v^Vm,/ x:^ TN*.. \., 'K:->> V y I \ k -*D»- J^v^V^c A TREATISE ON Elementary Geometry WITH APPENDICES CONTAINING A COLLECTION OF EXERCISES FOR STUDENTS AND AN INTRODUCTION TO MODERN GEOMETRY. BY WILLIAM CHAUVENET, LL.D., Pbofessor 07 Matbematios ans Astronoxt vx Washinqton Universitt. PHILADELPHIA J. B. LIPPINCOTT & CO. 1876. Entered aaording to Act of Congress, in the year 1870, by J. B. LIPPINCOTT & CO^ In cue Clerk's OfBce of the Districf Court of the United States for the Eastern Dunxtci of Pennsylvania. Ct7?IllC0TT'8 PRBSS, P»ii.A».i.raiA. (^ \ PREFACE The invention of Analytic Geometry by Descartes in the early part of tiie seventeenth century, quickly followed by that of the Infinitesimal Calculus by Newton and Leibnitz, pro- duced a complete revolution in the mathematical sciences them- selves and accelerated in an astonishing degree the progress of all the sciences in which mathematics are aji{)lied, but arreste4Hington University, St. Louis, June 1, 1869. J« CONTENTS INTRODUCTION ^ 9 PLANE GEOMETRY. BOOK I. Rectiuneab Figubes ~ 12 BOOK IL The Cibcle o2 BOOK III. Pboportional Lines. Similab Figubes 91 BOOK ly. CJOMPABISON AND MeASUBEMENT OP THE SUBFACES OP BeCTILINEAB Figures 126 BOOK V. Regulab Polygons. Measurement op the Circle. Maxima and Minima op Plane Figures 142 GEOMETRY OF SPACE. BOOK VI. The Plane. Polyedral Angles 17J 7 8 CONTENTS. BOOK VII. PAOI POLTEDBONS 196 BOOK VIII. The Three Round Bodies. The Cylindeb. The Cone. The SpdEBE « 238 BOOK IX. Measubement op the Theee Round Bodies 271 APPENDIX I. 4 Exebcises in Elementaby Geometby 291 APPENDIX II. IWTBODUCTION TO MODEBN GeOMETBY 338 universittI Elements of Geometry. INTRODUCTION. 1. Every person possesses a conception of space indefinitely ex- tended in all directions. Material bodies occupy finite, or limited, portions of space. The portion of space which a body occupies can be conceived as abstracted from the matter of which the body is composed, and is called a geometrical solid. The material body filling the space is called 2i physical solid. A geometrical solid is, therefore, merely the form, or figure, of a physical solid. In this work, since only geometrical solids will be considered, we shall, for brevity, call them simply solids. 2. Definitions. In geometry, then, a solid is a limited, or bounded, portion of space. The limits, or boundaries, of a solid are surfaces. The limits, or boundaries, of a surface are lines. The limits of a line are points. 3. A solid has extension in all directions ; but for the purpose of measuring its magnitude, it is considered as having three specific dimensions, called length, breadth and thickness. A surface has only two dimensions, length and breadth. A line has only one dimension, namely, length. The intersection tf two surfaces is a line. A point has no extension, and therefore neither length, breadth nor thickness. The intersection of two lines is a point. 4. Although our first notion of a surface, as expressed in the defi- nition above given, is that of the boundary of a solid, we can suppose 10 GEOMETRY. such boundary to be abstracted and considered separately from the solid. Moreover, we may suppose a surface of indefiuite extent as to length and breadth ; such a surface has no limits. Similarly, a line may be considered, not only as the limit of a surface, but as abstracted from the surface and existing separately in space. Moreover, we may suppose a line of indefinite length, or without limits. Finally, a point may be considered, not merely as a limit of a line, but abstractly as having only position in space. 5. Definitions. A straight line is the shortest line between two points ; as AB. Since our first conception of a straight line may be regarded as derived from a comparison of all the lines that can be imagined to exist between two points, i.e., of lines of limited length, this definition (which is the most common one) may be admitted as expressing such a first conception ; but since we can suppose straight lines of indefi- nite extent, a more general definition is the following: A straight line is a line of which every portion is the shortest line between the points limiting that portion. A broken line is a line composed of difiTer- ent successive straight lines; as ABCDEF. A curved line, or simply a curve, is a line no portion of which is straight; as ABC. If a point moves along a line, it is said to de^a^ibe the line. 6. Definitions. A plane surface, or simply a plane, is a surface in which, if any two points are taken, the straight line joining these points lies wholly in the surface. A curved surface is a surface no portion of which is plane. 7. Solids are classified according to the nature of the surfaces w hich limit them. The most simple are bounded by planes. 8. Definitions. A geometrical figure is any combination of points, lines, surfaces, or solids, formed under given conditions. Figures formed by points and lines in a plane are called plane figures. Those formed by straight lines alone are called rectilinear, or right-lined^ figures ; a straight line being otlten called a right line. INTRODUCTION. 11 9. Definitions. Geometry may be defined as the science of extension and position. More specifically, it is the science which treats of the construction of figures under given conditions, of their rneasuremeni. and of their properties. Plane geometry treats of plane figures. The consideration of all other figures belongs to the geometry oj space, also called the geometry of three dimensions. 10. Some terms of frequent use in geometry are here defined. A theorem is a truth requiring demonstration. A lemma is an auxiliary theorem employed in the demonstration of another theo- rem. A problem is a question proposed for solution. An axiom is a truth assumed as self-evident. A postulate (in geometry) assumes the possibility of the solution of some problem. Theorems, problems, axioms and postulates are all called propo- sitions. A corollary is an immediate consequence deduced from one or more propositions. A scholium is a remark upon one or more propositions, pointing out their use, their connection, their limitation, or their extension. An hypothesis is a supposition, made either in the enun- ciation of a proposition, or in the course of a demonstration. PLANE GEOMETRY BOOK I. RECTILINEAR FIGURES. THE STRAIGHT LINE. 1. Ax TOM. There can be but one straight line between the same two points. 2. Podulate. A straight line can be drawn between any two points ; and any straight line can he produced (i. e., prolonged) indefinitely. 3. Axiom. If two indefinite straight lines coincide in two points, they coincide throughout their whole extent, and form but one line. Hence two points determine a straight line; and a straight line may be designated by any two of its points. 4. Different straight lines drawn from the same point are said to have different directions ; as OA, OD, etc. The ^ c _ point from which they are drawn, or at which they commence, is often called the origin. If any one of the lines, as OA, be produced through 0, the portions OA, OB, on opposite sides of 0, may be regarded as two different lines having opposite directions reckoned from the common origin 0. Hence, also, every straight line AB has two opposite directions, namely, from A toward B (J. being regarded as its origin) expressed by AB, and from B toward A (B being regarded as its origin) expressed by BA. If a line AB is to be produced through B, that is, toward C, we should express this by saying that ^ — + J — - AB is to be produced; but if it is to be 12 BOOK I. 13 produced through A^ that is toward D, we should express this by saying that BA is to be produced. ANGLES. 5. Definition. An angle is a figure formed by two straight lines drawn from the same point; thus OA, OB form an angle at 0. The lines OA, OB are called the sides of the angle ; the common point 0, its vertex. An isolated angle may be designated by the letter at its vertex, as "the angle 0;" but when several angles are formed at the samf point by different lines, as OA, OB, OC, we desig- nate the angle intended by three letters ; namely, by one letter on each of its sides, together with the one at its vertex, which must be written between the other two. Thus, with these lines there are formed three different angles, which are distinguished as A OB, BOC and AOC. Two angles, such as A OB, BOC, which have the same vertex and a common side OB between them, are called adjacent. 6. Definition. Two angles are equal when one can be placed upon the other so that they shall coincide. Thus, the angles A OB and A' O'B' sire equal, i{ A' O'B' can be superposed upon A OB so that while O'A' coin- cides with OA, 0' B' shall also coincide with OB. The equality of the two angles is not affected by producing the sides ; for the coincident sides con- tinue to coincide when produced indefinitely (3).* Thus the magnitude of an angle is independent of the length of its sides. 7. A clear notion of the magnitude of an angle will be obtained by supposing that one of its sides, as OB, was at first coincident with the other side OA, and that it has revolved about the point (turning upon as the leg of a pair of dividers turns upon its hinge) until it has arrived at the position OB. During this revolution the movable side makes with the fixed side a varying angle, which increases by insensible degrees, that is, continuously; and the revolving line is * An Arabic numeral alone refers to an article in the same Book ; but in refer- ring to articles in another Book, the number of the Book is also given. 14 G EOMETR Y. said to describe, or to generate, the angle A OB. By continuing the revolution, an angle of any magnitude may be generated. It is evident from this mode of generation, as well as from the defi- nition (6), that the magnitude of an angle is independent of the length of its sides. PERPENDICULARS AND OBLIQUE LINES. 8. Definition. When one straight line meets another, so as to make two adjacent angles equal, each of these angles is called a right angle ; and the first line is said to be perpendicular to the second. Thus, \i AOG and BOC are equal angles, ^ each is a riglit angle, and the line CO is per- pendicular to AB. Intersecting lines not perpendicular are said to be oblique to each other. PROPOSITION L— THEOREM. 9. At a given point in a straight line one perpendicular to the line can he drawn^ and but one. Let be the given point in the line AB. Suppose a line 02), constantly passing through 0, to revolve about ^ 0, starting from the position OA In any one of its successive positions, it makes two different angles with the line AB; one, AODy with the portion OA; and another, ^0Z>, with the por- tion OB. As it revolves from the position OA around to the posi- tion OB, the angle AOD will continuously increase, and the angle BOD will continuously decrease. There will therefore be one posi- tion, as OC, where the two angles become equal ; and there can evi- dently be but one. 10. Corollary. All right angles are equal. That is, the right angles AOC, BOC made by a line CO meeting AB, are each equal to each of the right anglesJ^'O'C", B'O'C, made by a line CO' meeting any other line A'B'. For, the line A'O'B' can be ap- ^ - plied to the line A OB, so that 0' shall Cf Bf BOOK 15 fall upon 0, and then O C will fall upon OC, unless there can be two perpendiculars to AB at 0, which by the preceding proposition is impos^sible. The lines will therefore coincide and the angles will be equal (6). ; PKOPOSITION II.— THEOREM. 11. The two adjacent angles which one straight line makes with another are together equal to two right angles. If the t'.vo angles are equal, they are right angles by the definition (8), and no proof is necessary. If they are not equal, as AOD and BOD, still the sum o^ AOD and BOD is equal to two right angles. For, let OCbe drawn at perpendicular to AB. The angle AOD is the ^ sum of the two angles J^OC and COD. Adding the angle BOD, the sum of the two angles A OD and BOD is the sum of the three angles AOC, COD and BOD. The first of these three is a ^ ^ right angle, and the other two are together equal to the right angle BOC; hence the sum of the angles AOD and BOD is equal to two right angles. 12. Corollary I. If one of the two adjacent angles which one line makes with another is a right angle, the other is also a right angle. 13. Corollary II. If a line CD is perpen- dicular to another line AB, then, reciprocally, the line AB is perpendicular to CD. For, CO being perpendicular to AB at 0, AOC ^ is a right angle, hence (Cor. I.) AOD is a right angle, and A or AB is perpendicular to CD, ^ 14. Corollary III. The sum of all the consecutive angles, A OB^ BOCf COD, DOE, formed on the same side of a straight line AE, at a common point 0, is equal to two right angles. For, their sum is equal to the sum of the two adjacent angles A OB, BOE, which by the proposition is equal to two right angles. 16 GEOMETRY. 15. Corollary TV. The sum of all the cousecutive angles AOB BOC, COD, DOE, EOA, formed about a point 0, is equal to four right angles. For, if two straight lines are drawn through 0, perpendicular to each other, the sum of all the consecutive angles formed about will be equal to the four right angles formed by the perpendiculars. 16. Scholium. A straight line revolving from the position OA around to the position OB describes the two right angles AOC and COB; hence OA and OB, regarded as two different lines having opposite directions (4), are frequently said to make an angle with each other equal to two right angles. A line revolving from the position OA from right to left, that is, successively into the positions OC, OB, OD, when it has arrived at the position OD will have described an angle greater than two right angles. On the other hand, if the position OD is reached by revolving from left to right, that is, successively into the positions OE, OD, then the angle AOD is less than two right angles. Thus, any two ^ straight lines drawn from a common point make two different angles with each other, one less and the other greater than two right angles. Hereafter the angle which is less than two right angles will be understood, unless otherwise expressly stated. 17. Definitions. An acute angle is an angle ^ less than a right angle; as AOD. An obtuse angle is an angle greater than a right angle; as BOD. 18. When the sum of two angles is equal to a right angle, each is called the complement of the other. Thus DOC is the complement of AOD, and AOD is the complement of DOC. 19. When the sum of two angles is equal to two right angles, each is called the supplement of the other. Thus BOD is the supplement of AOD, and AOD is the supplement of BOD. BOOK I. 17 20. It is evident that the complements of equal angles ar*' equal to each other; and also that the supplements of equal angles are equal to each other. PROPOSITION III.— THEOREM. 21. Conversely, if the sum of two adjacent angles is equal to two right angles, their exterior sides are in the same straight line. Let the sum of the adjacent angles AOD, BOD, be equal to two right angles ; then, 0^ and OB are in the same straight line. For BOD is the supplement of AOD (19), and is therefore identical with the angle which OD makes with the prolongation of ^0 (11). Therefore OB and the prolongation of A are the same line. 22. Every proposition consists of an hypothesis and a conclusion. The converse of a proposition is a second proposition of which the hypothesis and conclusion are respectively the conclusion and hy- pothesis of the first. For example, Proposition II. may be enun- ciated thus : Hypothesis — if two adjacent angles have their exterior sides in the same straight line, then — Conclusion— i\\Q sum of these adjacent angles is equal to two right angles. And Proposition III. may be enunciated thus : Hypothesis — if the sum of two adjacent angles is equal to two right angles, then — Conclusion — these adjacent angles have their exterior sides in the same straight line. Each of these propositions is therefore the converse of the other. A proposition and its convei-se are however not always both true. PROPOSITION IV.— THEOREM. 23. If two straight lines intersect each other, the opposite (or vertical) angles are equal. Let AB and CD intersect in 0; then will the B c opposite, or vertical, angles AOC and BOD be equal. For, each of these angles is the supple- ment of the same angle BOC, or AOD, and hence they are equal (20). 2* B 18 GEOM.ETRY. In like manner it is proved that the opposite angles A OD and BOC are equal. 24. Corollary I. The straight line EOF which bisects the angle AOG also bisects its vertical angle BOD, For, the angle FOD is equal to its vertical angle EOC, and FOB is equal to its vertical angle EOA , therefore if I EOC ahd EOA are equal, FOD and FOB are equal. 25. Corollary II. The two straight lines EOFf HOG, which bisect the two pairs of vertical angles, are perpendicular to each J other. For, HOC = HOB and COE = BOF; hence, by addition, HOC + COE = HOB + BOF; that is, HOE = HOF; therefore, by the definition (8), HO is perpen- dicular to FE. PROPOSITION v.— THEOREM. 26. From a given point without a straight line, one perpendicular can be drawn to that line, and hut one. Let AB be the given straight line and P the given point. The line AB divides the plane in which it is situated into two portions. Let the por- tion containing P, which we suppose to be the upper portion, be revolved about the line AB {i.e., folded over) until the point P comes into the lower portion; and let P' be that point in the plane with which P coincides after this revolution. Restoring P to its original position, join PF\ cutting AB in C, and again revolve the upper portion of the plane about AB until P again coincides with P'. Since the line AB is fixed during the revolution, the point Cis fixed; therefore PC will coincide with P'C, aid the angle PCD with the angle P' CD. These angles are therefore equal (6), and BC is perpendicular to PP' (8), or PC perpendicular to AB (13). There can therefore be one perpendicular from the point P to the line AB. Moreover, PC is the only perpendicular. Let PD be any other BOOK I. 19 line drawn from P to AB^ and join P'D. Then, when the upper portion of the plane is revolved until P coincides with P', D being fixed, PD coincides with P'D, and consequently the angle Pi) with the angle P'DC. Hence the angles PDC and P'DC are equal. Now PP' being the only straight line that can be drawn from P to P' (1), PDP' is not a straight line ; and if PD is produced to E, PDE and DP' are different straight lines. Hence the angle PDP' is less than two right angles, and its half, PDC, is less than one right angle; that is, PD is an oblique line. Therefore PC is the only perpendicular. . 27. Corollary. Of the two angles which any oblique line drawn from P makes with AB, that one is acute within which the perpeu dicular from Pupon AB falls ; thus, PDC is acute. PROPOSITION VI.— THEOREM. 28. The perpendicular is the shortest line that can be drawn from a point to a straight line. Let PC be the perpendicular, and PD any oblique line, from the point P to the line AB. Then PC < PD. For, produce PC to P', making CP' = CPf and join P'D. When the portion of the plane which contains P is revolved about AB, as in the preceding proposition, until P coincides with P', PD also coincides with P'D; and hence PD = P'D. But the straight line PP', being the shortest distance between the points P and P', is less than the broken line PDP'. Therefore PC, the half of the straight line, is less than PD, the half of the broken line. 29. Definition. By the distance of a point from a line is always understood the shortest distance. By the preceding proposition, therefore, the perpendicular measures the distance of a point from a straight line. Also, by the distance of one point from another is understood the diortest distance, that is, the straight line between the points. 20 GEOM ETIl Y. PROPOSITION VII.— THEOREM. 30. Two oblique lines drawn from the same point to a straight line^ cutting off equal distances from the foot of the perpendicular , are equal. Let the oblique lines PD, FE, meet the line AB in the points D and Ef cutting off the equal distances CD and CE from the foot of the perpendicular. Then PB = PE. For, DCE being perpendicular to PC, and CD = CE, the figure PCD may be re- volved about PC into coincidence with PCE; and yince the point D will fall on E, PD will coincide with PE. Therefore PD = PE. 31. Corollary. The angles PDC, and PEC are equal ; that is, two equal straight lines from a point to a straight line make equal acute angles with that line. 32. Definition. A broken line, as ABCDE, is called convex, when no one of its component straight lines, if pro- duced, can enter the space enclosed by the b ^ — --— ..^/> broken line and the straight line joining its extremities. CH PROPOSITION VIII.— THEOREM. 33. A convex broken line is less than any other line which envelops it and has the same extremities. Let the convex broken line AFGE have the same extremities -4, E, as the line ABCDE, and be enveloped by it; that is, wholly included within the space bound-ed by ABCDE and the straight line AE. Then AFGE < ABCDE. For, produce AF and EG to meet the en- veloping line in H and K. Imagine ABCDE to be the path of a point moving from A to E. If the straight line AH be substituted for ABCH, the path AHDE mil be shorter than the path ABCDE the portion HDE being common to both. If, further, the straight line FK be substituted for FHDK, the path AFKE will be a still shorter path from A to E. And if, finally, GE be substituted for BOOK I. 21 GKE, AFGEviWl be a still shorter path. Therefore, AFGE is less than any enveloping line. 34. Scholium. The preceding demonstration applies when the en- veloping line is a curve, or any species of line whatever. PKOPOSITION IX.— THEOREM. 35. Oj two oblique lines drawn from the same point to the suine straight line, that is the greater which cuts off upon the line the greater distance from the perpendicular. Let FC be the perpendicular from F to AB, and suppose CE > CD; then FE > FD. For, produce FC to PVraaking CF' = CF, and join DF', EF'. Then, as in Pro- position VI., we have FD = F'D, and FE ---= F'E. But (33), the broken line FDF' is less than the enveloping line FEF' ; therefore FD, the half of FDF', is less than FE, the half of FEF'. If the two oblique lines are on opposite sides of the perpendicular, as FE and FD', and if CE > CD', take CD = CD', and join FD. Then, as above FE^ FD; and, by Proposition VII., FD = FD' ; hence FE > FD'. 36. Corollary I. (Converse of Proposition VII.). Two equal ob- lique lines cut off equal distances from the perpendicular. 37. Corollary II. (Converse of Proposition IX.). Of two unequal oblique lines, the greater cuts off the greater distance from the per- pendicular. . PEOPOSITION X.— THEOREM. 38. If a perpendicular is erected at the middle of a straight line, then, 1st. Every point in' the perpendicular is equally distant from the extremities of the line ; 2d. Every point withou. the perpendicular is unequally distant from the extremities of the line. 22 GEOMETRY. Let AB be a finite straight line, and C its middle point ; then, 1st. Every point P in the perpendicular erected at C is equally distant from A and B. For, since CA = CB, we have (30) PA =^PB. 2d. Any point Q without the perpendicular is unequally distant from A and B. For, Q being on one side or the other of the perpendicular, one of the lines QAy QB must cut the perpendicular ; let it be QA and let it cut in P; join PB. The straight line QB is less than the broken line QPB, that is, QB <: QP -\- PB. But PB = PA; therefore QB PROPOSITION XIX.— THEOREM. 74. The angle contained by two straight lines drawn from any point within a triangle to the extremities of one of the sides is greater than the angle contained by the other two sides of the triangle. From any point D, within the triangle ABC, let ^ DB, DC be drawn; then, the angle BDC is greater than the angle J5J.C For, produce BD to meet AC in E. We have the angle BDC>BEC(i6d), and the angle BEC> BA C; hence BDC> B AC 75. Definition. Equal triangles, and in general equal figures, are those which can exactly fill the same space, or which can be applied to each other so as to coincide in all their parts. 4 PROPOSITION XX.— THEOREM. 76. Two triangles are equal when two sides and the included angle of the one are respectively equal to two sides and the included angle of the other. In the triangles ABC, DEF, let AB be equal to DE, BC to EF, and the included angle B fl equal to the included angle ^ ^ ^mm E; then, the triangles are equal. For, the triangle ABC may be superposed upon the triangle DEF, by applying the angle B to the equal angle E, the side BA upon its equal ED, and the side BC upon its equal EF. The points A and C then coinciding with the points D and F, the side A C will coincide with the side DF, and the triangles will coin- cide in all their parts ; therefore they are equal (75). " 77. Corollary. If in two triangles ABC, DEF, there are given B = E, AB = DE Sind BC = EF, there will follow A = D,C = F, and AC =DF, r- /■ BOOK I. 31 PROPOSITION XXI.— THEOREM. 78. Two triangles are equal when a side and the two adjacent angles of the one are respectively equal to a side and the two adjacent angles of the other. In the triangles ABC, DBF, let ^C be equal to EF, and let th^ angles B and C adjacent to BGhQ respectively equal to the angles E and F adja- cent to EF; then, the tri- angles are equal. For, the triangle ABC may be superposed upon the triangle DEF, by applying BC to ita equal EF, the point B upon E, and the point C upon F. The angle B being equal to the angle E, the side BA will take the direction of ED, and the point A will fall somewhere in the line ED. The angle C being equal to the angle F, the side CA will take the direction of FD, and the point A will fall somewhere in the line FD. Hence the point A, falling at once in both the lines ED and FD, must fall at their intersection D. Therefore the triangles will coincide through- out, and are equal. 79. Corollary. If in two triangles ABC, DEF, there are given B = E, C = F, and BC = EF, there will follow A = D, AB = DE,aiidAC=DF. C E f F PROPOSITION XXII.— THEOREM. 80. Two triangles are equal when the three sides of the one are re- spectively equal to the three sides of the other. In the triangles ABC, DEF, let AB be equal to DE, AC to DF, and BC to EF; then, the triangles are equal. ^^ ^^^ For, suppose the triangle ABC to ^^ \ y^ h\ \ be placed so that its base BC coin- ^ ^ ^\^ \7 cides with its equal EF, but so that ^^G the vertex A falls on the opposite side of EF from Z), as at 6?; and join DG which intersects EF in H. 32 GEOMETRY. Then, by hypothesis, EG = ED and EG = ED ; therefore, E and F being two points equally distant from D and G, the line EE is per- pendicular to DG Sit its middle point -H'(41).^ Hence, if the figure DEE be revolved upon the line EF, H being a fixed point, HD will fall (ipon its equal HG, and the triangle DEE will coincide entirely with the triangle GEE. Therefore, the tri- angle DEE is equal to the triangle GEE, or to the triangle ABC. 81. Corollary. If in two triangles ABC, DEE, there are given AB = DE,AC=DE,BC= EE, there will follow A = D,B=E, C=E. 82. Scholium. In two equal triangles, the equal angles lie opposite to the equal sides. PROPOSITION XXIII.— THEOREM. 83. Two right triangles are equal, 1st, when the hypotenuse and a side of the one are respectively equal to the hypotenuse and a side of the other ; or, 2d, when the hypotenuse and an acute angle of the one are respectively equal to the hypotenuse and an acute angle of the other. 1st. In the right triangles ABC, DEE, let the hypotenuse AB be equal to DE, and the side AC to DE; then, the triangles are equal. For, applying AC to its equal DE, the angles C and E being c^qual, the side CB will take the direction FE, and B will fall somc- ;\liere in the line EE. But AB being equal to DE, will cut off on FE the same distance from the perpendicular (36), and hence B will fall at E. The triangles will therefore coincide, and are equal. 2d. Let AB = DE, and the angle ABC = the angle DEE', then, the triangles are equal. For, the third angles BAC and EDF are equal (73), and hence iho triangles are equal by (78). B O O K 1 . 33 _; PKOPOSITION XXIV.—TITEOKEM. 84. If two sides of a triangle are respectively equal to two sides oj another, hut the included angle in the first triangle is greater than the included angle in the second, the third side of the first triangle is greater than the third side of the second. Let ABC and ABD be the two triangles in which the sides A 7?, A C are respectively equal to the sides AB, AD, but the included angle BA C is greater than the included angle BAD ; then, BC is greater than BD. For, suppose the line AE to be drawn, bisect- ing the angle CAD and meeting BC in E; join DE. The triangles AED, AEC are equal (76), and therefore ED = EC. But in the triangle BDE we have BE^EDy BD, and substituting EC for its equal ED, . BE-\-EG> BD, or BC > BD. 85. Corollary. Conversely, if in two triangles ABC, DEE, we have AB = DE,AC = DF, hut BC> EF; then, A > D. For, if A were equal to D, we should ^ have BC =EF (76) ; and if A were less than D, we should have ^C < EF (by the above proposition) ; but as both these conclu- sions are absurd, being contrary to the hy- pothesis, we can only have J. > Z>. PEOPOSITION XXV.— THEOKEM. 86 In an isosceles triangle, the angles opposite to the equal sides are equal. Let AB and AC he the equal sides of the isosceles triangle ABC. then, the angles B and Care equal. ^ For, let D be the middle point of BC, and draw AD. a The triangles ABD and ADC are equal (80) ; therefore / : \ the angle ABD = the angle ACD (82). / \ \ 87. Corollary 1. From the equality of the triangles ^ — ^ — ^, ABD and ACD, we also have the angles ADB = ADC, 3** 34 GEOMETRY. and BAD = CAD ; that is, the straight line joining the vertex and the middle of the base of an isosceles triangle is perpendicular to the hose and bisects the vertical angle. ^ Hence, also, the straight line which bisects the vertical angle of an isosceles triangle bisects the base at right angles. H%.^ Corollary II. Every equilateral triangle is also equiangular; and by (68), each of its angles is equal to one-third of two right angles, or to two-thirds of one right angle. PROPOSITION XXVI.— THEOREM. 89. If two sides of a triangle are unequaly the angles opposite to them are unequal, and the greater angle is opposite to the greater side. In the triangle ABC, let AB be greater than AC\ then, the angle A CB is greater than the angle B. For, from the greater side AB cut off a part AD = AC, and join CD. The triangle ADC is isosceles, and therefore the angles ADC and ACD are equal (86). But the whole angle ACB is greater than its part ACD, and therefore greater than ADC; and ADC, an exterior angle of the triangle BDC, is greater than the angle B (69) ; still more, then, is A CB greater than B. PROPOSITION XXVII.— THEOREM. 90. If two angles of a triangle, are equal, the sides opposite to them are equal. In the triangle ABC, let the angles B and C be equal ; then, the sides AB and A C are equal. For, if the sides AB and A C were unequal, the angles B and C could not be equal (89). 91. Corollary. Every equiangular triangle is also equilateral. 1 • B O O K 1 . 35 PROPOSITION XXVIII.— THEOREM. 92. If two angles of a triangle are unequal, the sides opposite to them are unequal, and the greater side is opposite to the greater angle. In the triangle ABC let the angle Cbe greater than the angle JB; then, AB is greater than AC. For, suppose the line CD to be drawn, cutting off i'roni the greater angle a part BCD = B. Then BDC is an isosceles triangle, and DC= DB. But in the triangle ADC, we have AD + DC '> AC', or, putting DB for its equal DC, AD ^ DB> AC', ov AB> A C. POLYGONS. 93. Definitions. A polygon is a portion of a plane bounded by straight lines; as ABCDE. The bounding lines are the sides; their sura is the perimeter of the polygon. The angles which the adjacent sides make with each other are the angles of the polygon ; anS the vertices of these angles are called the vertices of the polygon. Any line joining two vertices not consecutive is called a diagonal; as J-C 94. Definitions. Polygons are classed according to the number of their sides : A triangle is a polygon of three sides. A quadrilateral is a polygon of four sides. A pentagon has five sides ; a hexagon, six ; a heptagon, seven ; an octagon, eight; an enneagon, nine; a decagon, ten; a dodecagon, twelve; etc. An equilateral polygon is one all of whose sides are equal; an equiangular polygon, one all of whose angles are equal. 95. Definition. A convex polygon is one no side of which when produced can enter within the space enclosed by the perimeter, as ABCDE in (93). Each of the angles of such a polygon is less than two right angles. It is also evident from the definition that the perimeter of a convex 36 GEO M ETR Y. polygon canuot be intersected by a straight line in more than two |X)ints. A concave polygon is one of which two or more sides, when produced, will enter the space enclosed by the perimeter; as MNOPQ, of which, OP and QP when produced will enter within the polygon. The angle OPQ, formed i)y two adjacent re-entrant sides, is called a re- entrant angle; and hence a concave polygon is sometimes called a re-entrant polygon. All the polygons hereafter considered will be understood to be convex. 96. A polygon may be divided into triangles by diagonals drawn from one of its vertices. Thus the pentagon ABODE is divided into three triangles by the diagonals drawn from A. The number of triangles into which any polygon can thus be divided is evi- dently equal to the number of its sides, less two. The number of diagonals so drawn is equal to the number of sides, Tess three. 97. Two polygons ABODE, A'B'G'D'E', are equal when they can be divided by diagonals into the same number of triangles, equal each to each, and similarly arranged ; for the polygons can evidently be super- posed, one upon the other, so as to coincide. 98. Definitions. Two polygons • are mutually equiangular when the angles of the one are re- spectively equal to the angles of the other, taken in the same order; as ABOD, A'B'O'D', in which A = A', B = B', etc. The equal angles are called homologous angles; the sides containing equal - angles, and similarly placed, are homologous sides ; thus A and A' are homologous angles, AB and A'B' are homologous eides, etc. BOOK I. 37 Two polygons are mutually equilateral when the sides of the one are respectively equal to the sides of the other, taken in the same order ; as MNPQ, M'N'P'q\ in which MN = M'N\ NP = N'P', etc. The equal sides are homologous ; and angles contained by equal sides simi- larly placed, are homologous ; thus MN and M'N' are homologous sides ; M and M' are homologous angles. Two mutually equiangular polygons are not necessarily also mu- tually equilateral. Nor are two mutually equilateral polygons necessarily also mutually equiangular, except in the case of tri- angles (80). If two polygons are mutually equilateral and also mutually equi- angular, they are equal ; for they can evidently be superposed, one upon the other, so as to coincide. PEOPOSITION XXIX.— THEOREM. 99. The sum of all the angles oj any polygon is equal to two right angles taken as many times less two as the polygon has sides. For, by drawing diagonals from any one vertex, the polygon can be divided into as many triangles as it has sides, less two (96). The sum of the angles of all the triangles is the same as the sum of the angles of the polygon, and the sum of the angles of each triangle is two right angles (68). Therefore, the sum of the angles of the polygon is two right angles taken as many times less two as the polygon has sides. 100. Corollary I. If N denotes the number of the sides of the polygon, and B a right angle, the sum of the angles is 2B X (N—2) = (_2N— 4:)E = 2NB — 4B; that is, twice as many right angles as the polygon has sides, less four right angles. For example, the sum of the angles of a quadrilateral is four right angles ; of a pentagon, six right angles ; of a hexagon, eight right angles, etc. 4 38 GEOMETRY. 101. Corollary II. If all the sides of any polygon ABODE, be /5roduced so as to form one exterior angle at each vertex, the sum of these exterior angles, a, b, c, d, e, is four right angles. For, the sum of each interior and its adjacent exterior angle, as 4. + <*j is two right angles (11); therefore, the sum of all the angles, both interior and exterior, is twice as many right angles as the polygon has sides. But the sum of the interior angles alone is twice as many right angles as the polygon has sides, less four right angles (100) ; therefore the sum of the exterior angles is equal to four right angles. This is also proved in a very simple manner, by drawing, from any point in the plane of the polygon, a series of lines respectively parallel to the sides of the polygon and in the same directions as their prolongations. The consecutive angles formed by these lines will be equal to the exterior angles of the polygon (60), and their sum is four right angles (15). QUADEILATERALS. 102. Definitions. Quadrilaterals are divided into classes as follows : 1st. The trapezium (J.) which has no two of its sides parallel. 2d. The trapezoid (J5) which has two sides par- allel. The parallel sides are called the bases, and the perpendicular distance between them the alti- tude of the trapezoid. 3d. The parallelogram (C) which is bounded by two pairs of parallel sides. The side upon which a parallelogram is supposed to stand and the opposite side are called its lower and upper bases. The perpendicular distance between the bases is the altitude. 103. Definitions. Parallelograms are divided into species, ju" follows : \ \Z^ BOOK I. 1st. The rhomboid (a), whose adjacent sides are not equal and whose angles are not right angles. 2d. The rhombus, or lozenge (6), whose sides are all equal. 3d. The rectangle (c), whose angles are all equal and therefore right angles. 39 4th. The square (d), whose sides are all equal and whoise angles are all equal. The square is at once a rhombus and a rectangle. PEOPOSITION XXX.— THEOEEM. 104. In every parallelogram, the opposite angles are equal, ana cfie opposite sides are equal. Let ABCD be a parallelogram. 1st. The opposite angles B and D, contained An by parallel lines lying in opposite directions, are equal (61) ; and for the same reason the opposite angles A and C are equal. 2d. Draw the diagonal A C. Since AD and BC a.re parallel, the alternate angles CAD and ACB are equal (49), and since DC and AB are parallel, the alternate angles ACD and CAB are equal. Therefore, the triangles ADC and CBA are equal (78), and the sides opposite to the equal angles are equal, namely, AD = BC,2indiDC=: AB. 105. Corollary I. A diagonal of a parallelogram divides it into two equal triangles. 106. Corollary II. If one angle of a parallelogram is a riglil angle, all its angles are right angles, and the figure is a rectangle. 40 GEOMETRY. PEOPOSITION XXXI.— THEOREM. 107. If the opposite angles of a quadrilateral are equals or if tXs opposite sides are equal, the figure is a parallelogram. 1st. Let the opposite angles of the quadrilateral ABCD be equal, or ^ -== O and B = D. Then, by adding equals, we have s- --. A + B=0 + D; \ \^ therefore, each of the sums A -{- B and C -\- D is equal to one-half the sum of the four angles. But the sum of the four angles is equal to four right angles (100) ; therefore, A -{- B is equal to two right angles, and the lines AD and ^Care parallel (56). In like manner it may be proved that AB and CD are parallel. Therefore the figure is a parallelogram. 2d. Let the opposite sides of the quadrilateral ABCD be equal, \j. _^L = AD and AB = DC. Then, drawing the diagonal AC, the triangles ABC, A CD are equal (80) ; therefore, the angles CAD and ACB are equal, and the lines AD and BC Sive parallel (54). Also since the angles CAB and ACD are equal, the lines AB and DC are parallel. Therefore ABCD is parallelogram. PROPOSITION XXXII.— THEOREM. 108. If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. Let the opposite sides BC and AD of the quadrilateral ABCD be equal and parallel. Draw the diagonal AC. The alternate angles CAD and ACB are equal (49), and hence the triangles ADC and CBA are equal (76). There- fore, the sides AB and CD are equal and the figure is a parallelo- gram (107). BOOK I. 41 / 7 PROPOSITION XXXIII.— THEOREM. 109. The diagonals of a parallelogram bisect each other. Let the diagonals J.C, BD of the parallelogram ABCD intersect in E) then, AE = ^Cand ED = EB. For, the side AD and the angles EAD, ADE, of the triangle EAD, are respectively equal to / the side CB and the angles ECB, EBC of the f^ Y triangle ECB\ hence these, triangles are equal (78), and the sides respectively opposite the equal angles are equal, namely, AE = EC and ED = EB. 110. Corollary I. The diagonals of a rhombus ABCD bisect each other at right angles in E. For, since AD = CD and AE = EC, ED is perpendicular to AC (41). 111. Corollary II. The diagonals of a rhombus bisect its opposite angles. For, in each of the isos- celes triangles ADC, ABC, BCD, DAB, the line drawn from the vertex to the middle of the base bisects the vertical angle (87). PROPOSITION XXXIV.— THEOREM. 112. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Let the diagonals of the quadrilateral ABCD bisect each other m E. Then, the triangles AED and CEB are equal (76), and the angles EAD, ECB, respect- ively opposite the equal sides, are equal. There- fore AD and BC are parallel (54). In like manner AB and DC are shown to be parallel, and the figure is a parallelogram. 113. Corollary. If the diagonals of a quadrilateral bisect each other at right angles, the figure is a rhombus. 4 * A /) />• 42 GEOMETRY. PROPOSITION XXXV.— THEOREM. 114. The diagonals of a rectangle are equal. ^ lyet ABCD be a rectangle; then its diagonals, AC and J5Z), are equal. For, the right triangles ABC and DCB are equal (76) ; therefore, AC = BD. 115. Corollary I. The diagonals of a square are equal, and, since the square is also a rhombus, they bisect each other at right angles (110), and also bisect the angles of the square (111). 116. Corollary II. A parallelogram is a rectangle if its diagonals are equal. 117. Corollary III. A quadrilateral is a square, if its diagonals are equal and bisect each other at right angles. 118. Scholium. The rectangle, being a species of parallelogram, has all the properties of a parallelogram. The square, being at once a ^rallelogram, a rectangle and a rhombus, has the properties of all these figures. PROPOSITION XXXVI.— THEOREM. 119. Two parallelograms are equal when two adjacent sides and the included angle of the one are equal to two adjacent sides and the included angle of the other. Let AC, A'C\ have AB = A'B\ ^ ^ j!! £" AD = A'D', and the angle BAD = I / / J B'A'D'\ then, these parallelograms a b a' B' are equal. For they may evidently be applied the one to the other so as to coincide throughout. 120. Corollary. Two rectangles are equal when they have equal bases and equal altitudes. BOOK I. 43 APPLICATIONS. PROPOSITION XXXVII.— THEOREM. 121. If a straigJit line draivn parallel to the base of a triangle bisects one of the sides, it also bisects the other side; and the portion of it intercepted between the two sides is equal to one-half the base. Let DE be parallel to the base BC of the triangle ABC, and bisect the side AB in D; then, it bisects the side ^Oin ^, and JDE = hBG. 1st. Through D suppose DF to be drawn parallel to AC. In the triangles ADE, DBF, we have AD -—z DB, and the angles adjacent to these sides equal, namely DA.E = BDF, and ADE = DBF (51) ; therefore these triangles are equal (78), and AE = DF. Also, since DEGF is a parallelogram, DF = EC (104) ; and hence AE = EC. 2d. The triangles ADE and BDF being equal, we have DE = BE, and in the parallelogram DECF we have DE = FC; therefore BE = FC. Hence F is the middle point of BG, and DE = hBC. 122. Corollary I. The straight line DE, joining the middle points of the sides AB, AC^ of the triangU ABC, is parallel to third side BC, and is equal to one-half of BC. For, the straight line drawn through D parallel to BC, passes through E (121), and is therefore identical with DE Consequently, also, DE= iBC. 123. Corollary 11. The straight line dravm parallel to the bases of a trapezoid, bisecting one of the non-parallel sides, also bisects the opposite side. A D Let ABCD be a trapezoid, BC and AD its y^ 1 parallel bases, E the middle point of AB, and ^/L. -X^ 1 ^ let EF be drawn parallel to -BC or AD ; then, / F is the middle of DC. For, draw the diago- nal AC, intersecting EF in H. Then in the triangle ABC, EH is drawn through the middle of AB parallel to ^C; therefore H is the middle of AC. In the triangle ACD, HE is drawn through the middle of AC parallel to AD ; therefore F is the middle of DC. 124. Corollary III. In a trapezoid, the straight line joining the middle points of the non-parallel sides is parallel to the bases, and ii equal to one-half their sum. A }) e/ /\. \ / ^\ 44 GEOMETRY. Let EF join the middle points, E and Fy of ^J5 and DC. Then, 1st, £i^ is parallel to BC. For, by Cor. II. the straight line drawn through E parallel to BC passes through i^and is therefore identical with EF. 2d. Drawing the diagonal AC^ intersecting EF in H, we have, in the triangle ABC, EH=iBQ and in the triangle ACD, HF=iAD, the sum of which gives EF=^^BC + AD), PROPOSITION XXXVIIL— THEOREM. 