin b sinl c Adding 1-to each member, we have, cos a + sin b sin c - cos b cos c I + cos A = sin b sin c But, 1 + cos A = 2 cos2 ~A (Plane Trig., Art. 85), and, sin b sin c cos b cos c = - cos (b + c) (Art. 73); cos a - cos (b + c) hence 2 cos2 A = sinb sin sin b sin c or, cos2 -A = sin~(a + b + c) sin (b + c - a) (Art. 85). sin b sin c Putting s = a + b + c, we shall have, s= =(a + + + ) and -- a = (+ c - a): hence, A sin I (s) sin ( S - a) sin b sin c os ~ = \/s,(s) sin ( - b) si a sin c sin (s) sin (~S - sin a sin b 9. Had we subtracted each member of the first equa. tion in the last article, from 1, instead of adding, we should, by making similar reductions, have foilnd, sin ~A = /sin ~ (a b - b-c) sin ~ (a + c - b) -sin b sin c sin B = /sin (a + b-c) sin 1 (b + c - a) (5) sin a sin c sin = sn (a + c - b) sin (b + c - a) sin a sin b 326 SPHERICAL TRIGONOMETRY. Putting s = a + b + c, we shall have, - (b+c-a), 1 s-b=~ (a+c-b), and - s-c= (a+b-c); hence, sin A = V/sin (-s- c) sin (2s - b) sin b sin c sin B = /sin (~ s -c) sin (~s -a) (6) sin a sin c ~;n 1 sinsin b) sin (s-a) sin a sin b 10. From equations (4) and (6) we obtain, tan ~.A = V/sin (~ s - c) sin (~s - b) /sin s - c) sin (2s - a) tan I B si () sin ( S-b)' (7) tan GC = /sin (~s-b) sin (ls-a), sin (s) sin ( s- c) 11. We may deduce the value of the side of a triangle in terms of-the three angles by applying equations (5), to the polar triangle. Thus, if a', b', c', A', B', C', represent the sides and angles of the polar triangle, we shall have (B. IX., P. 6), A = 180~-a', B =180~-b', = 180 - c'; a= 180~ - A', b = 180~ - B', and c = 180 - C'; hence, omitting the', since the equations are applicable to any triangle, we shall have, cos a =.os-(A+B -C) cos~(A+ C —B) cos aa-i/ sin 2B sin 2 cos c os b = /s -- (A + B- C) cos (B+ a-A), (8) sin B sin C /cos (A aC- a B) cos (B + a- A), cos ~~ ~ sin A sin B Putting S = A + B + C, we shall have SPHERICAL TRIGONOMETRY. 327 S-a = ( + B- A), - B- -B= (/+ C — B), and, S-a = (A +B- C); cos 1 a cos (2 S- C) cos ( - B) hence, cos2a=V / s sin B sin C os~- /s (Is- C) cos (~ s - A) Cos lb / 2 i n (9) 2 W ~' sin A sin C 1sc \cos (~S - B) cos ( S - A) sin A sin B 12. All the formulas necessary for the solution of spherical triangles, may be deduced from equations marked (1). If we substitute for cos b in the third equation, its value taken from the second, and substitute for cos2 a its value 1 - sin2 a, and then divide by the common factor, sin a, we shall have, cos c sin a = sin c cos a cos B + sin b cos C. sin B sin ~ But equations (3) give sin b sin B sn hence, by substitution, sin B cos C sin c cos c sin a = sin c cos a cosB+ s i siln Dividing by sin c, we have, cos c. sin B cos C s a sin a = a cos B + - sin C sin c sin C COS But, - =cot (Art. 55). sin Therefore, cot c sin a = cos a cos B + cot C sin B. Hence we may write the three symmetrical equations, cot a sin b = cos b cos C + cot A sin C, cot b sin c = cos c cos A + cot B sin A, (10) cot c sin a = cos a cos B + cot C sin B. J That is: In every spherical triangle, the cotangent of one o, the sides into the sine of a second side, is equal to the cosine of the second side into the cosine of the included angle, plus the cotangent of the angle opposite the first side into the sine of the included angle. 328 SPHERICAL TRIGONOMETRY. NAPIER'S ANALOGIES. 13. If from the first and third of equations (1), cos c be eliminated, there will result, after a little reduction,. cos A sin c = cos a sin b - cos C sin a cos b. By a simple permutation, this gives, cos B sin c = cos b sin a - cos C sin b cos a. Hence, by adding these two equations, and reducing, we shall have, sin c (cos A + cos B) = (1 - cos C) sin (a + b). sin c sin a sin b But since,. = = siA, we shall have, sin c (sin A + sin B) = Sin C (sin a + sin b), land, sin c (sin A - sin B) = sin C (sin a - sin b). Dividing these two equations, successively, by the preceding; we shall have, sin A + sin B sin C sin a + sin b cos A + cos B 1 -cos C sin (a + b) sin A - sin B sin C sin a - sin b cos A + cos B- — cos C X sin (a + b) reducing these by the formlulas (Plane Trig., Arts. 85, 86), we have, Oos ~ ~- a ] tang 1 (A + B) = cot 2 X s (a - 6) cos I (a + b) sin (a - b) tang - (A B) = cot 1 C X 2 sin ~a + b) Hence, two sides, a and b, with the included angle C being given, the two other angles A and B may be found by the proportions, cos~ (a + b): cos (a - b):: cot a: tang (A + B), sin (a + b): sin (a - b):: cot ~ C: tang~(A - B). SPHERICAL TRIGONOMETRY. 329 We may apply the same proportions to the triangle, polar to ABC by putting 1800 - A', 1800 - B', 180~ - a', 1800~- b', 180~- c', instead of a, b, A, B, C, respectively; and after reducing and omitting the accents, we shall have, cos I(A + B): cos (A - B): tang: tang: tang(a + b), sin I(A + B): sin ~ (A - B):: tang c: tang (a - b); by means of which, when a side c and the two adjacent angles A and B are given, we are enabled to find the two other sides a and b. These four proportions are known by the name of Napier's Analogies. 14. In the case in which there are given two sides and an angle opposite one of them, there will in general be two solutions corresponding to the two results in Case II., of rectilineal triangles. It is also plain, that this ambiguity will extend itself to the corresponding case of the polar triangle, that is, to the case in which there are given two angles and a side opposite one of them. In every case we shall avoid all false solutions by recollecting, 1st. That every angle, and every side of a spherical triangle is less than 180~. 