125. If a series of parallels cutting any two straight lines intercept equal distances on one of these lines, they also intercept equal distances on the other line. Let MN, M'N', be two straight lines cut by a series of parallels AA', BB', CC, DD'-, then, \i AB, BC, CD are equal, A'B', B'C, CD' are also equal. For, through the points A, B, C, draw Ah, Be, Cd, parallel to M'N'. In the triangles ABh, BCc, CDd, we h^ve AB = BC = CD; and the corresponding angles adjacent to these sides are equal (51), namely, BAb = CBc = DCd,&ndABb = BCG = CDd ; therefore, these triangles are equal to each other (78), and Ab = Be = Cd. But, the figures A'b, B'c, C'd, being parallelo- grams, we have Ab = A'B', Be = B'C, Cd = CD'; therefore, A'B' ^-^ B'C = CD'. PROPOSITION XXXIX.— THEOREM. 126. Every point in the bisector of an angle is equally distant from the sides of the angle; and every point not in the bisector, but icithin the angle, is unequally distant from the sides of the angle. BOOK I 45 1st. Let AD be the bisector of the angle BA Q P any point in it, and PEj PF, the per- pendicular distances of Pfrom AB and AC] then, PE = PF. For, the right triangles APE, APE, having the angles PAE and PAF equal, and AP com- mon, are equal (83) ; therefore, PE = PF. 2d. Let Q be any point not in the bisector, but within the angle ; then, the perpendicular distances §^and QH are unequal. For, suppose that one of these distances, as QE, cuts the bisector in some point P: from P let PP be drawn perpendicular to AC, and join QF. We have QH < QF; also QF < QP -{- PF, or QF < §P -f PE, or QF< QE; therefore, QH When the angle BAC is obtuse, the point Q, not in the bisector, may be so situated that the perpendicular on one of the sides, as AB, will fall at the vertex A ; the perpendicular QH is then less than the oblique line QA. Or, a point Q' may be so situated that the perpendicular Q'E', let fall on one of the sides, as AB, will meet that side produced through the vertex A; this perpendicular must cut the side A C in some point, K, and we then have Q'H' < Q'K < Q'E\ 127. Corollary. The bisector of an angle is the locus (40) of all the points within the angle which are equally distant from its sides. QF PEOPOSITION XL.— THEOKEM. 128. The three bisectors of the three angles of a triangle meet in the same point. Let AD, BE, CF, be the bisectors of the angles A, B, C, respectively, of the triangle ABC. Let the two bisectors AD, BE, meet in 0. The point 0, being in AD, is equally dis- tant from AB and AC (126); and being in BE, it is equally distant from AB and BC; 46 GEOMETRY. therefore, the point is equally distant from -4 Cand BC, and must lie in the bisector of the angle C (127). That is, the bisector CF also passes through 0, and the three bisect- ors meet in the same point. 129. Corollary. The point in which the three bisectors of the angles of a triangle meet is equally distant from the three sides of the triangle. PROPOSITION XLI.— THEOREM. 130. The three perpendiculars erected at the middle points of the sides of a triangle meet in the same point Let DGy EHj FK, be the perpendiculars erected to BC, CA, AB, respectively, at their middle points, D, E, F, It is first necessary to prove that any two of these perpendiculars, as DO, EH, meet in some point. If they did not meet, they would be parallel, and then CB and CA being perpen- diculars to these parallels from the same point C, would be in one straight line, which is impossible, since they are two sides of a tri- angle. Therefore, DG and EH are not parallel, and must meet in some point, as 0. Now the point being in the perpendicular i) G^ is equally distant from B and C (38), and being also in the perpendicular EH, it -is equally distant from A and C\ therefore it is equally distant from A iftid B, and must lie in the perpendicular FK (39). That is, the perpendicular FK passes through 0, and the three perpendiculars meet in the same point. 131. Corollary. The point in which the three perpendiculars meet is equally distant from the three vertices of the triangle. BOOK I. 47 PKOPOSITION XLII.— THEOREM. 182. The three perpendiculai's from the vertices of a triangle to the opposite sides meet in the same point. Let AD, BE, CF, be the perpen- ?:.... 4 J?' diculars from the vertices of the tri- BC and angle ABC to the opposite sides, re- spectively. Through the three vertices, A, B, C, draw the lines B'C, A'B', A'C, re- spectively parallel to BC, AB, AC, ^' Then the two quadrilaterals ABCB' and ACBC are parallelograms, and we have AB' AC = BC; therefore AB' = AC, or A is the middle of B' C But AD being perpendicular to ^C is perpendicular to the parallel B'C', therefore AD is the perpendicular to B'C erected at its middle point A. In like manner, it is shown that BE and CFare the perpendiculars to A' C and A'B' at their middle points; there- fore, by (130), the three perpendiculars meet in the same point. 133. Definition. A straight line drawn from any vertex of a tri- angle to the middle point of the opposite side is called a medial line of the triangle. Thus, D being the middle point of BC, AD is the medial line to BC. PEOPOSITION XLIII.— THEOREM. 134. The three medial lines of a triangle meet in the samejyoint. Let D, E, F, be the three middle points of the sides of the triangle ABC', AD, BE, CF, the three medial lines. Let the two medial lines, AD and BE, meet in 0. Let G be the middle point of OA, a^nd H the middle point of QB; join OH, HD, DE, EG. In the triangle A OB, OH is par- allel to J.^, and OH = ^AB: and in the triangle ABC,"ED is parallel to ^^, and ED = ^AB (122). ' Therefore, HO and ED, being parallel to AB,nre parallel to each other; and each being 48 GEOMETRY. equal to hAB, they are equal to each other ; consequeutly, EGHD is a parallelogram (108), and its diagonals bisect each other (109). Therefore OB = OG = GA, or OD = ^AD ; that is, the medial line BE cuts the medial line AD at a point whose distance from D is one-third of AD. In the ^me way it is proved that the medial line OF cuts AD at a point whose distance from D is one-third of AD, that is, at the same point 0; and therefore the three medial lines meet in the same point. SYMMETRICAL FIGURES. a. Symmetry with 7'espect to an axis. 135. Definition. Two points are symmetrical with respect to a fixed straight line, called the axis of symmetry, when this axis bisects at right angles the straight line joining the two points. Thus, A and A' are symmetrical with respect to the axis 3fN, if 3IN bisects AA' at right angles at a. If the portion of the plane containing the point A on one side of the axis MN, is revolved about ^ '^* this axis (or folded over) until it coincides with the portion on the other side of the axis, the point A' at which A ftills is the symmetrical point of A. 136. Definition. Any two figures are symmetrical with respect to an axis when every point of one figure has its symmetrical point on the other. Thus, A'B' is the symmetrical figure of the straight line AB, with respect to (lie axis MN, every point, as C, of the one, ^~ having its symmetrical point C" in the other. The symmetrical figure of an indefinite straight line, AB, is an indefinite straight line, A'B\ which intersects the first in the axis and makes the same angle with the axis as tlie first line. BOOK I. 49 137. Definition. In two symmetrical figures the corresponding symmetrical lines are called homologous. Thus, in the symmetrical figures ABODE, A'B'C'D'E', the homologous lines are AB and A'B\ BCsmd B'C, etc. In all cases, two figures, symmetrical with I'C'spect to an axis, can be brought into coin- cidence by the revolution of either about the axis. 6. Symmetry with respect to a centre. 138. Definition. Two points are symmetrical with respect to a fixed point, called the centre of symmetry, when this centre bisects the straight line joining the two points. Thus, A and A' are symmetrical with respect ...'■^, to the centre 0, if the line AA' passes through ,.••'* and is bisected at 0. ,..-''' The distance of a point from the centre is called a its radius of symmetry. A point A is brought into coincidence with its symmetrical point A\ by revolving its radius OA through two right angles (16). 139. Definition. Any two figures are symmetrical with respect to a centre, when every point of one figure has its symmetrical point on the other. Thus, A'B' is the symmetrical -i figure of the straight line AB with respect to the centre 0. Since the triangles AOB, A' OB', are equal (76), the angle B is equal to the angle B' ; therefore, AB and A'B' are parallel. the homologous lines of two figures, cymmetrical with respect to a centre, are parallel. Thus, in the symmetri- cal figures ABOD, A'B'O'D', the homologous lines J.^ and A'B' are parallel, BO and B' O' are parallel, etc. Twd figures symmetrical with respect to a centre can be brought 5 D ,'-' O In general, 50 GEOMETRY. into coincidence by revolving one of them, in its own plane, about the centre ; every radius of symmetry revolving through two right angles at the same time. 140. Definition. Any single figure is called a symmetrical figure, either when it can be divided by an axis into two figures symmetri- cal Y>^ith respect to that axis, or when it has a centre such that every strafght line drawn through it cuts the figure in two points which are symmetrical with respect to this centre. Thus, ABCDC'B' is a symmetrical figure with respect to the axis MN, being divided by MN into two figures, ABCD and AB'C'D, which are sym- metrical with respect to MN. Also, the figure ABCDEF is symmetrical with respect to the centre 0, its vertices, taken two and two, being symmetrical with respect to 0. In this case, any straight line KL drawn /^ "■--->.v::l yo through the centre and terminated by the perimeter, is called a diameter. PROPOSITION XLIV.— THEOREM. 141. IJ a figure is symmetrical with respect to two axes perpendicular to each other, it is also symmetrical with respect to the interseciion of these axes as a centre of symmetry. Let the figure ABCDEF GH be symmetrical with respect to the two perpendicular axes MNy PQ, which intersect in 0\ then, the point is also the centre of symmetry of the figure. For, let T be any point in the perimeter of the figure; draw TRT' perpendicular to MN, and TSt per- pendicular to P§; join T' 0, Ot and i^/S. Since the figure is symmetrical with respect to MN, we have RT' = ET; and since RT = OS, it follows that RT' = OS; therefore, 1 3 G "- \f J I'''' n 7V\ a - D BOOK I. 51 RT' OS is a parallelogram (108), and RS is equal and parallel to or. Again, since the figure is symmetrical with respect to P§, we have St = ST = OR; therefore, SROt is a parallelogram, and RS is equal and parallel to Ot Hence, T\ and t, are in the same straight line, since there can be but one parallel to RS drawn through the same point 0. Now we have OT' = jRaS and 0< = RS, and consequently OT' = Ot; therefore, any straigh^: line T'Ot, drawn through 0, is bisected at C ; that is, is the centre of symmetry of the figure. BOOK II, THE CIRCLE. 1. Definitions. A circle is a portion of a plane bounded by a curve, all the points of which are equally distant from a point within it called the centre. The curve which bounds the circle is called its circumference. Any straight line drawn from the centre to the circumference is called a radius. Any straight line drawn through the centre and terminated each way by the circumfer- ence is called a diameter. In the figure, is the centre, and the curve ABGEA is the cir- cumference of the circle ; the circle is the space included within the circumference; OA, OB, OC, are radii; AOCis a diameter. By the definition of a circle, all its radii are equal ; also all its diameters are equal, each being double the radius. If one extremity, 0, of a line OA is fixed, while the line revolves in a plane, the other extremity, J., will describe a circumference, whose radii are all equal to OA. 2. Definitions. An arc of a circle is any portion of its circumfer- ence ; as DEF. A (Aord is any straight line joining two points of the circum- ference; as DF. The arc DEF is said to be subtended by its chord DF. Every chord subtends two arcs, which together make up the whole circumference. Thus DF subtends both the arc DEF and the arc DCBAF. Whei an arc and its chord are spoken of, the arc less than 52 BOOK II. 53 a semi-circumference, as DEF, is always understood, unless otherwise stated. A segment is a portion of the circle included between an arc and its chord ; thus, by the segment DEF is meant the space included between the arc DF and its chord. A sector is the space included between an arc and the two radii dra^vn to its extremities ; as A OB. 3. From the definition of a circle it follows that every point within the circle is at a distance from the centre which is less than the radius; and every point without the circle is at a distance from the centre which is greater than the radius. Hence (I. 40), the locus of all the points in a plane which are at a given distance from a given point is the circumference of a circle described with the given point as a centre and with the given distance as a radius. 4. It is also a consequence of the definition of a circle, that two circles are equal when the radius of one is equal to the radius of the other, or when (as we usually say) they have the same radius. For if one circle be superposed upon the other so that their centres coin- cide, their circumferences will coincide, since all the points of both are at the same distance from the centre. If when superposed the second circle is made to turn upon its centre as upon a pivot, it must continue to coincide with the first. 5. Postulate. A circumference may be described with any point as r a centre and any distance as a radius. Ql j^ 4; AKCS AND CHOKDS. PROPOSITION I.— THEOEEM. 6. A straight line cannot intersect a circumference in more iiian two points. For, if it could intersect it in three points, the three radii drawn - " to these three points would be three equal straight lines drawn from the same point to the same straight line, which is impossible (I. 36). 5* 64 GEOMETRY. PEOPOSITION II.— THEOREM. 7. Every diameter bisects the circle and its circumference. J jet AMBN be a circle whose centre is ; then, any diameter A OB bisects the circle and its cirjcuraference. For, if the figure ANB be turned about AB as an axis and superposed upon the figure AMB, the curve ANB will coincide with the curve AMB, since all the points of both are equally distant from the centre. The two figures then coincide throughout, and are therefore equal in all respects. Therefore, J.^ divides both the circle and its circumference into equal parts. 8. Definitions. A segment equal to one half the circle, as the seg- ment AMB, is called a semi-circle. An arc equal to half a circum- ference, as the arc AMB, is called a semi-drcwnference. PEOPOSITION III.— THEOREM. 9. A diameter is greater than any other chord. Let AC he any chord which is not a diame- ter, and A OB a diameter drawn through A : then AB> AG. For, join OC. Then, AO -{r OC > AC (I. 66) ; that is, since all the radii are equal, AO-^ OB > AC, or AB> AC. PROPOSITION IV.— THEOREM. 10. In equal circles, or in the same circle, equal angles at the centre intercept equal arcs on the circumference, and conversely. Let 0, 0', be the centre of equal ^ A T> A' D circles, and AOB, A'O'B', equal angles at these centres ; then, the intercepted arcs, AB, A'B', are equal. For, one of the angles, together with its arc, may be superposed upon the other; and when BOOK II. 55 the equal angles coincide, their intercepted arcs will evidently coin- cide also. Conversely, if the arcs AB, A'B' are equal, the angles A OB, A' O'B' are equal. For, when one of the arcs is superposed upon its equal, the corresponding angles at the centre will evidently coincide. If the angles are in the same circle, the demonstration is siniilar. 11. Definition. A fourth part of a circum- ference is called a quadrant. It is evident from the preceding theorem that a right angle at the centre intercepts a quadrant on the circum- ference. Thus, two perpendicular diameters, AOCy BOD, divide the circumference into four quad- rants, AB, BQ CD, DA. PROPOSITION v.— THEOREM. 12. In equal circles, or in the same circle, equal arcs are subtended by equal chords, and conversely. Let 0, O^he the centres of equal circles, and AB, A'B\ equs,\ arcs; then, the chords AB, A'B', are equal. For, drawing the radii to the extremi- ties of the arcs, the angles and 0' are equal (10), and consequently the triangles A OB, A' O'B', are equal (I. 76). Therefore, AB == A'B'. Conversely, if the chords AB, A'B', are equal, the triangles AOB, A' O'B' are equal (I. 80), and the angles 0, 0' are equal. There- fore (10), arc ^^ = arc A'B'. If the arcs are in the same circle, the demonstration is similar. PROPOSITION VI.— THEOREM. 13. In equal circles, or in the same circle, the greater arc is subtended by the greater chord, and conversely ; the arcs being both less than a semi-circumference 56 GEOMETRY. Let the arc A C be greater than the arc AB; then, the chord AC is greater than the chord AB. For, draw the radii OA, OB, 00. In the triangles AOC, A OB, the angle A 00 is obviously greater than the angle A OB; therefore, (I. 84), chor?lJLO> chord ^^. Conversely, if chord J.(7 > chord AB, then, arc J.C > arc AB. For, in the triangles A 00, A OB, the side J.C > the side AB; therefore (I. 85), angle AOC > angle A OB; and consequently, arc AC ^ arc AB. 14. Scholium. If the arcs are greater than a semi -circumference, the contrary is true ; that is, the arc AMB, which is greater than the arc AMCy is subtended by the less chord ; and conversely. PROPOSITION VII.— THEOREM. 15. The diameter perpendicular to a chord bisects the chord and the arcs subtended by it Let the diameter DOD' be perpendicular to the chord AB at C; then, 1st, it bisects the chord. For, the radii OA, OB being equal oblique lines from the point to the line AB, cut off equal distances from the foot of the per- pendicular (I. 36); therefore, AC = BC. 2d. The subtended arcs ADB, AD'B, are bisected at D and D' , respectively. For, evei-y point in the per- pendicular DOD' drawn at the middle of AB being equally distant from its extremities A and B (I. 38), the chords AD and BD are equal; therefore, (12), the arcs AD and BD are equal. For the same reason, the arcs AD' and BD' are equal. 16. Corollary I. The perpendicular erected upon the middle of a chord passes through the centre of the circle, and through the middle of the arc subtended by the chord. Also, the straight line drawn through any two of the three points O, C, D, passes through tlie third and is perpendicular t'» the chord AB. BOOK It 67 17. Corollary II. The middle points of any number of parallel cliords all lie in the same diameter perpendicular to the chords. In other words, the locus of the middle points of a system of parallel chords is the diameter perpendicular to these chords. PKOPOSITION VIII.— THEOREM. 18. In the same circle^ or in equal circles, equal chords are equally distant from the centre; and of two unequal chords, the less is at the greater distance from the centre. 1st. Let AB, CD, be equal chords ; OE, OF, the perpendiculars which measure their distances from the centre 0; then, OE = OF. , For, since the perpendiculars bisect the chords (15), AE=CF; hence (I. 83), the right triangles AOE and COF are equal, and OE = OF. 2d. Let CG, AB, be unequal chords; OE, OH, their distances from the centre ; and let CG be less than AB ; then, OH > OE. For, since chord AB > chord CG, we have arc AB > arc CG ; BO that if from C we draw the chord CD = AB, its subtended arc CD, being equal to the arc AB, will be greater than the arc CG. Therefore the perpendicular OH will intersect the chord CD in some point J. Drawing the perpendicular OF to CD, we have, by the first part of the demonstration, OF = OE. But OH > 01, and 01 > 0F(1. 28); still more, then, is 0H> OF, or 0H> OE. If the chords be taken in two equal circles, the demonstration is the same. 19. Corollary I. The converse of the proposition is also evidently true, namely : in the same circle, or in equal circles, chords equally distant from the centre are equal ;' and of two chords unequally distant from the centre, that is the greater ivhose distance from the centre is the less. 5** 58 G E O M P: T R Y. 20. Corollary II. The least chord that x;an be drawn in a circle through a given point P is the chord, AB, perpendicular to the line OP joining the given point and the centre. For, if CD is any other chord drawn through P, the perpen- dicular OQ to this chord is less than OP; there- fore, by the preceding corollary, CD is greater than AB. PKOPOSITION IX.— THEOREM. 21. Through any three points, not in the same straight line, a eireum.' ference can he made to pass, and hut one. Let A, B, C, be any three points not in the same straight line. 1st. A circumference can be made to pass through these points. For, since they are not in the same straight line, the lines AB, BC, AC, joining them two and two, form a triangle, and the three perpendiculars DE, FG, HK, erected at the middle points of the sides, meet in a point which is equally distant from the three points A, B, C, (I. 131). Therefore a circumference described from as a centre and a radius equal to any one of the three equal distances OA, OB, OC, will pass through the three given points. 2d. Only one circumference can be made to pass through these points. For the centre of a7iy circumference passing through the three points must be at once in two perpendiculars, as DE, FG, and therefore at their intersection ; but two straight lines intersect in only one point, and hence is the centre of the only circumference that can pass through the three points. 22. Corollary. Two circumferences can intersect in but two points; for, they could not have a third point in common without having the same centre and becoming in fact but one circumference. BOOK II, 59 TANGENTS AND SECANTS. 23. Definitions. A tangent is an indefinite straight line which has but one point in common with the cir- cumference; as ACB. The common point, 0, is called the point of contact, or the point of tangency. The circum- ference is also said to be tangent to the line AB at the point C. A secant is a straight line which meets the circumference in two points ; as^i^. 24. Definition. A rectilinear figure is said to be circumscribed about a circle when all its sides are tangents to the circumference. In the same case, the circle is said to be inr scribed in the figure. PEOPOSITION X.-.THEOKEM. 25. A straight line oblique to a radius at its extremity cuts the dr- cumference. Let AB be oblique to the radius OC at its extremity C; then, AB cuts the circumfer- ence at (7, and also in a second point D. For, let OE be the perpendicular from upon AB; then OE < OC, and the point E is within the circumference. Therefore AB cuts the circumference in C, and must evi- dently cut it in a second point D. 60 txEOMETRY. PROPOSITION XL— THEOREM. 26. A straight line perpendicular to a radius at its extremity ia a tangent to the circle. Let AB be perpendicular to the radius 00 at its extremity 0; then, AB is a tangent to the circle at the point O. For, from the centre draw the oblique line OD to any point of AB except 0. Then, OD "> 00, and Z) is a point without the cir- cumference. Therefore AB having all its points except without the circumference, has but the point in common with it, and is a tangent at that point (23). 27. Corollary. Conversely, a tangent AB at any point is perpen- dicular to the radius 00 drawn to that point. For, if it were not perpendicular to the radius it would cut the circumference (25), and would not be a tangent. 28. Scholium. If a secant EF, passing through a point of the <5ircumference, be supposed to revolve upon this point, as upon a pivot, its second point of intersection, D, will move along the circumference and ap- proach nearer and nearer to 0. AVhen the second point comes into coincidence with 0, the revolving line ceases to be strictly a secant, and becomes the tan- gent AB; but, continuing the revolution, the revolving line again becomes a secant, as E'F', and the second point of intersection reappears on the other side of 0, as at D'. If, then, our revolving line be required to be a secant m the stridi sense imposed by our definition, that is a line meeting the circum- ference in two points, this condition can be satisfied only by keeping the second point of intersection, D, distinct from the first point, C, however near these points may be brought to each other ; and, there- fore, under this condition, the tangent is often called the limit of the secants drawn through the point of contact; that is to say, a limit toward which the secant continually approaches, as the second point BOOK II. 61 of intersection (on either side of tlie first) continually approaches the first, but a limit which is never reached by the secant as such. On the other hand, as the tangent is but one of the positions of our revolving line, it has properties in common with the secant ; and in order to exhibit such common properties in the most striking manner, it is often expedient to regard the tangent as a secant whose tivo points of intersection are coincident. But it is to be observed that we then no longer consider the secant as a cutting line, but simply as a line drawn through two points of the curve ; and we include the tangent as that special case of such a line in which the two points are coincident. In this, we generalize in the same way as in algebra, when we say that the expression x = a — b signifies that x is the difference of a and b, even when a = b, and there is really no differ- ence between a and b. PKOPOSITION XII.— THEOREM. 29. Two parallels intercept equal arcs on a circumference, "VVe may have three cases : 1st. When the parallels AB, CD, are both secants ; then, the intercepted arcs J.Cand BD are equal. For, let OM be the radius drawn perpendicular to the parallels. By Prop. VII. the point M is at once the middle of the arc AMB and of the arc CMD, and hence we have AM = BM and CM=DM, whence, by subtraction, AM — CM = BM — DM; AC = BD. that is, 2d. When one of the parallels is a secant, as AB, and the other is a tangent, as EF at 3f, then, the intercepted arcs A3f and B3I are equal. For, the radius OM draw^n to the point of contact is per- pendicular to the tangent (27), and consequently perpendicular also 10 its parallel AB; therefore, by Prop. VII., AM = BM. 3(1. Wlien both tlie parallels are tangents, as ^i^at M, and GH 62 G E O M E 1 R Y. if at N; then, the intercepted arcs MAN and MEN are equal. For, drawing any secant AB parallel to the tangents, we have by the second case, AM = BM and AN = BN, whence, by addition, that is. AM-\-AN = BM-{-B]Sr, c/ \2) H V ^ V MAN=MBN\ and each of the intercepted arcs in this case is a semi-circumference. 30. ScJiolium 1. The straight line joining the points of contact of two parallel tangents is a diameter. 31. Scholium 2. According to the principle of (28), the tangent being regarded as a secant whose two points of intersection are coin- cident, the demonstration of the first case in the preceding theorem embraces that of the other two cases. RELATIVE POSITION OF TWO CIRCLES. 32. Definition. Two circles are concentric^ when they have the same centre. 33. Definition. Two circumferences are tangent to each other, or touch each other, when they have but one point in common. The common point is called the point of contact, or the poi7it of tangency. Two kinds of contact are distinguished : external contact, when each circle is outside the other ; internal contact, when one circle is within the other. PROPOSITION XIII.— THEOREM. 34. When two circumferences intersect, the straight line, joining their centres bisects their common chord at right angles. Let and 0' be the centres of two circumferences which intersect in the points A, B; then, the straight line 00' bisects their common chord AB at right angles. For, the perpendicular to AB erected BOOK II 63 at its middle point (7, passes through both centres (16) ; and there can be but one straight line drawn between the two points and 0', 35. Corollary. When, two circumferences are tangent to each other, their point of contact is in the straight line joining their centres. It has just been proved that when two circumferences intersect, the two points of intersection lie at equal distances from the line joining the centres and on opposite sides of this line. Now let the circles be supposed to be moved so as to cause the points of intersection to approach each other; these points will ultimately come together on the line joining the centres, and be blended in a single point C, common to the two cir- cumferences, which will then be their point of contact. The perpendicular to 00' erected at (7 will then be a common tangent to the two circumferences and take the place of the common chord. PKOPOSITION XIV.— THEOKEM. 36. When two circumferences are wholly exterior to each other, the distance of their centres is greater than the sum of their radii. Let 0, 0' be the centres. Their dis- tance 00' is greater than the sum of the radii OA, O'B, by the portion AB interposed between the circles. PEOPOSITION XV.— THEOREM. 37. When two circumferences are tangent to each other externally, the distance of their centres is equal to the sum of their radii. Let 0, 0', be the centres, and C the point of contact. The point C being in the line joining the centres (35), we have 00' = 00-\- O'C. 64 GEOMETRY. PROPOSITION XVI.— THEOREM. 88. When tivo circumferences intersect, the distance of their centres is less than the sum of their radii and greater than the difference of their radii. Let aiid 0' be their centres, and A one of their points of intersection. The point A is not in the line joining the centres (34) ; and consequently there is formed the triangle AOO', in which we have 00' < OA -^ O'A, and also 00' > OA - O'A (I. 67). PROPOSITION XVII-THEOREM. 39. When two circumferences are tangent to each other interrially, the distance of their centres is equal to the difference of their radii. Let 0, 0', be the centres, and (7 the point of contact. The point C being in the line joining the centres (35), we have 00' = 00— 0' C. PROPOSITION XVIII.— THEOREM. 40. When one cirmimference is wholly within another, the distance of their centres is less than the difference of their radii. Let 0, 0', be the centres. We have the dif- ference of the radii OA — O'B == 00' -f AB. Hence 00' is less than the difference of the radii by the distance AB. 41. Corollary. The converse of each of the preceding five propo- sitions is also true : namely — 1st. When the distance of the centres is greater than the sum of the radii, the circumferences are wholly exterior to each other. BOOK IT. 65 2d. \YheD the distance of the centres is equal to the sum of the radii, the circumferences touch each other externally. 3d. When the distance of the centres is less than the sum of the radii, but greater than their difference, the circumferences intersect. 4th. When the distance of the centres is equal to the difference of the radii, the circumferences touch each other internally. 5th. When the distance of the centres is less than the difference of the radii, one circumference is wholly within the other. MEASURE OF ANGLES. As the measurement of magnitude is one of the principal objects of geometry, it will be proper to premise here some principles in regard to the measurement of quantity in general. 42. Definition. To measure a quantity of any kind is to find how many times it contains another quantity of the same kind called the unit. Thus, to measure a line is to find the number expressing how many times it contains another line called the unit of length, or the linear unit. The number which expresses how many times a quantity contains the unit is called the numerical measure of that quantity. 43. Definition. The ratio of two quantities is the quotient arising from dividing one by the other ; thus, the ratio of A to B is — To find the ratio of one quantity to another is, then, to find how many times the first contains the second; therefore, it is the same thing as to measure the first by the second taken as the unit (42). It is implied in the definition of ratio, that the quantities compared are of the same kind. Hence, also, instead of the definition (42), we may say that to measure a quantity is to find its ratio to the unit. The ratio of two quantities is the same as the ratio of their numerical measures. Thus, if P denotes the unit, and if Pis con- tained m times in A and n times in B, then, A mP m B~ nP~ n 44. Definitioii. Two quantities are commensurable when there ia 6 * E 66 GEOMETRY.- some third quantity of the same kind which is contained a whole number of times in each. This third quantity is called the common measure of the proposed quantities. Thus,. the two lines, A and B, are commensurable, if there is some line, C, which is contained a whole num- ber of times in each, as, for example, ' ' ' ' ' ' ' ' 7 times in A, and 4 times in B. ^' — • — « — ' — » The ratio of two commensurable quan- q, — , tities can, therefore, be exactly expressed by a number whole or fractional (as in the preceding example by -1, and is called a commensurable ratio. 45. Definition. Two quantities are incommensurable when they have no common measure. The ratio of two such quantities is called an incommensurable ratio. If A and B are two incommensurable quantities, their ratio is still expressed by — 46. Problem. To find the greatest common measure of two quantities. The well-known arithmetical process may be extended to quantities of all kinds. Thus, suppose AB and CD are two straight lines whose common measure is required. Their greatest common meas- ure cannot be greater than the less line CD. Therefore, let CD be applied to AB ^' ' ' J"^ as many times as possible, suppose 3 times, ^' ""^^ with a remainder EB less than CD. Any common measure of AB and CD must also be a common measure of CD and EB ; for it will be contained a whole number of times in CD, and in AE, which is a multiple of CD, and therefore to measure AB it must also measure the part EB. Hence, the greatest common measure of AB and CD must also be the greatest common measure of CD and EB. This greatest common measure of CD and EB cannot be greater than the less line EB ; therefore, let EB be applied as many times as possible to CD, suppose twice, with a remainder FD. Then, by the same reasoning, the greatest common measure of CD and EB, and consequently also that of AB and CD, is the greatest common measure of EB and FD. Therefore, let FD be applied to EB as many times as possible : suppose it is contained B O O K 1 1 . 67 exactly twice in EB without remainder ; the process is then com- pleted, and we have found FD as the required greatest common measure. The measure of each line, referred to FD as the unit, will then be as follows : we have EB = 2FD, CD = 2EB -{- FD = 4FD -\- FD = 6FD, AB =^ ZCD -i- EB = IdFD -\- 2FD = IIFD. The proposed lines are therefore numerically expressed, in terms of 17 the unit FDy by the numbers 17 and 5 ; and their ratio is — 5 47. When the preceding process is applied to two quantities and no remainder can be found which is exactly contained in a pre- ceding remainder, however far the process be continued, the two quantities have no common measure; that is, they are incommen- surable, and their ratio cannot be exactly expressed by any number whole or fractional. 48. But although an incommensurable ratio cannot be exactly expressed by a number, it may be approximately expressed by a number within any assigned measure of precision. A Suppose — denotes the incommensurable ratio of two quantities B A and B] and let it be proposed to obtain an approximate numeri- cal expression of this ratio that shall be correct within an assigned measure of precision, say Let B be divided into 100 equal parts, and suppose A is found to contain 314 of these parts with a remainder less than one of the parts ; then, evidently, we have A 314 ... 1 — = withm — -» B 100 100 that, is, is an approximate value of the ratio — within the as- 100 ^^ B signed measure of precision. A . To generalize this, — denoting as before the incommensurable B ratio of the two quantities A and jB, let B be divided into n equal 68 GEOMETRY. parts, and let A contain m of these parts with a remainder less than one of the parts ; then we have — ; = — within - ; * B n n and, since n may be taken as great as we please, - may be made less •, n 7)1/ than any assigned measure of precision, and — will be the approxi- n A mate value of the ratio — within that assigned. measure. 49. Theorem, Two ineommensurahle ratios are equals if their approxi- mate 7iumerical values are always equals when both are expressed within the same measure of precision however small. A A' Let — and — be two incommensurable ratios whose approximate numerical values are always the same when the same measure of precision is employed in expressing both ; then, we say that B~ B' For, let - be any assumed measure of precision, and in accordance with the hypothesis of the theorem, suppose that for any value of 1 A A' ->the ratios—' — j have the same approximate numerical expres- sion, say — » each ratio exceeding — by a quantity less than -; n n n then, these ratios cannot differ' /rom each other by so much as - n But the measure - may be assumed as small as we please, that is less A A' than any assignable quantity however small; hence — and — cannot B B differ by any assignable quantity however small, and therefore they must be equal. The student should study this demonstration in connection with that of Proposition XIX., which follows. BOOK II. 69 50. Definition. A proportion is an equality of ratios. Thus, if the A . A' . ratio -- is equal to the ratio —y the equality B B B~ B' is a proportion. It may be read : " Ratio of A to B equals ratio of A' to B\" or "A is to B as A' is to B':' A proportion is often written as follows : A:B = A':B' where the notation A : B is equivalent to ^ -^ ^. When thus written, A and B' are called the extremes, B and A' the means, and B' is called a fourth proportional to A, B and A' ; the first terms J. and A\ of the ratios are called the antecedents — the second terms, B and B', the consequents. When the means are equal, as in the proportion A:B = B: Q the middle term B is called a mean proportional between A and C, and G is called a third proportional to A and B, PKOPOSITION XIX.— THEOREM. 51. In the same circle, or in equal circles, two angles at the centre are in the same ratio as their intercepted arcs. Let A OB and AOChe two angles at the centre of the same, or at the centres of equal circles; AB and J. C, their intercepted arcs; then, A OB AB AOC~ AC 1st. Suppose the arcs to have ^' a common measure which is con- tained, for example, 7 times in the arc AB and 4 times in the arc AC; so that if AB is divided into 7 parts, each equal to the common measure, A C will contain 4 of these parts. Then the ratio of the arcs AB and AC is 7 : 4 ; Uiat is, 70 GEOMETRY. AG~ 4: /^vfv\ Drawing radii to the several ^O / i 1 \ \^^ points of division of the arcs, ^'^^^~c^^ the partial angles at the centre subtended by the equal partial arcs will be equal (10) ; therefore the angle AOB will be divided into 7 equal parts, of which the angle AOG will contain 4 ; hence the ratio of the angles A OB and ^0Cis7 : 4; that is, AOB 7 AOC~ 4 Therefore, we have AOB ^AB AOC~ AG or, AOB:AOG=AB:Aa 2d. If the arcs are incommensurable, suppose one of them, as A C, to be divided into any number n of equal parts ; then AB will con- tain a certain number m of these parts, plus a remainder less than AB one of these parts. The numerical expression of the ratio will ^ C 77l< 1 then be — , correct within - (48). Drawing radii to the several points of division of the arcs, the angle JL 0(7 will be divided into n equal parts, and the angle A OB will contain m such parts, plus a remainder less than one of the parts. Therefore, the numerical expression of the ratio will also be —> correct within - ; that ^ AOG . n n' . , . AOB, , . . , . IS, the ratio has the same approximate numerical expression as AB ... the ratio —r-f however small the parts into which AG is divided; therefore these ratios must be absolutely equal (49), and we have for incommensurable, as well as for commensurable, arcs, AOB AB AOG~ AG ar, AOB:AOC=AB:AC. BOOK II. 71 PKOPOSITION XX.— THEOEEM. 62. The numerical measure of an angle at the centre of a circle is the same as the numerical measure of its intercepted arc, if the adopted unit of angle is the angle at the centre which intercepts the adopted unit of arc. Let A OB be an angle at the centre 0, and AB its intercepted arc. Let AOC be the angle which is adopted as the unit of angle, and let its intercepted arc AG ho. the arc which is adopted as the unit of arc. By Proposition XIX. we have AOB_AB AOG~ AG But the first of these ratios is the measure (42) of the angle A OB referred to the unit AOC', and the second ratio is the measure of the arc AB referred to the unit A G. Therefore, with the adopted units, the numerical measure of the angle A OB is the same as that of the arc AB. 53. Scholium I. This theorem, being of frequent application, is usually more briefly, though inaccurately, expressed by saying that an angle at the centre is measured by its intercepted arc. In this con- ventional statement of the theorem, the condition that the adopted units of angle and arc correspond to each other is understood ; and the expression " is measured by" is used for "has the same numerical measure as." 54. Scholium II. The right angle is, by its nature, the most simple unit of angle ; nevertheless custom has sanctioned a different unit. The unit of angle generally adopted is an angle equal to -^^^th part of a right angle, called a degree, and denoted by the symbol °. The corresponding unit of arc is -^ih part of a quadrant (11), and is also called a degree. A right angle and a quadrant are therefore both expressed by 90°. Two right angles and a semi-circumference are both expressed by 180°. Four right angles and a whole circumference are both ex- pressed by 360°. The degree (either of angle or arc) is subdivided into minutes and 72 GEOMETRY. seconds, denoted by the symbols ' and " : a minute being -^^^th part of a degree, and a second being -^th part of a minute. Fractional parts of a degree less than one second are expressed by decimal parts of a second. An angle, or an arc, of any magnitude is, then, numerically ex- pressed by the unit degree and its subdivisions. Thus, for example, an angle equal to ^th of a right angle, as well as its intercepted arc, will be expressed by 12° 51' 25". 714 55. Definition. When the sum of two arcs is a quadrant (that is, 90°), each is called the complement of the other. When the sum of two arcs is a semi-circumference (that is, 180°), each is called the supplement of the other. See (I. 18, 19). 56. Definitions. An inscribed angle is one whose vertex is on the circumference and whose sides are chords ; as BA C. In general, any rectilinear figure, as ABC, is said ta be inscribed in a circle, when its angular points are on the circumference; and the circle is then said to ho, circumscribed about the figure. An angle is said to be inscribed in a segment when its vertex is in the arc of the segment, and its sides pass through the extremities of the sub- tending chord. Thus, the angle BAC is inscribed in the segment BAC. PROPOSITION XXI.— THEOREM. 57. An inscribed angle is measured by one-half its intercepted arc. There may be three cases : hi. Let one of the sides AB of the inscribed angle BAC be a diameter; then, the measure pf the angle BACis one-half the arc BC. For, draw the radius OC. Then, AOC being an isosceles triangle, the angles OAC and OCA are equal (I. 86). The angle BOC, an exterior angle of the triangle AOC, is equal to the sum of the interior angles OAC Siud OCA (I. 69), and therefore double BOOK II. 73 either of them. But the angle BOC, at the centre, is measured by the arc J5C'(53); therefore, the angle OAC is measured by one-half the arc BC. 2d. Let ^he centre of the circle fall within the inscribed angle BAC; then, the measure of the angle BAC is one-half of the arc BC. For, draw the diameter AD. The measure of the angle BAD is, by the first case, one-half the arc BD; and the measure of the angle CAD is one-half the arc CD; therefore, the measure of the sum of the angles BAD and CAD is one-half the sum of the arcs BD and CD; that is, the measure of the angle BAC is one-half the arc BC. 3d. Let the centre of the circle fall without the inscribed angle BAC; then, the measure of the angle BAC is one-half the arc BC. For, draw the diameter AD. The measure of the angle BAD is, by the first case, one-half the arc BD ; and the measure of the angle CAD is one-half the arc CD; therefore, the measure of the difference of the angles BAD and CAD is one-half the difference of the arcs BD and CD; that is, the measure of the angle BAC is one-half the arc BC. 58. Corollary 1. All the angles BAC, BDC, etc., inscribed in the same segment, are equal. For eacr is measured by one-half the same ^ arc BMC. •59. Corollary IL Any angle BAC, inscribed in a semicircle is a right angle. For it is measured by half a semi-circumference, or by a quad- rant (54). 74 GEOMETRY. 60. Corollary III. Any angle BA C, inscribed ^^^ — ^^ in a segment greater than a semicircle, is acute; / .^■^"''^^ \\ for it is measured by half the arc BD C, which / ^^ V\ is less than a semi-circumference. ^K" ---pMO Any angle BDC, inscribed in a segment less ^^^^^^^ than a semicircle, is obtuse ; for it is measured d by lialf the arc BA C, which is greater than a semi-circumference. 61. Corollary IV, The opposite angles of an inscribed quadrilateral ABDC, are supplements of each other. For the sum of two oppo- site angles, as BAC and BDC, is measured by one-half the circum- ference, which is the measure of two right angles, (54) and (I. 19). PROPOSITION XXII.— THEOREM. 62. An angle formed by a tangent and a chord is measured by one- half the intercepted arc. Let the angle BAC be formed by the tangent AB and the chord AC; then, it is measured by one-half the intercepted arc AMC. For, draw the diameter AD. The angle BAD being a right angle (27), is measured by one-half the semi-circumference AMD ; and the angle CAD is measured by one-half the arc CD; therefore, the angle BA C, which is the difference of the angles BAD and CAD, is measured by one-half the difference of AMD and CD, that is, by one-half the arc AMC. Also, the angle B'AC is measured by one-half the intercepted ar(; ANC. For, it is the sum of the right angle B'AD and the angle CA D, and is measured by one-half the sum of the semi-circumference AND and the arc CD; that is, by one-half the arc ANC. 63. Scholium. This proposition may be treated as a particular case of Prop. XXI. by an application of the principle of (28). For, con- sider the angle CAD which is measured by one-half the arc CD. Let the side AC remain fixed, while the side AD, regarded as a secant, revolves about A until it arrives at the position of the tangent BOOK II 75 AB'. The point D will move along the circumference, and will ultimately coincide with A, when the line AD has become a tangent and the intercepted arc has become the arc CNA. PROPOSITION XXIII.— THEOREM. 64. An angle formed by two chords, intersecting within the circum- ference, is measured by one-half the sum of the arcs intercepted between its sides and between the sides of its vertical angle. Let the angle AEC be formed by the chords AB, CD, intersecting within the circumference; then will it be measured by one-half the sum of the arcs AC and BD, intercepted between the sides of AEC and the sides of its vertical angle BED. For, join AD. The angle AEC is equal to the sum of the angles EDA and EAD (I. 69), and these angles are measured by one-half of J.Oand one-half of BD, respectively'; therefore, the angle AEC is measured by one-half the sum of the arcs A C and BD. PROPOSITION XXIV.— THEOREM. 65. An angle formed hy two secants, intersecting without the circum- ference, is measured by one-half the difference of the intercepted arcs. Let the angle BAC be formed by the secants AB and AC; then, will it be measured by one- half the difference of the arcs BC and DE. For, join CD. The angle BDC is equal to the sum of the angles DA C and A CD (I. 69) ; there- fore, the angle A is equal to the difference of the angles BD C and A CD. But these angles are meas- ured by one-half of BC and one-half of DE re- spectively ; hence, the angle A is measured by one-half the differ- ence of BC and DE. 76 GEOMETRY. 66. Corollary. The angle BAE, formed by a tangent AB and a secant AE, is measured by one-half the difference of the intercepted arcs BE and BC. For, the tangent AB may be regarded as a secant whose two points of intersection are coincident at B (28). For, the same reason, the angle BAD, formed by two tangents AB and AD, is measured by one-half the difference of the intercepted arcs BCD and BED. A proof may be given, without using the principle of (28), by drawing EB and BC. PROBLEMS OF CONSTRUCTION. Heretofore, our figures have been assumed to be constructed under certain conditions, although methods of constructing them have not been given. Indeed, the precise construction of the figures was not necessary, inasmuch as they were only required as aids in following the demonstration of principles. We now proceed, first, to apply these principles in the solution of the simple problems necessary for the construction of the plane figures already treated of, and then to apply these simple problems in the solution of more complex ones. All the constructions of elementary geometry are effected solely by the straight line and the circumference, these being the only lines treated of in the elements ; and these lines are practically drawn^ or described, by the aid of the ruler and compasses, with the use of which the student is supposed to be familiar. PROPOSITION XXV.— PROBLEM. 67. To bisect a given straight line. Let AB be the given straight line. With the points A and B as centres, and with a radius greater than the half of AB, describe arcs intersecting in the two points D and E. Through these points draw the straight line DE, A'hich bi- sects AB at the point C. For, D and E being '.F E B O O K I 1 . 77 ecjually distant from A and B, the straight line DE is perpendicular to AB Sit its middle point (I. 41). PKOPOSITION XXVI.— PEOBLEM. 68. At a given point in a given straight line, to erect a perpendicular to that line. Let AB be the given line and C the given point. Take two points, D and E, in the line and at equal distances from C. With D and E a.s cen- — ^ A D c E B tres and a radius greater than DO or CE de- scribe two arcs intersecting in F. Then CF is the required perpen- dicular (I. 41). 69. Another solution. Take any point 0, without the given line, as a centre, and with a radius equal to the distance from to G describe a circumference intersecting AB in C ^\ /'^ "^ and in a second point D. Draw the diameter DOE, and join EO. Then EG will be the re- quired perpendicular : for the angle EGD, inscribed in a semicircle, is a right angle (59). This construction is often preferable to the preceding, especially when the given point is at, or near, one extremity of the given line, and it is not convenient to produce the line through that extremity. The point must evidently be so chosen as not to lie in the required perpendicular. PKOPOSITION XXVII.— PEOBLEM. 70. From a given point without a given straight line, to let fall a per- pendicular to that line. Let AB be the given line and the given point. With C as a centre, and with a radius suf- ficiently great, describe an arc intersecting ^'^ AB in D and E. With D and E as centres and a radius greater than the half of DE, 7* ' .--- E 78 GEOMETRY'. describe two arcs intersecting in F. The line CF is the required perpendicular (I. 41). 71. Another solution. With any point in the line ^liS as a centre, and with the radius OC, describe an arc CDE intersecting AB ^ o in D. With Z) as a centre and a radius equal to the distance DC describe an arc intersecting the arc CDE in E. The line CJ^is the required perpen- dicular. For, the point D is the middle of the arc CDE, and the radius OD drawn to this point is perpendicular to the chord CE (16). /E PROPOSITION XXVIII.— PROBLEM. 72. To bisect a given arc or a given angle, 1st. Let AB be a given arc. Bisect its chord AB by a perpendicular as in (67). This perpendicular also bisects the arc (16). 2d. Let BA be a given angle. With A as a centre and with any radius, describe an arc intersecting the sides of the angle in D and E. With D and E as centres, and with equal radii, describe arcs intersecting in F. The straight line AF bisects the arc DE^ and consequently also the angle J5^0 (12). 73. Scholium. By the same construction each of the halves of an arc, or an angle, may be bisected ; and thus, by successive bisections, an arc, or an angle, may be divided into 4, 8, 16, 32, etc., equal parts. BOOK II 79 PKOPOSITION XXIX.— PEOBLEM. 74. At a given point in a given straight line,to construct an angle fqual to a given angle. Let A be the given point in the straight line AB, and the given angle. With as a centre and with any radius describe an arc MN terminated by the sides of the angle. With ^ as a centre and with the same radius, OM, describe an indefinite arc BC. With J5 as a centre and with a radius equal to the chord of MN describe an arc intersecting the indefinite arc ^O in D. Join AD. Then the angle BAD is equal to the angle 0. For the chords of the arcs MX und BD are equal ; therefore, these arcs are equal (12), and consequently also the angles and A (10). PKOPOSITION XXX.— PEOBLEM. 75. Through a given point, to draw a parallel to a given straight line. Let A be the given point, and BC the given line. From any point B in ^C draw the straight -P line BAD through A. At the point A, by / the preceding problem, construct the angle / \ DAE equal to the angle ABC. Then AE is A. parallel to BC {I. 55). L j c 76. Scholium. This problem is, in practice, more accurately solved by the aid of a triangle, constructed of wood or metal. This triangle has one right angle, and its acute angles are usually made equal to 30° and 60°. Let A be the given point, and BC the given line. Place the triangle, EFD, with one of its sides in coinci- dence with the given line BC. Then place the straight edge of a ruler MN .N 80 GEOMETRY. against, the side EF of the triangle. Now, keeping the ruler firmly fixed, slide the triangle along its edge until the side ED passes through the given point A. Trace the line EAD along the edge ED of the triangle ; then, it is evident that this line will be parallel to BG. One angle of the triangle being made very precisely equal to a right angle, this instrument is also used in practice to construct per- [)endiculars, with more facility than by the methods of (68) and (70). PROPOSITION XXXI.— PROBLEM. 77. Two angles of a triangle being given, to find the third. Let A and B be the given angles. Draw the indefinite line QM. From any y-oint in this line, draw ON making the angle MON = A, and the line OP making the angle NOP = B, Then POQ is the required third angle of the triangle (I. 72). PROPOSITION XXXII.— PROBLEM. 78. Two sides of a triangle and their included angle being given, to construct the triangle. Let b and c be the given sides and A their y' * included angle. ^ c Draw an indefinite line AE, and construct p ihesing\eEAF=A. On AE take AC = b, and on AF take AB = c ; join BC. Then ABC is the triangle required; for it is a formed with the data. With the data, two sides and the included angle, only one triangle can be constructed ; that is, all triangles constructed with these data are equal, and thus only repetitions of the same triangle (I. 76). 79. Scholium. It is evident that one triangle is always possible, whatever may be the magnitude of the proposed sides and their in- cluded angle. BOOK II. 81 PEOPOSITION XXXIII.— PKOBLEM. ■ 80. One side and two angles of a triangle being given, to construct the triangle. Two angles of the triangle being given, ^^ ^-^ the third angle can be found by (77) ; and we shall therefore always have given the c two angles adjacent to the given side. Let, thp^, c be the given side, A and B the angles adjacent to it. Draw a line AB = c; at A make an angle BAD = A, and at B an angle ABE = B. The lines AD and BE intersecting in C, we have ABC as the required triangle. With these data, but one triangle can be constructed (I. 78). 81. Scholium. If the two given angles are together equal to or greater tlian two right angles, the problem is impossible; that is, no triangle can be constructed with the data; for the lines AD and BC will not intersect on that side of AB on which the angles have been constructed. PEOPOSITION XXXIV.— PKOBLEM. 82. The three sides of a triangle being given, to construct the triangle. Let a, 6 and c be the three given sides. a Draw BC = a; with C as a centre and a ^ ^ radius equal to b describe an arc ; with B as a centre and a radius equal to c describe a second arc intersecting the first in A. Then, ABC is the required triangle. With these data but one triangle can be con- structed (I. 80). 83. Scholium. The problem is impossible when one of the given Bides is equal to or greater than the sum of the other two (I. 66). 7** F c- 82 G E O M E T K Y. PKOPOSITION XXXV.— PEOBLEM. 84. Two sides of a triangle and the angle opposite to one of them being given, to construct the triangle. AVe shall consider two cases. y ^ ls|i. When the given angle A is acute, " and Ihe given side a, opposite to it in the triangle, is less than the other given side c. Construct an angle DAE = A. In one of its sides, as AD, take AB = c; with ^ as a centre and a radius equal to a, describe an arc which (since a < c) will intersect AE in two points, C and C", on the same side of A. Join BC and BC". Then, either ABC or ABC" is the required tri- angle, since each is formed with the data ; and the problem has two solutions. There will, however, be but one solution, even with these data, when the side a is so much less than the side c as to be just equal to the perpendicular from B upon AE. For then the arc described from B as a centre and with the radius a, will touch AE in a single point C, and the required triangle will be ABC, right angled at C. 2d. When the given angle A is either acute, right or obtuse, and the side a opposite to it is greater than the other given side c. The same construction being made as in the first case, the arc described with B as a centre and with a. radius equal to a,' will intersect AE in only one point, C, on the same side of A. Then ABC will be the triangle required, and will be the only possible triangle with the data. The second point of intersection, C", will fall in EA produced, and the triangle ABC thus formed will not contain the given angle. 85. Scholium. The problem is impossible when the given angle A is acute and the proposed side opposite to it is less than the perpen- dicular from B upon AE) for then the arc described from B will not intersect AE, The problem is also impossible when the given angle is right, or C'\ B O O K I 1 . 83 obtuse, if the given side opposite to the angle is less than the other given side ; for either the arc described from B would not intersect AE, or it would intersect it only when produced through A. More- over, a right or obtuse angle is the greatest angle of a triangle (I. 70), and the side opposite to it must be the greatest side (I. 92). PEOPOSITION XXXVI.— PROBLEM. 86. The adjacent sides of a parallelogram and their included angle being given, to construct the parallelogram. Construct an angle A equal to the given angle, and take A C and AB respectively equal to the given sides. With jB as a centre and a radius equal to A C, describe an arc ; with (/ as a centre and a radius equal to AB, describe another arc, intersect- ing the first in D. Draw BD and CD. Then ABDCis a parallelo- gram (I. 107), and it is the one required, since it is formed with tlie data. Or thus: through B draw BD parallel to J.C, and through C draw CD parallel to AB. PROPOSITION XXXVII.— PROBLEM. 87. To find the centre of a given circumference, or of a given are. Take any three points, A, B and C, in the given circumference or arc. Bisect the arcs AB, BC, by perpendiculars to the chords AB, B C (72) ; these perpendiculars intersect in the required centre (16). 88. Scholium. The same construction serves to describe a circum- ference which shall pass through three given points A, B, C; or to circumscribe^Si circle about a given triangle ABC, that is, to describe % circumference in which the given triangle shall be inscribed (56), (S4 GEOMETRY. PEOPOSITION XXXVllI.— PROBLEM. 89. At a given point in a given circumjerence, the circumference. Let A be the given point in the given circum- ference. Draw the radius OA, and at A draw BAD perpendicular to OA; BC will be the re- quired tangent (26). If the centre of the circumference is not given, it may first be found by the preceding problem, or we may proceed more directly as follows. Take two points D and E equidistant from A ; draw the chord DE, and through A draw BAC parallel to DE. Since A is the middle point of the arc DE, the radius drawn to A will be perpendicular to DE (16), and con- sequently also to BC' therefore ^Cis a tangent at A. to draw a tan^ ent t/t PROPOSITION XXXIX.— PROBLEM. 90. Through a given point without a given circle to draw a tangmt to the circle. Let be the centre of the given circle and P the given point. Upon OF, as a diameter, describe a circumfer- ence intersecting the circumference of the given circle in two points, A and A'. Draw PA and PA, both of which will be tangent to the given circle. For, drawing the radii OA and OA', the angles OAP and OA'P are right angles (59) ; therefore PA and PA' are tangents (26). In practice, this problem is accurately solved by placing the straight edge of a ruler through the given point and tangent to the given circumference, and then tracing the tangent by the straight edge. The precise point of tangency is then determined by drawing a perpendicular to the tangent from the centre. 91. Scholium. This problem always admits of two solutions. More- over, the portions of the two tangents intercepted between the given B O O K I I . 85 point aud the points of tangency are equal, for the right triangles POA and FOA' are equal (I. 83) ; therefore, FA = FA'. PKOPOSITION XL.— PKOBLEM. y2. To draw a commoyi tangent to two given circles. Let and 0' be the centres of the given circles, and let the radius of the first be the greater. 1st. To draw an exterior , will pass through the three points Z>, E, Fj will be tangent to the three sides of the triangle at these points (26), and will therefore be inscribed in the triangle. 95. Scholium. If the sides of the triangle are produced and the exterior angles are bisected, the intersections 0', 0", 0"\ of the H BOOK II. 87 bisecting lines, will be the centres of three circles, each of which will touch one side of the triangle and the two other sides produced. In general, therefore, /our circles can he drawn tangent to three inter- secting straight lines. The three circles which lie without the triangle liave been named escribed circles. PROPOSITION XLII.— PEOBLEM. 96. Upon a given straight line, to describe a segment which shall contain a given angle. Let ABhe the given line. At the point B construct the angle J.^(7 equal to the given angk. Draw BO per- pendicular to BCy and DO perpendicular to AB at its middle point D, intersecting jBO in 0. With as a centre, and radius OB describe the circumference AMBN. The segment AMB is the required segment. For, the line BC, being perpendicular to the radius OB, is a tangent to the circle; therefore, the angle ABC is meas- ured by one-half the arc ANB (62), which is also the measure of any angle AMB inscribed in the segment AMB (57). Therefore, any angle inscribed in this segment is equal to the given angle. 97. Scholium. If any point P is taken within the segment AMB, the angle APB is greater than the inscribed angle AMB (I. 74) ; and if any point Q is taken without this segment, but on the same side of the chord AB as the segment, the angle AQB is less than the in- scribed angle AMB. Therefore, the angles whose vertices lie in the arc AMB are the only angles of the given magnitude whose sides pass through the two points A and B ; hence, the arc AMB is the locns of the vertices of all the angles of the given magnitude whose sides pass through A and B. If any point M' be taken in the arc AM'B, the angle AMB is the supplement of the angle AM'B (61) ; and if BM' be produced to J5', the angle AM'B' is also the supplement oi AM'B', therefore AM'B' = AMB. Hence the vertices of all the angles of the given magnitude whose sides, or sides produced, pass through A and B, lie 88 GEOMETRY. in the circumference AMBM' ; that is, the locus of the vertices of all the angles of a given magnitude whose sides, or sides produced, pass through two fixed points, is a circumference passing through these points, and this locus may be constructed by the preceding problem. It may here be remarked, that in order to establish a certain line as a locus of points subject to certain given conditions, it is necessary nof -only to show that every point in that line satisfies the conditions, but also that no other points satisfy them ; for the asserted locus must be the assemblage of all the points satisfying the given condi- tions (I. 40). INSCRIBED AND CIRCUMSCRIBED QUADRILATERALS. 98. Definition. An inscriptible quadrilateral is one which can be inscribed in a circle ; that is, a circumference can be described pass- ing thiiough its four vertices. I PROPOSITION XLIIL— THEOREM. i 99. A quadrilateral is inscriptible if two opposite angles in it are supplements of each other. Let the angles A and G, of the quadrilateral ^ ABCD, be supplements of each other. De- /^ ^^ \\ scribe a circumference passing through the b^- -^2) three vertices B, C, D; and draw the chord l\ ^^ j BD. The angle A, being the supplement of \\y-^ J C, is equal to any angle inscribed in the seg- ^^ raeut BMD (61) ; therefore the vertex A must be on the arc BMD (97), and the quadrilateral is inscribed in tlie circle. 100. Scholium. This proposition is the converse of (61). BOOK II 89 PKOPOSITION XLIV.— THEOEEM. 101. In any circumscribed quadrilateral^ the sum of two opposite sides '.s equal to the sum of the other two opposite sides. Let ABCD be circumscribed about a circle; then, AB^ DC=AD-j- BC. For, let U, Fj G, H, be the points of contact of the sides ; then we have (91), AE=AH, BE= BF, CG = CF, DG = DH. \dding the corresponding members of these equalities, we have AE-^ BE-\r CG-{^ DG = AH-\- DH^ BF-{- CF, that is, AB^DC=AD-\-BC. PROPOSITION XLV.— THEOREM. 102. Conversely, if the sum of two opposite sides of a quadrilateral is equal to the sum of the other two sides, the quadrilateral may be cir- cumscribed about a circle. In the quadrilateral ABOD, let AB-\- DC = AD -\- BC; then, the quadrilateral can be cir- cumscribed about a circle. Since the sura of the four angles of the quad- rilateral is equal to four right angles, there must be two consecutive angles in it whose sum is not greater than two right angles ; let B and C be these angles. Let a circle be described tangent to the three sides AB, BC, CD, the centre of this circle being the intersection of the bisectors of the angles B and C; then it is to be proved that this circle is tangent also to the fourth side AD. From the point A two tangents can be drawn to the circle (90). One of these tangents being AB, the other must be a line cutting CD (or CD produced) ; for, the sum of the angles B and C being not greater than two right angles, it is evident that no straight line a* 90 GEOMETRY. can be drawn from A, falling on the same side of BA with Ci), and not cutting the circle, which shall not cut CD. This second tangent, then, must be either AD \ or some other line, AM, cutting CD in a point M H^^^^^^^^^^^^^d differing from D. If now AM is a tangent, K ^ AB CM is a circumscribed quadrilateral, and by I \ the "preceding proposition we shall have |\ / \ AB + CM= AM-\- BC. i ""^^ — c But we also have, by the hypothesis of the present proposition, AB -\- DC=AD-\-Ba Taking the difference of these equalities, we have DM=AM—AD; that is, one side of a triangle is equal to the difference of the other two, which is absurd. Therefore, the hypothesis that the tangent drawn from J. and cutting the line CD, cuts it in any other point than D, leads to an absurdity ; therefore, that hypothesis must be false, and the tangent in question must cut CD in D, and consequently coincide with AD. Hence, a circle has been described which is tangent to the four sides of the quadrilateral ; and the quadrilateral is circum- scribed about the circle. 103. Scholium. The method of demonstration employed above is called the indirect method, or the reductio ad absurdum. At the outset of a demonstration, or at any stage of its progress, two or more hypotheses respecting the quantities under consideration may be admissible so far as has been proved up to that point. If, now, these hypotheses are such that one must be true, and only one can be true, then, when all except one are shown to be absurd, that one must stand as the truth. While admitting the validity of this method, geometers usually prefer the direct method whenever it is applicable. There are, how- ever, propositions, such as the preceding, of which no direct proof is known, or at least no proof sufficiently simple to be admitted into elementary geometry. We have already employed the reductio ad absurdum in several cases without presenting the argument in full ; Bee (I. 47), (I. 85), (27). BOOK III. PROPORTIONAL LINES. SIMILAR FIGURES. THEOEY OF PROPORTION. 1. Definition. One quantity is said to be proportional to another when the ratio of any two values, A and B^ of the first, is equal to the ratio of the two corresponding values, A' and B'y of the second ; 60 that the four values form the proportion AiB = A':B\ A A' B B' This definition presupposes two quantities, each of which can have various values, so related to each other that each value of one cor- responds to a value of the other. An example occurs in the case of an angle at the centre of a circle and its intercepted arc. The angle may vary, and with it also the arc ; but to each value of the angle there corresponds a certain value of the arc. It has been proved (II. 51) that the ratio of any two values of the angle is equal to the ratio of the two corresponding values of the arc ; and in ac- cordance with the definition just given, this proposition would be briefly expressed as follows : " The angle at the centre of a circle is proportional to its intercepted arc." 2. Definition. One quantity is said to be reciprocally proportional to another when the ratio of two values, A and B, of the first, is equal to the reciprocal of the ratio of the two corresponding values, A! and B\ of the second, so that the four values form the proportion A'.B = B':A\ A B' ^ A' 92 GEOMETRY. For example, if the product j) of two numbers, x and y, is given so that we have xy=p, then, a; and 2/ may each have an indefinite number of values, but as X increases y diminishes. If, now, A and B are two values of x, while A' and B' are the two corresponding values of y, we must have A X A' =p, BxB'=p, whence, by dividing one of these equations by the other, and therefore B A' ~ B' = 1, B' ~A'' that is, two numbers whose product is tional. constant are reciprocally propor- 3. Let the quantities in each of the 5 couplets of the proportion | = ||, ovA:B = A':B', [1] be measured by a unit of their own kind, and thus expressed by numbers (11. 42) ; let a and h denote the numerical measures of A and jB, a' and 5' those of A' and J5' ; then (11. 43), A a J/ a' B~h 'b'~V and the proportion [1] may be replaced by the numerical proportion, a a' I. f i.t h h' 4. Conversely, if the numerical measures a, 6, a', b\ of four quan- tities A, B, A\ B\ are in proportion, these quantities themselves are in proportion, provided that A and B are quantities of the same kind, and A' and B' are quantities of the same kind (though not neces- sarily of the same kind as A and B) ; that is, if we have a : h =^ a' : h' , 4 BOOKIII. 93 we may, under these conditions, infer the proportion A:B = A': B'. 5. Let us now consider the numerical proportion a'.h = a'^h'. Writing it in the form a a' and multiplying both members of this equality by bb \ we obtain ab' = a'b, whence the theorem : ilie product of the extremes of a (numericaV) proportion is equal to the product of the means. Corollary. If the means are equal, as in the proportion a : 6 = 6 : c, we have b"^ = ac, whence b = Vac] that is, a mean proportional be- tween two numbers is equal to the square root of their product. 6. Conversely, if the product of two numbers is equal to the product of two others, either two may be made the extremes, and the other *wo the means, of a proportion. For, if we have given ab' = a'b, then, dividing by bb', we obtain - = — > or a : = a : . b b' ».. Corollary. Tlie terms of a proportion may be written in any order which will make the products of the extremes equal to the product of the means. Thus, any one of the following proportions may be inferred from the given equality ab' = a'b: a : b = a' : b', a : a' = b : b', b : a = b' : a', b :b' = a '. a, b' : a' = b : a, etc. Also, any one of these proportions may be inferred from any other. 7. Definitions. When we have given the proportion a : b = a' : b', 94 GEOMETRY. and infer the proportion a:a' = h'.h', the second proportion is said to be deduced fty alternation. "When we infer the proportion 6 : a = 6' : a', this proportion is said to be deduced hy inversion, 8. It is important to observe, that when we speak of the products^ of the extremes and means of a proportion, it is implied that at least two of the terras are numbers. If, for example, the terms of the proportion are all lines^ no meaning can be directly attached to the products A y^ B'y B y^ A', since in a product the multiplier at least must be a number. But if we have a proportion such as A : B = m : n. 4 in which m and n are numbers, while A and B are any two quanti- ties of the same kind, then we may infer the equality nA = mB. M Nevertheless, we shall for the sake of brevity often speak of the product of two lines, meaning thereby the product of the numbers which represent those lines when they are measured hy a common unit. 9. If J. and B are any two quantities of the same kind, and m any number whole or fractional, we have, identically, mA A^ MB'~^ B' 4 that is, equimultiples of two quantities are in the same ratio as the quanfMies themselves. Similarly, if we have the proportion A : B = A' : B\ and if m and n are any two numbers, we can infer the proportions mA : mB = nA' ! nB\ mA : nB = mA' : nB'. B ' B' > 1 ■^> or, reducing, A-\- B A' + B' B B' and dividing this by [1], A-]- B -4L + -B' B o o K 1 1 r . 95 10. Composition and division. Suppose we have given the propor- tion - = -, [1] B B' *- -^ in which A and B are any quantities of the same kind, and A' and B' quantities of the same kind. Let unity be added to both mem- bers of [1] ; then i+1 ^'.. [2] A A' results which are briefly expressed by the theorem, if four quantities are in proportion, they are in proportion by composition; the term composition being employed to express the addition of antecedent and consequent in each ratio. If we had subtracted unity from both members of [1], we should have found A — B A'^B' B ~ B' ' [3] A — B _ A' — B' ' •■ -" A ~ ■ A' results which are briefly expressed by the theorem, if four quantities are in proportion,they are in proportion by division; where the term division is employed to express the subtraction of consequent from antecedent in each ratio, this subtraction being conceived to divide, or to separate, the antecedent into parts. The quotient of [2] divided by [3] is A-\- B _ A' -\- B' A — B~A' — B''' that is, if four quantities are in proportion, they are in proportion hy composition and, divimon. 1 96 GEOMETKY. 11. Definition. A continued proportion is a series of equal ratios, as A:B = A':B'=^A":B" = A!" : B'" = etc. 12. Let r denote the common value of the ratio in the continued proportion of the preceding article ; that is, let A A' A" A" r = = — = — = — = etc. : B B' B" B'" then, we have A = Br, A=B'r, A!' = B"r, ^'" = jB'"r, etc., and adding these equations, A-^A-\-A'-\- A'" + etc. = {B ^ B' -\- B" + B'" + etc.) r, whence A^ A' ^ A" -^ A'" + etc. A A' ! ! ! =1 r = — = — = etc • ^ + ^' + ^" + jB"' + etc. B B' that is, the sum of any number of the antecedents of a continued pro- portion is to the sum of the corresponding consequents as any antecedent is to its consequent. . al If any antecedent and its corresponding consequent be taken with the negative sign, the theorem still holds, provided we read algebraic sum for sum. In this theorem the- quantities A, B, C, etc., must all be quantities of the same kind. 13. If we have any number of proportions, as a : b =^ c : d, a':b' = c':d\ a":6" = c":t^",etc.; then^ writing them in the form, a c a' ^_c' a" c" b~d' b'~d' 'b"~J''^^'' and multiplying these equations together, we have I BOOK III. 97 hb' h" ... dd'd" ... or aa' a" ... '.hh' h" ... = c c' c" . . . : d d' d" , . . , that is, if the corresponding terms of two or more proportions are mul- tiplied together, the products are in proportion. If the corresponding terms of the several proportions are equal, that isy if a = a' = a", b = b' =^ b", etc., then the multiplication of two or more proportions gives a' :b' = c': d\ a':b' = c':d'; that is, if four numbers are in propor-tion, like powers of these number, are in proportion. 14. If A, B and C are like quantities of any kind, and if A , B — = m, and — = n, B ' G then A ~ = mn. If Ay B and C were numbers, this would be proved, arithmetically, by simply omitting the common factor B in the multiplication of the two fractious ; but when they are not numbers we cannot regard B as a factor, or multiplier, and therefore we should proceed more strictly as follows. By the nature of ratio we have J. = J5Xm, B = CX n, therefore, putting C X n for B, we have A= C X n X m= CX mn, thai is, A — = mn: C a result usually expressed as follows : the ratio of the first of three quantities to the third is compounded of the ratio of the first to the second and the ratio of the second to the third. 9 G 98 GEOMETRY. PROPORTIONAL LINES. PROPOSITION L— THEOREM. Id. A parallel to the base of a triangle divides the other two sides ^rwortionally. Let DE he a parallel to the base, BC, of the triangle ABC; then, AB:AD = AC:AE. ^ « 1st. Suppose the lines AB, AD, to have a /-\ M common measure which is contained, for exam- / "\ ^ pie, 7 times in AB, and 4 times in AD ; so that J \ „ if AB is divided into 7 parts each equal to the common measure, AD will contain 4 of these parts. Then the ratio of AB to AD is 7 : 4 ^ o (II. 43) ; that is Jl — = - J l AD 4* ^^■1 Through the several points of division of AB, draw parallels to the base; then J. will be divided into 7 equal parts (I. 125), of which AE will contain 4. Hence the ratio of AC to AE is 7 : 4 ; that is,^ AC_7_ AE~ A Therefore, we have AB AC AD ~ AE or AB:AD = AC:AE. 2d. If AB and AD are incommensurable, suppose one of them, as AD, to be divided into any number n of equal parts; then, AB will contain a certain number m of these parts pltis a remainder less than one of these parts. The numerical expression of the ratio will then be —> correct within - (II. 48). Drawing parallels to AD n n BC, through the several points of division of AB, the line ^^will be divided into n equal parts, and the line AC will contain m such parts plus a remainder less than one of the parts. Therefore, the BOOKIII. 99 numerical expression of the ratio — — will also be — > correct within — AE n n Since, then, the two ratios always have the same approximate nu- merical expression, however small the parts inio which AD is dVided, these ratios must be absolutely equal (II. 49), and we have, as before, AB_AC^ AD ~ AE or AB:AD = ACiAE, [1] 16. Corollary I. By division (10), the proportion [1] gives AB — AD'.AB = AC—AE'.AC, or • DB:AB = EC:Aa Also, if the parallel DE intersect the sides BA and CA produced through A, we find, as in the preceding demonstration, AB:AD = AC:AE, from which, by composition (10), AB-\-AD:AB = AC-\-AE:AOy or DB:AB = EC:Aa 17. Corollary II. By alternation (7), the preceding proportions give AB:AC=AD:AE, DB:EC=AB:AC, which may both be expressed in one continued proportion, AB AD DB AC~ AE~ EC This proportion is indeed the most general statement of the proposi- tion (15), which may also be expressed as follows : if a straight line is drawn j^'^rallel to the base of a triangle, the corresponding segments on the two sides are in a constant ratio. ■ 100 GEOMETRY. ^ 18. Corollary III. If two straight lines MN, M'N\ are intersected by any number of parallels AA\ BB\ CC\ etc., the corresponding segments of the two lines are proportional. For, let the two lines meet in 0; then, by * Corollary II., OA AB OB BC DC CD ^ etc., o'l'- A'B' OB' B'C OC CD' whence, by (11), AB A'B' BC B'C CD CD' AC A'C BD ^ -B'D'^'' If ilifiVand M'N' were parallel, this proportion would still hold, since we should then have AB = A'B', BC^B'Cy etc. PROPOSITION II.— THEOREM. 19. Conversely, if a straight line divides two sides of a triangle pro-' portionally, it is parallel to the third side. Let DE divide the sides AB, A C, of the triangle ABC, proportionally; then, DE is parallel to BC, For, if DE is not parallel to B^C, let some other line DE', drawn through D, be parallel to BC, Then, by the preceding theorem, AB:AD = AC:AE'. But, by hypothesis, we have AB:AD = AC:AE, whence it follows that AE' = AE, which is impossible unless DE' coincides with DE. Therefore, DE is parallel to BC. 20. Scholium. The converse of (18) is not generally true. PROPOSITION III.— THEOREM. 21. In any triangle, the bisector of an angle, or the bisector of Us exterior angle, divides the opposite side, internally or externally, into mgments which are projyortional to the adjacent sides. B O O K I I 1 . 101 1st. Let AD bisect the angle A of the triangle ABC; then, DB:DC=AB:Aa For, through B draw BE parallel to DA, meeting CA produced in E. The angle ABE = BAD (I. 49), and the angle AEB = CAD (I. 51) ; and, by hypothesis, the angle BAD = CAD ; therefore, the angle ABE = AEB, and AE = AB (I. 90). Now, in the triangle CEB, AD being parallel to EB, we have (17), DB:DC=AE:Aa or DB: DC=AB:AC; that is, the side BC is divided by AD internally into segments pro- portional to the adjacent sides AB and A C. 2d. Let AD' bisect the exterior angle BAE; then, D'B:D'C=AB:AC For, draw J5jE^' parallel to D'A; then, ^J5^' is an isosceles tri- angle, and AE' = AB. In the triangle CAD', we have (17), D'B:D'C=AE' :AC, or D'B'.D'C=AB:AC{ that is, the side BC is divided by AD' externally into segments pro- portional to the adjacent sides AB and A C. 22. Scholium. When a point is taken on a given finite line, or on the line produced, the distances of the point from the extremities of the line are called the segments, internal or external, of the line. The given line is the sum of two internal segments, or the difference tjf two external segments. 23. Corollary. If a straight line, drawn from the vertex of any angle of a triangle to the opposite side, divides that side internally in the ratio of the other two sides, it is the bisector of the angle ; if it divides the opposite side externally in that ratio, it is the bisector of the exterioi angle. (To be proved). 9* 102 Gi:OMETRY. I SIMILAR POLYGONS. 24. Definitions. Two polygons are similar^ when they are mutually equiangular and have their homologous sides proportional. In similar i)olygons, any points, angles or lines, similarly situated in each, are called homologous. The ratio of a side of one polygon to its homologous side in the ether is called the ratio of similitude of the polygons. PROPOSITION IV.— THEOREM. J 25. Two triangles are similar, when they are mutually equiangular. Let ABCy A'B'C'y be mutually equiangular triangles, in which A = A\ B = B\ O = C; then, these triangles are similar. For, place the angle A' upon its equal angle A, and let B' fall at b and C" at c. Since the angle Abe is equal to B, be is parallel to BC (I. 55), and we have (15), AB:Ab = AC:Ac, or AB:A'B' = AC:A'C\ In the same manner, it is proved that AB:A'B' = BC:B'C'; and, combining these proportions, 4B _ AC _ BC A'B'~ A'C'~ B'C' [1] Therefore, the homologous sides are proportional, and the triangles are similar (24). 26. Corollary. Two triangles are similar when two angles of the one are respectively equal to two angles of the other (I. 73). 27. Scholium I. The homologous sides lie opposite to equal angles. 28. Scholium II. The ratio of similitude (24) of the two similar triangles, is any one of the equal ratios in the continued propor- tion [1]. BOOK III, 103 29. Scholium III. In two similar triangles, any two homologoua lines are in the ratio of similitude of A A' the triangles. For example, the per- pendiculars AD, A'D\ drawn from the homologous vertices A, A\ to the op- posite sides, are homologous lines of the two triangles; and the right tri- angles ABD, AB'D\ being similar (25), we have AD _ AB _ AC A'D'~ A'B''' A'C BG B'C In like manner, if the lines AD, A'D\ were drawn from Aj A! , to the middle points of the opposite sides, or to two points which divide the opposite sides in the same ratio in each triangle, these lines would still be to each other in the ratio of similitude of the two triangles. PROPOSITION v.— THEORExM. 30. Two triangles are similar, when their homologous sides are pro- portional. In the triangles ABC, A' B'C, let AB__A^^ BC , A'B'~ A'C'~ B'C' then, these triangles are similar. For, on AB take Ah = A'B', and draw he parallel to BC. Then, the triangles Ahe and ABC are mutually equiangular, and we have (25), AG_BG Ac be [1] AB AB — or Ah A'B' Comparing this -with the given proportion [1], we see that the first ratio is the same in both ; hence the second and third ratios in each are equal respectively, and, the numerators being the same, the denominators are equal; that is, A,C = Ac, and B'C = be. Therefore, the triangles A' B'C and Abe are equal (I. 80) ; and since Abo is similar to ABC, A'B' C is also similar to ABC. 104 GEOMETRY. 31. Scholium. In order to establish the similarity of two polygons according to the definition (24), it is necessary, in general, to sho\A that they fulfill two conditions : 1st, they must be mutually equi- angular, and 2d, their homologous sides must be proportional. In the case of triangles, however, either of these conditions involves the other ; and to establish the similarity of two triangles it will be suf- ficient to show, either that they are mutually equiangular, or that their homologous sides are proportional. I PROPOSITION VI.— THEOREM. '^^■1 32. Trvo triangles are similar^ when an angle of the one is equal to an angle of the other, and the sides including these angles are propor- portional. In the triangles ABC, A'B'C, let a a' A = A\ and yy // A'B'~A'C'' b^-.-./c ^, ^ then, these triangles are similar. ^ ^ ^^H For, place the angle Al upon its ^^^ equal angle A ; let B' fall at 6, and C at c. Then, by the hy- pothesis, AB_AC Ab~ Ac' Therefore, be is parallel to BC (19), and the triangle Abe is simi to ABC (25). But Abe is equal to A'B'C; therefore, A'B'C is also similar to ABC. 4 liar I PROPOSITION VII.— THEOREM. 33. Two triangles are similar, when they have their sides parallel each to each, or perpendicular each to each. Let ABC, abc have their sides par- allel each to each, or perpendicular each to each ; then, these triangles are similar. For, when the sides of two angles BOOK III 106 are pamllel each to each, or perpen- dicular each to each, these angles are either equal, or supplements of each other, (I. 60, 62, 63). In the present case, therefore, three hypotheses may be made, namely, denoting a right angle hyE, 1st hyp. A + a = 2B, B -\- b = 2R, C+ c = 2B; 2d " A = a. B -{-b = 2B, C-{- c = 2B; 3d " A = a, B = b, whence C = c. The 1st and 2d hypotheses cannot be admitted, since the sum of all the angles of the two triangles would then exceed four right angles (I. 68). The 3d hypothesis is therefore the only admissible one; that is, the two triangles are mutually equiangular and consequently similar. 34. Scholium. Homologous sides in the two triangles are either two parallel sides, or two perpendicular sides; and homologous, or equal, angles, are angles included by homologous sides. PKOPOSITION VIII.— THEOREM. 