2d. That the greater angle lies opposite the greater side, and the least angle opposite the least side, and reciprocally. NAPIER'S CIRCULAR PARTS. 15. Besides the analogies of Napier already demonstrated, that Geometer invented rules for the solution of all the cases of right-angled spherical triangles. In every right-angled spher- C ical triangle BAGC, there are six parts: three sides and three angles. If we omit the consideration of the right angle, which is always known, there B~ 0. A are five remaining parts, two G of which must be given before the others can be determined. 330 SPHERICAL TRIGONOMETRY. The circular parts, as they are called, are the two sides c C and b, about the right angle, 09 the complements of the oblique angles B and C, and the comnplement of the hypothenuse a. B A Hence, there are five circular parts. The right angle A not being a circular part, is supposed not to separate the circular' parts c and b, so that these parts are considered "as lying adjacent to each other. If any two parts of the triangle are given, their corresponding circular parts are also known, and these, together with a required part, will make three parts under consideration.. Now, these three parts'will all lie together, or ooze of them will be separated fronmi both of the others. For example, if B and c were given, and a required, the three parts considered would lie together. But, if B and C were given, and b required, the parts would not lie together; for, B would be separated from- C by the part a, and from b by the part c. In eit~her case, B is the middle part. Hence, when there are three of the circular parts, under consideration, the middle part is that one of them to which both of the pthers are adjacent, or from which both of them are separated. In the former case, the parts are said to be adjacent, and, in the latter case, the parts are said to b]e opposite. This being premised, we are now to prove the following theorems for the solution of right-angled spherical triangles, which, it must'be remembered, apply to the circular parts, as. already defined. 1st. Radius into the sine of t4e- middle part is equal to the rectangle of the tangents of the adjacent Tparts. 2d. Radius into the sine of the middle part is equal to the rectangle of the cosines of the opposite parts. These theorems are proved by assuming each of the five circular parts, in succession, as the middle part, and by taking the extremes first opposite, and then adjacent. Having thus fixed the three'parts which are to be consid SPHERICAL TRIGONOMETRY. 331 ered, take that one of the general equations for obliqueangled triangles, that will contain the three corresponding parts of the triangle, together with the right angle; then make A = 90~, and after making the reductions corresponding to this supposition, the resulting equation will prove the rule for that particular case. For example, let comp. a, be the middle part and the extremes opposite. The'equation to'be applied in this case must contain a, b, c, and A. The first of equations (1) contains these four quantities: cos a = cos b cos c + sin b sin c cos A. If A = 90~ cos A = 0; hence, cos a = cos b cos c; that, is, radius, which is 1, into the sine of the middle part, (which is the complement of a,) is equal to the rectangle of the cosines of the opposite parts. Suppose, now, that the comple- C ment of a were the middle part and the extremes adjacent. The \e equation to be applied must contain the four quantities a., B, 6, and B A A. It is the first of equations (2): c cos A = sin B sin C cos a - cos B cos C. Making A = 90~, we have, sin B sin C cos a = cos B cos C or, cos a = cot B cot C; that is, radius, which is 1, into the sine of the middle part is equal to the rectangle of the tangent of the com-'plement of B, into the tangent of the complement of C, that is, to the' rectangle of the tangents of the adjacent circular parts. Let us now take the comp. B, for the middle part and the extremes opposite. The two other parts under consideration will then be the perpendicular b and the angle C. The equation to be applied must contain the four parts A, B, C, and b: it is the second of equations (2), cos B = sin A sin C cos b - cos A cos C. 332 SPHERICAL TRIGONOMETRY. Making A = 90~, we have, cos B = sin C cos b. Let comp. B be still the middle part and the extremes adjacent. The equation to be applied must then contain the four parts a, B, c, and A. It is similar to equations (10); cot a sin c = cos c cos B + cot A. sin B. But, if A=90~, cot A = 0; hence, cot a sin c = cos c cos B: or, cos B = cot a tang c. By pursuing the same method of demonstration when each circular part is made the middle part, and making the terms homogeneous, when we change the radius from 1 to P (Plane Trig., Art. 87), we obtain the five following equations, which embrace all the cases. B cos a = cos b cos c = cot B cot C, Rcos B = cos b sin C = cot a tang c, R cos C= cos c sin B = cot a tang b, (11) R sin b = sin a sin B= tang c cot U, R sin c = sin a sin C = tang b cot B. We see from these equations that, if the middle part is required we must begin the proportion with radius; and when one of the extremes is reguired we imust begin the proportion with the other extreme. We also conclude, from the first of the equations, that when the hypothenuse is less than 90~, the sides b and c are of the same species, and also that the angles B aid C are likewise of the same species. When a is greater than 90~, the sides b and c are of different species, and the same is true of the angles B and C. We also see from the two last equations that a side and its opposite angle are always of the same species. These properties are proved by considering the algebraic signs which have been attributed to the trigonometrical functions, and by remembering that the two members of an equation must always have the same algebraic sign. SPHERICAL TRIGONOMETRY. 