35. If three or more straight lines drawn through a common point intersect two parallels, the corresponding segments of the parallels are in proportion. Let OA, OB, OC, OB, drawn through the common point 0, intersect the parallels AD and ad, in the points A, B, C, D and a, b, c, d, respectively ; then, 4B_BC_CD ab be cd For, the triangle OAB is similar to the tri- angle Oab (25) ; OB C is similar to Obc ; and CD to Ocd ; therefore, we have 4B_qB__BC_0C_CD ab Ob be Oc cd \<. c fh A i c \f // ~x '/■/ c V' vvhich includes the proportion that was to be proved. 106 G E O M E T K Y. I 36. Scholium. The demonstration is the same whether the parallels cut the system of diverging lines on the same side, or on opposite sides, of the point 0. Moreover, the demonstration extends to any corresponding segments, as AC and ac, BD Sind bd^ etc. ; and the ratio of any two corresponding segments is equal to the ratio of. the distances of the parallels from the point 0, measured on any one of the'diverging lines. PROPOSITION IX.— THEOREM. 37. Conversely, if three or more straight lines divide two parallels pro2yortionally, they pass through a common point. Let Aa, Bb, Cc, Dd, divide the parallels AD and ad proportionally ; that is, so that AB_BC_CD ~ ab be cd [1] then, Aaj Bb, etc., meet in a common point. For, let Aa and Cc meet in 0; join Ob. Then, in order to prove that Bb passes through 0, we have to prove that Ob and Bb are in the same straight line. Now, if they are not in the same straight line, Ob produced cuts AD in some point P differing from B ; and by the preceding theorem, we have AP ab AG ac But, from the hypothesis [IJ, wB have by (12), AC AB ab ac whence, AP = AB, which is impossible unless P coincides with B, and Ob produced coincides with Bb. Therefore, Bb passes through 0. In the same way, Dd is shown to pass through O. BOOK III. 107 PROPOSITION X.— THEOREM. 38. If two polygons are composed of the same number of triangles similar each to each and similarly placed, the polygons are similar. Let the polygon ABCD, etc., be composed of the triangles ABC, A CD, etc.; and let the polygon A'B' C'D\ etc., be com- posed of the triangles A'B'C, A'C'D\ etc., similar to ABC, A CD, etc., respectively, and similarly placed ; then, the polygons are similar. 1st. The polygons are mutually equiangular. For, the homolo- gous angles of the similar triangles are equal ; and any two corre- sponding angles of the polygons are eithv homologous angles of two similar triangles, or sums of homologous angles of two or more similar triangles. Thus B = B' ; BCD = BCA -{- ACD = B'C'A' -f A'C'D' = B'C'D'', etc. 2d. Their homologous sides are proportional. For, from the simi- lar triangles, we haye AB A'B' BG B'C AC A'C CD AD DE = — - — : = etc. CD' A'D' D'E' Therefore, the polygons fulfill the two conditions of similarity (24). PROPOSITION XI.— THEOREM. 39. Conversely, two similar polygons may he decomposed into the same number of triangles similar each to each and similarly placed. Let ABCD, etc., A'B' CD', etc., be two similar polygons. From two homologous vertices, A and A', let diagonals be drawn in each polygon ; then, the polygons will be decomposed as required. For, 1st. We have, by the definition of similar polygons, 108 G E O M E T K Y. Angle B = B', and = BG_- B'C" therefore, the triangles AB C and JL'J5'C" are similar (32). 2d. Since ABC and A'B'C are similar, the angles BCA and B'^'A' are equal; subtracting these equals from the equals ^CZ) and B' C'D\ respectively, there remain the equals AGD and A' C'D\ Also, from the similarity of the triangles ABC and A'B'C, and from that of the polygons, we have BC CD . AG_^ ^ A'C'~ B'C'~ CD'' therefore, the triangles ACD and A' CD' are similar (32). Thus, successively, each triangle of one polygon may be shown be similar to the triangle similarly situated in the other. 40. Scholium. Two similar polygons may be decomposed into simi- lar triangles, not only by diagonals, but by lines drawn from any two homologous points. Thus, let be any arbitrarily assumed point in the plane of the polygon ABCD, etc.; and draw OA, OB, OC, etc. In the similar polygon A'B'CD', etc., draw A' 0' making the angle B'A'O' equal to BAO, and B' 0' making the angle A'B' 0' equal to AB 0. The intersection 0' of these lines, regarded as a point belonging to the polygon A'B' CD', etc., is homologous to the point of the polygon ABCD, etc.; and the lines O'A', O'B', O'C, etc., being drawn, the triangles O'A'B', O'B'C, etc., are ghown to be similar to OAB, OBC, etc., respectively, by the same method as was employed in the preceding demonstration. If the point is taken without the polygon, and its homologcus BOOKIII. 109 point 0' found as before by constructing the triangle O'AIB' similar a O^.fliff::: to OAB, the polygons will be decomposed into triangles partly addi- tive and partly subtractive. Thus the polygon ABODE is equal to the sum of the two triangles 0J5C and OCD, diminished by the triangles OBA, OAE and OED-, and the polygon A'B' C'D'E' \9, similarly decomposed. Homologous lines in the two polygons are lines joining pairs of homologous points, such as OA and 0'A\ OB and O'B', etc., the diagonals joining homologous vertices, etc. ; and it is readily shown that any two such homologous lines are in the same ratio as any two homologous sides, that is, in the ratio of similitude of the poly- gons (24). 41. Oorollary. Two similar polygons are equal when any line in one is equal to its homologous line in the other. PROPOSITION XII.— THEOREM. 42. Tlie perimeters of two similar polygons are in the same ratio as any two homologous sides. For, we have (see preceding figures), AB _ BO _ _„ A'B'~ B'0'~~ — ' — ^^•' etc. whence (12), AB -f ^(7 -f CD + etc. _ AB_ _ BC^ A'B' J^ B'O' ^ O'D' H- etc. ~" AB' ~ B' 0' CD' AB \ 43. Corollary. The perimeters of two similar polygons are in the same ratio as any two homologous lines ; that is, in the ratio of "similitude of the polygons (40). 10 no GEOMETRY. APPLICATIONS. PROPOSITION XIII.— THEOREM. 44. If a perpendicular is drawn from the vertex of the right an to the hypotenuse of a right triangle : Jst. The two triangles thus formed are similar to each other and to ffl the whole triangle ; 2d. The perpendicular is a mean proportional between the segments ^ of the hypotenuse; M 3d. Each side about the right angle is a mean proportional between the hypotenuse and the adjacent segment. Let C be the right angle of the triangle ABC, and CD the perpendicular to the hy- potenuse ; then, 1st. The triangles J. C-Z) and CBD sire simi- lar to each other and to ABC. For, the triangles A CD and ABC have the angle A common, and the right angles, ADC, ACB, equal; therefore, they are similar (26). For a like reason CBD is similar to AB C, and consequently also to A CD. ■■ 2d. The perpendicular CD is a mean proportional between the segments AD and DB. For, the similar triangles, A CD, CBD, give AD: CD = CD: BD. 3d. The side J. is a mean proportional between the hypotenuse AB and the adjacent segment AD. For, the similar triangles, ACD, ABC, give AB:AC=AC:AD. In the same way, the triangles CBD and ABC give, AB:BC = BC:BD. i 45. Corollary I. If all the lines of the figure are supposed to be expressed in numbers, being measured by any common unit, the preceding proportions give, by (5), CD' = ADX BD, AC'=ABX AD, BG'= AB X BD; BOOK III. Ill u'liere we employ the notation CD^, as in algebra, to signify the pro- duct of CD multiplied by itself, or the second power of CD\ ob- serving, however, that this is but a conventional abbreviation for "second power of the number representing CD" (8). It may be read " the square of CD," for a reason that will appea r hereaf ter. 46. Corollary II. By division, the last two ceding corollary give /"' "^^"'of the that is, the squares of the sides including the right angle are propor- tional to the segments oj the hypotenuse. 47. Corollary III. If from any point C in the circumference of a circle, a perpendicular CD is drawn to a diameter AB, and also the chords CJ., CB ; then, since A CB is a right angle (II. 59), it follows that the perpendicular is a mean proportional between the segments of the diameter ; and each chord is a mean proportional be- tween the diameter and the segment adjacent to that chord. PKOPOSITION XIV.— THEOKEM. 48. The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. Let ABC be right angled at C; then, '^ AB' = AC'' + ^C\ For, by the preceding proposition, we have AC' = ABX AD, and BC' = AB X BD, the sum of which is ACP -i:BC' = ABx iAD + BD) = AB X AB = AB\ 49. Corollary I. By this theorem, if the numerical measures of two sides of a right triangle are given, that of the third is found. For example, if^C=3, ^C = 4; then, AB = -,/[3' 4 4^] = 5. 112 GEOMETRY. I If the hypotenuse, AB, and one side, AC, are given, we have BC' = AB' — AC'; thus, if there are given AB = 5, AC = 3, then, we find BC= x/\b' — 3^] == 4. 50. Corollary II. If JL C is the diagonal of a square ^ ABCD, we have, by the preceding theorem, wh AC' = AB' + BC' = 2AB\ ence, Tc' AB = % and extracting the square root, AC AB = V2 = 1.41421 + ad inf. Since the square root of 2 is an incommensurable number, it follows that the diagonal of a square is incommensurable with its side. 51. Definition. The projection of a point A upon an indefinite straight line XI^ is the foot P of the perpendicular let fall from the point upon the line. . ... ^ ^ ^ The projection of a finite straight line AB upon the line XY is the distance PQ between the projections of th extremities of AB. If one extremity B of the line AB is in the line XY, the distance from B to P (the projec- tion of A) is the projection of AB on XY; for the point J5 is in this case its own projection. PKOPOSITION XV.— THEOREM. H 52. In any triangle, the square of the side opposite to an acute angle is equal to the sum of the sqiLares of the other two sides diminished by twice the product of one of these sides and the projection of the other upon that side. Let C be an acute angle of the triangle ABC, P the projection of A upon BC by the perpen- dicular APf PC the projection of J. C upon BC; i>^ then, Fig. 1. . BOOK III. 113 AB'=BC''-^AG' — 2BC^ PC, A Fig. 2. For, if P falls on the base, as in Fig. 1, we X'^--^ have F~ li C PB = BG—PC, and if P falls upon the base produced, as in Fig. 2, we have PB = PC—BC, but in either case the square of PB is, by a theorem of algebra, "^ PB^ = BC' + PC" — 2BC X PC. Adding AP^ to both members of this equality, and observing that by the preceding theorem, PB'' + AP' = AB\ and PC' -{- AP'' = A C , we obtain AB'=BC'-}'AX!' — 2BCXPC PEOPOSITION XVI.— THEOREM. 53. In an obtuse angled triangle, the square of the side opposite to the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the prelection of the other upon that side. Let C be the obtuse angle of the triangle ABC, ^ P the projection of A upon BC (produced) ; then. AB'=BC' + AC + 2BC X PC. For, since P can only fall upon BC produced, ACB being an obtuse angle, we shall in all cases have PB = BC-^PC, and the square of PB will be, by an algebraic theorem, f PB'=BC' + PC'-{-2BCX PC. Adding AP^ to both members, we obtain IB' = BC' -{- AG' -i- 2BC X PC. * (x — yy or (y — x)^ = x^ + ^/^ — 2xy. f {x + yY =^x^ + y2 _^ 2ly. 10* ^ 114 GEOMETRY. 54. Corollary. From the preceding three theorems, it follows that an angle of a triangle is acute, right or obtuse, according as the square of the side opposite to it is less than, equal to, or greater than, the sum of the squares of the other two sides. PROPOSITION XVII.— THEOREM. i 55. If through a fixed point within a circle any chord is drawn, the product of its two segments has the same value, in whatever direction the chord is drawn. Let P be any fixed point within the circle 0, AB and A'B' any two chords- drawn through P; then, PAX PB = PA' X PB'. For, join AB' and A'B. The triangles APB', A'PBy are similar, having the angles at P equal, and also the angles A and A' equal (II. 58) ; therefore, whence (5), PA : PA' = PB' : PP, PAX PB = PA' X PB'. 56. Corollary. If AB is the least chord, drawn through P (II. 20), then, since it is perpendicular to OP, we have PA = PB (II. 15), and hence PA^ = PA' X PB ' ; that is, either segment of the least chord drawn through a fixed point is a mean proportional between the segments of any other chord drawn through that point. 57. Scholium. If a chord constantly passing through a fixed point P, be conceived to revolve upon this point as upon a pivot, one seg- ment of. the chord increases while the other decreases, but their product being constant (being always equal to the square of half the least chord), the two segments are said to vary reciprocally, or to be recip'^oeally proportional (2). BOOK III 116 PROPOSITION XVIII.— THEOREM. 58. If through a fixed point without a circle a secant is drawn, the product of the whole secant and'its external segment has the same value, in whatever direction the secant is drawn. Let P be any fixed point without the circle 0, PAB and PA'B' any two secants drawn through P; then, PA XPB = PA' X PB\ For, join AB' and A'B. The triangles APB\ A'PB, are similar, having the angle at P common, and also the angles B and B' equal (II. 58) ; there- fore, PA : PA' = PB' : PB, whence (5), PAX PB = PA' X PB'. 59. Corollary. If the line PAB, constantly passing through the fixed point P, be conceived to revolve upon P, as upon a pivot, and to approach the tangent PT, the two points of intersection, A and B, will approach each other ; and when the line has come into coinci- dence with the tangent, the two points of intersection will coincide in the point of tangency T. The whole secant and its external seg- ment will then both become equal to the tangent PT; therefore, regarding the tangent as a secant whose two points of intersection are coincident (II. 28), we shall have PT' = PA' X PB'; that is, if through a fixed point without a circle a tangent to the circle is drawn, and also any secant, the tangent is a mean proportional be- tween the whole secant and its external segment. 60. Scholium I. When a secant, constantly passing through a fixed point, changes its direction, the whole secant and its external seg- ment vary reciprocally, or they are reciprocally proportional, since their product is constant (2). 61. Scholium II. The analogy between the two preceding proposi- tions is especially to be remarked. They may, indeed, be reduced to a single proposition in the following form : If through any fixed 116 GEOMETRY. point in the plane of a circle a straight line is drawn Intersecting the circumference, the product of the distances of the fixed point from the two points of intersection is constant. PKOPOSITION XIX.— THEOREM. 62. In any triangle, if a medial line is drawn from the vertex to the 1st. The sum of the squares of the two sides is equal to twice the square of half the base increased by twice the square of the medial line; 2d. The difference of the squares of the two sides is equal to twice the product of the base by the projection of the medial line on the base. In the triangle ABC, let D be the middle point of the base BC, AD the medial line from A to the base, P the projection of A upon the base, DP the projection of AD upon the base ; then, 1st. AB' + AG' = 2BD' + 2AD' ; 2d. AB'—AG' = 2BG X DP, For, if AB> AC, the angle ADB will be obtuse and ADO will be acute, and in the triangles ABD, ADC, we shall have, by (53) and (52). AB' = BD' + AD' + 2BD X DP, AC'=DC' + ^^' — 2DC X DP. Adding these equations, and observing that BD = DC, we have 1st. IB' + AG' = 2BD' + 2AD\ Subtracting the second equation from the first, we have AB' — AG' :-■= 2 {BD ^ DC) X DP; that is, 2d. AB' — AC' = 2BG X DP. BOOK III. 117 63. Corollary I. In any quadrilateral, the sum of the squares of the four sides is equal to the sum of the squares of the diagonals plus four times the square of the line joining the middle points of the diag- onals. For, let E and F be the middle points of the diagonals of the quadrilateral ABCD; join EF, EJj, ED. Then, by the preceding theorem, we have in the triangle ABC, IE' + BG" = 2AE' + 2BE\ and in the triangle ADCy CD'-^DA' = 2AE' + 2DE\ whence, by addition, AB' -\-BC'-{- GD'-\- DT = 4AE' + 2 (BE* -f DE'). Now, in the triangle BED, we have BE' + im'=.2BF'+2EF''; therefore, AB' -i- BC" + CD' -\- DA' = 4AE' -f 4BF' + AEF\ But AAE' = (2AEy = AU\ and ABF' = (2BFy = BD' ; hence, finally, AB' -f BC' -f CD'-^DA' = AC' + BD' + 4EF\ 64. Corollary II. In a parallelogram, the sum of the squares of the four sides is equal to the sum of the squares of the diagonals. For if the quadrilateral in the preceding corollary is a parallelo- gram, the diagonals bisect each other, and the distance EF is zero. PKOPOSITION XX.— THEOREM. 65. In any triangle, the product of two sides is equal to the product of the diameter of the circumscribed circle by the perpendicular let fall upon the third side from the vertex of the opposite angle. 118 GEOMETRY. Let AB, A C, be two sides of a triangle ABC, AD the perpendicular upon BC, AE the di- ameter of the circumscribed circle ; then, ABXAC=AEXAD. For, joining CE, the angle ACE is a right angle (II. 59), and the angles E and B are equal (II. 58) ; there- fore, the right triangles AEC, ABD, are similar, and give AB:AE=AD:AQ whence, AB X AC = AE X AD. . PROPOSITION XXI.— THEOREM. 66. In any triangle, the product of two sides is equal to the proi of the segments of the third side formed by the bisector of the opposite angle plus the square of the bisector. Let AD bisect the angle A of the triangle ABC; then, ABXAC=DBX DC-i- DT. For, circumscribe a circle about ABC, produce AD to meet the cir- cumference in E, and join CE. The triangles ABD, AEC, are similar, and give AB:AE=DA:AC, whence AB X AC = AE X DA = (DE + DA) X DA = DEX l>A-\- DA\ Now, by (55), we have DE X DA = DB X DC, and hence ABX AC=DBX DC^DT. 67. Corollary. If the exterior angle BAF is bisected by AD', the same theorem holds, except that plus is to be changed to minus. BOOK III 119 For, producing D'A to meet the circumference in E\ and joining CE', the triangles ABD\ AE'C, are similar, and give AB : AE' = AD' : AC, whence AB X AC =^ AE' X AD' = {D'E' — D'A) X D'A = D'E' X D'A — WA\ or, by (58), AB X AC=D'B X D'C— D'A' PROBLEMS OF CONSTRUCTION. PROPOSITION XXII.— PROBLEM. 68. To divide a given straight line into parts proportional to given straight lines. Let it be required to divide AB into parts proportional to 31, N and P. From A draw an indefinite straight line AX, upon which lay- off AC =31, CD = N, DE = F, join EB, and draw CF, DG, parallel to EB; then AF, FG, GB, are proportional to 31, N, F (18). 69. Corollary. To divide a given straight line AB into any num- ber of equal parts, draw an indefinite line AX, upon which lay ofi* the same number of equal distances, each distance being of any convenient length ; through 3f the last point of division on AX draw 3IB, and through the other points of division of AX draw par- allels to 3IB, which will divide AB into the required number of equal parts. This follows both from the theory of proportional lines and from (I. 125). 120 GEOMETRY. PEOPOSITION XXIII.— PROBLEM. 70. To find a fourth proportional to three given straight lines. Let it be required to find a fourth propor- tional to My N and P. Draw the indefinite ^v n lines AX, AY, making any angle with each otRer. Upon AX lay off AB = M, AD = N; and upon AY lay off J. (7= P; join BC, and draw DE parallel to BC; then AE is the re- quired fourth proportional. For, we have (15), AB:AD = AC: AE, or M: N=PxAE. 71. Corollary. If AB = M, and both AD and J. Care made equal to N, AE will be a third proportional to Jf and N] for we shall have M:N=N:AE. PEOPOSITION XXIV.— PROBLEM. 72. To find a mean proportional between two given straight lines. Let it be required to find a mean proportional between M and N. Upon an indefinite line lay off AB = 3I,BC = N; upon AC describe' a semi-circumference, and at B erect a perpen- dicular, BD, to A C. Then BD is the required mean proportional (47). Second method. Take AB equal to the greater line 3f, and upon it lay off BC = N. Upon AB describe a semi-circumference, erect CD per- pendicular to AB and join BD. Then BD is the required mean proportional (47). 73. Definition. When a given straight line is divided into two segments such that one of the segments is a mean proportional between the given line and the other segment, it is said to be divided in extreme and mean ratio. Thus AB is divided in extreme and mean ratio at C, if AB : AC = ACiCB. cf BOOK III. 121 If C is taken in BA produced so that AB : AC = AC : CB then AB is divided at C", externally, in extreme and mean ratio. PROPOSITION XXV.— PROBLEM. 74. To divide a given straight line in extreme and mean ratio. Let AB be the given straight line. At B erect the perpendicular BO equal to one half of AB. With the centre and radius / ^X/?' OB J describe a circumference, / ^^^^"j and through A and draw A ^.^^\ J cutting the circumference first I; -^f— — i^^ ^^ in D and a second time in i)'. Upon AB lay off ^C = AD, and upon BA produced lay off AC ^= AD'. Then AB is divided at G internally, and at C exter- nally, in extreme and mean ratio. For, 1st, we have (59), AD' :AB = AB:AD or AC, [1] whence, by division (10), AD' — AB:AB = AB — AC:AQ or, since DD' = 20B = AB, and therefore AD' — AB = AD' — DD'=:AD = AC, AC : AB = CB : AC, and, by inversion (7), AB'.AC=AC'.CB; that is, AB is divided at C, internally, in extreme and mean ratio. 2d. The proportion [1] gives by composition (10), AD' + AB : AD' = AB + AD : AB, jT, since AD' = AC, AD' + AB = CB, AB -^ AD = DD' -f AD = AD' = AC, CB:AC = AC:AB, and, by inversion, AB : AC = AC : CB; m .that is, AB is divided at C, externally, in extreme and mean ratio. 11 122 GEOMETRY. 75 /Scholium, Since OD = OD' AB ~ 2 ' we have AC = A0 AB 2 ' AC' = ■■A0 + ' AB 2 ■ But the right triangle A OB gives whence, extracting the square root, AO=^AB. Therefore, AC=AB. t/5 — 1 VI '- .^ t/5 + 1 AC = AB. 2 2 76. Definitions. When a straight line is divided internally and externally in the same ratio, it is said to be divided harmonically. Thus, AB is divided harmonically at G and A if CA : CB = DA : DB; I 1 1 that is, if the ratio of the distances of C from A and B is equal to the ratio of the distances of D from A and B. Since this proportion may also be written in the form AC:AD = BC:BD, the ratio of the distances of A from C and D is equal to the ratio of the distances of B from C and D ; consequently the line CD is divided harmonically at A and B. The four points A, B, C, D, thus related, are called harmonic points, and A and B are called conjugate points, as also C and D. PROPOSITION XXVL— PROBLEM. 77. To divide a given straight line harr^onically in a given ratio. Let it be required to divide AB harmonically in the ratio of il^f to N. Upon the indefinite line AX, lay oflf ^i; = M, and from E lay off EF and EG, each equal to N\ join FB, GB; and draw EC paraUel to FB, ED parallel to GB. J a I book: III. 123 Then, by the construction we have (17), M_ CA DA N~ CB~ DB' therefore, by the definition (76), AB is divided harmonically at C and Z), and in the given ratio. 78. Scholium. If the extreme points A and D are given, and it is required to insert their conjugate harmonic points B and C, the har- monic ratio being given = 31 : N, we take on AX, as before, AE = M and EF = EG = iV, join ED, and draw GB parallel to ED, which determines B) then, join FB and draw jE^C parallel to i^^, which determines C Also if, of four harmonic points J., B, C, D, any three are given, the fourth can be found. PEOPOSITION XXVII.— PROBLEM. 79. To find the locus of all the points whose distances from two given points are in a given ratio. Let A and B be the given points, and let the given ratio be M ; N, Suppose the problem solved, and that P is a point of the required locus. Divide AB internally at C and externally at D, in the ratio M'. N, and join PA, PB, PC, PD. By the condition imposed upon P we must have PA:PB = 31: N= CA : CB = DA : DB; therefore, PC bisects the angle APB, and PD bisects the exterior angle BPE (23). But the bisectors PC and PD are perpendicular to each other (I. 25) ; therefore, the point P is the vertex of a riglit angle whose sides pass through the fixed points C and D, and the locus of P is the circumference of a circle described upon CD as a diameter (II. 59, 97). Hence, we derive the following Construction. Divide AB harmonically, at C and D, in the given ratio (77), and upon CD as a diameter describe a circumference. This circumference is the required locus. 124 GEOMETRY. PROPOSITION XXVIII.— PROBLEM. 80. On a given straight line, to construct a polygon similar 'o a given polygon. Let it be required to construct u^on A'B' a polygon similar to ABCDEF. Divide ABCDEF into tri- angles by diagonals drawn from A. Make the angles B'A'C and A'B'C equal to jBJ.(7and J.jB (7 respectively; then, the triangle A'B'C will be similar to ABC (25). In the same manner construct the triangle A'D'C similar to ADC, A'E'D' similar to AED, and A'E'F' similar to AEF. Then, A'B'C'D'E'F' is the required polygon (38). PROPOSITION XXIX.— PROBLEM. 81. To construct a polygon similar to a given polygon, the ratio similitude of the two polygons being given. Let ABCDE be the given polygon, and let the given ratio of similitude be M : N. Take any point 0, either within or without the given polygon, and draw straight lines from through each of the vertices of the polygon. Upon any one of these lines, as OA, take OA' a fourth proportional M, Nf and OA, that is, so that Jf:N= OA: 0A\ N\- to In the angle AOB draw A'B' parallel to AB; then, in the angle BOC, B'C parallel to BC, and so on. The polygon A'B'C'D'E' will be similar to ABODE) for the two polygons will be composed BOOK III. 125 of the same number of triangles, additive or subtractive, similarly placed; and their ratio of similitude will evidently be the given ratio if: iV.' (40). 82. Scholium. The point in the preceding construction is called the centre of similitude of the two polygons. 11 » BOOK IV. COMPARISON AND MEASUREMENT OF THE SURFACES OF RECTILINEAR FIGURES. 1. Definition. The area of a surface is its numerical measure, referred to some other surface as the unit ; in other words, it is the ratio of the surface to the unit of surface- (II. 43). The unit of surface is called the superficial unit. The most con- venient superficial unit is the square whose side is the linear unit. 2. Definition. Equivalent figures are those wliose areas are equal. PEOPOSITION I.— THEOREM. 3. Two rectangles having equal altitudes are to each other as their Let ABCD, AEFD, be two rectangles hav- ing equal altitudes, AB and AE their bases ; then, ABCD AB AEFD ~~ AE-^ -I 1-— r— E Suppose the bases to have a common meas- ure which is contained, for example, 7 times in AB, and 4 times in AE\ so that if AB is divided into 7 equal parts, AE will contain 4 of these parts ; then, we have AB_1 AE~~ A If, now, at the several points of division of the bases, we erect [)erpendicu]ars to them, the rectangle ABCD will be divided into 7 126 I BOOK IV 127 equal rectangles (I. 120), of which AEFD will contain 4; conse- quently, we have ABCD 7 AEFD ~ 4 and therefore ABCD AB AEFD ~ AE The demonstration is extended to the case in which the bases are incommensurable, by the process already exemplified in (II. 51) and (III. 15). 4. Corollary. Since AD may be called the base, and AB and AE the altitudes, it follows that tioo rectangles having equal bases are to each other as their altitudes. Note. In these propositions, by "rectangle" is to be understood "surface of the rectangle." PKOPOSITION II.— THEOKEM. 5. Any two rectangles are to each other as the products of their bases by their altitudes. Let B and B' be two rectangles, k and k' their bases, h and A' their h altitudes; then. B^ B' k X h k' X h'' For, let /S be a third rectangle having the same base k as the rec- tangle B, and the same altitude h' as the rectangle B' \ then we have, by (4) and (3), B _h S__k S~h'* B'^F and multiplying these ratios, we find (III. 14), B_ _ kX h B'~k'Xh'' 6. Scholium. It must be remembered that by the product of two 128 GEOMETRY. lines, is to be understood the product of the numbers which represent them when they are measured by the linear unit (III. 8). PROPOSITION III.— THEOREM. 7. The area of a rectangle is equal to the product oj its base and altiiude. Let H be any rectangle, k its base and h its altitude numerically expressed in a terms of the linear unit; and let Q be the square whose side is the linear unit ; then, by the preceding theorem, B _ JcX h I 1 XI k X h. B But since Q is tlie unit of surface, — = the numerical measure, area, of the rectangle B (1); therefore, Area of B= k X h. 8. Scholium I. When the base and altitude are exactly divisible by the linear unit, this proposition is rendered evident by dividing the rectangle into squares each equal to the superficial unit. Thus, if the base contains 7 linear units and the altitude 5, the rec- tangle can obviously be divided into 35 squares each equal to the superficial unit ; that is, its area =: 5 X 7. The proposition, as above demonstrated, is, however, more general, and includes also the cases in which either the base, or the altitude, or both, are incommensurable with the unit of length. 9. Scholium II. The area of a square being the product of two equal sides, is the second power of a side. Hence it is, tliat in arith- metic and algebra, the expression "square of a number" has been adopted to signify "second power of a number." AVe may also here observe that many writers employ the expres- sion " rectangle of two lines" in the sense of " product of two lines,'* because the rectangle constructed upon two lines is measured by the product of the numerical measures of tlie lines. BOOK IV. 129 PROPOSITION IV.— THEOREM. 10. The area of a parallelogram is equal to the product of its base and altitude. Let ABCD be a parallelogram, k the numerical measure of its base AB, h that of its altitude AF; and denote its area by S; then, S=k X A. For, let the rectangle ABEF be con- structed having the same base and alti- tude as the parallelogram ; the upper bases of the two figures will be in the same straight line FG (I. 58). The right triangles AFD and BEG are equal, having AF = BE, and AD = BG (I. 83). If from the whole figure ABGFwe take away the triangle AFD, there remains the parallelogram ABGD; and if from the whole figure we take away the triangle BEG, there remains the rectangle ABEF; therefore the surface of the parallelogram is equal to that of the rectangle. But the area of the rectangle is Jc X^ h (7) ; therefore that of the parallelogram is also k X h; that is S := k X h. 11. GorollaryJ. Parallelograms having equal bases and equal alti- tudes are equivalent. 12. Gorollary II. Parallelograms having equal altitudes are to each other as their bases ; parallelograms having equal bases are to each other as their altitudes; and any two parallelograms are to each other as the products of their bases by their altitudes. PROPOSITION v.— THEOREM. 13. The area of a triangle is equal to half the product of its base and altitude. Let ABGhe a triangle, k the numerical raeas- ^ ^ ure of its base BG, h that of its altitude AD; and 8 its area ; then, S=ik X h. For, through A draw AE parallel to CB, and through B draw BE 11** 130 GEOMETllY. parallel to CA. The triangle ABC is one-half the parallelogram AEBG (1. 105) ; but the area of the parallelogram =^k y^h) there- fore, for the triangle, we have S = hk X h. 14. Corollary I. A triangle is equivalent to one-half of any par allelogram having the same base and the same altitude. 15. Corollary II. Triangles having equal bases and equal altitudes are equivalent. 16. Corollary III. Triangles having equal altitudes are to each other as the'ir bases ; triangles having equal bases are to each other as their altitudes ; and any two triangles are to each other as the products of their bases by their altitudes. I 1 PKOPOSITION VI.— THEOREM. 17. 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; JfiV= A, its al- ^ ^f ^ n titude; AD = a, BC = b, its parallel bases; and let 8 denote its area ; then, i S=l{a-\-b) X h. N For, draw the diagonal A C. The altitude of each of the triangles J.i)Cand ABC is equal to /t, and their bases are respectively a and b ; the area of the first is I a X h, that of the second is 2 6 X /i ; and the trapezoid being the sum of the two triangles, we have S=laXh + H X h = h(a-^ b) X h. 18. Corollary. The straight line EF, joining the middle points of AB and i)(7, being equal to half the sum of AD and BC (I. 124), the area of the trapezoid is equal to the product MN X EF. 19. Scholium. The area of any polygon may be found by finding the areas of the several triangles into which it may be decomposed l)y drawing diagonals from any vertex. The following method, however, is usually preferred, especially in ^surveying. Draw the longest diagonal AD of tiie proposed polygOD I BOOK IV. 131 ABGDEF] and upon AD let fall the per- pendiculars BM, CJSr, JEP, FQ. The poly- gon is thus decomposed into right triangles and right trapezoids, and by measuring the lengths of the perpendiculars and also of the distances AM, MN, ND, AQ, QP, PD, the bases and altitudes of these triangles and trapezoids are known. Hence their areas can be computed by tbe preceding theorems, and the sum of these areas will be the area of the polygon. PKOPOSITION VII.— THEOREM. 20. Similar triangles are to each other as the squares of their homolo- gous sides. Let ABC, A'B'C be similar tri- angles; then, ABC BC' A'B'C'~ WV" Ijet AD, A'D', be the altitudes. By (16), we have BCX AD ^BG^ AD ABC A'B'C B'C'XA'D' B'C'^A'D' But the homologous lines AD, AID' , are in the ratio of similitude of the triangles (III. 29) ; that is, AD therefore, ABC BC AID' B'C BC ^ BC__ BG^ — - - - I rit^ A'B'C B'C B'C B'C 21. Corollary. If we had put the ratio ^42) : A'D' in the place'of the ratio BC' B^C, we should have found ABC AD' A'B'C A'D' 132 GEOMETRY. li and in general, we may conclude that the surfaces of two similar tri angles are as the squares of any two homologous lines; or, again, the ratio of the surfaces of two similar triangles is the square of the ratio of similitude of the triangles. PROPOSITION VIII.— THEOREM. 22. Two triangles having an angle of the one equal to an angle o the other are to each other as the products of the sides including the equal angles. Two triangles which have an angle of the one equal to an angle of the other may be placed with their equal angles in coincidence. Let ABG, ADE, be the two triangles having the common angle A ; then, ABC^ ABX AG ADE~~ ADX AE I For, join BE. The triangles ABC, ABE^ having the common vertex B, and their bases A (7, AE, in the same straight line, have the same altitude; therefore (16), ABC AG ABE~ AE 4 The triangles ABE, ADE, having the common vertex E, and their bases AB, AD, in the same straight line, have the same altitude ; therefore, ABE AB ADE~ ad' Multiplying these ratios, we have (III. 14), ABG ABXAO ADE~ AD X AE BOOK IV. 188 PKOPOSITION IX.— THEOKEM. 23. Similar polygons are to each other as the squares of their homolo- gous sides. Let ABCBEF, A'B' G'D'E'F', be two similar polygons; and denote their surfaces by S and 8'; then, b^ S' ~ aw' ^""••-"" ^"^ For, let the polygons be de- composed into homologous tri- angles (III. 39). The ratio of the surfaces of any pair of homolo- gous triangles, as ABC and A'B'C\ A CD and A' CD', etc., will be the square of the ratio of two homologous sides of the polygons (20) ; therefore, we shall have ABC ACD ADE AEF AB^ A'B'C A' CD' A'D'E' A'E'F' A'B" Therefore, by addition of antecedents and consequents (III. 12), ABC + ACD + ADE + AEF S AB' A'B'C ^ A' CD' -{- A'D'E' -\- A'E'F' S' A'B 72 24. Corollary. The ratio of the surfaces of two similar polygons is the square of the ratio of similitude of the polygons ; that is, the square of the ratio of any two homologous lines of the polygons. PROPOSITION X.— THEOREM. 25. The square described upon the hypotenuse of a right triangle is eqxiivalent to the sum of the squares described on the other two sides. 12 134 GEOMETRY. Let the triangle ABC be right angled at C\ then, the square AK^ described upon the hypotenuse, is equal in area to the sum of the squares AF and J5Z), described on the other two sides. For, from Cdraw CF perpendicular to *AB and produce it to meet KFl in L. Join CK, EG. Since ACF and ACB are right angles, CF and CB are in the same straight line (I. 21) ; and for a similar reason A C and CD are in the same straight line. In the triangles CAK, GAB, we have AK equal to AB, being sides of the same square ; A C equal to A G, for the same reason ; and the angles CAK, GAB, equal, being each equal to the sum of the angle CAB and a right angle; therefore, these triangles are equal (I. 76). The triangle CAK and the rectangle AL have the same base AK; and since the vertex C is upon LF produced, they also have the same altitude; therefore, the triangle CAK in equivalent to' one-half the rectangle AL (14). The triangle GAB and the square AF have the same base A G ; and, since the vertex B is upon FC produced,, they also have the same altitude ; therefore, the triangle GAB is equivalent to one- half the square AF (14). But the triangles CAK, GAB, have been shown to be equal; therefore, the rectangle AL is equivalent to the square AF. In the same way, it is proved that the rectangle BL is equivalent to the square BD. Therefore, the square AH, which is the sum of the rectangles AL and BL, is equivalent to the sum of the squares ^i^and BD. 26. Scholium. This theorem is ascribed to Pythagoras (born about 600 B. C), and is commonly called the Pythagorean Theorem. ■ The preceding demonstration of it is that which was given by Euclid in his Elements (about 300 B. C). It is important to observe, that we may deduce the same result from the numerical relation AB^ = AC^ -{- BC^, already established in rill. 48) For, since the measure of the area of a square is the « BOOK IV. 135 second power of the number which represents its side, it follows directly from this numerical relation that the area of which AB^ is the gaeasure is equal to the sum of the areas of which AC^ and BG^ are the measures. In the same manner, most of the numerical rela- tions demonstrated in the articles (III. 48) to (III. 67) give rise to theorems respecting areas by merely substituting, for a product, the area represented by that product. This may be called a transition from the abstract (pure number) to the concrete (actual space). On the other hand, we may pass from the concrete to the abstract. For example, in the above figure it has been proved that the areas of the rectangles AL, BL, are respectively equal to the areas of the squares AF, BD. But the rectangles, having the same altitude, are to each other as their bases J.P, PB', and the squares are to each other as their numerical measures AC^^ BC^; hence, we infer the numerical relation AC' :BC' = AP: PB, which was otherwise proved in (III. 46). Henceforth, we shall employ the equation AB^ = AC' -f- PC', as the expression of either one of the theorems (III. 48) and (IV. 25). 27. Corollary. If the three sides of a right triangle be taken as the homologous sides of three similar polygons constructed upon them, then the polygon constructed upon the hypotenuse is equivalent to the sum of the polygons constructed upon the other two sides. For, let P, Q, E, denote the areas of the polygons constructed upon the sides AC, PC, and upon the hypotenuse AB, respectively. Then, the polygons being similar, we have Q BG' Q ~ BG' from the first of which we derive, by composition, P+ Q ^ AC' ^PG' ^ AP^^ Q PC' ~ PC' which compared with the second gives at once P = P+ Q. 136 GEOMETRY. PKOBLEMS OF CONSTRUCTION. PROPOSITION XI.— PROBLEM. 28. To construct a triangle equivalent to a given polygon. Let ABCDEFhe the given polygon. ,Take any three consecutive vertices, as A, By C, and draw the diagonal A C. Through B draw BP parallel to ^0 meeting DC pror duced in P; join AF. The triangles APQ ABQ have the same base AC; and since their vertices, Pand B, lie on the same straight line BP parallel to AG, they also have the same altitude; therefore they are equivalent. Therefore, the penta- gon APDEF is equivalent to the hexagon ABCDEF. Now, taking any three consecutive vertices of this pentagon, we shall, by a pre- cisely similar construction, find a quadrilateral of the same area; and, finally, by a similar operation upon the quadrilateral, we shall find a triangle of the same area. Thus, whatever the number of the sides of the given polygon, a series of successive steps, each step reducing the number of sides by one, will give a series of polygons of equal areas, terminating in a triangle. PROPOSITION XII.— PROBLEM. 29. To construct a square equivalent to a given parallelogram or to a aiven triangle. 1st. Let J. C be a given parallelogram ; k its base, and h its altitude. Find a mean proportional x between h and k, by (III. 72). The square constructed upon x will be equivalent to the parallelogram, since x' = hX k. 2d. Let ABC be a given triangle; a its base and h its altitude. Find a mean proportional x between a and ^h; the square constructed upon a; will be equivalent to the triangle, since a;'' = a X ^^ r= \ah. \Zj^ BOOK IV 137 30. Scholium. By means of this problem and the preceding, a square can be found equivalent to any given polygon. PKOPOSITION XIII.— PROBLEM. 31. To construct a square equivalent to the sum of two or more given squareSj or to the difference of two given squares. 1st. Let m, n, p, q, be^the sides of given squares. Draw AB = m, and BC = n, perpendicular to each other at B; join AC. Then (25), AG^ = m^ -\- n\ Draw CD = p, perpendicular to A C, and join AD. Then AD' = AG' -\-p'=m' + n'-{-p\ Draw DE = q perpendicular to ADy and join AE. Then, AE' = AD' -{- q' = m' + n' + p'^ -{- q^; therefore, the square constructed upon AE will be equivalent to the sura of the squares constructed upon m, w, p, q. In this manner may the areas of any number of given squares be added. 2d. Construct a right angle ABC, and lay off BA = n. With the centre A and a radius = m, describe an arc cutting -BC in C. Then BC' = AC' — AB'= m^ — n"^; therefore, the square con- ''' structed upon ^Cwill be equivalent to the difference of the squares constructed upon m and n. 32. Scholium I. By means of this problem, together with the pre- ceding ones, a square can be found equivalent to the sum of any number of given polygons^ or to the difference of any two given polygons. 33. Scholium II. If m, n, p, q, in the preceding problem are ho- mologous sides of given similar polygons, the line AE in the first figure is the homologous side of a similar polygon equivalent to the sum of the given polygons (27). And the line BC, in the second figure, is the homologous side of a similar polygon, equivalent to the difference of two given similar polygons. 138 GEOMETRY. One side of a polygon, similar to a given polygon, being known,! the polygon may be constructed by (III. 80). PROPOSITION XIV.— PROBLEM. 34. Upon a given straight line to construct a rectangle equivalent to a given rectangle. Let k ' be the given straight line, and A C the given rectangle whose base is k and altitude h. Find a fourth proportional h\ to k\ k and A,, by (III. 70). Then, the rectangle constructed upon the base k' with the altitude^A' is equiva- lent to AC] for, by the construction, k' : k = h:h\ whence, k' X h' = k X h (7). k' PROPOSITION. XV.— PROBLEM. 35. To construct a rectangle^ having given its area and the sum of two adjacent sides. Let MN be equal to the given sum of the adjacent sides of the required rectangle; and let the given area be that of the square whose side is AB. Upon 3IN as a diameter describe a semi- circle. At 31 erect MP = AB perpendicular to MN, and draw PQ parallel to 3fN, intersecting the circumference in Q. From Q let fall QE perpendicular to 3fN; then, MB and RN are the base and altitude of the required rectangle. For, )\y (IIL 47), MBX BN=QB' = FM' = AB' PROPOSITION XVI.— PROBLEM. 36. To construct a rectangle, having given its area and the difference of two adjacent side.^. BOOK IV. 139 Let MN be equal to the given difference of the adjacent sides of the required rectangle; and let the given area be that of the square described on AB. Upon MN as a diameter describe a circle. At M draw the tangent MP = AB^ and from P, draw the secant PQR through the centre of the circle ; then, PR and PQ are the base and altitude of the required rectangle. For, by (III. 59), PR X PQ = PM'' = AB\ and the difference of PR and PQ is QR = MN. PROPOSITION XVII.— PROBLEM. 37. To find two straight lines in the ratio of the areas of two given polygons. Let squares be found equal in area to the given polygons, respectively (30). Upon the sides of the right angle. ^C'-B, take CA and CB equal to the sides of these squares, join AB and let fall CD perpendicular to AB. Then, by (III. 46), we have AD : DB = CA' : CB'] therefore, AD, DB, are in the ratio of the areas of the given polygons. PROPOSITION XVIII.— PROBLEM. 38. To find a square which shall he to a given square in the ratio of two given straight lines. Let AB' be the given square, and M : N I | ^^J" the given ratio. Upon an indefinite straight line CL, lay ofi' CD = M, DE = N; upon CE as a i\'i diameter describe a semicircle ; at D erect fy^ p^ the perpendicular DE cutting the circum- ^ del ference in F\ join EC, EE; lay off i^'jS' =^J5, and through H draw HO parallel to EC; then, EG is the side of the required square. For, by (III. 15), we have EG:EH=EC:EE, 140 GEOMETRY. whence (III. 13), FO^ : FH' = FG' : FE\ Also, by (III. 46), FG':FE'=CD: DE=M: K Hence, FG':FH' = MiN. But FFL = AB^ therefore the square constructed upon FG is to the square upon AB in the ratio M : N. PROPOSITION XIX.— PROBLEM. 39. To construct a polygon similar to a given polygon and whose area shall be in a given ratio to that of the given polygon. Let P be the given polygon, and let a be one of its sides; let 31: Nhe the given ratio. Find, by the preceding problem, the side a' of a square which shall be to a"^ in the ratio M : N; ^^ upon a', as a homologous side to a, construct the polygon P' similar to P (III. 80) ; this will be the \ ^' polygon required. For, the polygons being similar, their areas are in the ratio a'^ : a^ or if : iV, as required. PROPOSITION XX.— PROBLEM. 40. To construct a polygon similar to a given polygon P and equiva- lent to a given polygon Q. Find 31 and JV, the sides of squares respectively equal in area to P and Q, (30). Let a be any side of P, and find a jjf fourth proportional a' to 31, N and a: ^_____^ upon a', as a homologous side to a, con- \ P' | struct the polygon P' similar to P; this ^ — —, — ' will be the required polygon. For, by construction, M_ a_^ N~ a'' BOOK IV. 141 therefore, taking the letters P, Q and P\ to denote the areas of the polygons, but, the polygons P and P' being similar, we have, by (23), P' a"' and comparing these equations, we have P' = Q. Therefore, the polygon P' is similar to the polygon Pand equiva- lent to the polygon Q, as required. T 1 . ^7^. "t . -^--tT » BOOK V. REGULAR POLYGONS. MEASUREMENT OF THE CIRCLE, MAXIMA AND MINIMA OF PLANE FIGURES. REGULAR POLYGONS. 1. Definition. A regular polygon is a polygon which is at once equilateral and equiangular. The equilateral triangle and the square are simple examples of regular polygons. The following theorem establishes the possibility of regular polygons of any number of sides. PROPOSITION L— THEOREM. 2. If the circumference of a circle be divided into any number of equal parts, the chords joining the successive points of division form a regular polygon inscribed in the circle ; and the tangents drawn at the points of division form a regular polygon circumscribed about the circle. Let the circumference be divided into the equal arcs AB, BC, CD, etc. ; then, 1st, draw- ing the chords AB, BC, CD, etc., ABCD, etc., is a regular inscribed polygon. For, its sides are equal, being chords of equal arcs; and its angles are equal, being inscribed in equal segments. 2d. Drawing tangents at A, B, C, etc., the polygon GHK, etc., is a regular circumscribed polygon. For, in the triangles AGB, BHC, CKD, etc., w^e have AB = BC= CD, etc., and the angles GAB, GBA, HBC, HCB, etc., are equal, since each is formed by a tangent and chord and is measured by half of one of the equal parts of the circumference 142 BOOK V. 143 (II. 62) ; therefore, these triangles are all isosceles and equal to each other. Hence, we have the angles G'= H = K, etc., and AG = GB = BH = EC = CK, etc., from which, by the addition of equals, it follows that GH = HK, etc. 3. Corollary I. Hence, if an inscribed polygon is given, a circum- scribed polygon of the same number of sides can be formed by drawing tangents at the vertices of the given polygon. And if a circumscribed polygon is given, an inscribed polygon of the same number of sides can be formed by joining the points at which the sides of the given polygon touch the circle. It is often preferable, however, to obtain the circumscribed polygon from the inscribed, and reciprocally, by the following methods : 1st. Let ABCD . .. . be a given inscribed polygon. Bisect the arcs ABj BC, CD, etc., in the points E^F, G, etc., and draw tangents, A'B', B' C', C'D\ etc., at these points ; then, since the arcs EF, EG, etc., are equal, the polygon A'B' C'D' is, by the preceding propo- 'tion, a regular circumscribed polygon of the same number of sides as ABCD .... Since the radius OE is perpendicular to AB (II. 16) as well as to A'B', the sides A'B\ AB, are parallel; and, for the same reason, all the sides of A'B' CD' are parallel to the sides of ABCD respectively. Moreover, the radii OA, OB, OC, etc., when produced, pass through the vertices A',BIC', etc. ; for since B'E = B'F, the point B' must lie on the line OB which bisects the angle EOF (I. 127). 2d. If the circumscribed polygon A'B' CD' .... is given, we have only to draw OA', OB', OC, etc., intersecting the circumference in A, B, C, etc., and then to join AB, BC, CD, etc., to obtain the in- scribed polygon of the same number of sides. 4. Corollary II. If the chords AE, EB, BE, EC, etc., be drawn, a regular inscribed polygon will be formed of double the number of sides of ^^ CD.... If tangents are drawn at A, B, C, etc., intersecting the tangents A'B', B'C, CD', etc., a regular circumscribed polygon will be formed of double the number of sides of A'B' CD'. . . . It is evident that the area of an inscribed polygon is less than B' F C As,.^"'^ 1 ^"^v^/v j^.A/a ^A /\ \? \A~~.^ ^— -A/ 144 GEOMETRY. that of the inscribed polygon of double the number of sides ; and the area of a circumscribed polygon is greater than that of the cir- cumscribed polygon of double the number of sides. •: PROPOSITION II.— THEOREM. 5. A circle may he circumscribed about any regular polygon ; and a circle may also be inscribed in it. « Let ABCD. . . be a regular polygon ; then, ^^^Z^^^^ n 1st. A circle may be circumscribed about //^'- 1 •' \\ ■ it. For, describe a circumference passing I/I \\/ \M through three consecutive vertices A, B, C \\ /6\ J/j (11.88); let be its centre, draw OJT per- VW'' ^\/// Mi pendicular to -SCand bisecting it at jET, and \^_J^^^^ j | join OA, OD. Conceive the quadrilateral -^^hI A OHB to be revolved upon the line OH (i. e., folded over), iS^i HB falls upon its equal HC. The polygon being regular, the angle HBA = HCD, and the side BA = CD; therefore the side BA will take the direction of CD and the point A will fall upon D. Hence OD = OA, and the circumference described with the radius OA and passing through the three consecutive vertices A, B, C, also passes through the fourth vertex D. It follows that the circumfer- ence which passes through the three vertices B, C, D, also passes through the next vertex E, and thus through all the vertices of the polygon. The circle is therefore circumscribed about the polygon. ■■ 2d. A circle may be inscribed in it. For, the sides of the polygon ■ being equal chords of the circumscribed circle, are equally distant from the centre ; therefore, a circle described with the centre and the radius OH will touch all the sides, and will consequently be in- scribed in the polygon. 6. Definitions. The centre of a regular polygon is the common cen- tre, 0, of the circumscribed and inscribed circles. The radius of a regular polygon is the radius, OAy of the circum- scribed circle. The apothem is the radius, OH, of the inscribed circle. The angle at the centre is the angle, A OB, formed by radii drawn to the extremities of any side. BOOK 145 7. The angle at the centre is equal to four right angles divided by the number of sides of the polygon. 8. Since the angle ABC is equal to twice ABO, or to ABO -f BAO, it follows that the angle ABC of the polygon is the supple- ment of the angle at the centre (I. 68). PROPOSITION III.— THEOREM. 9. Regular polygons of the same number of sides are similar. Let ABODE, A'B'C'D'E', be regular polygons of the same num- ber of sides ; then, they are similar. For, 1st, they are mutually equi- angular, since the magnitude of an angle of either polygon de- pends only on the number of the sides (7 and 8), which is the same in both. 2d. The homologous sides are proportional, since the ratio AB : A'B' is the same as the ratio BC : B'C, or CD : CD', etc. Therefore the polygons fulfill the two conditions of similarity. 10. Corollary. The perimeters of regular polygons of the same num' her of sides are to each other as the radii of the circumscribed circles, or as the radii of the inscribed circles ; and their areas are to each other as the squares of these radii. For, these radii are homologous lines of the similar polygons (III. 43), (IV. 24). PROPOSITION IV.— PROBLEM. 11. To inscribe a square in a given circle. Draw any two diameters AC, BD, perpen- licular to each other, and join their extremities )y the chords AB, BC, CD, DA ; then, ABCD an inscribed square (II. 12), (II. 59). (/ \ ^ °A 12. Corollary. To circumscribe a square about the circle, draw mgents at the extremities of two perpendicular diameters A C, BD. 13 146 GEOMETRY. 13. Scholium. In the right triaDgle ABO, we have AB* = OA^ + OB' = 20A\ whence AB = OA. V% by which the side of the inscribed square can be computed, the radius being given. PROPOSITION v.— PROBLEM. 14. To inscribe a regular hexagon in a given circle. Suppose the problem solved, and let ABCDEFhe a regular inscribed hexagon. Join BE and AD; since the arcs AB, BG, CDy etc., are equal, the lines BE, AD, bisect the circumference and are diameters inter- secting in the centre 0. The inscribed angle ABO is measured by one-half the arc AFE, that is, by AF, or one of the equal divisions of the circumference ;"■ the angle A OB at the centre is also measured by one division, that IS, hj AB; and the angle BAO == ABO; therefore the triangle ABO \& equiangular, and AB =^ OA. Therefore the side of tho^g inscribed regular hexagon is equal to the radius of the circle. ^| Consequently, to inscribe a regular hexagon, apply the radius six times as a chord. 15. Corollary. To inscribe an equilateral triangle, ACE, join the alternate vertices of the regular hexagon. fli 16. Scholium. In the right triangle ACD, we have J-0' = AD' — DC' = (2 A Of — To' = SAO'; whence, AC = AO.V% by which the side of the inscribed equilateral triangle can be com- puted, the radius being given. The apothem, OH, of the inscribed equilateral triangle is equal to one-half the radius OB; for the figure AOCB is a rhombus and its diagonals bisect each other at right angles (I. 110). The apothem of the inscribed regular hexagon is equal to one-half the side of the inscribed equilateral triangle, that is, to l/3 ; for the perpendicular from upon AB is equal to the perpendicular from A upon OB, that is, to AH. The angle at the centre of the regular inscribed hexagon is ^ of 4 right angles, that is, f of one right angle = 60°. BOOK V, 147 The angle of the hexagon, or ABC, is f of a right angle = 120°. The angle at the centre of the inscribed equilateral triangle is f of one right angle = 120°. PROPOSITION VI.— PROBLEM. 17. To inscribe a regular decagon in a given circle. Suppose the problem solved, and let ABC X, be a regular inscribed decagon. Join AF, BG; since each of these lines bi- sects the circumference, they are diameters and intersect in the centre 0. Draw J5^ intersect- ing OA in 31. The angle A3fB is measured by half the sum of the arcs KF and AB (II. 64), that is, >y two divisions of the circumference; the inscribed angle 3IAB is measured by half the arc BF, that is, also by two divisions ; there- fore A3IB is an isosceles triangle, and MB = AB. Again, the inscribed angle MB is measured by half the arc KG, that is, by one division, and the angle 3I0B at the centre has the same measure ; therefore 03IB is an isosceles triangle, and 031 = MB = AB. The inscribed angle MBA, being measured by half the arc AK, khat is, by one division, is equal to the angle AOB. Therefore the isosceles triangles A31B and AOB are mutually equiangular and nmilar, and give the proportion whence OA'.AB = ABi AM, OA X A3I=AB'= 031'; i at is. the radius OA is divided in extreme and mean ratio at 31 (III. 73) ; and the greater segment 031 is equal to the gide AB of the inscribed regular decagon. Consequently, to inscribe a regular decagon, divide the radius in extreme and mean ratio (III. 74), and apply the greater segment ten times as a chord. 148 GEOMETRY. 18. Corollary. To inscribe a regular penta- gon, ACEGK, join the alternate vertices of the regular inscribed decagon. 19. JScholium, By (III. 75), we have AB = OA. VI by which the side of the regular decagon may be computed from the radius. The angle at the centre of the regular decagon is f of one right angle = 36° ; the angle at the centre of the regular pentagon is \ of one right angle = 72°. The angle ABC of the regular decagon is f of one right angle = 144° ; the angle A CE of the regular pentagon is -f of one right angle = 108°. PROPOSITION VII.-PROELEM. 20. To inscribe a regular pentedecagon in a given circle. Suppose AB is the side of a regu- lar inscribed pentedecagon, or that the arc AB is ^ of the circumfer- ence. Now the fraction -^-^ :z^ ^ — J^ ; therefore the arc AB \s the dif- ference between ^ and y\j- of the circumference. Hence, if we inscribe the chord AC equal to the side of the regular inscribed hexagon, and then CB equal to that of the regular inscribed decagon, the chord AB will be the side of the regular inscribed pentedecagon required. 21. Scholium. Any regular inscribed polygon being given, a regu- lar inscribed polygon of double the number of sides can be formed by bisecting the arcs subtended by its sides and drawing the chords of the semi-arcs (4). Also, any regular inscribed polygon being given, a regular circumscribed polygon of the same number of sides lan be formed (3). Therefore, by means of the inscribed square, we can inscribe and circumscjibe, successively, regular polygons of 8, 16, 32, etc., sides ; by means of the hexagon, those of 12, 24, 48, etc., sides ; by means of the decagon, those of 20, 40, 80, etc., sides ; and. BOOK V. 149 finally, by means of the pentedecagon, those of SO, 60, 120, etc., sides. Until the beginning of the present century, it was supposed that these were the only polygons that could be constructed by elementary geometry, that is, by the use of the straight line and circle only. Gauss, however, in his Disquisitiones Arithrneticce, LipsiWy 1801, proved that it is possible, by the use of the straight line and circle only, to construct regular polygons of 17 sides, of 257 sides, and in general of any number of sides which can be expressed by 2" -j- 1, n being an integer, provided that 2^* + 1 is a prime number. PROPOSITION VIII.— THEOREM. 22. The area of a regular polygon is equal to half the product oj its perimeter and apothem. For, straight lines drawn from the centre to the vertices of the polygon divide it into equal triangles whose bases are the sides of the polygon and whose common altitude is the apothem. The area of one of these triangles is equal to half the product of its base and altitude (IV. 13) ; therefore, the sum of their areas, or the area of the polygon, is half the product of the sum of the bases by the common altitude, that is, half the product of the perimeter and apothem. PROPOSITION IX.— THEOREM. 23. The area of a regular inscribed dodecagon is equal to three times the square of the radius. Let AB, BC, CD, DE, be four consecu- tive sides of a regular inscribed dodeca- gon, and draw the radii OA, OE; then, the figure ABODE is one-third of the dodecagon, and we have only to prove that the area of this figure is equal to the square of the radius. Draw the radius OD ; at A and D draw the tangents AF and GDF meeting in F; join AC and CE, and let J.Oand OEhe produced to meet the tan- 13 « 150 GEOMETRY, gent GF in H and G. The arc AD, containing three of the sidea of the dodecagon, is one fourth of the cir- cumference ; therefore the angle A OD is a right angle, and OF is a square de- scribed on the radius. Since A C and CE are sides of the regu- lar inscribed hexagon, each is equal to the radius ; therefore OA CE is a paral- lelogram. Hence also GO^lTand GECH are parallelograms. The triangles D^Oand BCA are equal (I. 80). The area of the triangle DEC is one-half that of the parallelogram EH (IV. 14) ; therefore the two triangles DEC and BCA are together equivalent to the parallelogram EEC. Adding the parallelogram OC to these equals, we have the figure OABCDE equivalent to the parallelogram OH. But the parallelogram OH is equivalent to the square OF (IV. 11); therefore the .figure OABCDE, or one-third the dodecagon, is equivalent to the square OF, that is, to the square of the radius. Therefore, the area of the whole dodecagon is equal to three times the square of the radius. 24. Scholium. The area of the circumscribed square is evidently equal to four times the square of the radius. The area of the circle is greater than that of the inscribed regular dodecagon, and less than ihat of the circumscribed square; therefore, if the square of the radius is taken as the unit of surface, the area of a circle is greater than 3 and less than 4. I PROPOSITION X.— PROBLEM. 25. Given the perimeters of a regular inscribed and a similar cir- rumscribed polygon, to compute the perimeters of the regular inscribed and circums&rihed polygons of double the number of sides. Let AB be a side of the given inscribed polygon, CD a side of the similar circumscribed polygon, tangent to the arc AB at its middle point E: Join AE, and at A and B draw the tangents AF and BG; then AE is a side of the regular inscribed polygon of double BOOK V. 151 the number of sides, and FO is a side of the circumscribed polygon of double the number of sides' (4). Denote the perimeters of the given \ \ \ inscribed and circumscribed polygons \ \ ; by p and P, respectively ; and the pe- \ \ i / rimeters of the required inscribed and '%' circumscribed polygons of double the number of sides by p' and P', respectively. Since 0(7 is the radius of the circle circumscribed about the poly- gon whose perimeter is P, we have (10), P_qc qc p~ OA^^ OE* and since OF bisects the angle COE, we have (III. 21), qc_ CF^ 0E~ FE therefore, P__CF p ~ FE whence, by composition, P_±jp _ CFJ-JE _ CE 2p ~ 2FE "FG Now FG is a side of the polygon whose perimeter is P', and is con- tained as many times in P' as CE is contained in P, hence (III. 9), CEP^ FG~ P'' and therefore, p+p P 2p P' P' = - 2pP whence P' = ^-\-P [1] Again, the right triangles AEH and EFNsire similar, since their 152 GEOMETRY. acute angles ^^^and FEN are equal, and give AH EN AE ~ EF Since AH and AE are contained the same number of times in p and p\ re- spectively, we have o AH p AE~P'' and since EN and EF are contained the same number of times in jo' and P', respectively, we have EF~ P'' therefore, we have P' F'' whence __^___ p' = Vp X P', [2] % Therefore, from the given perimeters p and P, we compute P' by the equation [1], and then with p and P' we compute p' by the equation [2]. 26. Definition. Two polygons are isoperimetric when their peri] ters are equal. PROPOSITION XI.— PROBLEM. 27. Given the raditis and apothem of a regular polygon, to compute the radius and apothem of the isoperimetric polygon of double the num- ber of sides. Let AB be a side of the given regular polygon, the centre of this polygon, OA its radius, OB its apothem. Produce DO to meet the circumference of the circumscribed circle in O'; join O'A, O'B; let fall OA^ perpendicular to O'A, and through A' draw J.'P' parallel to AB. Since the new polygon is to have twice as many sides as the given polygon, the angle at its centre must be one-half the angle AOP>\ B O O K V. 158 therefore the angle AO'B, which is equal to one-half of AOB (II. 57), is equal to the angle at the centre of the new polygon. Since the perimeter of the new polygon is to be equal to that of the given polygon, but is to be divided into twice as many sides, each of its sides must, be equal to one-half of AB\ therefore A'B', which is equal to one-half of AB (I. 121), is a side of the new polygon ; O'A' is its radius, and O'D' its apothem. If, then, we denote the given radius OA by R, and the given apothem OD by r, the required radius O'A' by jB', and the apothem O'D' by r'i we have 0'/>' = ^ = -^-^-±-^ 2 2 ^ ' or r' = ^- [1] In the right triangle OA'O'y we have (III. 44), WA'=00' X O'D', or ■ E' = VRy:^'\ [2] therefore, r' is an arithmetic mean between R and r, and jR' is a geometric mean between R and r'. MEASUREMENT OF THE CIKCLE. The principle which we employed in the comparison of incommen- surable ratios (II. 49) is fundamentally the same as that which we are about to apply to the measurement of the circle, but we shall now state it in a much more general form, better adapted for subse- quent application. 28. Definitions. I. A variable quantity, or simply, a variable, is a quantity which has different successive values. II. When the successive values of a variable, under the conditions imposed upon it, approach more and more nearly to the value of some fixed or constant quantity, so that the difference between the variable and the constant may become less than any assigned quan- tity, without becoming zero, the variable is said to approach indefi- 13=^* 154 GEOMETRY. nitely to the constant ; and the constant is called the limit of the variable. Or, more briefly, the limit of a variable is a constant quantity to which the variable, under the conditions imposed upon it, approaches indefinitely. As an example, illustrating these definitions, let a point be re- quii"fid to move from A to B under the fol- lowing conditions : it shall first move over | | i i i one-half of AB, that is to C; then over ^ ^ one-half of CB, to C" ; then over one-half of C'B, to G" ; and so on indefinitely ; then the distance of the point from -4 is a variable^ and this variable approaches indefinitely to the constant AB, as its lirnit, without ever reaching it. As a second example, let A denote the angle of any regular poly- gon, and n the number of sides of the polygon ; then, a right angle being taken as the unit, we have (8), ^ = 2-1 n The value of J. is a variable depending upon n; and since n may be 4 taken so great that - shall be less than any assigned quantity how- ever small, the value of ^ approaches to two right angles as its limit, but evidently never reaches that limit. 29. Principle of Limits. Theorem. If two variable quantities are always equal to each other and each approaches to a limit, the two limits are iiecessarily equal. For, two variables always equal to each other present in fact but one value, and it is evidently impossible that one variable value shall at the same time approach indefinitely to two unequal limits. 30. Theorem. The limit of the product of two variables is the pro- duct of their limits. Thus, if x approaches indefinitely to the limit a, and y approaches indefinitely to the limit b, the product xy must approach indefinitely to the product ab ; that is, the limit of the pro- duct xy is the product ab of the limits of x and y. 31. Theorem. If two variables are in a constant ratio and each approaches to a limit, these limits are in the same constant ratio. Let X and y be two variables in the constant ratio m, tliat is, let BOOK V. 155 X = my; and let their limits be a and b respectively, Since y ap- proaches indefinitely to b, my approaches indefinitely to mb ; there- fore we have x and my, two variables, always equal to each other, whose limits are a and mb, respectively, whence, by (29), a = mb; that is, a and b are in the constant ratio m. PROPOSITION XII.— THEOREM. 32. An arc of a circle is less than any line which envelops it and has the same extremities. Let A KB be an arc of a circle, AB its chord j and let ALB, AMB, etc., be any lines enveloping it and terminating at A and B. Of all the lines AKB, ALB, AMB, etc. (each of which includes the segment, or area, AKB, be- tween itself and the chord AB), there must be at least one minimum, or shortest line.* Now, no one of the lines ALB, AMB, etc., envelop- ing AKB, can be such a minimum ; for, drawing a tangent CKD to the arc AKB, the line A CKDB is less than A CLDB ; therefore ALB is not the minimum ; and in the same way it is shown that no other enveloping line can be the minimum. Therefoi*, the arc AKB is the minimum. 33. Corollary. The circumference of a circle is less than the perimeter of any polygon circumscribed about it. 34. Scholium. The demonstration is applicable when AKB is any convex curve whatever. PROPOSITION XIII.— THEOREM. 35. If the number of sides of a regular polygon inscribed in a circle be increased indefinitely, the apothem of the polygon will approach to the radius of the circle as its limit. * If we choose to admit the possibility of two or more equal shortest lines, still we say that of all the lines, AKB, ALB, etc., there must be one which is either {he minimum line, or one of the minimum lines. 156 GEOMETRY. Let AB be a side of a regular polygon inscribed in the circle whose radius is OA ; and let OD be its apothem. In the triangle OAD we have (I. 67), OA— 0D< AD. •N'ow, by increasing the number of sides of the polygon, the length of a side AB may evidently be made as small as we please, or less than any quantity that may be assigned. Hence AD, or ^AB, and still more OA — OD, which is still less than AD, may become less than any assigned quantity ; that is, the apothem OD approaches to the radius OA as its limit (28). PROPOSITION XIV.— THEOREM. 36. The circumference of a circle is the limit to which the perimeters of regular inscribed and circumscribed polygons approach when the number of their sides is increased indefinitely; and the area of the circle is the limit of the areas of these polygons. Let AB and CD be sides of a regular inscribed and a similar circumscribed polygon ; let r denote the apothem OE, E the radius OF, p the perimeter of the inscribed polygon, P the perimeter of the cir- cumscribed polygon. Then, we have (10), £== — Pi: r' whence, by division (III. 10), Now we have seen, in the preceding proposition, that by increasing the number of sides of the polygons, the difference B — r may be made less than any assigned quantity ; consequently the quantity p, may also be made less than any assigned |x(iJ r), or P quantity. But P being always greater, and p always less, than the circumference of the circle, the difference between this circumference BOOK V w: and either Por^ is less than the difference P — p, and consequently may also be made less than any assigned quantity. Therefore, the circumference is the common limit of P and p. Again, let s and S denote the areas of two similar inscribed and circumscribed' polygons. The difference between the triangles COD and A OB is the trapezoid CABD, the measure of which is \{CD -\- AB) X EF) therefore, the difference between the areas of the polygons is ^-5 = i(P + i>) X (i^-r); consequently, /S — s < P X (P — r). Now by increasing the number of sides of the polygons, the quantity P X {B — r), and consequently also S — s, may be made less than any assigned quantity. But S being always greater, and s always less, than the area of the circle, the difference between the ariea of the circle and either ^ or s is less than the difference S — s, and conse- quently may also be made less than any assigned quantity. There- fore, the area of the circle is the common limit of /S' and s. PKOPOSITION XV.— THEOKEM. 37. The circumferences of two circles are to each other as their radii j and their areas are to each other as the squares of their radii. Let B and P' be the radii of the circles, C and C their cir- cumferences, 8 and S' their areas. Inscribe in the two circles simi- lar regular polygons; let P and P' denote the perimeters, A and A' the areas of these polygons; then, the polygons being similar, we have (10), P__B A — 31 F'~~B'' A'~B"' These relations remain the same whatever may be the number of Bides in the polygons, provided there is the same number in each (9). When this number is indefinitely increased, P approaches C as its 158 GEOMETRY. limit, and P' approaches C as its limit (36) ; and since Pand P' are in the constant ratio of R to R\ their limits are in the same ratio (31); therefore C R^ R' [1] And since the limit of A is S, and the limit of A' is S\ it follows in the same manner that S_ R' R'^ [2] «S8. Corollary I. The proportion [1] is by (III. 9) the same as 0' 2R 2R' and the proportion [2] is the same as S'~ 4.R"~~ (2R'y'' therefore, the circumferences of circle are to each other as their diame- ters, and their areas are to each other as the squares of their diameters, ■I 39. Corollary II. Similar arcs, as AB, A'B'j are those which subtend equal an- \ /' gles at the centres of the circles to which they belong; they are therefore like parts of their respective circumferences, and are in the same ratio as the circumferences. Also the similar sectors A OB and A'O'B' are like parts of the circles to which they belong. Therefore, similar arcs are to each other as their radii, and similar sectors are to each other as the squares of their radii. 40. Corollary III. The ratio of the circumference of a circle to its diameter is constant; that is, it is the same for all circles. For, from the proportion [3] we have Yl _0 2R 2R'' BOOK V. 169 This constant ratio is usually denoted by r, so that for any circle whose diameter is 2R and circumference 0, we have — = TT, or 0= 2t.R. 2R 41. Scholium. The ratio ;r is incommensurable (as can be proved by the higher mathematics), and can therefore be expressed in num- bers only approximately. The letter ;r, however, is used to symbolize its exact value. PROPOSITION XVI.— THEOREM. 42. The area of a circle is equal to half the product of its circum- ference by its radius. Let the area of any regular polygon circum- scribed about the circle be denoted by J., its perimeter by P, and its apothem which is equal to the radius of the circle by B ; then (22), A = iPXR,orj = iE. Let the number of the sides of the polygon be continually doubled, then A approaches the area S of the circle as its limit, and P ap- proaches the circumference C as its limit ; but A and P are in the constant ratio ^B ; therefore their limits are in the same ratio (31), and we have |=iiJ,or^ iCxB. [IJ 43. Corollary I. The area of a circle is equal to the square of its radius multiplied by the constant number tt. For, substituting for C its value 27rB in [1], we have S=7:B\ 44. Corollary II. The area of a sector is equal to half the product of its arc by the radius. For, denote the arc ab of the sector aOb hy c, and the area of the sector by s ; then, since c and s are lilie parts of G and S, we have (IIL 9), 160 G E O M E T li Y. 8_ c_ ^c X B S~ C~~ ^G X B I But S=iCX B; therefore s = -J-c X i?. - 45. Scholium. A circle may be regarded as a regular polygon of an infinite number of sides. In proving that the circle is the limit to- wards which the inscribed regular polygon approaches when the number of its sides is increased indefinitely, it was tacitly assumed that the number of sides is always finite. It was shown that the dif- ference between the polygon and the circle may be made less than any assigned quantity by making the number of sides sufficiently great; but an assigned difference being necessarily a finite quantity, there is also scjme finite number of sides sufficiently great to satisfy the imposed condition. Conversely, so long as the number of sides is finite, there is some finite difference between the polygon and the circle. But if we make the hypothesis that the number of sides of the inscribed regular polygon is greater than any finite number, that is, infinite, then it must follow that the difference between the poly- gon and the circle is less than any finite quantity, that is, zero ; and consequently, the circle is identical with the inscribed polygon of an infinite number of sides. This conclusion, it will be observed, is little else than an abridge statement of the theory of limits as applied to the circle ; the abridg- ment being effected by the hypothetical introduction of the infinite into the statement. Ji PEOPOSITION XVII.— PEOBLEM. 46. To compute the ratio of the^circumference of a circle to its diame- ter, approximately. First Method, called the Method of Perimeters. In this method, we take the diameter of the circle as given and compute the perimeters of some inscribed and a similar circumscribed regular polygon. We then compute the perimeters of inscribed and circum- scribed regular polygons of double the number of sides, by Propo- sition X. Taking the last-found perimeters as given, wo compute the perimeters of polygons of double the number of sides by the same method ; and so on. As the number of sides increases, the lengths BOOK V. 161 of the perimeters approach to that of the circumference (36) ; hence, their successively computed values will be successive nearer and nearer approximations to. the value of the circumference. Taking, then, the diameter of the circle as given = 1, let us begin by inscribing and circumscribing a square. The perimeter of the inscribed square = 4 X -j X V^2 = 2i/2 (13) ; that of the circum- scribed square = 4 ; therefore, putting P=4, p = 2|/2 = 2.8284271, we find, by Proposition X., for the perimeters of the circumscribed and inscribed regular octagons, r>. __ 2p X P P-hP = 3.3137085, y = |/p X P' =-3.0614675. Then taking these as given quantities, we put P = 3.3137085, p = 3.0614675, and find by the same formulae for the polygons of 16 sides P' = 3.1825979, / = 3.1214452. Continuing this process, the results will be found as in the followmg TABLE.* Number Perimeter of Perimeter of of sides. circumscribed polygon. inscribed polygon. 4 4.0000000 2.8284271 8 3.3137085 3.0614675 16 3.1825979 3.1214452 32 3.1517249 3.1365485 64 3.1441184 3.1403312 128 3.1422236 3.1412773 256 3.1417504 3.1415138 512 3.1416321 3.1415729 1024 3.1416025 3.1415877 2048 3.1415951 3.1415914 4096 3.1415933 3.1415923 8192 3.1415928 3.1415926 * The computations have been carried out with ten decimal places in order to ensure the accuracy of the seventli place as given in the table. 14- L 162 GEOMETRY. 1 From the last two numbers of this table, we learn that the cir- cumference of the circle whose diamtter is unity is less than 3.1415928 and greater than 3.1415926 ; and since, when the diame- ter =^ 1, we have C = tt, (40), it follows that' :r = 3.1415927 within a unit of the seventh decimal place. Second Method, called the Method of Isoperimeteks. This method is based upon Proposition XI. Instead of taking the diame- ter as given and computing its circumference, we take the circum- ference as given and compute the diameter ; or we take the semi- circumference as given and compute the radius. Suppose we assume the semi-circumference ^C=l; then since C = 27ri?, we have E R' that is, the value of tt is the reciprocal of the value of the radius o the circle whose semi-circumference is unity. Let ABCD be a square whose semi-perimeter = 1 ; then each of its sides ^= ^. Denote its radius OA by R, and its apothem OE by r ; then we have r = \ =-0.2500000, i2=il/2^= 0.3535534. Now, by Proposition XI., we compute the apothem r' and the radius R' of the regular polygon of 8 sides having the same pe- rimeter as this square ; we find r' = "^^tr ^ 0.3017767, 2 R' = i/R X r' = 0.3266407. Again, taking these as given, we put r = 0.3017767, R = 0.3266407, and find by the same formulae, for the apothem and radius of the isoperimetric regular polygon of 16 sides, the values r' = 0.3142087, R' = 0.3203644. BOOK V. 163 Continuing this process, the results are found as in the following TABLE. Number of sides. Apothem. Radiiis. 4 0.2500000 0.3535534 8 0.3017767 0.3266407 16 0.3142087 0.3203644 32 0.3172866 0.3188218 64 0.3180541 0.3184376 128 0.3182460 .0.3183418 256 0.3182939 0.3183179 512 0.3183059 0.3183119 1024 0.3183089 0.3183104 2048 0.3183096 0.3183100 4096 0.3183098 0.3183099 8192 0.3183099 0.3183099 Now, a citcumference described with the radius r is inscribed in the polygon, and a circumference described with a radius R is circum- scribed about the polygon ; and the first circumference is less, while the second is greater, than the perimeter of the polygon. Therefore the circumference which is equal to the perimeter of the polygon has a radius greater than r and less than R ; and this is true for each of the successive isoperimetric polygons. But the r and R of the polygon of 8192 sides do not differ by so much as .0000001 ; there- fore, the radius of the circumference which is equal to the perimeter of the polygons, that is, to 2, is 0.3183099 within less than .0000001 ; and we have 1 0.3183099 = 3.141593 within a unit of the sixth decimal place. 47. Scholium I. Observing that in this second method the value of r = -J, for the square, is the arithmetic mean of and ^, and that R = ^t/2 is the geometric mean between ^ and J^, we arrive at the following proposition : The value of - is the, limit approached by the successive numbers ob- tained by starting from the numbers and ^ and taking alternately the arithmetic mean and the geometric mean between the two which precede. 164 GEOMETRY. 1 I taS 48. Scholium II. Archimedes (born 287 B. C.) was the first, assign an approximate value of -. By a method similar to the above " first method," he proved that its value is between Z\ and 3ff, or, in decimals, between 3.1428 and 3.1408 ; he therefore assigned its value correctly within a unit of th^ third decimal place. The number 3|, or %f^ usually cited as Archimedes' value of n- (although it is but one of the two limits assigned by him), is often used as a sufficient approximation in rough computations. Metius (A. D. 1640) found the much more accurate value f-ff» which correctly represents even the sixth decimal place. It is easily remembered by observing that the denominator and numerator writ- ten consecutively, thus 113|355, present the first three odd numbers each written twice. More recently, the value has been found to a very great number of decimals, by the aid of series demonstrated by the DiflTerential Calculus. Clausen and Dase of Germany (about A. D. 1846), com- puting independently of each other, carried out the value to 200 decimal places, and their results agreed to the last figure. The mutual verification thus obtained stamps their results as thus far the best established value to the 200th place. (See Schumacher's Asironomische Nachrichten, No. 589.) Other computers have car- ried the value to over 500 places, but it does not appear that their faults have been verified. The value to fifteen decimal places is .r = 3.141592653589793. d For tlie greater number of practical applications, the value tz = 3.1416 is sufficiently accurate. MAXIMA AND MINIMA OF PLANE FIGUEES. 49. Definition. Among quantities of the same kind, that which is greatest is called a maximum ; that which is least, a minimum. Thus, the diameter of a circle is a maximum among all straight lines joining two points of the circumference ; the perpendicular is a minimum among all the lines drawn from a given point to a given straight line. BOOK V. 166 An enclosed figure is said to be a maximum or a minimum, when its area is a maximum or a minimum. 60. Definition. Any two figures are called isoperimetrie when their perimeters are equal. PROPOSITION XVIII.— THEOREM. 51. Of all triangles having the same base and equal areas^ that which is isosceles has the minimum perimeter. Let ABC be an isosceles triangle, and A'BC any other triangle having the same base and an y equal area ; then, AB -\- AC < A'B -\- A' C. ^ ZX\ For, the altitudes of the triangles must be /y^ \ equal (IV. 15), and their vertices A and A lie ^ ^ in the same straight Jine MN parallel to BC. Draw CND perpendicular to MN^ meeting BA produced in D ; join A!D. Since the angle NAC = ACB =- ABC = DAN, the right triangles ACN, ADN, are equal ; therefore, AN is perpendicular to CD at its middle point N, and we have AD = AC, A'D = A' C. But BD < A'B + A'D] that h, AB -^ AG < A'B + A' C. 52. Corollary. OJ all triangles having the same area, that which is equilateral has the minimum perimeter. For, the triangle having the minimum perimeter enclosing a given area must be isosceles which- ever side is taken as the base. PROPOSITION XIX.— THEOREM. 53. Of all triangles having the same hose and equal perimeters, thai which is isosceles is the maximum. Let ^J50 be an isosceles triangle, and let A'BC, ^ ^ ^ standing on the same base BC, have an equal /S^^^^^\\ perimeter ; that is, let Al B -\- A! C =^ AB -{- A O; /^'^"'^^ ^ then, the area of ^j5C is greater than the area of A'BC. For, the vertex A' must fall between BC and the parallel MN drawn through A ; since, if it fell upon MN, we should have, as in the preceding demonstration, A'B -\- A' C y> AB -\- AC. and if it 166 GEOMETRY. fell above MN, the sum A'B -{- A' C would be still greater. There- fore the altitude of the triangle ABC is greater than that of A'BC, and hence also its area is the greater. 54. Corollary, Of all isoperimetric triangles, that which is equilai eral is the maximum. For, the maximum triangle having a given perimeter must be isosceles whichever side is taken as the base. PEOPOSITION XX.— THEOREM. 55. Of all triangles formed with the same two given sides, that in which these sides are perpendicular to each other is the maximum. Let ABC, A'BC, be two triangles having the a sides AB, BC, respectively equal to A'B, BC; then, if the angle ABC is a right angle, the area of the triangle ABC is greater than that of the triangle A'BC. For, taking BC as the common base, the altitude AB of the tri- angle ABC is evidently greater than the altitude A'D of the triangle A'BC. I PROPOSITION XXI.— THEOREM. i 56. Of all isoperimetric plane figures, the circle is the maximum. 1st. With a given perimeter, there may be constructed an hifinite ^ : number of figures of different forms and various areas. The areaW may be made as small as we please (ly. 35), but obviously cannot be increased indefinitely. Therefore, among all the figures of the same perimeter there must be .one maximum figure, or several maxi- mum figures of diflTerent forms and equal areas. 2d. Every closed figure of given perimeter containing a maximum urea must necessarily be convex, that is, such that any straight line joining two points of the perimeter lies wholly within the figure. Let ACBNA be a non-convex figure, the straight line AB, joining two of the points in its /^ "^ perimeter, lying without the figure ; then, if the / X re-entrant portion ACB be revolved about the \^ ^l:)^' line AB into the position ACB, the figure \ ^ ACBNA has the same perimeter as the first ^~ -^ BOOK V. 167 figure, but a greater area. Therefore, the non-convex figure cannot be a maximum among figures of equal perimeters. 3d. Now let ACBFA be a maximum figure formed with a given perimeter; then we say that, taking any point A in its perimeter and drawing AB ^,^"" ^7"^-^' so as to divide the perimeter into two equal y^^ / YsN„ parts, this line also divides the area of the f / / ^y^^'^\ figure into two equal parts. For, if the \\ /i^^^^^"^^^ W)^' area of one of the parts, as AFB, were ^\" '";^ greater than that of the other part, A CB, — q-^"^^ then, if the part AFB were revolved upon the line AB into the position AF'B, the area of the figure AF'BFA would be greater than that of the figure A CBFA, and yet would have the same perimeter; thus the figure ACBFA would not be a maximum. Hence also it appears that, ACBFA being a maximum figure, AF'BFA is also one of the maximum figures, for it has the same perimeter and area as the former figure. This latter figure is sym- metrical with respect to the line ABy since by the nature of the revo- lution about ABj every line FF' perpendicular to ABj and termi- nated by the perimeter, is bisected by AB (1. 140).* Hence F and F' being any two symmetrical points in the perimeter of this figure, the triangles AFB and AF'B are equal. Now the angles AFB and AF'B must be right angles ; for if they were not right angles the areas of the triangles AFB and AF'B could be increased without varying th'e lengths of the chords AF^ FB, AF', F'B (55), and then (the segments AGF, FEB, AG'F', F'E'B, still standing on these chords), the whole figure would have its area increased without changing the length of its perimeter; con- sequently the figure AF'BFA would not be a maximum. There- fore, the angles i^and F' are right angles. But jPis any point in the curve AFB; therefore, this curve is a semi-circumference (H. 59, 97). Hence, if a figure ACBFA of a given perimeter is a maximum, its half AFB, formed by drawing AB from any arbitrarily chosen point A in the perimeter, is a semicircle. Therefore the whole figure is a circle.'