333 SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES BY LOGARITHMS. 16. It is to be observed, that when any part of a triangle becomes known by means of its sine only, there may be two values for this part, and consequently two triangles that will satisfy the question; because, the same sine which corresponds to an angle or an arc, corresponds likewise to its supplement. This will not take place, when the unknown quantity is determined by means of its cosine, its tangent, or cotangent. In all these cases, the sign will enable us to decide whether the part in question is less or greater than 90~; the part is less than 90~, if its cosine, tangent, or cotangent, has the sign +; it is greater if one of these quantities has the sign -. In order to discover the species of the required part of the triangle, we shall annex the minus sign to the logarithms of all the elements whose cosines, tangents, or cotangents, are negative. Then, by recollecting that the product of the two extremes has the same sign as that of the means, we can at once determine the sign which is to be given to the required element, and then its species will be known. It has already been observed, that the tables are calculated to the radius R, whose logarithm is 10 (Plane Trig., Art. 100); hence, all expressions involving the circular functions, must be made homogeneous, to adapt them to the logarithmic formulas. EXAMPLES. 1. In the right-angled spherical o triangle BAC, right-angled at A, there are given a = 640 40' and b = 42~ 12': required the remaining parts. B A First, to find the side c. The hypothenuse a corresponds to the middle part, and the extremes are opposite: hence,'. cos a = cos b cos c, or, 334 SPHERICAL TRIGONOMETRY. As cos b 42~ 12' ar. comp. log. 0.130296..... 10.000000:: cos a 64~ 40'.... 9.631326 cos c 54~ 43' 07".... 9.761622 To find the angle B. The side b is the middle part and the extremes opposite: hence, R sin b = cos (comp. a) X cos (comp. B) = sin a sin B. As sin a 64~ 40' ar. comp. log. 0.043911 sin b 42~ 12'.... 9.827189: P...... 10.000000 sin b 48~ 00' 14"... 9.871100 To find the angle C. The angle C is the middle part and the extremes adjacent: hence, R cos C = cot a tang b. As R.. ar. comp. log. 0.000000: cot a 640 40'.... 9.675237:: tang b 42~ 12'... 9.957485: cos C6 64~ 34' 46".... 9.632722 2. In a right-angled triangle BAC, there are given the hyppthenuse a = 105~ 34', and the angle B= 80~ 40': required the remaining parts. To find the angle C. The hypothenuse is the middle part and the extremes adjacent: hence, R cos a = cot B cot C. As cot B 80~ 40' ar. comp. log. 0.784220 +: cos a 105~ 34'.... 9.428717 -: P R.......o10.00.0000 + cot C 1480 30' 54".... 10.212937Since the' cotangent of C is negative, the angle C is greater than 90~, and is the supplement of the are which would correspond to the cotangent, if it were positive. SPHERICAL TRIGONOMETRY. 335 To find the side c. The angle B corresponds to the middle part, and the extremes are adjacent: hence, R, cos B = cot a tang c. As cot a 1050 34' ar. comp. log. 0.555053 -: R...... 10.000000 +:: cos B 80~ 40'.... 9.209992 +: tang c 149~ 47' 36"... 9.765045 - To find the side b. The side b is the middle part and the extremes are opposite: hence, R sin b = sin a sin B. As R. ar. comp. log.. 0.000000: sin a 105~ 34'.. 9.983770:: sin B 80~ 40'... 9.994212: sin b 71~ 54' 33"... 9.977982 OF QUADRANTAL TRIANGLES. 17. A quadrantal spherical triangle is one which has one of its sides equal'to 90~. Let BA C be a quadrantal tri- C angle of which the side a = 90~. If we pass to the corresponding polar triangle, we shall have A' = 180~ - a = 90~, B' = 180 -b, B d A C' = 180 - c, a' 1800 - A, b' = 180' - B, c' =180~- C; — "D from which'we see, that the polar triangle will be rightangled at A', and hence, every case may be referred to a right-angled triangle. But we can solve the quadrantal triangle by means of the right-angled triangle in a manner still more simpleo 336 SPHERICAL TRIGONOMETRY. Let the side BC of the quad- C rantal triangle BA C be equal to 90~; produce the side CA till C'D is equal' to 90~, and conceive the arc of a great circle to be drawn A through B and D. B d 16 Then C will be the pole of the arc BD, and the angle C a- -- will be measured by BD (B. IX., P. 4), and the angles CBD and D will be right angles. Now before the remaining parts of the quadrantal triangle can be found, at least two parts must be given in addition to the side BC= 90~; in which case two parts of the right-angled triangle BDA, together with the right angle, become known. Hence, the conditions which enable us to determine one of these triangles, will enable us also to determine the other. EXAMPLES. 1. In the quadrantal triangle BCA, there are given CB = 90~, the angle C = 420 12', and ihe angle A = 1!5~ 20'; required the remaining parts. Having produced CzA to D, making CD = 90~, and drawn the arc BD), there will then be given in the rightangled triangle.BAD, the side a = C = 42~ 12', and the angle BAD = 180~ - BA C = 180~ - 115~ 20' = 64~ 40', to find the remaining parts. To find the side d. The side a is the middle part, and the extremes o!posite: hence, R sin a = sin A sin d. As sin A 64~ 40' ar. comp. log. 0.043911: -2R. ~.. i... 10.000000:: sin a 420 12'.... 9.827189: sin d 48~ 00' 14". 9.871100 TO find the angle B. The angle A corresponds to the middle part, and the extremes aro opposite: hence, SPHERICAL TRIGONOMETRY. 337 R cos A = sin B cos a. As cos a 42~ 12' ar. comp. log. 0.130296..... 10.000000:: cos A 64~ 40'.. 9.631326: sin B 35~ 16' 53"... 9.761622 To find the side b. The side b is the middle part, and the extremes are adjacent: hence, R sin b = cot A tang a. As R.. ar. comp. log. 0.000000: cot A 64~ 40'.... 9.675237:: tang a 42~ 12'.... 9.957485: sin b 25~ 25' 14"... 9.632722 Hence, CA = 90~ - b = 90~ - 25~ 25' 14" = 64~ 34' 46"' CBA = 90 - ABD = 900 - 350 16' 53" = 540 43' 07" BA = d..= 48~ 00' 14" 2. In the right-angled triangle BA(C right-angled at A, there are given a = 115~ 25', and c = 60~ 59': required the remaining parts. B = 1480 56' 45" Ans. U= C 75~ 30' 33" b = 152~ 13' 50" 3. In the right-angled spherical triangle BAC rightangled at A, there are given c = 116~ 30' 43", and b = 29~ 41' 32": required the remaining parts. C= 1030 52' 46" Ans. B = 320 30' 22" a = 1120 48' 58" 4. In a quadrantal triangle, there are given the quadrantal side = 90~, an adjacent side = 115~ 09', and the included angle =- 115~ 55': required the remaining parts. side, 1130 18' 19" Ans. anles 1170 33' 52" 0 7' 41010 40' 07" 22 338 SPHERICAL TRIGONOMETRY. SOLUTION OF OBLIQUE-ANGLED TRIANGLES BY LOGARITHMS. 18. There are six cases which occur in the solution of oblique-angled spherical triangles. 1. Having given two sides, and an angle opposite one of them. 2. Having given two angles, and a side opposite one of them. 3. Having given the three sides of a triangle, to find the angles. 4. Having given the three angles of a triangle, to find the sides. 5. Having given two sides and the included angle. 6. Having given two angles and the included side. CASE I. Given two sides, and an angle opposite one of them, to find the remaining parts. 19. For this case, we employ equations (3); sin a: sin b:: sin A sin B. Ex. 1. Given the side a= C 440 13' 45", b= 84~ 14' 29", a and the angle A = 320 26' 07": b required the remaining parts. A B To find the angle B. B' D As sin a 440 13' 45" ar. comp. log. 0.156437: sin b 84~ 14' 29".... 9.997803:: sin A 32~ 26' 07".... 9.729445: sin B 49~ 54' 38", or sin B' 130~ 5' 22" 9.883685 Since the sine of an arc is the same as the sine of its supplement, there are two angles corresponding to the logarithmic sine 9.883685, and these angles are supplements of each other. It does not follow, however, that both of them will satisfy all the other conditions of the question. If they do, there will be two triangles ACGB', A CB; if not, there will be but one. SPHERICAL TRIGONOMETRY. 339 To determine the circumstances under which this ambiguity arises, we will consider the 2d of equations (1) cos b = cos a cos c + sin a sin c cos B, from which we obtain, cos b - cos a cos c cos _B -- sin a sin c ANow, if cos b be greater than cos a, we shall have, COS b > cos a cos c, or, the sign of the second member of the equation will depend on that of cos b. HIence, cos B and cos b will have the same sign, or B and b will be of the same species, and there will be but one triangle. But when cos b > cos a, then sin b < sin a: hence, If the sine of the side opposite the required angle be less than the sine of the other given side, there will be but one triangle. If, however, sin b > sin a, the cos b will be less than cos a, and it is plain that such a value may then be given to c, as to render cos b < cos a cos c, or, the sign of the second member may be made to depend on cos c. We can therefore give such values to c as to satisfy the two equations, cos b - cos a cos c +cos B= - sin a sin c cos b- cos a cos c - cos B=: sin a sin c hence, if the sine of the side opposite the required angle be greater than the sine of the other given side, there will be two triangles which will fulfil the given conditions. Let us, however, consider the triangle A CB, in which we are yet to find the base AB and the angle C. We can find these parts by dividing the triangle into two rightangled triangles. Draw the are CD perpendicular to the base AB: then, in each of the triangles there will be given the hypothenuse and the angle at the base. And generally, 340 SPHERICAL TR-IGONOMETRY. when it is proposed to solve an oblique-angled triangle by meansof the right-angled triangle, we must so draw the perpendicular, that it shall pass through the extremity of a given side, and lie opposite to a given angle. To find the angle C, in the triangle A CD. As cot A 32~ 26' 07" ar. comp. log. 9.803105 R...... 10.000000:: cos b 840 14' 29"... 9.001465: cot A CD 86 21' 06"... 8.804570 To find the angle C in the triangle _DCB. As cot B 49~ 54' 38" ar. comp. log. 0.074810.R... 10.000000:: cos a 44~ 13' 45". 9.855250: cot D CB490 35' 38"... 9.930060 Hence, A CB- 135~ 56' 44". To find the side AB. As sin A 32~ 26' 07" ar. comp. log. 0.270555: sin C 135~ 56' 44"... 9.842198:: sin a 44~ 13' 45"... 9.843563 sin c 115~ 16' 12"... 9.956316 The arc 64~ 43' 48", which corresponds to sin c is not the value of the side AB: for the side AB must be greater than b, since it lies opposite to a greater angle. But b = 840 14' 29": hence, the side AB must be the supplement of 64~ 43' 48", or, 115~ 16' 12". Ex. 2. Given b = 91~ 03' 25", a = 40~ 36' 37", and A = 35~ 57' 15": required the remaining parts, when the obtuse angle B is taken. B = 1150 35' 41" Ans. G = 58~ 30' 57" c- 70~ 58' 52" SPHERIICAL TRIGONOMETRY. 341 CASE II. tHaving given two angles and a side opposite one of them, to -find the remaining parts. 