*' * This demonstration is due to Steiner, Crelle^s Journal fur die reine uwf angewandte Mathematik, vol. 24. (Berlin, 1842.) 168 GEO M ET il Y. PROPOSITION XXIL— THEOREM. 57. Of all plane figures containing the same area, the circle has the minimum perimeter. Let (7 be a circle, and A any- other figure having the same area f \ / \ as C\ then, the perimeter of is less than that of A. For, let ^ be a circle having the same perimeter as the figure A ; then, by the preceding theo- rems A <:^ B, OY G < B. Now, of two circles, that which has the less area has the less perimeter ; there- fore, the perin^eter of Cis less than that of B, or less than that of ^. PROPOSITION XXIII.— THEOREM. J 58. OJ all the poly go7is constructed with the same given sides, thai is the maximum which can he inscribed in a circle. Let P be a polygon constructed with the sides a, b, c, d, e, and inscribed in a circle S, and let P' be any other polygon constructed with the same sides and not inscrip- tible in a circle.; then, P ^ P'. For, upon the sides a, b, c, etc., of the polygon P' construct cir- cular segments equal to those stand- ing on the corresponding sides of P. The whole figure S' thus formed has the same perimeter as the circle >S; therefore, area of S^ area of S' (56); subtracting the circular segments from both, we have P>P'. i BOOK V. 169 PROPOSITION XXIV.— PROBLEM. 59. Oj all isoperimetric polygons having the same number of sides^ the regular polygon is the mojcimum. 1st. The maximum polygon P, of all the isope- rimetric polygons of the same number of sides must have its sides equal ; for if two of its sides, as AB', B' C, were unequal, we could, by (53), substitute for the triangle AB'C the isosceles triangle ABC having the same perimeter as ^^'Oand a greater area, and thus the area of the whole polygon could be increased with out changing the length of its perimeter or the number of its sides. 2d. The maximum polygon constructed with the same number of equal sides must, by (^5S), be inscriptible in a circle ; therefore it must be a regular polygon. PROPOSITION XXV.-THEOREM. 60. Of all polygons having the same number of sides and the same area, the regular polygon has the minimum perimeter. Let Pbe a regular polygon, and i/ any irregular polygon having the same number of sides and the same area as P; then, the perimeter of P is less than that of M. For, let iV be a regular polygon having the same perimeter and the same number of sides as ilf ; then, by (59), if < iV, or P < K But of two regular polygons having the same number of sides, that which has the less area has the less perimeter ; therefore the perimeter of P is less than that of N, or less than that of if. 15 170 GEOMETRY, PROPOSITION XXVI.— THEOREM. Ql, Jf a regular polygon be constructed with a given perimeter j its area will be the greater^ the greater the number of its sides. Let F be the regular polygon of three sides, and Q the regular polygon of four sicles, constructed with the same given perimeter. In any side AB of P take any arbitrary point D ; the polygon P may be regarded as an irregular poly- gon of four sides, in which the sides AD, DB, make an angle with each other equal to two right angles (I. 16); then, the irregular polygon P of four sides is less than the regular isoperimeti'ic polygon Q of four sides (59). In the same manner it follows that Q is less than the regular isoperimetric polygon of five sides, and so on. PROPOSITION XXVII.— THEOREM. 62. If a regular polygon be constructed with a given area, its perim-i eter will be the less, the greater the number of its sides. Let P and Q be regular polygons having the same area, and let Q have the greater number of sides; then, the perimeter of P will be greater than that of §. For, let P be a regular polygon having the same perimeter as Q and the same number of sides as P; then, by (61), Q > P, or P '> R\ therefore the perimeter of P is greater than that of 11, or greater than that of Q. GEOMETRY OF SPACE. I BOOK VI. THE PLANE. POLYEDRAL ANGLES. 1. DEFimrioN. A plane has already been defined as a surface such that the straight line joining any two points in it lies wholly in the surface. Thus, the surface MN is a plane, if, A and B being any two points in it, the straight line AB lies wholly in the surface. The plane is understood to be indefinite in extent, so that, however far the straight line is produced, all its points lie in the plane. But to represent a plane in a diagram, we are obliged to take a limited portion of it, and we usually represent it by a parallelogram supposed to lie in the plane. DETERMINATION OF A PLANE. PROPOSITION I.— THEOREM. 2. Through any given straight line an infinite number .of planes may he passed. Let AB be a given straight line. A straight line may be drawn in any plane, and the position of that plane may be changed until the line drawn in it is brought into coincidence with AB. We shall then have one plane 171 ^' / '^ 172 GEOMETRY. passed through AB ; and this plane may be turned upon AB as an axis and made to occupy an infinite number of positions. 3. Scholium. Hence, a plane subjected to the single condition that it shall pass through a given straight line, is not fixed, or deter- minate, in position. But it will become determinate if it is required to pass through an additional point, or line, as shown in the next pr()position. A plane is said to be determined by given lines, or points, when it is the only plane which contains such lines or points. PROPOSITION II.— THEOREM. 4. A plane is determined, 1st, by a straight line and a point ivithout that line; 2d, by two intersecting straight lines; 3d, by three points not in the same straight line ; 4th, by two parallel straight lines. 1st. A plane MN being passed through a given straight line ABy and then turned upon this line as an axis until it contains a given point C not in the line ABy is evidently determined ; for, if it is then turned in either direction about AB, it will cease to contain the point C The plane is therefore determined by the given straight line and the point without it. 2d. If two intersecting straight lines ABy A C, are given, a plane passed through AB and any point C (other than the point A) of ACy contains the two straight lines, and is determined by these lines. 8d. If three points are given, J., By C, not in the same straight line, any two of them may be joined by a straight line, and then the plane passed through this line and the third point, contains the three points, and is thus determined by them. 4th. Two parallel lines, AB, CD, are by ^*' '" definition (I. 42) necessarily in the same plane, and there is but one plane containing e them, since a plane passed through one of them, AB, and any point E of the other, is determined in position. 5. Corollary. The intersection of two planes is a straight line. For, the intersection cannot contain three points not in the same straight line, sii ce only one plane can contain three such points. I AC. For, upon PD take PB = PC, and join yIP: then ^P>^P (L 35); but AB = AC; therefore, AD>AC. 11. Corollary I. Conversely, equal oblique lines from a point to a plane meet the plane at equal distances from the perpendicular; and of two unequal oblique lines, the greater meets the plane at the greater distance from the perpendicular. BOOK VI. 175 12. Corollary II. Equal straight lines from a point to a plane meet the plane in the circumference of a circle whose centre is the foot of the perpendicular from the point to the plane. Hence we derive a method of drawing a perpendicular from a given point A to sl given ~ plane il/iV^ find any three points, B, 0, JE, in the plane, equidistant from Ay and find the centre P of the circle passing through these points; the straight line AP will be the required perpendicular. PROPOSITION v.— THEOREM. IS. If a draight line is perpendicular to each of two straight lines at their point of intersection, it is perpendicular to the plane of those lirfes. Let AP be perpendicular to PB and PC, at their intersection P; then, AP is perpen. dicular to the plane MN which contains those lines. For, let PD be any other straight line drawn through P in the plane MN. Draw any straight line BDC intersecting PB, PC PD, in B, C, D; produce AP to A' making ! /// PA' ~= PA, and join A and A' to each of ''j'i'' the points B, C, D. . / Since BP is perpendicular to AA', at its middle point, we have BA = BA', and for a like reason CA = CA' ; therefore, the triangles ABC, A'BC, are equal (I. 80). If, then, the triangle ABC is turned about its base PC until its plane coin- cides with that of the triangle A'BC, the vertex J. will fall upon A'^, and as the point D remains fixed, the line AD will coincide with A'D] therefore, D and Pare each equally distant from the extremi- ties 0^ AA', and DP is perpendicular to AA' or AP(J. 41). Hence -4P is perpendicular to any line PD, that is, to every line, passing through its foot in the plane 3IN, and is consequently perpendicular to the plane. 14. Corollary I. At a given point P of a straight line AP, a plane can be passed perpendicular to that line, and but one. For, two per- pendiculars, PB, PC, being drawn to AP in any two different planes APB, APC, passed through AP, the plane of the lines PB, PC, will 176 GEOMETRY. I u Q JV be perpendicular to the line AP. Moreover, no other plane passed through P can be perpendicular to AP; for, any other plane not con- taining the point (7 would cut the oblique line^C in a point C different from C, and we should have the angle APC different from APC, and therefore not a right angle. 15. Corollary II. All the perpendiculars ^ PB, PC, PDj etc., drawn to a line AP at the same point, lie in one plane perpendicular to ylP. Hence, if an indefinite straight line PQ, perpendicular to AP, be made to revolve, always remaining perpendicular to AP, it is said to generate the plane 3IN perpendicular. to AP; for the line PQ passes successively, during its revolution, through every point of this plane. 16. Corollary III. Through any point C without a given straight line AP, a plane can be passed perpendicular to AP, and but one. For, in the plane determined by the line AP and the point C, the perpendicular CP can be drawn to AP, and then the plane generated by the revohition of PC about AP as an axis will, by the preceding coroliary, be perpendicular to AP; and it is evident that there can be but one such perpendicular plane. PROPOSITION VI.— THEOREM. 17. If from the foot of a perpendicular to a plane a straight line is drawn at right angles to any line of the plane, and its intersection with that line is joined to any point of the perpendicular, this last line will he perpendicular to the line of the plane. Let AP be perpendicular to the plane ^ MN; from its foot P let PD be drawn at right angles to any line BC of the plane; then, A being any point in AP, the straight line ^Z) is perpendicular to J5C For, lay off DB = DC, and join PB, PC, AB, AC. Since DB = DC, we have PB = PC (I. 30), and hence AB = AC (10). Therefore, A and D being each equally distant from~ B and C, the line AD is perpendicular to BC (I. 41). I BOOK VI. 177 iV PAKALLEL STRAIGHT LINES AND PLANES. 18 Definitions. A straight line is parallel to a plane Vihen it can- not meet the plane though both be indefinitely produced. In the" same case, the plane is said to be parallel to the line. Two planes are parallel when they do not meet, both being indefi • nite in extent. PEOPOSITION VII.— THEOREM. 19. If two straight lines are parallel, every plane passed through one of thein and not coincident with the plane of the parallels is parallel to the other. Let AB and CD be parallel lines, and MN any plane passed through CD; then, the line AB and the plane 3/iVare parallel. For, the parallels AB, CD, are in the same plane, A CDB, which intersects the plane 3IN in the line CD ; and if AB could meet the plane 3IN, it could meet it only in some point of CD; but AB cannot meet CD, since it is parallel to it ; there- fore AB cannot meet the plane MN. 20. Corollary I. Through any given straight line HK, a plane can be passed parallel to any other given straight line AB, For, in the plane determined by AB and any ^ ~'^ ^ point H of UK, let HL be drawn parallel to AB ; then, the plane MN, determined by HK and HL, is parallel to AB. 21. Corollary II. Through any given point 0, a plane can be passed parallel to any two given straight lines AB, CD, in space. For, in the plane determined by the given point and the line AB let aOb be drawn through parallel to AB ; and in the plane determined by the point O and the line CD, let cOc^ be drawn through parallel to CD; then, the plane determined by the lines ab and cd is parallel to each 01 the lines AB and CD. 15= M 178 G E O M E T R Y. • PKOPOSITION VIII.— THEOREM. 22. If a straight line and a plane are parallel, the inter sections 0/ the plane with planes passed through the line are parallel to that line and to each other. Let the line AB be parallel to the plane MNy and let CD, EF, etc., be the intersec- tions of MN with planes passed through AB] then, these intersections are parallel to AB and to each other. For, the line AB cannot meet CD, since it cannot meet the plane in which CD lies ; and since these lines are in the same plane, AD, and cannot meet, they are parallel. For the same reason, EF, GH, are parallel to AB. Moreover, no two of these intersections, as CD, EF, can meet ; for if they met, their point of meeting and the line AB would be at once in two different planes, AD and AF, which is impossible (4). 23. Corollary. If a straight line AB is parallel to a plane MN, a parallel CD to the line AB, drawn through any point C of the plane, lies in the plane. -«| For, the plane passed through the line AB and the point C inter- f | sects the plane MN in a parallel to AB, which must coincide with CD, since there cannot be two parallels to AB drawn through the same point C. I PROPOSITION IX.— THEOREM. 24. Planes perpendicular to the same straight line are parallel tc each other. The planes MN, PQ, perpendicular to the same straight line ^5, cannot meet; for, if they met, we should have through a point of their intersection two planes perpendicular to the same straight line, which is impossible (16) ; therefore these planes are parallel. .1/ / - ^^ / /' iV / 1 I" / Q BOOK VI. 179 J l-KOPOSITION X.— THEOEEM. 25. The intersections oj tivo parallel planes with any third plane are parallel. Let JOT and PQ be parallel planes, and ^ b AD any plane intersecting them in the V /\ \ lines AB and CD ; then, AB and CD are \ / \ parallel. ^\~ V ^ For, the lines AB and CD cannot meet, p \ \i> since the planes in which they are situ- \ \ / \ ated cannot meet, and they are lines in \ -^- ^ the same plane AD\ therefore they are parallel. 26. Corollary. Parallel lines A (7, BDy intercepted between parallel planes 3IN, PQ, are equal. For, the plane of the parallels A C, BD, intersects the parallel planes i/iV, PQ, in the parallel lines AB, CD; therefore, the figure ^^DCis a parallelogram, and AC == BD. PROPOSITION XL— THEOREM. 27. A straight line perpendicular to one of two parallel planes is perpendicular to the other. Let JfiV and PQ be parallel planes, and let the ^f I straight line AB be perpendicular to PQ ; then, I ~Jj / it will also be perpendicular to MN. For, through A draw any straight line AC m the plane MN, pass a plane through AB and AC, and let BD be the intersection of this plane with PQ. Then jiC and BD are parallel (25); but AB IS perpendicular to BD (6), and consequently also to AC) therefore AB, being perpendicular to any line AC which it meets in the plane MN, is perpendicular to the plane MN. 28. Corollary. Through any given point A, one pla/^e can be passed parallel to a given plane PQ, and but one. For, iiuni A a perpen- dicular AB can be drawn to the plane PQ (7), and then through A 180 GEOMETRY. a plane MN can be passed perpendicular to AB (14) ; the plane MN is parallel to PQ (24). No other plane can be passed through A parallel to PQ\ for every plane parallel to PQ must be perpendicular to tlie line AB (27), and there can be but one plane perpendicular to AB passed through the same point A (14). PROPOSITION XII.— THEOREM. 29. The locus of all the straight lines drawn through a given point parallel to a given plane, is a plane passed through the point parallel to thz given plane. Let A be the given point, and PQ the given plane ; then, every straight line AB, drawn through A parallel to the plane PQ, lies in the plane MN passed through A parallel to PQ. For, pass any plane through AB, intersecting the plane PQ in a straight line CD ; then AB is paral- lel to CD (22). But CD is parallel to the plane MN, since it is in the parallel plane PQ and can- not meet MN-, therefore, the line AB drawn through the point A parallel to CD lies in the plane MN (23). 30. Scholium. In the geometry of space, the term loctis has the same general signification as in plane geometry (I. 40) ; only it is not limited to lines, but may, as in this proposition, be extended to a surface. In the present case, the locus is the assemblage of all the points of all the lines which satisfy the two conditions of passing through a given point and being parallel to a given plane. 31. Corollary. Since two straight lines are sufficient to determine a plane (4), if two intersecting straight lines are each parallel to a given plane, the plane of these lines is parallel to the given plane. PROPOSITION XIII.— THEOREM. 32. If two angles, not in the same plane, have their sides respectively parallel and lying in the same direction, they are equal and their planes are parallel. I I 3f P ^V 1 < -«/ BOOK VI. 181 Let BA C,B'A' C, be two angles lying in the planes MN, M'N'-, and let AB, A (7, be parallel respectively to A'B', AC\ and in the same directions.* 1st. The angles BA C and B'A ' C are equal. For, through the parallels AB, A'B', pass a plane AB\ and through the parallels AC A'C, pass a plane AC, intersecting the first in the line AA'. Let BC be any plane parallel to AA', intersecting the planes AB', AC, in the lines BB', CC, and the planes 3IN, 31' N', in the lines BC, B'C, respectively. Since AA' is parallel to the plane BC, the intersections BB', CC, are parallel to AA' SLud. to each other (22) ; hence, the quadrilaterals AB' and AC are parallelograms, and we have AB = A'B', AC= ^'C, and BB' = AA' = CC. Therefore. BB' and CC are equal and parallel, and the quadrilateral .BC" is a parallelogram, and we have BC=B'C. The triangles ABC, A'B' C, therefore, have their three sides equal each to each, and consequently the angles BA C and B'A'C are equal. 2d. The planes of these angles are parallel. For, each of the lines AB, A C, being parallel to a line of the plane M'N', is parallel to that plane, therefore the plane MN of these lines is parallel to the plane M'N' (31). PKOPOSITION XIV.— THEOREM. 33. If one of two parallel lines is perpendicular to a plane, the other is also perpendicular to that plane. Let AB, A'B', be parallel lines, and let AB be perpendicular to the plane MN] then, A'B' is also perpendicular to MN. For, let A and A' be the intersections of these lines with the plane ; through A' draw ' " jv any line A' C in the plane MN, and through A draw J. parallel to ^'C and in the same direction. The angles * Two parallels AB, A^B^, lie in the same direction when they lie on the same side of the line ^^^ joining their origins A and A\ Compare note (I. 60). 16 182 GEOMETRY. BAG, B'A'C, are equal (32) ; but BAG is a right angle, since BA is perpendicular to the plane; hence, B'A'G' is a right angle; that is, B'A' is perpendicular to any line A' G' drawn through its foot in the plane MN, and is consequently perpendicular to the plane. 34. Gorollary I. Two straight lines AB, A'B', perpendieular to the same plane MN, are parallel to each other. For, if through any point of *A'B' a parallel to AB is drawn, it will be perpendicular to the plane MN, since AB is perpendicular to that plane ; but through the same point there cannot be two perpendiculars to the plane ; there- fore, the parallel drawn to AB coincides with A'B\ 35. Gorollary II. If two straight lines A and B are parallel to a third G, they are parallel to each other. For, let MN be a plane perpendicular to C; then (33), A and B are each perpendicular to this plane and are parallel to each other (34), 36. Gorollary III. Two parallel planes are everywhere equally dis- tant All perpendiculars to one of two parallel planes are also- per- pendicular to the other (27) ; and since they are parallels (34) inter- cepted between parallel planes, they are equal (26). M PROPOSITION XV.— THEOREM. 37. If two straight lines are intersected by three parallel planes, their corresponding segments are proportional. Let AB, GD, be intersected by the parallel planes MN, PQ, RS, in the points A, E, B, and G,F,D; then, AE_CF EB ~ FD M J 1 ^ ^^^C P \ \ ^ E V.^ a For, draw AB cutting the plane P§ in G, and join EG and EG. The plane of the lines AB^ AD, cuts the parallel planes PQ and RS in the lines EG and BD; therefore, EG and BD are parallel (25), ani we have (III. 15), AE_AG EB ~ GD i BOOK VI. 183 Thi plane of the lines DA and DO cuts the parallel planes MN and PQ in the lines AC and GF; therefore, AC and GF are parallel, and we have AG CF GD ~ FD Comparing these two proportions, we obtain AE_CF EB ~~ FD DIEDRAL ANGLES.— ANGLE OF A LINE AND PLANE, ETC. 38. Definition. When two planes meet and are terminated by their common intersection, they form a diedral angle. Thus, the planes AE, AF, meeting in AB, and ter- minated by ABj form a diedral angle. The planes AE, AF, are called the faces, and the line AB the edge, of the diedral angle. A diedral angle may be named by four letters, one in each face and two on its edge, the two on the edge being written between the other two ; thus, the angle in the figure may be named DABC. When there is but one diedral angle formed at the same edge, it may be named by two letters on its edge ; thus, in the preceding figure, the diedral angle DABC may be named the diedral angle AB. 39. Definition. The angle CAD formed by two straight lines -4 C, AD, drawn, one in each face of the diedral angle, perpendicular to its edge AB at the same point, is called the plane angle of the diedral angle. The plane angle thus formed is the same at whatever point of the edge of the diedral angle it is constructed. Thus, if at B, we draw BE and BE in the two faces respectively, and perpendicular to AB, the angle EBF is equal to the angle CAD, since the sides of these angles are parallel each to each (32). It is to be observed that the plane of the plane angle CAD is perpendicular to the edge AB (13) ; and conversely, a plane perpen- (li'^ular to the edge of a diedral angle cuts its faces in lines which 184 GEOMETRY. I be A T) A' C "^ C B <<;- — F Bt ^. — Ft Et are perpendicular to the edge and therefore form the plane angle ol the diedral angle. 40. A diedral angle DABG may be conceived to be generated by a plane, at first coincident with a fixed plane "^E", revolving upon the line AB as an axis until it comes into the position AF. In this revolution, a straight line CA^ perpendicular to AB^ generates the plane angle CAD. 41. Definition. Two diedral angles are equal when they can placed so that their faces shall coincide. Thus, the diedral angles CABD, C'A'B'D', are equal, if, when the edge A'B' is applied to the edge AB and the face A'F' to the face AF, the face A'E' also coincides with the face AF. Since the faces continue to coincide when produced indefinitely, it is apparent that the magnitude of the diedral angle does not depend upon the extent of its faces, but only upon their relative position. Two diedral angles are evidently equal when their plane angles are equal. 42. Definition. Two diedral angles CABD, DABE, which have a common edge AB and a common plane BD between them, are called adjacent. Two diedral angles are added together by placing them adjacent to each other. Thus, the diedral angle CABE is the sum of the two diedral and DABE. 43. Definition. When a plane CAB meets another MN, forming two equal adjacent die- dral angles, CABM and CABN, each of these angles is called a right diedral angle, and the plane CAB is perpendicular to the plane MN. It is evident that in this case the plane angles CDF, CDF, of the two equal diedral angles, are right angles. Through any straight line AB in a plane MN, a plane CAB can be passed perpendicular to the plane MN. The proof is similar to that of the corresponding proposition in plane geometry (I. 9). « angles CABD N BOOK VI, 185 PROPOSITION XVI.— THEOREM. 44. Tivo diedral angles are in the same ratio as their plane angles. Let CABD and GEFHhe two die- dral angles; and let CAD and GEH be their plane angles. Suppose the plane angles have a common measure, contained, for exam- ple, 5 times in CAD and 3 times in GIlH; the ratio of these angles is then 5 : 3. Let straight lines be drawn from the vertices of these angles, dividing the angle DA C into 5 equal parts, and the angle HEG into 3 equal parts, each equal to the common measure; let planes be passed through the edge AB and the several lines of division of the plane angle CAD, and also planes through the edge EF and the several lines of division of the plane angle GEH. The given diedral angles are thus divided into partial diedral angles which are all equal to each other since their plane angles are equal. The diedral angle CABD contains 5 of these partial angles, and the diedral angle GEFH contains 3 of them ; therefore, the given diedral angles are also in the ratio 5:3; that is, they are in the same ratio as their plane angles. The proof is extended to the case in which the given plane angles are incommensurable, by the method exemplified in (II. 51). 45. Corollary I. Since the diedral angle is proportional to its plane angle (that is, varies proportionally with it), the plane angle is taken as the meamire of the diedral angle, just as an arc is taken as the mea- sure of a plane angle. Thus, a diedral angle will be expressed by 45° if its plane angle is expressed by 45°, etc. 46. Corollary II. The sum of two adjacent die- dral angles, formed by one plane meeting another, is equal to two right diedral angles. For, the sum of the plane angles which measure them is equal to two right angles. In a similar manner, a number of properties of diedral angles can be proved, which are analo- gous to propositions relating to plane angles. The student can establish the following : 16 «- 186 GEOMETRY. Opposite or vertical diedral angles are equal; as CABN and DABMj in the preceding figure. When a plane intersects two parallel planes, the alternate diedral angles are equal, and the cofresponding diedral angles are equal; (the terms alternate and corresponding having sig- nifications similar to those given in plane geometry.) Two diedral angles which have their faces respectively parallel, or (if their edges are parallel) respectively perpendicular to each other, are either equal or supplementary. PROPOSITION XVII.— THEOREM. 47. If a straight line is perpendicular to a plane, every plane passedl through the line is also perpendicular to that plane. Let AB be perpendicular to the plane MN; then, any plane PQ, passed through AB, is also perpendicular to MN. For, at B draw BC, in the plane MN, perpen- dicular to the intersection BQ. Since AB is per- pendicular to the plane MN, it is perpendicular to BQ and BC; therefore, the angle ABC is the plane angle of the diedral angle formed by the planes PQ and MN\\ and since the angle ABCh a right angle, the planes are perpendicu- lar to each other. 48. Corollary. If AG, BO and CO, are three straight lines perpendicular to each other at a common point 0, each is per- pendicular to the plane of the other two, and the three planes are perpendicular to each other. BOOK VI. 187 and PKOPOSITION XVIII.— THEOKEM. 49. IJ two planes are perpendicular to each other, a straight line drawn in one of them, perpendicular to their intersection, is perpendicu- lar to the other. Let the planes PQ and MN be perpendicular to each other at any point B of their intersection BQ, let BA be drawn, in the plane PQ, perpendicular to BQ) then, BA is perpendicular to the plane MN. For, drawing BC, in the plane MN, perpendicu- lar to BQ, the angle ABC is a right angle, since it is the plane angle of the right diedral angle formed by the two planes ; therefore, AB, perpendicular to the two straight lines BQ, BC, is perpendicular to their plane i£i\^(13). 50. Corollary I. If two planes, PQ and MN, are perpendicular to each other, a straight line BA drawn through any point B of their intersection perpendicular to one of the planes MN, will lie in the other plane PQ (8). 51. Corollary II. If two planes, PQ and MN, are perpendicular to each other, a straight line drawn from any point A of PQ, per- pendicular to MN, lies in the plane PQ (7). PKOPOSITION XIX.— THEOREM. 52. Through any given straight line, a plane can be passed perpen- dicular to any given plane. Let AB be the given straight line and MN the given plane. From any point A of AB let AC be drawn perpendicular to MN, and through AB ' and A C pass a plane AD. This plane is perpen- / dicular to MN (47). ^ Moreover, since, by (51), any plane passed through AB perpendicular to MN must contain the perpendicular A C, the plane AD is the only plane perpendicular to MN that can be passed through AB, unless AB is itself perpendicular to MN, in which case an infinite number of planes can be passed through it perpendicular to MN (47). N 188 GEOMETRY. PROPOSITION XX.— THEOREM. 63. Ij two intersecting planes are each perpendicular to a third planCt their intersection is also perpendicular to that plane. Let the planes FQ, ES, intersecting in the .line AB, be perpendicular to the plane MN; then, AB is perpendicular to the plane MN. For, if from any point A of AB, a perpen- dicular be drawn to MN,_ this perpendicular will lie in each of the planes PQ and BS (51), and must therefore be their intersection AB. 54. Scholium. This proposition may be otherwise stated as follows: Tf a plane (MN) is perpendicular to each of two intersecting planes {PQ and BS), it is perpendicular to the intersection {AB) of those planes. PROPOSITION XXI.— THEOREM. 55. Every point in the plane which bisects a diedral angle is e^ distant from the faces of that angle. Let the plane AM bisect the diedral angle CABD; let P be any point in this plane ; PE and PF the perpendiculars from P upon the planes ABC and ABD; then, PE = PF. For, through PE and PF pass a plane, intersecting the planes ABC and ABD in OE and OF; join PO. The plane PEF is per- • pendicular to each of the planes ABC, ABD (47), and consequently perpendicular to their intersection AB (54). Therefore the angles FOE and POF measure the diedral angles MABC and MABD, which by hypothesis are equal. Hence the right triangles FOE and POi^are equal (L 83), and PE = PF. BOOK VI 189 66. Definitions. The projection of a point A upon a plane MN, is the foot a of the perpen- dicular let fall from A upon the plane. The projection of a line ABODE , . . , upon a plane MN, is the line abcde . . . formed by the projections of all the points of the line ABCDE. . . upon the plane. PROPOSITION XXII.— THEOREM. 57. The projection of a straight line upon a plane is a straight line. Let AB be the given straight line, and MN the given plane. The plane Ab, passed through AB perpendicular to the plane MN, contains all the perpendiculars let fall from points of AB upon MN (50) ; therefore, these perpendiculars all meet the plane MN in the intersection ah of the perpendicular plane with 3IN. The projection of AB upon the plane MN is, consequently, the straight line ah. 58. Scholium. The plane Ah is called the projecting plane of the straight line AB upon the plane MN. PROPOSITION XXIII.— THEOREM. 59. The acute angle which a straight line makes with its own pro- jection upon a plane, is the least angle which it makes with any line of that plane. A Let Ba be the projection of the straight line BA upon the plane MN, the point B being the point of intersection of the line BA with the plane; let J5C be any other straight line drawn through B in the plane ; then, the angle ABa is less than the angle ABC. For, take BC= Ba, and join AC. In the triangles ABa, ABC, we have AB common, and Ba = BC, but Aa <^ A C, since the perpendicular is less than any oblique line ; therefore, the angle ABa is less than the angle ABC (I. 85). 190 GEOMETRY. M : 1 ^ ^ / ■■■ A \ 1 l_ . yo . 1 60. Definition. The acute angle which a straight line makes wil its own projection upon a plane is called the inclination of the line to, the plane, or the angle of the line and plane. 61. Definition. Two straight lines ABj CD, not in the same plane, are regarded aj3 making an angle with each other which is equal to the angle between two straight lines Ob, Od, drawn through any point in space, parallel respectively to the two lines and in the Same directions. Since every straight line has two oppo- site directions (I. 4), the angle which one line makes with another is either acute or obtuse, according to the directions considered. Thus, if Oh is drawn in the direction expressed by AB (that is, on the same side of a straight line joining A and 0), and if Od is drawn in the direction expressed by CD, then dOb is equal to the angle which CD makes with AB ; but if Oa is drawn in the direction ex- pressed by BA (which is the opposite of AB), while Od is still in the direction of CD, then dOa is equal'to the angle which CD makes , with BA. M If 1/iVis any plane parallel to the two lines AB, CD (21), then the angle of these lines is the same as the angle of their projections ab, cd, upon this plane. II 62. From the preceding definition, it follows that when a straight '^ line is perperidicidar to a plane, it is perpendicular to all the lines of the plane', whether the lines pass through its foot or not. For, let AB be perpendicular to the plane MN, and CD any line of the plane. At any point B' in CD, let A'B' be drawn perpendicu- lar to the plane; then, A'B' being parallel to AB, the right angle ^'J5'C is equal to the angle of the lines AB and CD, that is, AB is perpen- dicular tk) CD. \ I ^ BOOK VI. 191 PROPOSITION XXIV.— THEOEEM. 63. Two straight lines not in the same plane being given: 1st, a common perpendicular to the two lines can he drawn ; 2d, hut one such common perpendicular can he drawn ; 3d, the common perpendicular is the shortest distance hetween the two lines. Let AB and CD be the given straight lines. 1st. Through one of the given lines, say ABj pass a plane 3fN, parallel to the other (20) ; let cd be the projection of CD upon this plane. Then, cd \yill be parallel to CD (22), and therefore not parallel to AB; hence it will meet AB in some point c. At c dra\y cC perpendicular to cd in the pro- jecting plane Cd ; then Cc is a common perpendicular to AB and CD. For, CD and cd being parallel, Cc drawn perpendicular to cd is perpendicular to CD. Also, since Cc is the line which projects the ])oint C upon the plane 3IN, it is perpendicular to that plane, and therefore perpendicular to AB. 2d. The line Cc is the only common perpendicular. For, if an- other line EF, drawn between AB and CD, could be perpendicular to AB and CD, it would be perpendicular also to a line FG drawn parallel to CD in the plane il/iV, and consequently perpendicular to the plane 3IN; but EH, drawn in the plane Cd, parallel to Cc, is perpendicular to the plane 3IN; hence we should have two perpen- diculars from the point E to the plane 3£N, which is impossible. 3d. The common perpendicular Cc is the shortest distance between AB and CD. For, any other distance EF is greater than the per- pendicular EH, or than its equal Cc. 64. Scholium. The preceding construction furnishes also the angle between AB and CD, namely, the angle Bed. 192 GEOMETRY. POLYEDRAL ANGLES. 65. Definition. When three or more planes meet in a common point, they form a polyedral angle. Thus, the figure S-ABCD, formed by the planes ASB, BSC, CSD, DSA, meeting in the tjommon point S, is a polyedral angle. The point S is the vertex of the angle ; the intersections of the planes, 8A, SB, etc., are its edges; the portions of the planes bounded by the edges are its faces; the angles ASB, BSC, etc., formed by the edges, are its /ace angles. A triedral angle is a polyedral angle having but three faces, which is the least number of faces that can form a polyedral angle. 66. Definition. Two polyedral angles are equal when they can be applied to each other so as to coincide in all their parts. Since two equal polyedral angles coincide however far their edges and faces are produced, the magnitude of a polyedral angle does not depend upon the extent of its faces ; but in order to represent the angle clearly in a diagram we usually pass a plane, as ABCD, cut- ting all its faces in straight lines AB, BC, etc. ; and by the face ASB is not meant the triangle ASB, but the indefinite surface included between the lines SA and SB indefinitely produced. 67. Definition, A polyedral angle S-ABCD is convex, when an section, ABCD, made by a plane cutting all its faces, is a convex polygon (I. 95). 68. Symmetrical polyedral angles. If we produce the edges ^.aS^ BS, etc., through the vertex S, we obtain another polyedral angle ^-^'jB' CD', which is symmetrical with the first, the vertex S being the centre of symmetry. If we pass a plane A'B'C'D', parallel to ABCD, so as to make SA' = SA, we shall also have SB' = SB, SC = SC, ^' etc. ; for we may suppose a third parallel plane passing through I 4 BOOK VI. 193 Sj and then AA', BB', etc., being divided proportionally by three parallel planes (37), if any one of them is bisected at S, the others are also bisected at that point. The points A'j B\ etc., are, then, symmetrical with A, B, etc., the definition of symmetry in a plane (I. 138), being extended to symmetry in space. The two symmetrical polyedral angles are equal in all their parts, for their face angles, ASB and A'SB', BSC and B'SC\ are equal, each to each, being vertical plane angles ; and the diedral angles at the edges SA and SA\ SB and SB', etc., are equal, being vertical diedral angles formed by the same planes. But the equal parts are arranged in inverse order in the two figures, as will appear more plainly, if we turn the polyedral angle S- A' B' CD' about, until the polygon A'B' CD' is brought into the same plane with ABCD, the vertex S remaining fixed ; the polygon A'B' CD' is then in the position ahcd, and it is apparent that while in the polyedral angle S-ABCD the parts ASB, BSC, etc., succeed each other in the order /ro9H right to left, their corresponding equal, jmrts aSb, bSc, etc., in the polyedral angle S-abcd succeed each other in the order from left to right. The two figures, therefore, cannot be made to coincide by superposition, and are not regarded as equal in the strict sense of the definition (I. 75), but are said to be equal by symmetry. PKOPOSITION XXV.— THEOREM. 69. The sum of a7iy two face angles of a triedral angle is greater than the third. The theorem requires proof only when the third angle considered is greater than each of the others. IjetS-ABC be a triedral angle in which the "^ face angle ASC is greater than either ASB or /^vX BSC; then, ASB + BSC > ASC. j \\\ For, in the face ASC draw SD making the ^i-;— X--v--N^ angle ASD equal to ASB, and through any point ' A' D of SD draw any straight line ADC cutting SA and SC; take SB == SD, and join AB, BC The triangles ASD and ASB are equ',1, by the construction (I. 76). whence AD — - AB. Now, in the triangle ABC, we have 17 N 194 GEOMETRY. AB-i- BC> AC, and subtracting the equals AB and AD, BC> DC; therefore, in the triangles BSC and DiSC, we have the angle BSC > ' i)SC (I. 85), and adding the equal angles ASB and ASD, we have ASB -{- BSC> ASC PROPOSITION XXVI.— THEOREM. i 70. The sum of the face angles of any convex polyedral angle is less than four right angles. Let the polyedral angle S be cut by a plane, a making the section ABCDE, by hypothe-sis, a // ;\\ convex polygon. From any point within this / / J \ \ polygon draw OA, OB, OC, OD, OE. /''i''^-\-\f> The sum of the angles of the triangles ASB, /\ / V^C' \ /\ BSC, etc., which have the common vertex S, is fF ^ | equal to the sum of the angles of the same num- * ber of triangles A OB, BOC, etc., which have the common vertex 0. But in the triedral angles formed at A, B, C, etc., by the faces of the polyedral angle and the plane of the polygon, we have (69). SAE + SAB > EAB, SBA -\- SBC> ABC, etc. ; hence, taking the sum of all these inequalities, it follows that the' sum of the angles at the bases of the triangles whose vertex is S is greater than the sum of the angles at the bases of the triangles whose vertex is ; therefore, the sum of the angles at S is less than the sum of the angles at 0, that is, less than four right angles. PROPOSITION XXVII.— THEOREM. 71. Two triedral angles are either equal or symmetrical, when the three face angles of one are respectively equal to the three face angles of the other. Id the triefl'ra. angles S and .^rea of a prism is equal to the product of the perim- eter of a right section of the prism by a lateral edge. Let AD' be a prism, an^^^IKL a right section of it ; then, the are jj^^B convex sur- face of the prism is equal^^The perimeter GHIKL multiplied by a lateral edge AA'. For,, the sides of the section GHIKL being perpendicular to the lateral edges AA'y BB ', etc., are the altitudes of the parallelo- grams which form the convex surface of the prism, if w^fkke as the bases of these paral- lelograms the lateral edges, AA' BB'y etc., which are all equal. Hence, the area of the sum of these parallelograms is (IV. 10), GH X AA' -{, HI X BB' + etc. = (GH-ir HI-{- etc.) X AA\ 17. Corollary. The lateral area of a right prism is equal to the product of the perimeter of its base by its altitude. PROPOSITION III.— THEOREM. 18. The four diagonals of a parallelepiped bisect each other. Let ABCD-G be a parallelopiped ; its four diagonals, A BH, DFj bisect each other. Through the opposite and parallel edges AE, CG, pass a plane which intersects the parallel faces ABCD, EFGH, in the parallel lines AC and EG. The figure ACGE is a parallelogram, and its diagonals A G and EC bisect each other in the point 0. In the same manner it is shown that A G and BH, A G and DF, bisect each other ; therefore, the four diagonals bisect each other in the point 0. a, EC, 200 GEOMETRY. 19. Scholium. The point 0, in which the four diagonals intersect, is called the centre of the parallelopiped ; and it is easily proved that any straight line drawn through and termiiiated by two opposite faces of the parallelopiped is bisected in that point. >-'-~"^. PKOPOSITION IV,— THEOEEM. 20. The sum of the squares of the four diagonals of a parallelopiped is equal to the S2im of the squares of its twelve edges. In the parallelogram A CGE we have (III. 64), ^ J^ ^=^^'AG' -^CE' = 2AE' + 2AC\ nnd in the parallelogram DBFH, BH^ + DF' = 2BF' + 2BD\ Adding, and observing that BF = AE, and also that in the parallelogram ABCD, 2AC'' + 2BD' = 4AB' + 4AD\ we have AG' +CE'-i- BE' + DF' = AAE'' + 4AB' + 4AD\ which proves the theorem. 21. Corollai'y. In a rectangular parallelopiped, the four diagona are equal to eS^eh oth&i^ ; and the square of a diagonal is equal to the sum of the squares of the three edges which meet at a common vertex. \ Thus, i? AG is a rectangular parallelopiped, we have, by dividing the preceding equation )3y/4, ) AG' = AE' + AB' + 3^'. 22. Scholium. If any three straight lines AB, AE, AD, not in the -^ame plane, are given, meeting in a common point, a parallelopiped can be constructed upon them. For, pass a plane through the extremity of each line parallel 'to the plane of the other two; these {)lanes, together with the planes of the given lines, determine the parallelopiped. In a rectangular parallelopiped, if the plane of two of the three edges which meet at a common vertex is taken as a base, the third edge is the altitude. These three edges, or the three perpendicular BOOK VII. 201 distances between the opposite faces of a rectangular parallelopiped, are palled its three dimensions. PROPOSITION v.— THEOREM. 23. Two prisms are equals if three faces including a triedral angle of tK^^ne are respectively equal to three faces similarly placed including a triedral angle of the other. Let the triedral angles ^ and a of the prisms ABODE- A\ abcde-a\ be contained by equal faces similarly placed, namely, ABODE equal to abcde, AB' equal to ah\ and AE' equal to ae' ] then, the prisms are equal. For, the triedral angles A and a are equal (VI. 71), and can be applied, the one to the other, so as to coincide; and then the bases ABODE, abcde, coinciding, the face AB' will coincide with ab\ and the face AE' with ae' ) therefore the sides A'B', A'E', of the upper base of one prism, will coincide with the sides a'b'y a'e', of the upper base of the other prism, and since these bases are equal they will coincide throughout; conse- quently also the lateral faces of the two prisms will coincide, each to each, and the prisms will coincide throughout ; therefore, the prisms are equal. 