20. For this case, we employ the equation (3). sin A sin B:: sin a sin b. Ex. 1. In a spherical triangle ABC, there are given the angle A = 50~ 12', B = 58~ 8', and the side a = 62~ 42'; to find the remaining parts. To find the side b. As sin A 50~ 12' ar. comp. log. 0.114478: sin B 58~ 08'.... 9.929050:: sin a 62~ 42'..... 9.948715: sin b 79~ 12' 10", or, 1000 47' 50" 9.992243 We see here, as in the last example, that there are two arcs corresponding to the 4th term of the proportion, and these arcs are supplements of each other, since they have the same sine. It does not follow, however, that both of them will satisfy all the conditions of the question. If they do, there will be two triangles; if not there will be but one. To determine when there are two triangles, and also when there is but one, let us consider the second of equations (2), cos B = sin A sin C cos b - cos A cos C, cos B + cos A cos C which gives, cos b = sin A sin C Now, if cos B be greater than cos A, we shall have, cos B > cos A cos C, and hence, the sign of the second member of the equation will depend on that of cos B, and consequently cos b and cos B will have the same algebraic sign, or b and B will be of the same species. But when cos B > cos A the sin B < sin A: hence, If the sine of the angle opposite the required side be less 342 SPHERICAL TRIGONOMETRY. than the sine of the other given angle, there will be but one solution. If, however, sin B > sin A, the cos B will be less than cos A, and it is plain that such a value may then be given'o cos C, as to render cos B < cos A cos C, or, the sign of the second member of the equation may be made to depend on cos C. We can therefore give such values to C as to satisfy the two equations, cos B + cos A cos C sin A sin C cos B + cos A cos C and -cos b = sin A sin C Hence, if the sine of the angle opposite the required side be greater than the sine of the other given angle, there will be two solutions. Let us first suppose the side b to be less than 90~, or, equal to 79~ 12' 10". If, now, we let fall from the angle C, a perpendicular on the base BA, the triangle wil be divided into two rightangled triangles, in each of which there will be two parts known besides the right angle. Calculating the parts by Napier's rules, we find, C= 1300 54' 28" c = 1190 03' 26" If we take the side b = 100~ 47' 50", we shall find, C = 1560 15' 06" c = 1520 14' 18" Ex. 2. In a spherical triangle ABC, there are given A = 1030 59' 57", B = 46~ 18' 07", and a = 42~ 08' 48"; required the remaining parts. There will be but one triangle, since sin B < sin A. b = 300 Ans. C= 360 07' 54" c= 240 03' 56" SPHERICAL TRIGONOMETRY. 343 CASE III. Having given the three sides of a spherical triangle, to find the angles. 21. For this case we use equations (4). cos A=R sin s sin (s- a) sin b sin c Ex. 1. In an oblique-angled spherical triangle, there are given a = 560 40', b = 83~ 13', and c = 114~ 30': required the angles. =(a + b + c) = s = 127~ 11' 30", -(b + c - a) = (s - a)= 700 31' 30". log sin i-s 1270 11' 30"... 9.901250 log sin (Is -- a) 70~ 31' 30"... 9.974413 - log sin b 830 13' ar. comp. 0.003051 - log sin c 1140 30' ar. comp. 0.040977 Sum... 19.919691 Half sum = log cos 2 A 24~ 15' 39".. 9.959845 Hence, angle A = 480 31' 18". The addition of twice the logarithm of radius, or 20, to the numerator of the quantity under the radical, just cancels the 20 which is to be subtracted on account of the arithmetical complements, so that the 20, in both cases, may be omitted. Applying the same formulas to the angles B and C, we find, B = 62~ 55' 46" C = 125~ 19' 02" -Ex. 2. In a spherical triangle there are given a = 40~ 18' 29", b = 67~ 14' 28", and c = 890 47' 06": required the three angles. A = 340 22' 16".Ans. B- 530 35' 16 " ( = 119~ 13' 32" 344 SPHERICAL TRIGONOMETRY. CASE IV. Raving given the three angles of a siherical triangle, to find the three sides. 22. For this case we employ equations (9). cos I a = R ios os S- sin ]3 sin C Ex. 1. In a spherical triangle ABC there are given A = 48~ 30', B = 125~ 20', and C= 62~ 54'; required the sides. (A + B + C)= ~S= 118~ 22' (2S-A). = 69~ 52' (2I -B) - -- 6~ 58' (2S - C). 55~ 28' log cos (~S- B) - 6~ 58'.. 9.996782 log cos (~S- - C) 550 28'.. 9.753495 - log sin B 1250 20' ar. comp. 0.088415 - log sin C 62~ 54' ar. comp. 0.050506 Sum. 19.889198 Half sum = log cos ~a = 280 19" 48". 9.944599 Hence, side a = 56~ 39' 36". In a similar manner we find, b= 114~ 29' 58" c- = 83~ 12' 06" Ex. 2. In a spherical triangle ABC, there are given A = 109~ 55' 42", B= 116~ 38' 33", and C= 120~ 43' 37"; required the three sides. a 980 21' 40" Ans. b -1090 50' 22" c = 1150 13' 26" CASE V. Havbig given in a spherical triangle, two sides and their included angle, to find the remaining parts. 23. For this case we employ the two first of Napier's Analogies. SPHERICAL TRIGONOMETRY. 345 cos ~(a + ): cos (a - b):: cot C: tang (A + B), sin (a + b): sin (a- b):: cot C: tang~(A - B). Having found the half sum and the half difference of the angles ii and B, the angles themselves become known; for, the greater angle is equal to the half sum plus the half difference, and the lesser is equal to the half sum minus the half difiference. The greater angle is then to be placed opposite the greater side. The remaining side of the triangle can be found by Case II. Ex. 1. In a spherical.triangle ABC, there are given a = 68 46" 02", b = 37 10', and C =39 23'; to find the remaining parts. (a + b) = 520 58' 1", -(a -b)=15~ 48' 01", IC=-190 41' 30". As cos *(a + b) 52~ 58' 01" log. ar. comp. 0.220210:.cos (a — b) 15~ 48' 01"... 9.983271:: cot C C 190 41' 30"... 10.446254: tang ~(A + B) 77~ 22' 25". 10.649735 As sin ~(a + b) 520 58' 01" log. ar. comp. 0.097840: sin ~(a —b) 15~ 48' 01"... 9.435016:: cot ~C 19~ 41' 30"... 10.446254: tang -(A - B) 43~ 37' 21"... 9.979110 Hence, A = 770 22' 25" + 430 37' 21" = 120~ 59' 46' B = 770 22' 25" - 430 37' 21"= 330 45' 04" side c.. = 43~ 37' 37" Ex. 2. In a spherical triangle ABC, there are given b = 830 19' 42", = 230 27' 46"; the contained angle A = 20~ 39' 48": to find the remaining parts. vB= 1560 30 16" Ans. C= 90 11' 48" (a= 61~ 32' 12" CASE VI. In a spherical triangle, having given two angles and the included side, to find the remaining parts. 24. For this case, we employ the second of Napier's Analogies. 346 SPHERICAL TRIGONOMETRY. cos 1-(A+B): cos-(A —B):: tang Ic: tang -(a+b), sin I (A + B): sin!-(A - B):: tang IC: tang -(a — b). From which a and b are found as in the last case. The remaining angle can then be found by Case I. Ex. 1. In a spherical triangle ABC, there are given A = 81~ 38' 20", B = 700 09' 38", c = 59~ 16' 23": to find the remaining parts. (A+B)=750 53' 59", (A-B)=50 44' 21", c=29~ 38' 11". As cos (A + B) 75~ 53' 59" log. ar. comp. 0.613287 cos ~-(A - B) 5~ 44' 21"... 9.997818 tang 2c 290 38' 11"... 9.755061: tang - (a + b) 660 42' 52".. 10.366156 As sin ~(A + B) 750 53' 59" log. ar. comp. 0.013286 sin ~(A - B) 50 14' 21"... 9.000000 tang I C 290 38' 11"... 9.755051: tang ~(a - b) 30 21' 25"... 8.768337 Hence, a = 660 42' 52" + 30 21' 25"= 70~ 04' 17" b = 66~ 42' 52" - 30 21' 25" = 630 21' 27" angle C... = 64~ 46' 33" Ex. 2. In a spherical triangle ABC, there are given A = 340 15' 03", B = 420 15' 13", and c = 760 35' 36": to find the remaining parts. a = 400 00' 10" Ans. b= 500 10' 30" C = 1210 36' 19" MENSURATION OF SURFACES. 1. WE determine the area, or contents of a surface, by finding how many times the given surface contains some other surface which is assumed as the unit of measure. Thus, when we say that a square yard contains 9 square feet, we should understand that one square foot is taken for the unit of measure, and that this unit is contained 9 times in the square yard. 2. The most convenient unit of measure for a surface, is a square whose side is the linear unit in which the linear dimensions of the figure are estimated. Thus, if the linear dimensions are feet, it will be most convenient to express the area in square feet; if the linear dimensions are yards, it will be most convenient to express the area in square yards, &c. 3. We have already seen (B. IV., P. 4, S. 2), that the term, rectangle or product of two lines, designates the rectangle constructed on the lines as sides; and that the numerical value of this product expresses the number of times which the rectangle contains its unit of measure. 4. To find the area of a square, a rectangle, or a parallelogram. Mfultiply the base by the altitude, and thle product will be the area (B. IV., P. 5). Ex. 1. To find the area of a parallelogram, the base being 12.25, and the altitude 8.5. Ans. 104.125. 2. What is the area of a square whose side is 204.3 feet? Ans. 41738.49 sq. ft. 3. What are the contents, in square yards, of a rectangle whose base is 66.3 feet, and altitude 33.3 feet? Ans. 245.31. 348 MENSURATION OF SURFACES. 4. To find the area of a rectangular board, whose length is 12~ feet, and breadth 9 inches. Ans. 93 sq. ft. 5. To find the number of square yards of painting in a parallelogram, whose base is 37 feet, and altitude 5 feet 3 inches. Ans. 217. 5. To find the area of a triangle. CASE I. When the base and altitude are given. Multiply the base by the altitude, and take half the product. Or, multiply one of these dimensions by half the other (B. IV., P. 6). Ex. 1. To find the area of a triangle, whose base is 625, and altitude 520 feet. Ans. 162500 sq. ft. 2. To find the number of square yards in a triangle, whose base is 40, and altitude 30 feet. Ans. 662. 3. To find the number of square yards in a triangle, whose base is 49, and altitude 251 feet. Ans. 68.7361. CASE II. 6. When two sides and their included angle are given. Add together the logarithms of the two sides and the logarithmic sine of their included angle; from this sum subtract the logarithm of the radius, which is 10, and the remainder will be the logarithm of double the area of the triangle. Find, from the table, the number answering to this logarithm, and divide it by 2; the quotienf will be the required area. Let BAC be a triangle, in which there are given BA, BC, and the included angle B. From the vertex A draw AD perpendicular to the base BC, and repre- B C sent the area of the triangle by Q. Then (Trig. Th. I.), R: sin B:: BA AD: BA X sinB hence, AD = BC x AD ~But, Q~= -2 (Art. 5): MENSURATION OF SURFACES. 349 hence, by substituting for AD its value, we have, BCx BA X sin B BC X BA X sinB Q2=2,or, 2Q= Taking the logarithms of both members, we have, log. 2 Q = log. BC + log. BA + log. sin B- log R. which proves the rule as enunciated. Ex. 1. What is the area of a triangle whose sides are, BC = 125.81, BA = 57.65, and the included angle B= 570 25'? + log. BC 125.81 2.099715 Then l 2 Q + log. BA 57.65 1.760799 og. 2 + log. sin B 57~ 25' 9.925626 - log. R.. -10. log. 2 Q.... 3.786140 and 2 Q = 6111.4, or Q = 3055.7, the required area. 2. What is the area of a triangle whose sides are 30 and 40, and their included angle 28~ 57'? Ans. 290.427. 3. What is the number of square yards in a triangle of which the sides are 25 feet and 21.25 feet, and their included angle 45~? Ans. 20.8694. CASE III. 7. When the three sides are known. 1. Add the three sides together, and takce half their sum. 2. From this half-sum subtract each side separately. 