24. Oorollary I. Two truncated prisms are equal, if three faces in- cluding a triedral angle of the one are respectively equal to three faces similarly placed including a triedral angle of the other. For, the pre- ceding demonstration applies whether the planes Jl'^'C"Z)'J&' and a'b'c'd'e' are parallel or inclined to the lower bases. 25. Oorollary II. Two right prisms are equal, if they have equal bases and equal altitudes. In the case of right prisms, 't is not necessary to add the condition that the faces shall be similarly placed ; for, if the two right prisms ABO-A' abc-a', cannot be made to coincide by placing the base ^jSCupon the equal base abc; yet, by inverting one of the 202 GEOMETRY. prisma and ap^ying the base ABC to the base a'b'c', they wilJ coincide. / ''/"\o/ PKOPOSiTiON VI.— theorem: 26. j^r/y mUq^e prism is equivalent to a right prism whose base is a right sd^ion of the^tque prism, and whose altitude is equal to a lateral ^dge of the oblique prism. I Let AB CDE-A' be the oblique prism. At any point F in the edge AA\ pass a plane perpendicular to A A' and forming the right section FGHIK. Produce AA' to F\ mak- ing FF' = AA', and through F' pass a second plane perpendicular to the edge AA', intersecting all the faces of the prism produced, and forming another right section F' G 'HTK' parallel and equal to the first. The prism FGHIK-F' is a right prism whose base is the right section and whose altitude FF' is equal to the lateral edge of the oblique prism. The solid ^5 GDjE^-jP is a truncated prism which is equal to the truncated prism A'B' C'D'E'-F' (24). Taking the first away from the whole solid J.^Ci)-E-i^', there remains the right prism; taking the second away from the same solid, there remains the oblique prism; therefore, the right prism and the oblique prism have the same/Tolume, that is, they are equivalent. PROPOSITION VII.— THEOREM. 27. The plane passed through two diagonally opposite edges of a paralldopiped divides it into two equivalent triangular prisms. X^ ABCD-A' be any parallelopiped ; the plane ACC'A', passed through its opposite edges J.^' and CC', divides it into two equiv- alent triangular prisms ABG-A' and ADC-A'. Let FGHI be any right section of the parallelopiped, made by a plane perpendicu- lar to the edge AA'. The intersection, FII, of this plane with the plane AC', is the di- BOOK VII. 203 agonal of the parallelogram FGHI, and divides that parallelogram into two equal triangles, FGI£ and FIH. The oblique prism ABC- A' is equivalent to a right prism whose base is the triangle FGII and whose altitude is A A' (26) ; and the oblique prism ADC-A' is equiva- lent to a right prism whose base is the triangle FIH and whose altitude is A A'. The two right prisms are equal (25) ; therefore, the oblique prisms, which are respectively equivalent to them, are equivalent, to^ each other. K PKOPOSITION VIII.— THEOEEM. V__\ \zzE ^ 2^. Two rectangular parallelopipeds having equal hoses are to each other as their altitudes. Let P and Q be two rectangular par- allelopipeds having equal bases, and let AB and CD be their altitudes. 1st. Suppose the altitudes have a com- mon measure, which is contained, for example, 5 times in AB and 3 times in CZ>, so that if AB is divided in 5 equal parts, CD will contain 3 of these parts ; then we have AB _b CD'~ Z If now we pass planes through the several points of division of AB and CD, perpendicular to these lines, the parallelopiped P will be divided into 5 equal parallelopipeds, and Q into 3 parallelopipeds, each equal to those in P ; hence, P_5 and, therefore, P AB Q~ CD 2d. If the altitudes are incommensurable, the proof may be given by the method exemplified in (II. 51) and (III. 15), or, according to the method of limits, as follows. Let CD be divided into any number of equal parts, and let cup of these parts be applied to AB as often as AB will contain it 204 GEOMETRY. p Q Since ^^ and CD are incommensurable, a certain number of these parts will extend from A to a point J5', leaving a remainder BB' less than one of the parts. Through B' pass a plane perpendicular to^:J.-B, and denote the parallelopiped whose base is the same as that of P or Qj and whose altitude is AB', by P'; then, since AB' and CD are commensur- able, P^ AB' Q ~ CD' Now, suppose the number of parts into which CD is divided to be continually increased ; the length of each part will become less and less,, and the point B' will approach nearer and nearer to B. The limit of AB' will be AB, md the limit of F' will be P (V. 28). P' P AB' AB The limit of — will therefore be —> and that of will be Q Q CD CD Since, then, the variables — and are constantly equal and q CD ^ ^ approach two limits, these limits are equal (V. 29), and we have AB CD 29. Scholium. The three edges of a rectangular parallelopiped which meet at a common vertex being called its dimensions, the pre- ceding theorem may also be expressed as follows : Two rectangular parallelopipeds which have two dimensions in com- mon are to each other as their third dimensions. PEOPOSITION IX.— THEOREM. 30. Two rectangular parallelopipeds having equal altitiides are to each other as their bases. Let a, b and c be the three dimensions of the rectangular par- allelopiped P; m,n and c those of the rectangular parallelopiped Q; the dimension c, or the altitude, being common. BOOK VII. 205 Let JR he Si third rectangular parallel- opiped whose dimensions are m, h and c; then, R has the two dimensions b and c in common with P, and the two dimensions )n and c in common with Q ; hence (29), P R a m R _b^ n and multiplyii ig these ratios together. P aXb Q m X ^ \ \ 1 But a X 6 is the area of the base of P, and m X ^ is the area of the base of Q ; therefore, P and Q are in the ratio of their bases. 31. Scholium. This proposition may also be expressed as follows: Two rectangular parallelopipeds which have one dimension in com- mon, are to each other in the products of the other two dimensions. p \ K C 1 k \ P ! \ ''N PROPOSITION X.— THEOREM. 32. Any two rectangular parallelopipeds are to each other as the pro- ducts of their three dimensions. Let a, b and c be the three dimensions 6 of the rectangular parallelepiped P; m, n and p those of the rectangular parallelepiped Q. Let P be a third rectangular paral- lelepiped whose dimensions are a, b and p ; then R has two dimensions in com- mon with P and one dimension in com- iu(m with Q\ hence, by (29) and (31), E — 1 p_o^ R p Q m X w wid multiplying these ratios together, P^ g X ^ X c Q m X n X p IS ■ ■ 206 GEOMETRY. PROPOSITION XI.— THEOREM. 33. The volume of a rectangular parallelopiped is equal to the pro- duct of its three dimensions, the unit of volume being the cube whose edge is the linear unit. ,i/et a, 6, c, be the three dimensions of the rectangular parallelopiped P; and let Q be the cube whose edge is the linear unit. The three dimensions of Q are each equal to unity, and we have, by the preceding proposition. P _. aXbXc Q IXlXl C I Q = aXbXc. Now, Q being taken as the unit of volume, — is the numerical mea- Q therefore the volume \ \ ~"\ "^-^ "^ \ ^. \ "^ ^^ \ \ \ sure, or volume of P, in terms of this unit (4) of P is equal to the product a X b X c. 34. Scholium I. Since the product aX b represents the base, when c is called the altitude, of the parallelopiped, this proposition may also be expressed as follows : The volume of a rectangular parallelopiped is equal to the product of its base by its altitude. 35. Sclwlium II. When the three dimensions of the parallelopiped are each exactly divisible by the linear unit, the truth of the propo- sition is rendered evident by dividing the solid into cubes, each of which is equal to the unit of volume. Thus, if the three edges which meet at a common vertex A are, respectively, equal to 3, 4 and 5, times the linear unit, these edges may be divided respectively into 3, 4 and & equal parts, and then planes passed through the several points of division at right angles to these edges will divide the solid into cubes, each equal to the unit cube, the number of which is evidently 3X4X5. But the more general demonstration, above given, includes also the. cases in which one of the dimensions, or two 3f them, or all three, are incommensurable with the linear unit. BOOK VII. v:.z:iyf^ 36. Scholium III. If the three dimensions of a rectangular paral- lelopiped are each equal to a, the solid is a cube whose edge is a, and its volume is a y\ a 'X a = a^; or, the volume of a cube is the third power of its edge. Hence it is that in arithmetic and algebra, the expression "cube of a number" has been adopii|du-to sig^ifv the third power of a number." /^^\^^ ^ " *^ 4p^ ' \^ OF THE ^^ PROPOSITION XII.— THEOREM. OIF- 37. The volume of any parallelopiped is equal to the product of its base by its altitude. Let ABCD-A' be any oblique parallelopiped, whose base is ABCD,2iVi([ altitude ^'0. Produce the edges AB, A'B\ DC, D' C \ in AB produced take FG== AB, and through jPand G pass planes, FFTI, GG'H'H, perpendicular to the produced edges, forming the right parallelopiped FGRI-F',vflth the base FF'I'I and altitude i^6r, equivalent to the given oblique parallelopiped ABCD-A' (26). From F'y draw F'X perpendicular to FI or F'F. Since AF is perpendicular to the plane FI', the plane of the base and the plane FI' are perpendicular to each other (VI. 47) ; therefore, F'K is perpendicular to the plane of the base (VI. 49) and is equal to B' 0, ' Now the three Vines F'G\ F'l' and F'K are perpendicular to each other ; consequently the parallelopiped KLMN-F' , constructed U})on them, is rectangular. The parallelopiped FGHI-F', regarded as an oblique prism whose base is FGG'F' and lateral edge F'I'f is equivalent to the right prism, or rectangular parallelopiped, KLMN-F', whose base is the right section F'L and whose altitude 208 GEOMETRY. is F'l' (26). Therefore, the given parallelepiped ABCD-A' is also equivalent to the rectangular parallelopiped KLMN-F'. The volume of this rectangular parallelopiped is equal to the product of its base KM by its altitude F'K\ its base KM is equal to F'H\ or FH, which is equivalent to AC, and its altitude F'K is equal to B' 0] therefore the volume of the parallelopiped ABCD-A' is equal to the* product of its base AChy its altitude B' 0. PEOPOSITION XIII.— THEOREM. 38. The volume of any prism is equal to the product^ of its base by its altitude. 1st. Let ABC- A' be a triangular prism. This prism is equivalent to one-half the par- allelopiped ABCD-A' constructed upon the edges AB, BC and BB' (27), and it has the same altitude. The volume of the parallelo- piped is equal to its base BD multiplied by its altitude ; therefore, the volume of the triangu- lar prism is equal to its base ABC^ the half of BD, multiplied by its altitude. 2d. Jjei AB CDF- A' be any prism. It may be divided into triangular prisms by planes passed through a lateral edge AA' and the sev- eral diagonals of its base. The volume of the given prism is the sum of the volumes of the triangular prisms, or the sum of their bases multiplied by their common altitude, which is the base ABCDE of the given prism multiplied by its altitude. 39. Corollary. Prisms having equivalent bases are to each other as their altitudes; prisms having equal altitudes ai*e to each other as their bases ; and any two prisms are to each other as the products of their bases and altitudes. Any two prisms having equivalent bases and equal altitudes are equivalent. BOOK VII. 209 PYEAMIDS. 40. Definitions. A pyramid is a polyedroii bounded by a poJvgoii and triangular faces formed by the intersections of planes passed through the sides of the poly- gon and a common point out of its plane; a? S-ABCDE. The polygon, ABODE, is the base of the pyra- mid ; the point, S, in which the triangular faces meet, is its vertex; the triangular faces taken to- gether constitute its lateral, or convex, surface ; the area of this surface is tjie lateral area ; the lines SA, SB, etc., in which the lateral faces intersect,are its lateral edges. The altitude of the pyramid is the perpendicular distance SO from the vertex to the base. A triangular pyramid is one whose base is a triangle; a quadrangvr tar pyramid, one whose base is a quadrilateral ; etc. A triangular pyramid, having but four faces (all of which are triangles), is a tetraedron ; and any one of its faces may be taken as its base. 41. Definitions. A regular pyramid is one whose base is a regular polygon, and whose vertex is in the perpendicular to the base erected at the centre of the polygon. This perpendicular is called the axis of the regular pyramid. From this definition and (VI. 10) it follows that all the lateral faces of a regular pyramid are equal iisosceles triangles. The slant height of a regular pyramid is the per- pendicular from the vertex to the base of any one of its lateral faces. 42. Definitions. A truncated pyramid is the portion of a pyra- mid included between its base and a plane cutting all its lateral edges. When the cutfing plane is parallel to the base, the truncated pyra- mid is called a frustum of a pyramid. The altitude of a friistum is the perpendicular distance between its bases. 18 * , O 210 GEOMETRY. In a frustum of a regular pyramid, the lateral faces are equal trapezoids ; and the perpendicular distance between the parallel sides of any one of these trapezoids is the §lant h'eight of the frustum. PROPOSITION XIV.— THEOREM. 43. If a pyramid is cut hy a plane parallel to its base: 1st, the edges and the altitude are divided proportionally ; 2d^ the section is a polygon similar to the base. Let the pyramid S-ABCDE, whose altitude is SO, be cut by the plane abcde parallel to the base, intersecting the lateral edges in the points a, by c, d, e, and the altitude in o ; then, 1st. The edges and the altitude are divided proportionally. For, suppose a plane passed through the ver- tex S parallel to the base ; then, the edges and altitude, being intersected by three parallel planes, are divided proportionally (VI. 37), and we have Et' Sa SA Sb SB Se SO So SO 2d. The section abcde is similar to the base ABCDE. For, the sides ab, be, etc., are parallel respectively to AB, BC, etc. (VI. 25), and in the same directions : therefore the angles of the two polygons are equal, each to each (VI. 32). Also, since ab is parallel to AB, and be parallel to BC, the tri- angles Sab and SAB are similar, and the triangles Sbc and SB Care similar; therefore, AB~ SB'^"" BC~ SB whence ab . be AB^BC' BOOK VII. mer we should find bo cd de BC~ CD~ DE~ ea EA 211 Therefore, the polygons ahede and ABODE are similar (III. 24). 44. Corollary 1. The polygons abcde and ABODE being similar, their surfaces are proportional to the squares of their homologous sides ; hence abcde ab Sa So ^ ABODE~AB' ~'M'~~SO'' that is, the surface of any section of a pyramid parallel to its base is proportional to the square of its distance from the vertex. 45. Oorollaryll. If two pyramids, S-ABODE and S'-A'B' 0'D\ having equal altitudes SO and S' 0\ are cut by planes parallel to their bases and at equal distances, So and S'o\ from their vertices, the sections abcde and a'b'c'd' will be proportional to the bases. For, by the preceding corollary, abdce ABODE SO' and 72 S'o s^" A'B'O'D' whence, since So = S'o' SO = S'0\ and abcde a'b'c'd' ABODE A'B'O'D' 46. Corollary III. If two pyramids have equal altitudes and equiva lent bases, sections made by planes parallel to their bases and at equal distances from their vertices are equivalent. 212 GEOMETRY. PROPOSITION XV.— THEOREM. 47. The lateral area of a regular pyramid is equal to the product i/ the perimeter of its base by one-half its slant height. ^or, let S- ABODE be a regular pyra- mid ; the lateral faces SAB, SBC, etc., be- ing equal isosceles triangles, whose bases are the sides of the regular polygon ABODE and whose common altitude is the slant height SH, the sum of their areas, or the lateral area of the pyramid, is equal to the sum of AB, BO, etc., multiplied by ^SH (IV. 13). 48. Oorollary. The lateral area of the frustum of a regular pyramid is equal to the half sum of the perimeters of its bases multiplied by the slant height of the frustum. For, this product is the measure of the sum of the areas of the trapezoids ABba, BOeb, etc., whose common altitude is the slant height hH (IV. 17). PROPOSITION XVI.— LEMMA. 49. A series of prisms may be inscribed in any given triangular pyramia whose total volume shall differ from the volume of the pyramid by less than any assigned volume. 'LQtS-ABOhe the given triangular pyramid, whose altitude is A T. Divide the altitude AT into any number of equal parts Ax, xy, etc., and denote one of these parts by h. Through the points of division x, y, etc., pass planes parallel to the base, cutting from the pyramid the sections DEF, GHI, etc. Upon the triangles DEF, GHI, etc., SLS upper bases, construct prisms whose lateral edges are parallel to >S'^, and whose altitudes are each equal to h. This is effected by passing BOOK VII. 213 planes through EF, HI, etc., parallel to SA. There will thus be formed a series of prisms DEF-A, GHI-D, etc., inscribed in the pyramid. Again, upon the triangles ABC, DEF, GUI, etc., as lower bases, construct prisms whose lateral edges are parallel to SA, and whose altitudes are each equal to h. This also is effected by passing planes through jBC, EF, HI, etc., parallel to SA. There will thus be formed a series of prisms ABC-D, DEF-G, etc., which may be said to be circumscribed about the pyramid. Now, the first inscribed prism DEF-A is equivalent to the second circumscribed prism DEF-G, since they have the same base DEF and equal altitudes (39) ; the second inscribed prism GHI-D is equivalent to the third circumscribed prism GHI-K; and so on. Therefore, the sum of all the inscribed prisms differs from the sum of all the cir- cumscribed prisms only by the first circumscribed prism ABC-D. But the pyramid is greater than the sum of the inscribed prisms and less than the sum of the circumscribed prisms ; therefore, the differ- ence between the total volume of the inscribed prisms anil the volume of the pyramid is less than the volume of the prism A BC-D. The volume of the prism ABC-D may be made as small as we })lease, or less than any assigned volume, by dividing the altitude A T into a sufiiciently great number of equal parts; for, if the as- signed volume is represented by a prism whose base is ABC and altitude Aa, we have only to divide ^T into a number of equal parts each less than Aa. Therefore, the difference between the total volume of the inscribed prisms and the volume of the pyramid may be made less than any assigned volume. 50. Corollary. If the number of parts into which the altitude is divided is increased indefinitely, the difference between the volume of the inscribed prisms and that of the pyramid approaches indefi- nitely to zero; and therefore the pyramid is the limit of the sum of the inscribed prisms, as their number is indefinitely increased (V. 28). 214 GEOMETRY. PROPOSITION XVII.— THEOREM. 51. Two triangular pyramids having equivalent bases and equal alti- tudes are equivalent. IjetS-ABC and S'-A'B'C be two triangular pyramids having equivalent bases, ABCy A'B'G'y in the same plane, and a common altitude .42! Divide the altitude AT into a number of equal parts Ax^ xy, yz, etc., and through the points of division pass planes parallel to the plane of the bases, intersecting the two pyramids. In the pyramid /S^^ 50 inscribe a series of prisms whose upper bases are the sections DEF, GHI, etc., and in the pyramid S'-A'B' C inscribe a series of prisms whose upper bases are the sections D'E'F\ G'H'I', etc. Since the corresponding sections are equivalent (46), the correspond- ing prisms, having equivalent bases and equal altitudes, are equiva- lent (39) ; therefore, the sum of the prisms inscribed in the pyramid S-ABC\s equivalent to the sum of the prisms inscribed in the pyra- mid S'-A'B'C \ that is, if we denote the total volumes of the two series of prisms by Fand F', we have F= V\ Now let the number of equal parts into which the altitude is divided be supposed to be indefinitely increased; the volume F approaches to the volume of the pyramid S-ABC as its limit, and the volume V approaches to the volume of the pyramid S'-A'B' C as its limit (50). Since, then, the variables F and V are always equal to each other and approach two limits, these limits are equal (V. 29) ; that is, the volumes of the pyramids are equal. 1 BOOK VII. 216 PEOPOSITION XYIII.— THEOKEM. 52. A triangular pyramid is one-third of a triangular prism of the %ame base and altitude. Let S-ABC be a triangular pyramid. Through one edge of the base, as A C, pass a plane A CDE parallel to the opposite lateral edge SB, and through S pass a plane SED parallel to the base ; the prism ABC-E has the same base and altitude as the given pyramid, and we are to prove that the pyramid is one- third of the prism. Taking away the pyramid S-ABC from the prism, there remains a quadrangular pyramid whose base is the parallelogram AGDE and vertex S. The plane SEC, passed through SE and SC, divides this pyramid into two triangular pyramids, S-AEC and S-ECD, which are equivalent to each other, since their triangular bases AEC and ECD are the halves of the parallelogram A CDE, and their common altitude is the perpendicular from S upon the plane A CDE (51). The pyramid S-ECD may be regarded as having ESD as its base and its vertex at C; therefore, it is equivalent to the pyramid S-ABC vfhich has an equivalent base and the same altitude. There- fore, the three pyramids into which the prism is divided are equiva- lent to each other, and the given pyramid is one-third of the prism. 53. Corollary. The volume of a triangular pyramid is equal to one- third of the product of its base by its altitude. PEOPOSITION XIX.— THEOKEM. 54. The volume of any pyramid is equal to one-third of the product of its base by its altitude. For, any pyramid, S-ABCDE, may be di- vided into triangular pyramids by passing planes through an edge SA and the diagonals AD, A C, etc., of its base. The bases of these pyramids ire the triangles which compose the base of the Iven pyramid, and their common altitude is the Ititude SO of the given pyramid. The volume the given pyramid is equal to the sum of the 216 GEOMETRY. volumes of the triangular pyramids, which is one-third of the sum of their bases multiplied by their common altitude, or one-third the product of the base ABODE by the altitude SO, 55. Corollary. Pyramids having equivalent bases are to each other as their altitudes. Pyramids having equal altitudes are to each other as tJy^ir bases. Any two pyramids are to each other as the products of their bases and altitudes. 56. Scholium. The volume of any polyedron may be found by dividing it into pyramids, and computing the volumes of these pyra- mids separately. The division may be effected by drawing all the diagonals that can be drawn from a common vertex ; the bases of the pyramids will be all the faces of the polyedron except those which meet at the common vertex. Or, a point may be taken within the polyedron and joined to all the vertices ; the polyedron will then be decomposed into pyramids whose bases will be the faces of the polyedron, and whose common vertex will be the point taken within it. PROPOSITION XX.— THEOREM. 57. Two tetraedrons which have a triedral angle of the one equal to a triedral angle of the other, are to each other as the products of the three edges of the equal triedral angles. Let ABCD, AB'C'D\ be the given tetraedrons, placed with their equal triedral angles in coincidence at A. From D and D', let fall DO and D'O' perpendicular to the face ABC. Then, taking the faces ABC, AB'C, as the bases of the triangu- lar pyramids D-ABC, D'-AB'C, and denoting the volumes by F and F', we have (55), J^ _ ABCX D O _ ABC ^ D0_ ~ ~ h'O'' V AB'C X D'O' AB'C By (IV. 22) and (III. 25), we have ABC __ ABX AC AB'C~ AB' X AC and DO AD D'C AD' ^ 3, li'C^Y S -{ ^/^-^^ therefore, "^ \ U "^B^OOK VII ■:i V AB X AO X AD V'~ AB' XAG'X AD'' 217 ^3 PKOPOSITION XXI.— THEOKEM. 58. A fy^jisium of a triangular pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean pro- portional between the bases, of the frustum. Let ABG-D be a frustum of a tri- angular pyramid, formed by a plane DEF parallel to the base ABC, Through the vertices A, E and C, pass a plane AEC;, and through the vertices E, D and G, pass a plane EDO, dividing the frustum into three pyra- mids. For brevity, denote the pyramid E-ABChy P, the pyramid E-DFChyp, and the pyramid E-ADC by §. The pyramids P and Q, regarded as having the common vertex C and their bases in the same plane BD, have a common altitude and are to each other as their bases AEB and AED (55). But the triangles AEB and AED, having a common altitude, namely, the altitude of the trapezoid ABED, are to each other as their bases AB and DE; hence we have P_AB ' Q~ DE The pyramids Q and p, regarded as having the common vertex E and their bases in the same plane AF, have a common altitude, and are to each other as their bases ADC and DCF. But the tri- angles ADC and DCF, having a common altitude, namely, the alti- tude of the trapezoid A CFD, are to each other as their bases A G and DF; hence we have Q^AG p DF ^Wf •f^ Moreover the section DEF being similar to ABC (43), we havp 19 i 218 GEOMETRY. DE~ DF and therefore ^=« Q p' whence Q' = FXp, Q^VFxTp; that is (III. 5), the pyramid Q is a mean proportional between the pyramids P and p. Now, denote the lower base ABC of the frustum by B, its upper ba^e by 6, and its altitude by h. The pyramid P, regarded as having its vertex at E, has the altitude h and the base B ; the pyramid p, regarded as having its vertex at C, has the altitude h and the base b ; hence (54), P=\hXB, p=.\hXb, and q = V\h XBXihXb = ih X VB X b; consequently, Q is equivalent to a pyramid whose altitude is Jjk and whose base is a mean proportional between the bases B and b ; and since the given frustum is the sum of P, p and §, the proposition is ' established. II ] f V denotes the volume of the frustum, the proposition is ex- pressed by the formula V=ihXB + ihXb + ihX VWxb, 59. Corollary A frustum of any pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the bases, of the frustum. For, let ABCDE-F be a frustum of any pyramid S-ABCDE. Let S'-A'B' C be a triangular pyramid, having the same altitude as the pyramid S-ABCDE, and a base A'B'C equivalent to the base ABCDE, and in the same plane with it. The volumes of the two pyramids are equivalent (55). Let the plane of the upper base of the given frustum be produced to cut the triangular pyramid. I BOOK VII. 219 The section i^'GT being equivalent to the section FGEIK (iS), the pyramid S'-F'GT is equivalent to the pyramid S-FGHIK; and taking away these pyramids from the whole pyramids, the frus- tums that remain are equivalent ; therefore, denoting by B the area of ABODE OT of A'B'C\ by b that of FGHIK or of F'GT, and by h the common altitude of the two frustums, we have for the vol- ume of the given frustum the same expression as for that of the tri- angular frustum ; namely, V=^h {B -\-h -^VTyTh). TEUNCATED TRIANGULAE PEISM, PEOPOSITION XXII.— THEOEEM. 60. A truncated triangular prism is equivalent to the mm of three pyramids whose common base is the base of theprismy and whose vertices are the three vertices of the iyiclined section. Let ABC-DEF be a truncated triangular prism whose base is ABC and inclined sec- tion DEF. Pass the planes AEC and DEC, dividing the truncated prism into the three pyramids, E-ABC, E-ACD and E-CDF. The first of these pyramids, E-ABC, has the base ABC a.nd the vertex E, 220 GEOMETRY. The second pyramid, E-ACD^ is equivalent to the pyramid B- A CD ; for they have the same base A CD, and the same altitude, since their vertices E and B are in the line EB parallel to this base. But the pyramid B-ACD is the same as D-ABC; that is, it has the base ABC and the vertex D. The third pyramid, E-CDF, is equivalent to the pyramid B-A CF; for they have equivalent bases CDF and ACF in the same plane, and also the same altitude, since their vertices E and B are in the line EB parallel to that plane. But the pyramid B-A CF is the same as F-ABC; that is, it has the base ABC and the vertex F. Therefore the truncated prism is equivalent to three pyramids whose common base is ABC and whose vertices are E, D and F. 61. Corollary I. The volume of a truncated right triangular prism is equal to the product of its base by one-third the sum of its lateral edges. For. the lateral edges AD, BE, CF, being perpendicular to the base, are the altitudes of the three pyramids to which the truncated prism has been proved to be equivalent ; therefore, the volume is ABC X iAD + ABC X iBE + ABC X iCF, ^ or ABCX AD-i-BE-i- CF 62. Corollary II. The volume of any truncated triangular prism is equal to the product of its right section by one-third the sum of its lateral edges. For, let AB C-A'B'C be any trun- cated triangular prism ; the right section DEF divides it into two truncated right prisms whose volumes are, by the preced- ing corollary. 3 and 3 the sum of which is 3 BOOK VII. 221 SIMILAR POLYEDRONS. 63. Definition. Similar polyedrons are those which &re bounded by the same number of faces similar each to each and similarly placed, and which have their homologous polyedral angles equal. Parts similarly placed in two similar polyedrons, whether faces, lines, or angles, are homologous. 64. Corollary I. Since homologous edges are in the ratio of simili- tude of the polygons of which they are homologous sides (III. 24), and every edge belongs to two faces, in each polyedron, it follows that the ratio of similitude of any two homologous faces is the same as that of any other two homologous faces, and this ratio may be called the ratio of similitude of the two polyedrons. Therefore, any two homologous edges of two similar polyedrons are in the ratio of similitude of the polyedrons ; or, homologous edges are proportional to each other, Qb. Corollary II. The ratio of the surfaces of any two homologous faces is the square of the ratio of similitude of the polyedrons (IV. 24) ; or, any two homologous faces are to each other as the squares of any two homologous edges. Hence, by the theory of proportions (III. 12), the entire surfaces of two similar polyedrons are to each other as the squares of any two homologous edges. PROPOSITION XXIII.— THEOREM. 66. If a tetraedron is cut by a plane parallel to one of its facesy the teiraedron cut off is similar to the first. Let the tetraedron ABCD be cut by the -^ plane B'C'D' parallel to BCD; then, the A tetraedrons AB' CD' and J.^ CD are simi- // \ lar. // \ For, since the edges AB, ACy AD, shq ^/\"f — l^^^ divided proportionally at B\ C, D\ the / \l/^ \ face AB' C is similar to the face ABC, / / \ AC'D' to ACD, and AB'D' to ABD) ^V"' 7 --^3 also, B'C'D' is similar to BCD (43). \ / ^^ Moreover, the homologous triedral angles, \y^ being contained by equal face angles simi- ^ 222 GEOMETRY. larly placed, are equal, each to each (VI. 71). Therefore, by the definition (63), the tetraedrons are similar. PROPOSITION XXIV.— THEOREM. •* 67. Two tetraedrons are similar^ when a diedral angle of the one is equal to a diedral angle of the other, and the faces including these angles are similar each to each, and similarly plaxied. Let ABCD, A'B'C'D', be two tetraedrons in which the diedral angle AB is equal to the diedral angle A'B', and the faces ABC and ABD are res- pectively similar to the faces A'B'O' and A'B'D') then, the tetraedrons are similar. The triedral angles A and A' are equal, since they may evidently be placed with their vertices in coincidence so as to coincide in all their parts. Therefore, the angles CAD and C'A'D' are equal. The given similar faces furnish the proportions AC A'C whence A'C'~ A'D'* therefore, the faces ACD and A' CD' are similar (III. 32). In like manner it is shown that the triedral angles B and B' are equal, and the faces BCD and B' CD' are similar. Finally, the triedral angles C and C are equal, since their face angles are equal each to each and are similarly placed (VI. 71) ; and the triedral angles D and D' are equal for the same reason. Therefore, the i\io tetraedrons are similar (63). AB AD AB A'B' A'D'~ ~ A'B' AC AD BOOK VII 223 PKOPOSITION XXV.— THEOEEM. 68. Two similar polyedrons may be decomposed into the same numh&i of tetraedrons similar each to each and similarly placed. Let ABCDEFGH and abcdefgh be similar polyedrons, of which A and a are homologous vertices. Let all the faces not adjacent to Ay in the first polyedron, be decomposed into triangles, and let straight lines be drawn from A to the vertices of these triangles ; the polyedron is then divided into tetraedrons having these triangles as bases and the common ver- tex A. Also decompose the faces not adjacent to a, in the second polye- dron, into triangles similar to those in the first polyedron and simi- larly placed (III. 39), and let straight lines be drawn from a to the vertices of these triangles ; the second polyedron is then divided into the same number of tetraedrons as the first, and it is readily proved that two tetraedrons similarly placed in the two polyedrons are similar. We leave the details of the proof to the student. See (III. 39). 69. Corollary. Homologous diagonals y and in general any two homol- ogous lineSy in two similar polyedrons^ are in the same ratio as any two homologous edges, that is, in the ratio of similitude of the polyedrons. PROPOSITION XXVI.— THEOREM. 70. Two polyedrons composed of the same number of tetraedrons^ similar each to each and similarly placed, are similar. The proof is left to the student. See (III. 38). 224 GEOMETRY. PROPOSITION XXVII.— THEOREM. 71. Similar polyedrons are to each other as the cubes of their homol- ogous edges. 1st. Let ABCDj abed, be two sirnjilar tetraedrons ; let the similar tnces BCD, bed, be taken as bases, and let AO, ao be their altitudes. Since the tetraedrons are simi- lar, they may be placed with their equal homologous polyedral angles A and a in coincidence, and the base bed will then be parallel to the base BCD, since their planes make equal angles with the plane of the face ABC. The perpen- dicular AO, to BCD, will also be perpendicular to bed, and Ao will be the altitude of the tetraedron Abed or abed. Denoting the volumes of the tetraedrons by V and v, we have (55), and by (69), we have BCDX AO BCD X AO bed X Ao bed Ao IT, we have BCD BC bed Te' AO AC Ao ac BC ~~ be ' hence Z_ ^ V — — — ' V W be be or, since any two homologous edges are in the same ratio as any other two, the two similar tetraedrons are to each other as the cubes of any two homologous edges. 2d. Two similar polyedrons may be decomposed into the same number of tetraedrons, similar each to each ; and any two homologous BOOK VII. 225 tetraedrons are to each other as the cubes of their homologous edges ; but the ratio of the homologous edges of the two similar tetraedrons is equal to ratio of any two homologous edges of the polyedron (69) ; therefore, any two homologous tetraedrons are to each other as the cubes of two homologous edges of the polyedron, and by the theory of proportion, their sums, or the polyedrons themselves, are in the same ratio, or as the cubes of their homologous edges. 72. Corollary I. Similar prisms or pyramids are to each other as the cubes of their altitudes. 73. Corollary II. Two similar, polyedrons are to each other as the cubes of any two homologous lines. N SYMMETKICAL POLYEDRONS. a. Symmetry with respect to a plane, 74. Definitions. Two points, A and A\ are sym- metrical with respect to a plane, 3IN, when this j^ plane bisects at right angles the straight Vine AA' I joining the points. / Two figures are symmetrical with respect to a plane, when every point of one figure has its sym- metrical point in the other. We leave the proof of the following simple theorems to the student. 75. Theorem. The symmetrical figure of a finite draight line, AB, is an equal straight line, A' B', ^N^^, 76. Theorem. The symmetrical figure of an indefinite straight line, AB, is another indefinite straight line, A'B\ which intersects the first in the plane of symmetry, and makes the same angle with the plane. V ^A 226 GEOMETRY. 77. Theorem. The symmetrical figure of a plane angle, BAC, is an equal 'plane angle, B'A' C (Fig. 1). Fig. 2. 78. Theorem. The symmetrical figure of a plane ABC, is a plane ABC ; and the two planes intersect in the plane of symmetry ABN, and make equal angles with it (Fig. 2). Corollary. If a plane is parallel to the plane of symmetry, its sym- metrical plane is also parallel to the plane of symmetry, and at the same distance from it. 79. Theorem. The symmetrical figure of a diedral angle, CABD, is an equal diedral angle, C'A'B'D' (Fig. 3). PROPOSITION XXVIII.— THEOREM. 80. If two polyedrons are symmetrical with respect to a plane, 1st, (heir homologous faces are equal; 2d, their homologous polyedral angles are symmetrical. 1st, lict Ay B, C, D, be the vertices of a face of one of the polyedrons; their symmetrical points. A', B', C, Z)', are in the same plane (78) ; tl^e homologous sides of the polygons ABCD, A'B'C'D', are equal (75), and their homologous angles are equal (77) ; therefore the homologous faces are equal. 2d. The homologous face angles of two polyedral angles, A and A ', are equal (77), and their homologous diedral angles are equal (79) ; but if one of the face angles as BOOK VIT. 22: BAD be applied to its equal B'A'D', so as to bring the other edges of the polyedral angles A and J.' on the same side of the common plane B'A'D', it will be apparent that the face angles succeed each other in inverse orders in the two figures ; therefore, the homologous polyedral angles of the two polyedrons are symmetrical (VI. 68). 81. Corollary. Two symmetrical polyedrons may he decomposed into the same number of tetraedrons symmetrical each to each. For one of the polyedrons being divided into tetraedrons by drawing diagonals from a common vertex, and the homologous diagonals being drawn in the other polyedron, any two corresponding tetraedrons thus formed will have their vertices symmetrical each to each, and will consequently be symmetrical tetraedrons. 82. Scholium. Two polyedrons whose faces are equal each to each and whose polyedral angles are symmetrical each to each, are called symmetrical polyedrons, whatever may be their position with respect to each other, since they admit of being placed on opposite sides of a plane so as to make their homologous vertices symmetrical with respect to that plane. PROPOSITION XXIX.— THEOREM. 83. Two symmetrical polyedrons are equivalent. Since two symmetrical polyedrons may be decomposed into the same number of tetraedrons symmetrical each to each, it is only necessary to prove that two symmetrical tetra- edrons are equivalent. Jjet SAB C be a tetraedron; let the plane of one of its faces, ABO, be taken as a plane of symmetry, and construct its symmetrical tetraedron S'ABC. The tetraedrons, having the same base ABC and equal altitudes SOy S' O, are equivalent (55). f 6. Symmetry mth respeet to a centre, 84. Definitions. Two points A and A' , are sym- metrical with respect to a fixed point, 0, called the centre of symmetry, when this point bisects the straight line, AA', joining the two points. a ./ 228 GEOMETRY. Any two figures suce symmetrical with respect to a centre, when every point of one figure has its symmetrical point on the othen These definitions are identical with those giveu in (I. 138), but are here extended to figures in space. The student can readily establish the following theorems on figur^ symmetrical with respect to a centre. 85. Theorem. The symmetrical figure of a finite straight line, AB^ is an equal straight line, A 'B,' parallel to the first (Fig. 1). Fig. 2. Fig. 1. ,'0 86. Theorem. The symmetrical figure of a plane angle, BA C,is ai equal plane angle, B'A'C (Fig. 2). 87. Theorem. The symmetrical figure of a plane, B AC, is a parallel plane, B'A'C (Fig. 2). 88. Theorem. The symmetrical figure of a diedral angle, DAB C, is an equal diedral angle, D'A'B'C. d 2 £t B' ; 89. Theorem. If two polye- drons are symmetrical with re- spect to a centre, 1st, their ho- mologous faces are equal; 2d, their homologous angles are sym- metrical. Corollary I. The symmetrical figure of a polyedron is the same, whether the symmetry be with respect to a plane or with respect to a centre. Corollary II. Two polyedrons, symmetrical with respect to a centre^ are equivalent. BOOK VII 229 c. Symmetry of a single figure. X 90. Definition. Any figure in space is called a symmetrical figure, 1st, if it can be divided by a plane into two figures which are sym- metrical with respect to that plane; 2d, if it has a centre which bisects ail straight lines drawn through it, and terminated by the sur- face of the figure ; 3d, if it has an axis which contains the centres of all the sections perpendicular to that axis. For example, 1st, the hexaedron SABCS' is symmetrical with respect to the plane ABC, which divides the solid into the two symmet- rical tetraedrons SABC, S'ABC. 2d. The intersection of the four diagonals of a parallelepiped is the centre of symmetry of the parallelepiped (18). 3d. The straight line 22', joining the cen- tres of the bases of a right parallelepiped AC, is an axis of symmetry of the figure, since it evidently contains the centre of any section abed perpendicular to it, or parallel to the bases. If the parallelepiped is rectangu- lar, it has three axes xx', yy', zz', perpendicu- lar to each other which intersect in its centre. We leave the demonstration of the following theorems to the student. 91. Theorem. If a figure has two planes of symmetry, MN and FQ, the intersection, xx', of these planes, is an axis of symmetry of the figure. See (I. 141). t/ ' . .r-4-?' .....^ /'."""^\ 7 230 GEOMETRY. 9^. Theorem. If a figure has three planes of symmetry perpendiculaf to each other (VI. 48), the intersections of these planes are three axes of symmetry, and the common intersection of thes&axes is the centre of symmetry of the figure. K THE REGULAR POLYEDRONS. 93. Definition. A regular polyedron is one whose faces are all equal regular polygons and whose polyedral angles are all equal to each other. PROPOSITION XXX.— PROBLEM. 