3. Multiply together the half-sum and each of the three re-'mainders, and the product will be the square of the area of the triangle. Then, extract the square root of this product, for the required area. Or, After having obtained the three remainders, add together the logarithm of the half-sum and thle logarithms of the respective remainders, and divide their sum by 2: the quotient will be the logarithm of the area. 350 MENSURATION OF SURFACES. Let A CB be a triangle: and denote C the area by Q: then, by the last case, we have, 6 a Q = - bc x sinA. Bu.t, we have (Plane Trig., Art. 78), A B sin A = 2 sin -A cos~JA; hence, Q = be sin'A cos i A. By substituting in this equation the values of sin WA, and cos ~ A, found in Arts. 92 and 93, Plane Trigonometry, we obtain, Q= s(s-a) (s-b) (s-c). Ex. 1. To find the area of a triangle whose three sides 20, 30, and 40. 20 45 45 45 half-sum. 30 20 30 40 40 25 ist rem. 15 2d rem. 5 3d rem. 2)90 45 half-sum. Then, 45 X 25 X 15 X 5 = 84375. The square root of which is 290.4737, the required area. 2. How many square yards of plastering are there in a triangle whose sides are 30, 40, and 50 feet? Ans. 663. 8. To find the area of a trapezoid. Add together the two parallel sides: then multiply their sum by the attitude of the trapezoid, and half the product will be the required area (B. IV., P. 7). Ex. 1. In a trapezoid the parallel sides are 750 and 1225, and the perpendicular distance between them is 1540; what is the area? Ans. 152075. 2. How many square feet are contained in a plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches? Ans. 132 sq. ft. 3. How many square yards are there in a trapezoid, whose parallel sides are 240 feet, 320 feet, and altitude 66 feet? Ans. 20533. MENSURATION OF SURFACES. 351 9. To find the area of a quadrilateral. Join two of the angles by a diagonal, dividing the quadrilateral into two triangles. Then, from each of the other angles let fall a perpendicular on the diagonal: then multiply the diagonal by half the sum of the two perpendiculars, and the product will be the area. Ex. 1. What is the area of the D quadrilateral ABCD, the diagonal A O being 42, and the perpendicA"" C ulars Dg, Bb, equal to 18 and 16 feet? Ans. 714. B 2. How many square yards of paving are there in the quadrilateral whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 and 331 feet? Ans. 222A-. 10. To find the area of an irregular polygon. Draw diagonals dividing the proposed polygon into trapezoids and triangles. Then find the areas of these figures separately, and add them together for the contents of the whole polygon. Ex. 1. Let it be required to deter- ID mine the contents of the polygon ABCDE, having five sides. E C Let us suppose that we have \ measured the diagonals and perpen- A diculars, and found AC= 36.21, (7 = 39.11, Bb = 4, Dd = 7.26, Aa = 4.18: required the area. Ans. 296.1292. 11. To find the area of a long and irregular figure, bounded on one side by a right line. 1. At the extremities of the right line measure the pe~pendicular breadths of the figujre; then divide the lilze into an!y number of equal parts, and measure the breadth at each point of division. 2. Add together the intermediate breadths and haly the sum of the extreme ones: then multiply this sum by one of the equal parts of the base line: the product will be the required area, very nearly. 3052 MENSURATION OF SURFACES. Let AEea be an irregular figure, d having for its base the right line a; AE. Divide AE into equal parts, and at the points of division A, B, C, D, and: X, erect the perpendiculars Aa, Bb, Cc, )Dd, Ee, to the base line AX, and designate them respectively by the letters a, b, c, d, and e. Then, the area of the trapezoid ABba = a 2 AB, b+c the area of the trapezoid BCcb 2 X BC, the area of the trapezoid CDdc= 2 X gCD d e and the area of the trapezoid _DEed = 2 X D)E; hence, their sum, or the area of the whole figure, is equal to 2,2 2 2 (a b~b~c+ c + /+)XAB, since AB, BC, &c., are equal to each other. But this sum is also equal to 2 J c c + d o 2t X AB, which corresponds with the enunciation of the rule. Ex. 1. The breadths of an irregular figure at five equidistant places being 8.2, 7.4, 9.2, 10.2, and 8.6, and the length of the base 40: required the area. 8.2 4)40 8.6 10 one of the equal parts. 2)16.8 8.4 mean of the extremes. 7.4 35.2 sum. 9.2 10 10.2 352 = area. 35.2 sum. 2. The length of an irregular figure being 84, and the breadths at six equidistant places 17.4, 20.6, 14.2, 16.5, 20.1, and 24.4; what is the area? Ans. 1550.64. -MENSURATION OF SURFACES. 353 12. To find the area of a regular polygon. Mfultiply half the perimeter of the polygon by the apothem, or Seepencdicular let fall from the centre onz one of the sides, and the product will be the area required (B3. v., p. 8). REMARK I.-The following is the manner of determining the perpendicular when one side and the number of sides of the regular polygon are known: First, divide 360 degrees by the number of sides of the polygon, and the quotient will be the angle at the centre; that is, the angle subtended by one of the equal sides. Divide this angle by 2, and half the angle at the centre wil'l then be known. Now, the line drawn from the centre to an angle of the polygon, the perpendicular let fall on one of the equal sides, and half this side, form a riglt-angled triangle, in which there are known the base, which is half the side of the polygon, and the angle at the vertex. Hence, the perpendicular can be determined. Ex. 1. To find the area of a reg- ----- ular hexagon, whose sides are 20 feet each. 6)360 \ 600 - A CB, the angle at the centre. 30~ = A CD, half the angle at centre. A Also, CAD = 90~ - A CD = 60~; and, AD = 10. Then, as sin AC). 30~, ar. comp. 0.301030: sin CAD. 60~,... 9.937531:AD.. 10,... 1.000000:CD. 17.3205.. 1.238561 Perimeter = 120, and half the perimeter = 60. Then, 60 X 17.3205 = 1039.23, the area. 2. What is the area of an octagon whose side is 20'! A 2. 1931.36886. REMARK II.-The area of a regular polygon of any number of sides is easily calculated by the above rule. 23 354 MENSURATION OF SURFACES. Let the areas of the regular polygons whose sides are unity, or 1, be calculated and arranged in the following TABLE. NAMt'S. SIDES. AREAS. NAMES. SIDES. AREAS. Triangle.. 3. 0.4330127 Octagon.. 8. 4.8284271 Square.. 4.. 1.0000000 Nonagon.. 9. 6.1818242 Pentagon. 5.. 1.7204774- Decagon..10. 7.6942088 Hexagon.. 6.. 2.5980762 Undecagon.11. 9.3656399 Heptagon... 3.6339124 Dodecagon.12 11.1961524 Now, since the areas of similar polygons are to each other as the squares of their homologous sides (B. IV., P. 27), we have, 12: any side squared:: tabular area: area. Hence, to find the area of any regular polygon, I. Square the side of the polygon. 2. Then multiply that square by the tabular area set opposite the polygon of the same number of sides, and the product will be the required area. Ex. 1. What is the area of a regular hexagon whose side is 20? 202 = 400, tabular area = 2.5980762. Hence, 2.5980762 x 400 = 1039.2304800, as before. 2. To find the area of a pentagon whose side is 25. Ans. 1075.298375. 3. To find the area of a decagon whose side is 20. Ans. 3077.68352. 13. To find the circumference of a circle when the diameter is given, or the diameter when the circumference is given. Multiply the diameter by 3.1416, and the product will be the circumference; or, divide the circumference by 3.1416, and the quotient will be the diameter. It is shown (B. V., P. 16, s. 1), that the circumference of a circle whose diameter is 1, is 3.1415926, or 3.1416. But, since the circumferences of circles are to each other as their MENSURATION OF SURFACES. 355 radii or diameters, we have, by calling the diameter of the second circle d, 1:d:: 3.1416: circumference, hence, d X 3.1416 = circumference. circumference Heence, also, d =.1416 3.1416 Ex. 1. What is the circumference of a circle whose diameter is 25? Ans. 78.54. 2. If the diameter of the earth is 7921 miles, what is the circumference? Ans. 24884.6136. 3. What is the diameter of a circle whose circumference is 11652.1904? Ans. 37.09. 4. What is the diameter of a circle whose circumference is 6850? Ans. 2180.41. 14. To find the length of an arc of a circle containing any number of degrees. Multiply the number of degrees in the given arc by 0.0087266, and the product by the diameter of the circle. Since the circumference of a circle whose diameter is 1, is 3.1416, it follows, that if 3.1416 be divided by 360 degrees, the quotient will be the length of an arc of 1 3.1416 degree: that is, 360 = 0.0087266 = are of one degree 360 to the diameter 1. This being multiplied by the number of degrees in an arc, the product will be the length of that arc in the circle whose diameter is 1; and this product being then multiplied by the diameter, the product is the length of the arc for any diameter whatever. REMARK.-When the arc contains degrees and minutes, reduce the minutes to the decimal of a degree, which is done by dividing them by 60. Ex. 1. To find the length of an arc of 30 degrees, the diameter being 18 feet. Ans. 4.712364. 2. To find the length of an arc of 12~ 10' or 12"~, the diameter being 20 feet. Ans. 2.123472. 3. What is the length of an arc of 10~ 15', or 10~", in a circle whose diameter is 68? Ans. 6.082396. 356 MENSURATION OF SURFACES. 15. To find the area of a circle. 1. Mialtiply the circumference by half the radius (B. v., P. 15). Or, 2. l]ziltiply the square of the radius by 3.1416 (B. v., P. 16). Ex. 1. To find the area of a circle whose diameter is 10, and circumference 31.416. Ans. 78.54. 2. Find the area of a circle whose diameter is 7, and circumference 21.9912. Ans. 38.4846. 3. How many square yards in a circle whose diameter is 31 feet? Ans. 1.069016. 4. What is the area of a circle whose circumference is 12 feet? Ans. 11.4595. 16. To find the area of a sector of a circle. 1. Multiply the arc of the sector by half the radius (B. V., P. 15, c). Or, 2. Compute the area of the whole circle: then say, as 360 degrees is to the degrees in the arc of the sector, so is the area of the whole circle to the area of the sector. Ex. 1. To find the area of a circular sector whose arc contains 18 degrees, the diameter of the circle being 3 feet. Ans. 0.35343. 2. To find the area of a sector whose are is 20 feet, the radius being 10. Ans. 100. 3. Required the area of a sector whose arc is 147~ 29', and radius 25 feet. Ans. 804.3986. 17. To find the area of a segment of a circle. 1. Find the area of the sector having the same arc, by the last problem. 2. Find the area of the triangle formed by the chord of the segment and the two radii of the sector. 3. Then add these two together for the answer when the segment is greater than a semicircle, and subtract the triangle from the sector when it is less. MENSURATION OF SURFACES. 357 Ex. 1. To find the area of the segment A CB, its chord AB being 12, and the radius EA, 10 feet. A/. As EA 10 ar. comp. 9.000000: AD 6.. 0.778151 iE:: sin D 90~.. 10.000000 * sin AED 36~ 52'= 36.87 9.778151 2 F 73.74 = the degrees in the are ACOB. Then, 0.0087266 X 73.74 X 20 = 12.87 = arc ABC nearly. 5 64.35 = area EA CB. Again, /EA-AD2 =