4 94. To construct a regular polyedron, having given one of its edges. There are five regular polyedrons, which we shall consider in their order. Construction of the regular tetraedron. Let AB be the given edge. Upon AB con- struct the equilateral triangle ABC. At the centre of this triangle erect a perpendicular, OD, to its plane, and take the point D so that AD = AB; join DA, DB, DC. The faces of the tetraedron ABCD are each equal to the face ABC (VI. 10), and its polyedral angles are all equal (VI. 71) ; therefore, ABCD is a regular tetraedron. Construction of the regular hexaedron. Upon the given edge AB, construct the square ABCD. The cube ABCDE, whose faces are each equal to this square, is a regular hexaedron, and the method of constructing it is obvious. Construction of the regular octaedron. Let AB be the given edge. Upon AB construct the square ABCD, and at the centre of the square erect the perpendicular FG to ite plane. In this perpendicular, take the points F and G so BOOK VII. 231 that 0F= OA and OG =- OA, and join FA, FB, FC, FD, GA, GB, GC, GD. These edges are equal to each other (VI. 10), and also to the edge AB, since A OF and A OB are equal triangles ; therefore, the faces of the figure are eight equal equilateral triangles. Since the triangles DFB and DAB are equal, DFBG is a square, and it is evident that the pyramid A-DFBG is equal in all its parts to the pyramid F-ABCD; therefore, the polyedral angles A and F are equal; whence, also, all the polyedral angles of the figure are equal to each other, and the figure is a regular octaedron. Construction of the regular dodecaedron. Upon the given edge AB, construct a regular pentagon ABODE; to each of the sides of this pentagon apply the side of an equal pentagon, and let the planes of these pentagons be so inclined to that of ABODE as to form triedral angles at A, B, C, D, E. There is thus formed a convex surface, FGHI, etc., composed of six regu- lar pentagons. Construct a second convex surface, F'G'HT, etc., equal to the first. The two surfaces may be combined so as to form a single con- vex surface. For, suppose the diagram to represent the exterior of the first surface and the interior of the second ; let the point P of the first be placed on F' of the second ; then the three equal angles OFF, P'F'A', A'F'G', can be united so as to form a triedral angle at F' equal to that at A\ since the diedral angle F'A' is already that which belongs to such a triedral angle. But when PF coin- cides with F'G\ there will be brought together at G' three angles PFA, AFG, F'G'H\ which will form a triedral angle equal to J' 232 GEOMETRY. since the diedral angles at the edges FA and F'G' are already those which belong to such an angle. Thus, it can be shown, successively, that all the edges PF, FO, etc., of the first figure, will coincide with the edges F'G\ G'H\ etc., of the second, and that all the polyedral angles of the whole convex surface thus formed are equal. This, fiurface is therefore a regular dodecaedron. Construction of the regular icosaedron. J Upon the given edge ABj construct a regular pentagon ABODE, and at its centre erect OS perpendicular to its plane, taking S so that SA = AB) then, joining SA, SB, etc., the pyramid S-ABCDE is regular, and each of its faces is an equilateral triangle. Now let tnc vertices A and B be taken (as in the second figure) as the vertices of two other pyramids, A-BSEFG and B-ASCHG, each equal to the first and having in common with it the faces ASB and ASE, ASB and BSC, respectively, and in common with each other the faces ASB and ABG. There is thus formed a convex surface CDEFGH, composed of ten equal equilateral triangles. Construct a second convex surface G'D'E'F'G'IF, equal in all re- spects to the first ; and let the figure represent the exterior of the first surface, and the interior of the second. Let the first surface be applied to the second by bringing the point Z), where two faces meet, upon tlie point C", where three faces meet. The edges DE and DC can then be brought into coincidence with the edges CD' and C'H\ re- spectively, to form a m^lyedral angle of five faces equal to S, without in any way changing the form of either surface, since the diedral angles at the edges SD, S'C', B'C', are those which belong to such a polyedral angle. But when DC has been brought into coincidence with C'H', there have been brought together, at the point H\ five BOOK VII. 233 equal faces having the necessary diedral inclinations to form another polyedral angle equal to S; and thus, in succession, it can be shown that all the outer edges of the first surface coincide with those of the second, and that all the polyedral angles of the entire convex sur- face thus formed are equal. This surface is therefore a regular icosaedron. yj y^^- PKOPOSITION XXXI.— THEOKEM. 95. Only jive regular {convex) polyedroris are possible. The faces of a regular polyedron must be regular polygons, and at least three faces are necessary to form a polyedral angle. 1st. The simplest regular polygon is the equilateral triangle. Three angles of an equilateral triangle can be combined to form a convex polyedral angle, and this combination, as shown in the pre- ceding proposition, gives the regular Jetraedron. The combination of four such angles gives the regular octaedron; and that of five gives the regular icosaedron. The combination of six or more (each being -f of a right angle) gives a sum equal to, or greater than, four right angles, and therefore cannot form a convex polyedral angle (VI. 70). Therefore, only three regular convex polyedrons are possible whose surfaces are composed of triangles. 2d. Three right angles can be combined to form a polyedral angle, ^nd this combination gives the regular hexaedron, or cube. Four or more right angles cannot form a convex polyedral angle (VI. 70) ; therefore, but one regular convex polyedron is possible whose surface is composed of squares. 3d. Three angles of a regular pentagon, being less than four right angles (each being f of a right angle), may form a polyedral angle, as in the case of the dodecaedron ; but four or more would exceed four right angles. Therefore, but one regular convex polyedron is possible with pentagonal faces. 4th. Three or more angles of a regular hexagon (each being f of a right angle) cannot form a convex polyedral angle ; nor can angles of any regular polygon of a greater number of sides form such a polyedral angle. Therefore, the five regular convex polyedrons constructed in the preceding proposition are the only ones possible. 20 -» 234 GEOMETRY. 96. Scholium. The student may derive some aid in comprehending the preceding discussion of the regular polyedrons by constructing models of them, which he can do in a very simple manner, and «,t the same time with great accuracy, as follows. Draw on card-board the following diagrams ; cut them out entire, ai\,d at the lines separating adjacent polygons cut the card-board half through ; the figures will then readily bend into the form of the respective surfaces, and can be retained in that form by glueing the Tetraedron. Hexaedron. {. Octaedron. Dodeoaedron. V:- Icosaedron. % - n GENERAL THEOREMS ON POLYEDRONS PROPOSITION XXXII.— THEOREM. 97. In any polyedron, the number of its edges increased 6y two is equal to the number of its vertices increased by the number (if its faces. BOOK VII. 235 Let E denote the number of edges of any polyedron, V the num* ber of its vertices, and F the number of its faces ; then we are to prove that E-\-2= V+F. In the first place, we observe that if we remove a face, as ABODE, from any con- vex polyedron GH, we leave an open sur- face, terminated by a broken line which was the contour of the face removed ; and in this open surface the number of edges and the number of vertices remain the same as in the original surface. Now let us form this open surface by putting together its faces successively, and let us examine the law of connection between the number of edges E, the number of vertices F, and the number of faces, at each successive step. Beginning with one face we have E= V. Annexing a second face, by applying one of its edges to an edge of the first, we form a surface having one edge and two vertices in common with the first ; therefore, whatever the number of sides of the new face, the whole number of edges is now one more than the whole number of vertices ; that is. For 2 faces. E= F+1. Annexing a third face, adjacent to each of the former, the new sur- face will have two edges and three vertices in common with the pre- ceding surface; therefore the increase in the number of edges is again one more than the increase in the number of vertices ; and we have For 3 faces, E = V -\- 2. At difierent stages of this process the number of common edges to two successive open surfaces may vary, but in all cases it is ap- parent that the addition of a new face increases E by one more unit than it increases V; and hence we have the following series of results : 236 In an open surfece of GEOMETRY. / 1 face, E=r, I 2 faces, E=r+i, f 3- ■' E--^:r+2, U " £=F+3, j etc. etc. V^— J faces, E= V+F 2; where the lato is, that, in the successive values of E, the number to be added to F is a unit less than the number of faces. The last line expresses the relation for the open surface of i^ — 1 faces, that is, for the open surface which wants but one face to make the closed sur- face of F faces. But the number of edges and the number of ver- tices of this open surface are the same as in the closed surface. Therefore, in a closed surface of F faces, we have or E= F+ F— 2, ^+2= F+F, as was to be proved. This theorem was discovered by Euler, and is called Fuler's Theo- rem on Polyedrons. PROPOSITION XXXin.— THEOREM. 98. The sum of all the angles of the fa^es of any polyedron is eqival to four righl angles taken as many times as the polyedron has vertices less two. Let F denote the number of edges, F the number of vertices, F the number of faces, and S the sum of all the angles of the faces, of any polyedron. If we consider both the interior angles of a polygon and the exterior ones formed by producing its sides as in (I. 101), the sum of all the angles both interior and exterior is 2R X w, where B denotes a right angle, and n is the number of sides of the polygon. If, then, F denotes the number of edges of the polyedron, 2E denotes the whole number of sides of all its feces considered as independent polygons, and the sum S of the interior angles of all the F faces plus the sum of their exterior angles is 2R X 2F. But the sura of BOOK VII. 237 * the exterior aDgles of one polygon is 4R, and the sum of the exterior angles of the F polygons is 4R X F; that is, S-{-UiXF=2Rx2E, ■)T, reducing, ^ iS=4i2x (Je;— F> But by Euler^s Theorem F-- F= F— ? : ^encei Boot Till. THE THREE ROUND BODIES. Of the various solids bounded by curved surfaces, but three are treated of in Elementary Geometry — namely, the cylindeVj the conef and the sphere^ which are called the three round bodies. THE CYLINDER 2. Definition. A cylindrical surface is a curved surface generated by a moving straight line which continually touches a given curve, and in all of its positions is parallel to a given fixed straight line not in the plane of the curve. Thus, if the straight line Aa moves so as continually to touch the given curve ABCDj and so that in any of its positions, as Bh, Cc, Ddy etc., it is parallel to a given fixed straight line ilifm, the surface ABCDdcba is a cylindrical surface. If the moving line is of indefinite length, a surface of indefinite extent is generated. The moving line is called the generatrix ; the curve which it touches is called the directrix. Any straight line in the surface, as Bh^ which represents one of the positions of the generatrix, is called an element of the surface. In this general definition of a cylindrical surface, the directrix may be any curve whatever. Hereafter we shall assume it to be a closed curve, and usually a circle, as this is the only curve whose properties are treated of in elementary geometry. 238 t BOOK VIII. 23y 3. Definition. The solid Ad bounded by a cylindrical surface and two parallel planes, ABD and abdy is called a cylinder; its plane surfaces, ABD, abd, are called its bases; the curved surface is some- times called its lateral surface; and the perpendicular distance be- tween its bases is its altitude. A cylinder whose base is a circle is called a eireular cylinder. 4. Definition. A right cylinder is one whose ele- ments are perpendicular to its base. 5. Definition. A right cylinder mth a circular base, as ABCa, is called a cylinder of revolution, be- cause it may be generated by the revolution of a rectangle A Opa about one of its sides, Oo, as an axis; the side Aa generating the curved surface, and the sides OA and oa generating the bases. The fixed side Oo is the axis of the cylinder. The radius of the base is called the radius of the cylinder. ^ .,4ifi PKOPOSITION I.— THEOKEM. 6. Every section of a cylinder made by a plane passing through an element is a parallelogram. Let Bb be an element of the cylinder Ac ; then, the section BbdD, made by a plane passed through Bb, is a parallelogram. 1st. The line Dd in which the cutting plane intersects the curved surface a second time is an element. For, if through any point D of this intersection a straight line is drawn paral- lel to Bb, this line by the definition of a cylindrical surface, is an element of the surface, and it must also lie in the plane Bd\ there- fore, this element, being common to both surfaces, is their inter- section. 2d. The lines BD and bd are parallel (VI. 25), and the elements Bb and Dd are parallel ; therefore, Bd is a parallelogram. 7. Corollary. Every section of a right cylinder made by a plane perpendicular to its base is a rectangle. •240 GEOMETRY. PROPOSITION II.— THEOREM. 8. The hoses of a cylinder are equal. Let BD be the straight line joining any two points of the perimeter of the lower base, and let a plane passing through BD and the element Bh cut the upper base in the line hd] then, BD = bd (6). Let A be any third point in the perimeter of the lower base, and Aa the corresponding element. Join AB^ AD, ah, ad. Then AB = ah and AD = ad (6) ; and the triangles ABD, ahd, are equal. Therefore, if the upper base be applied to the lower base with the line bd in coincidence with its equal BD, the triangles will coincide and the point a will fall upon A ; that is, any point a of the upper base will fall on the perimeter of the lower base, and consequently the perimeters will coincide throughout. Therefore, the bases are equal. 9. Corollary I. Any two parallel sections MPN, mpn, of a cylindrical surface Ah, are equal. For, these sections are the bases of the cylinder 3fn. 10. Corollary IL All the sections of a circular cylinder parallel to its bases are equal circles ; and the straight line joining the centres of the bases passes through the centres of all the parallel sections. This line is called the axis of the cylinder. 11. Definition. A tangent plane to a cylinder is a plane which passes through an element of the curved surface without cutting this surface. The element through which it passes is called the element of contact. r BOOK VIII. 241 PKOPOSITION III.— PROBLEM. 12. Through a given point, to pass a plane tangent to a given circular cylinder. 1st. When the given point is in the curved surface of the cylinder, in which case the element of ^^ c contact is given, since it must / ^ — 7^ / be the element passing through / s. V /^"^T/ / the given point. ^Ac-- Aj^s/y, // / Let the given point be a ^'-^^^'^v''"^i,^"7/b~ '/V / point in the element Aa. At /T / /^^^^// / A, in the plane of the base, i'4c" -/-.yfe.--// /s draw AT tangent to the base, ^"^"^"^^-^'T/o'"^^ and pass a plane lit through ^^"^^^z Aa and AT; this plane is tan- gent to the cylinder. For, let Pbe any point in this plane not in the element Aa, and through P pass a plane parallel to the base, in- tersecting the cylinder in the circle 3/iVand the plane Bt in the line MP. Let Q be the centre of the circle MN, and join QM. Since MP and 31 Q are parallel respectively to ^ T and A (VI. 25), the angle PMQ is equal to the angle TA 0, and PM is tangent to the circle MN at M; therefore, P lies without the circle MN and conse- quently without the cylinder. Hence the plane Pt does not cut the cylinder and is a tangent plane. 2d. When the given point is without the cylinder. Let P be the given point. Through P draw the straight line PT, parallel . to the elements of the cylinder, meeting the plane of the base in T. From Tdraw TA and TC tangents to the base (II. 90) ; through PT and the tangent TA pass a plane Bt, and through PT and TC pass a plane Ts. The plane Bt, passing through PTand the point A, must contain the element Aa, since ^a^is parallel to PT; and it is a tan- gent plane since it also contains the tangent A T. For a like reason the plane Ts is a tangent plane. 13. Corollary. The intersection of two tangent planes to a cylinder is parallel to the elements of the cylinder. 14. Scholium. Any straight line, drawn in a tangent plane and cutting the eleiient of contact, is tangent to the cylinder. . 21 Q 242 GEOMETRY. THE CONE. 15 DefinU.on. A conical surface is a curved surface generate* by a moving straight line which continually touches a given curve and passes through a given fixed point not in the plane of the onrve. Thus, if the straight line SA moves so IS continually to touch the given curve ABCD, and in all its positions, SB, SO, JSD, etc., passes through the given fixed point S, the surface S-ABCD is a conical surface. The moving line is called the generatrix ; the curve which it touches is called the directrix. Any straight line in the surface, as SB, which represents one of the positions of the generatrix, is called an element of the surface. The point S is called the vertex. If the generatrix is of indefinite length, as ASa, the whole surface" generated consists of two symmetrical portions, each of indefinite extent, lying on opposite sides of the vertex, sls S-ABCD and S-abcd, which are called nappes; one the upper, the other the lower nappe. 16. Definition. The solid S-ABCD, bounded by a conical surface and a plane ABD cutting the surface, is called a cone; its plane sur- face ABD is its base, the point S is its vertex, and the perpendicular distance SO from the vertex to the base is its altitude. A cone whose base is a circle is called a circular cone. The straight line drawn from the vertex ' of a circular cone to the centre of its base is the axis of the cone. 17. Definition. A right circular cone is a circular cone whose axis is perpendicular to its base, as S-ABCD. The right circular cone is also called a cone of revo- lution, because it may be generated by the revolution of a triangle, SAO, about one of its perpendicular Bides, SO, as an axis; the hypotenuse SA gener- ating the curved surface, and the remaining perpen- dicular T6' '^^ Since PP' passes through the centre o ^ ^\c\ of the circle, the distances Pa, Pb, Pc, are >blique lines from P to points a, b, c, equally /*/ BOOK VIII. 247 distant from the foot of the perpendicular, and are therefore equal (VI. 10). Hence, all the points of the circumference abed are equally- distant from the pole P. For .the same reason, they are equally dis- tant from the pole P'. 36. Corollary I. All the arcs of great circles drawn from a pole of a circle to points in its circumference, as the arcs Pa, Fb, Pc, are equal, since their chords are equal chords in equal circles. By the distance of two points on the surface of a sphere is usually understood the arc of a great circle joining the two points. The arc of a great circle drawn from any point of a given circle abc, to one of its poles, as the arc Pa, is called the polar distance of the given circle, and the distance from the nearest pole is usually understood. 37. Corollary II. The polar distance of a great circle is a quad- rant of a great circle ; thus PA, PB, etc., P'A, P'B, etc., polar dis- tances of the great circle ABCD, are quadrants; for, they are the measures of the right angles AOP, BOP, A OP', BOP', etc., whose vertices are at the centre of the great circles PAP', PBP', etc. In connection with the sphere, by a quadrant is usually to be understood a quadrant of a great circle. 38. Corollary III. If a point P on the surface of the sphere is at the distance of a quadrant from two points, B and C, of an arc of a great circle, it is the pole of that arc. For, the arcs PB and PC being quadrants, the angles POB and POC are right angles ; there- fore, the radius OP is perpendicular to each of the lines OB, OC, .and is consequently perpendicular to the plane of the arc BC (VI. 13) ; hence. Pis the pole of the arc BC. 39. Scholium. By means of poles, arcs of circles may be drawn upon the surface of a sphere with the same ease as upon a plane sur- face. Thus, by revolving the arc Pa about the pole P, its extremity a will describe the small circle abd ; and by revolving the quadrant PA about the pole P, the extremity A will describe the great circle ABD. If two points, B and C, are given on the surface, and it is required to draw the arc BC, of a great circle, between them, it will be neces- sary first to find the pole P of this circle ; for which purpose, take B and C as poles, and at a quadrant's distance describe two arcs on the surface intersecting in P. The arc PC can then be described with a pair of compasses, placing one foot of the compasses on P and 248 GEOMETRY. tracing the arc with the other foot. The opening of the compasses (distance between their feet) must in this case be equal to the chord of a quadrant ; and to obtain this it is necessary io know the radius of the sphere. ( PROPOSITION IX.— PROBLEM. 40. To find the radius of a given sphere. We here suppose that a material sphere is given, and that only measurements on the surface are possible. Fig. 3. 1st. With any point P (Fig. 1) of the given surface as a pole, and with any arbitrary opening of the compasses, describe a circum- ference ahc on the surface. The rectilinear distance Pa^ being the arbitrary opening of the compasses, is a known line. Take any three points, a, b, c, in this circumference, and with the compasses measure the rectilinear distances aft, he, ca. 2d. On a plane surface construct a triangle ahc (Fig. 2), with the three distances ah, he, ca, and find the centre o of the circle circum- scribed about the triangle (II; 87). The radius ao of this circle is the radius of the circle ahc of Fig. 1. 3d. With the radius ao as a side, and the known distance Pa as the hypotenuse, construct a right triangle aoP (Fig. 3). Draw aP' perpendicular to aP, meeting Po produced in P'. Then it is evident that PP', thus determined, is equal to the diameter of the given sphere, and its half PO is the required radius. 41. Definition A plane is tangent to a sphere when it has but one point in common with the surface of the sphere. 42. Definition. Two spheres are tangent to each ether when their surfaces have but one point in common. BOOK VIII. 249 )( PROPOSITION X.— THEOREM. ^ 43. A plane perpendicular to a radius of a sphere at its extremity ia tangent to the sphere. Let be the centre of a sphere, and ^ "\ let the plane MN be perpendicular to a / \ radius OA at its extremity A ; then, the [ *2 ] jv plane MN ia tangent to the sphere at the / Vn^ I "]"*-./ 7 point A. I ^vXJ/l,./ W For, taking any other point, as H, in / /2\. / the plane, and joining OH, the oblique ' L 1 £17 line OH is greater than the perpendicu- lar OA ; therefore the point H is without the sphere. Hence the plane MN has but the point A in common with the sphere, and is consequently tangent to the sphere. 44. Corollary. Conversely, a plane tangent to a sphere is perpen- dicular to the radius drawn to the point of contact. For, since every point of the plane except the point of contact is without the sphere, the radius drawn to the point of contact is the shortest line from the centre of the sphere to the plane, therefore it is perpendicular to the plane (VI. 9). 45. Scholium. Any straight line A T, drawn in the tangent plane through the point of contact, is tangent to the sphere. Any two straight lines, AT, A T\ tangent to the sphere at the same point A, determine the tangent plane at that point. PROPOSITION XI.— PROBLEM. 46. Through a given straight line without a given sphere, to pass a plane tangent to the sphere. Through the given straight line and the centre of the sphere, a plane can be passed which will cut the sphere in a great circle. Let the plane of the paper represent this plane ; let MN be the given line, the centre of the sphere, and aPcP' the great circle in which the plane passed through MN and the centre cuts the sphere. From any point M in the given line draw a tangent MaT to the great circle aPc, draw MO cutting the circumference of the circle 21** 250 GEOMETRY. in P and P'; let fall ao perpen- iicular to MO, and join Oa. Conceive the sphere to be gen- erated by the revolution of the semicircle PaP' about its diame- tei"t and let the tangent Ma re- volve with it. The line ao, per- pendicular to the axis, will gener- ate a small circle abc whose poles are P and P' ; the tangent MaT will generate a conical surface; and the portion of this surface between the point If and the circum- ference ahc is the surface of the cone whose vertex is M and whose base is the circle ahc. Every element of this cone as Mh is a tangent to the sphere, since it has the point 6, and that point only, in common with the sphere. Now, every plane which is tangent to this cone is also tangent to the sphere ; for any plane touching the cone in an element Mb, has the point b, and only the point 6, in common with the sphere. Therefore the solution of the present problem is reduced to passing a plane through the given line MN, tangent to the cone M-abe; which is done by Proposition VI. of this Book, observing the Scho- lium (24). Since there are two tangent planes to the cone, there are also two tangent planes to the sphere, passing through the given line 3IN. 47. Scholium. The indefinite conical surface generated by the revolution of the tangent MT is circumscribed about the sphere ; and the sphere is inscribed in this surface. The circle abc is called the circle of contact of the cone and sphere. PROPOSITION XII.— THEOREM. 48. The intersection of two spheres is a circle whose plane is perpen- dicular to the straight line joining the centres of the spheres, and ivhose centre is in that line. Through the centres and 0' of the two spheres, let any plane be passed, cutting the spheres in great circles which intersect each other in the points A and B ; the chord AB is bisected at C by the BOOK VIII. 251 line 00' at right angles (II. 34). If we now revolve the plane of these two circles about the line 00\ the circles will gener- ate the two spheres, and the point A will describe the line of intersection of their surfaces. Moreover, since the line AC will, during this revolution, remain perpendicular to 00\ it will generate a circle whose plane is perpendicular to 00' (VI. 15), and whose centre is C. 49. Scholium. Two spheres being given in any position whatever, if any plane is passed through their centres cutting them in two great circles, the spheres will intersect if these circles intersect, will be tangent to each other if these circles are tangent to each other, etc. For each of these positions, therefore, we shall have the same relations between the distance of the centres and the radii of the spheres, as have been established for the corresponding positions of two circles in Book II. PKOPOSITION XIII.— THEOKEM. 50. Through any four jjoints not in the same plane, a spherical sur- face can he made to pass, and bat one. Let A, B, C, D, be four given points not in the same plane. These four points may be taken as the vertices of a tetraedron ABCD. Let IE be the centre of the circle circum- scribed about the face ABC, and draw EM perpendicular to this face; every point in EM is equally distant from the points A, B and C (VI. 10). Let F be the centre of the circle circum- scribed about the face BCD, and draw FN perpendicular to this face ; every point in FN is equally distant from the points B, C and D. The two perpendiculars, EM and FN, intersect each other. For, let H be the middle point of ^0, and draw EH, FH. The lines EH and FH are each perpendicular to BC (II. 16); therefore, tlie i/^ 252 GEOMETRY. plane passed through EH and FH is perpendicular to BC (VI. 13) and consequently also to each of the faces ABGy BCD (VI. 47). Hence, the perpendiculars EM and FN lie in the Same plane EHF (VI. 50), and must meet unless they are parallel ; but they cannot be parallel unless the planes BCD and ABC are one and the same plajie, which is contrary to the hypothesis that the four given points are not in the same plane. The intersection of the perpendiculars EM and FN, being equally distant from A, B and C, and also equally distant from J5, C and D, is equally distant from the four points A, B, C and D, therefore, a spherical surface whose centre is and whose radius is the distance of from any one of these points, will pass through them all. Moreover, since the centre of any spherical surface passing through the four points A, B, C and D is necessarily in each of the perpen- diculars E3I, FN, the intersection is the centre of the only spheri- cal gurface that can be made to pass through the four given points. 51. Corollary I. The four perpendiculars to the planes of the faces of a tetraedron, erected at the centres of the faces, meet in the same point. 52. Corollary II. The six planes, perpendicular to the six edges of a tetraedron at their middle points, intersect in the same point. PROPOSITION XIV.— THEOREM. 53. A sphere may be inscribed in any given tetraedron. Let ABCD be the given tetraedron. Let the planes OAB, OBC, OAC, bisect the diedral angles at the edges AB, BC, AC, re- spectively. Every point in the plane OAB is equally distant from the faces ABC and ABD ^/r:::rr_.Jv-l----=)c'^ (VI. 55); every point in the plane OBC is equally distant from the faces ABC and DBC; and every point in the plane OAC is equally distant from thQ faces ABC and ADC; there- fore, the common intersection, 0, of these three planes is equally distant from the four faces of the tetraedron ; and a sphere described BOOK VIII. 253 with as a centre, and with a radius equal to the distance of from any face, will be tangent to each face, and will be inscribed in the tetraedron. 54. Corollary. The six planes, bisecting the six diedral angles of a tetraedron, intersect in the same point. SPHEEICAL ANGLES. 65. Definition. The angle of two eiirves passing through the same point is the angle formed by the two tangents to the curves at that point. This definition is applicable to any two intersecting curves in space, whether drawn in the same plane or upon a surface of any kind. Thus, in a plane, two circumferences inter- secting in a point A, make an angle equal to the angle TAT' formed by their tangents at A. In this case, the angle is also equal to the angle OAO' formed by the radii of the two circles drawn to the common point. In like manner, on a sphere, the angle formed by any two intersecting curves, AB, AB\ is the angle TAT\ formed by the lines AT, AT', tangents to the two curves, respectively, at their common point A. PKOPOSITION XV.— THEOEEM. 56. The angle of two intersecting curves on the surface of a sphere is equal to the diedral angle between the planes passed through the centre of the sphere and the tangents to the two curves at their point of in' ter^section. 254 GEOMETRY. Let the curves, AB and AB\ on the surface of a sphere whose centre is 0, in- tersect at A, and let A T and A T' be the tangents to the two curves, respectively. Since J. T and AT' do not cut the curves at 4> they do not cut the surface of the sphere, and are therefore tangents to the sphere. Hence they are both perpendicular to the radius OA drawn to the common point of contact, and consequently the angle T'ATy which is the angle of the two curves (55), measures the diedral angle of the planes OJjT, OAT'j passed through the radius OA and each of the tangents. \A>' PROPOSITION XVI.— THEOREM. 57. The angle of two arcs of great circles is equal to the angle of their planes, and is measured by the arc of a gi^eat circle described from its vertex as a pole and included between its sides {^produced if ne- Let ^5 and AB' be two arcs of great circles, ^ Tand AT' the tangents to these arcs at A, the centre of the sphere. The planes passing through the centre and the tangents AT, AT', are in this case the planes of the curves AB, AB', themselves ; consequently the angle BAB', or TAT', is equal to the angle of these planes (56), the edge of this "angle being the common diameter 'aod. Now let CC be the arc of a great circle described from Jl as a pole and intersecting the arcs AB, AB' (produced if necessary), in C and C". The radii OC and OC are perpendicular to AO, since the arcs AC, AC, are quadrants (37) ; therefore, the angle COG' is also equal to the diedral angle AO, or to the angle BAB', and it is measured by the arc CC. 58. Corollary. Any great circle arc J. C", drawn through the pole of a given great circle CC, is perpendicular to the circumference CC. For, the pole A being in the diameter A OD perpendicular to BOOK VIII. 255 the plane of CC, the plane of AC is perpendicular to the plane of CC (VI. 47), and hence the angle C is a right angle. Conversely, any great circle arc C'A perpendicular to the arc GC must pass through the pole A of CC. 59. Scholium. If it is required to draw a great circle B'C perpen- dicular to a given great circle CCE, through a given point B\ we have only to find the pole iV of the required arc by describing, from -B' as a pole and at a quadrant's distance, an arc cutting CCE in N; then, from JV as a pole, the perpendicular B'C can be de- scribed. SPHERICAL POLYGONS AND PYRAMIDS. 60. Definition. A spherical polygon is a portion .,^—~~~~~^c of the surface of a sphere bounded by three or y^ / /\ more arcs of erreat circles, as AB CD, ^ \ *<. / Since the planes of all great circles pass \ / / \ / through the centre of the sphere, the planes of \l{[ \J the sides of a spherical polygon form, at the cen- ^ ^ tre 0, a polyedral angle of which the edges are the radii drawn to the vertices of the polygon, the face angles are angles at the centre measured by the sides of the polygon, and the diedral angles are equal to the angles of the polygon (57). Since in a polyedral angle each face angle is assumed to be less than two right angles, each side of a spherical polygon will be as- sumed to be less than a semi-circumference. A spherical polygon is convex when its corresponding polyedral angle at the centre is convex (VL 67). A diagonal of a spherical polygon is an arc of a great circle join ing any two vertices not consecutive. 61. Definition. A spherical triangle is a spherical polygon of three sides. It is called right angled, isosceles, or equilateral, in the same cases as a plane triangle. 62. Definition. A spherical pyramid is a solid bounded by a spheri- cal polygon and the planes of the sides of the polygon ; as 0-ABCD. The centre of the sphere is the vertex of the pyramid ; the spherical polygon is its base. GEOMETRY. 63. Symmetrical spherical triangles and polygons. Let ABC be a spherical triangle, and the centre of the sphere. Drawing the radii OAy OBj OC, we form the triedral angle 0-ABC, at the centre. The sides AB, BC, AQ of the triangle are respectively the measures of the face angles AOB,BOC,AOC, of the triedral angle ; and the angles A, B, (7, of the triangle are respec- tively equal to the diedral angles at the edges OA, OB, OC, of the triedral angle (57). If the radii AO, BO, CO, are produced to meet the surface of the sphere in the points A', B', C, and if these points are joined by arcs of great circles A'B', B'C, A'C, a triedral angle O-A'B'C is formed symmetrical with 0-ABC (VI. 68), and its corresponding spherical triangle A'B' C is symmetrical with ABC. The spherical pyramid O-A'B'C is also symmetrical with the spherical pyramid 0-ABC In the same manner, we may form two symmetrical polygons of any number of sides, and corresponding symmetrical pyramids. 64. Two symmetrical spherical triangles, or polygons, are still called symmetrical in whatever position they may be placed on the surface of the sphere. If we place the symmetri- ^ cal triangles of the preceding figure with the ver- tices A' and B' in coincidence with their homolo- gous vertices A and B, their third vertices C and C will lie on opposite sides of the arc AB. In this position, it is apparent that the order of ar- rangement of the parts in one triangle is the reverse of that in the other, and that, in general, two symmetrical spherical triangles cannot be made to coincide by superposition. 65. There is, however, one exception to the last remark, namely, the case of symmetrical isosceles tri- angles. For, if ABC is an isosceles spherical triangle and AB = AC, then, in its symmetrical triangle we have A'B' = A'C, and consequently AB = A'C, AC = A'B', and since BOOK VIII 257 the angles A and A! are equal, if AB be placed on A! C\ AC will fall on its equal A'B' and the two triangles wijl coincide throughout. 66. In consequence of the relation established between polyedral angles and spherical polygons, it follows that from any property of polyedral angles we may infer an analogous property of spherical polygons. Reciprocally, from any property of spherical polygons we may infer an analogous property of polyedral angles. The latter is in almost all cases the more simple mode of proce- dure, inasmuch as the comparison of figures drawn on the surface of a sphere is nearly if not quite as simple as the comparison of plane figures. 67. Definition. If from the vertices of a spherical triangle as poles, arcs of great circles are described, these arcs form by their intersection a second triangle which is called the polar triangle of the first. Thus, if Ay B and C are the poles of the arcs of great circles, B' C\ A'C\ and A'B\ respec- tively, A'B'C is the polar triangle of ABC. Since all great circles, when completed, intersect each other in two points, the arcs B'C'^ A! C\ A!B\ if produced, will form three other triangles; but the triangle which is taken as the polar tri- dRgle is that whose vertex J.', homologous to A^ lies on the same siJe of the arc BC sls the vertex A ; and so of the other vertices. ^ PKOPOSITION XVII.— THEOKEM. 68. If A'B'C is the polar triangle of ABC, then, reeiprocally, iBC is the polar triangle of A'B'C, For, since A is the pole of the arc B'C, the point ^' is at a quadrant's distance from A ; and since C is the pole of the arc A'B', the point B' is at a quadrant's distance from C; therefore, B' is the pole of the arc AC (38). In the same man- ner, it is shown that A' is the pole of the arc BC, and C the pole of the arc AB. Moreover, A and A' are on the same side of B'C, B and B' on the same side of A'C, 22* R 258 GEOMETRY. C aud C on the same side of A'B' \ therefore, ABC is the polar triangle of J 'jB'C. . y^JtjK^^^^^^^^^ PROPOSITION XVIIL— THEOREM. dD. In two polar triangles, each angle of one is measured by the sup- plement of the side lying opposite to it in the other. Let ABC and A'B'C be two polar triangles. Let the sides AB and A C, produced if necessary, meet the side B'C in the points h and c. The vertex A being the pole of the arc 6c, the angle A is measured by the arc he (bl). Now, B' being the pole of the arc Ac and C the pole of the arc Ah, the arcs B'c and C'h are quadrants ; hence we have B' C -\- be = B'c + C'b :^ a semi-circumference. Therefore be, which measures the angle A, is the supplement of the BideJ5'0' (IL 55). In the same manner, it caa be shown that each angle of either triangle is measured by the supplement of the side lying opposite to it in the other triangle. 70. Scholium I. Let the angles of the triangle ABC be denoted by A, B and C, and let the sides opposite to them, namely, BC, AC and AB, be denoted by a, b and c, respectively. Let the cor- responding angles and sides of the polar triangle be denoted by A', B', C, a', b' and c'. Also let both angles and sides be expressed in degrees (I]. 54). Then, the preceding theorem gives the following relations : ^ + a' = J5 + 6' = -f c' = 180°, ^' + a = ^' + 6 = C" -f c = 180°, also A — a := A' — a', etc. 71. Scholium II. Two triedral angles at the centre of the sphere, corresponding to two polar triangles on the surface, are called s^ip- plementary triedral angles; for, it follows from the preceding BOOK VIII. 259 theorem, and from the relation between any spherical polygon and its corresponding polyedral angle (60), that the diedral angles of either of these triedral angles are respectively the supplements of the opposite face angles of the other. . + PROPOSITION XIX.— THEOREM. 72. Two triangles on the same sphere are either equal or symmetrical^ when two sides and the included angle of one are respectively equal to two sides and the included angle of the other. In the triangles ABC ojiH DEF, let the angle A be equal to the angle D, the side AB equal to the side DE, and the side A C equal to side DF. 1st. When the parts of the two triangles are in the same order, ABC can be applied to DEF, as in the corresponding case of plane triangles (I. 76), and the two triangles will coincide ; therefore, they are equal. 2d. When the parts of the two tri- angles are in inverse order, let DE'Fhe the symmetrical triangle of DEFy and therefore having its angles and sides equal, respectively, to those of DEF. Then, in the triangles ABC and DE'F, we ^ shall have the angle BAC equal to the angle E'DF, the side AB to the side DE\ and the side AC to the side DF, and these parts arranged in the same order in the two triangles; therefore, the triangle ABC is equal to the triangle DE'F, and consequently symmetrical with DEF. 73. Scholium. In this proposition, and in those which follow, the two triangles may be supposed on the same sphere, or on two equal spheres. PROPOSITION XX.— THEOREM. 74. Two triangles on the same sphere are either equal or symmetrical, when a side and the two adjacent angles of one are equal respectively to VL Kide and the two adjacent angles of the other. 260 GEOMETRY. For, one of the triaDgles may be applied to the other, or to ita symmetrical triaugle, as in the corresponding case of plane tri- angles (I. 78). V PROPOSITION XXI.— THEOREM. 75 Two triangles on the same sphere are either equal or symmetrical^ when the three sides oj one are respectively equal to the three sides of the other. For, their corresponding triedral angles at the centre of the sphere are either equal or symmetrical (VI. 71). PROPOSITION XXII.—THEOREM. 76. If two triangles on the same sphere are mutually equiangular, they are also mutually equilateral; and are either equal or sym- metrical. Let the spherical triangles M and N be mutually equian- gular. Let M' be the polar triangle of if, and N' the polar triangle of N. Since Jf and N are mutually equiangular, their polar triangles M' and N' are mutually equi- lateral (69) ; therefore, by the preceding proposition, the triangles M' and N' are mutually equiangular. But M' and N' being mutually equiangular, their polar triangles M and iVare mutually equilateral (69). Consequently, M&nd N are either equal or symmetrical (75). 77. Scholium. It may seem to the student that the preceding property destroys the analogy which subsists between plane and spherical triangles, since two mutually equiangular plane triangles are not necessarily mutually equilateral. But in the case of spheri- cal triangles, the equality of the sides follows from that of the angles only upon the condition that the triangles are constructed upon the same sphere or on equal spheres ; if they are constructed on spheres of different radii, the homologous sides of two mutually equiangular triangles will no longer be equal, but will be proportional to the radii of the sphere ; the two triangles will then be similar, as in the case of plane triangles. BOOK VIII. 261 d^r^ PROPOSITION XXIII.— THEOREM. 78. In an isosceles spherical triangle^ the angles opposite the eqtud sides are equal. In the spherical triangle ABC^ let AB=^AC; then, B= a For, draw the arc AD of a great circle, from the vertex A to the middle of the base EC. The tri- angles ABD and A CD are mutually equilateral, and in this case are symmetrical (75); therefore B=C. 79. Corollary. Since the triangles ABD and A CD are mutually equiangular, we have the angle BAD equal to the angle CAD^ and the angle ADB equal to the adjacent angle ADC; therefore, the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base is perpendicular to the base and also bisects the vertical angle. 80. Scholium. This proposition and its corollary may also be proved by applying the isosceles triangle to its symmetrical tri- angle (65). PROPOSITION XXIV.— THEOREM. 81. If two angles of a spherical triangle are equal, the triangle is In the triangle ABC let B =^ C; then, AB = AC. For, letA'B'C be the polar triangle of ^^C. Then, the sides A'B' and A'C are equal (69), and therefore the angles B' and C are equal (78). But since the angles B' and C are equal in the triangle A'B'C\ the sides AB and A C are equal in its polar triangle ABC, 262 " GEOMETRY. PEOPOSITION XXV.— THEOREM. 82. Any side of a spherical triangle is less than the sum of the other two. Let ABO be a spherical triangle; then, any side, as A C, is less than the sum of the other two, AB and BC. 1 For, in the corresponding tried ral angle / formed at the centre of the sphere, we / ,' ^*- have the angle AOG less than the supa of in-'^"^^' the angles AOB and BOG (VI. 69); and since the sides of the triangle measure these angles, respectively, we have^C<^^ + 5a c 83. Oorollary. Any side, AB^ of a spherical f/^ >v polygon ABODE is less than the sum of all the a^ ^-^ other sides. PROPOSITION XXVI.— THEOREM. 84. In a spherical triangle, the greater side is opposite the greater angle ; and conversely. 1st. In the triangle AB Csuppose AB 0^ AOB', ^ then, A0'> AB. For, draw the arc BD making the angle DBO= DOB; then, the triangle BDO is isosceles (81), and D0= DB. Adding DA to each of these equals we have A0= DB -\- DA. But DB-\-DA> AB (82) ; therefore, AO>AB. 2d. Conversely, in the triangle J.^ (7 suppose A0'> AB; then ABO> AOB. For, if ABO were equal to AOB, AO would be equal to AB (81), which is contrary to the hypothesis ; and if ABO were less than AOB, J. would be less than AB, which is also con- trary to the hypothesis; therefore, ABO must be greater than AOB. PEOPOSITION XXVIL— THEOREM. 85. If from the extremities of one side of a spherical triangle two area of great circles are drawn to a point within the triangle, the sum oj these arcs is less than the sum of the other two sides of the triangle. BOOK VIII 263 Id. the spherical triangle ABC, let the arcs BD and CD be drawn to any point D within the triangle; then, DB -[■ DC < AB ^ AC. For, produce BD to meet AC m E; then we have DCB t u t \ PROPOSITION XXVIII.— THEOREM. 86. The sum of the sides of a convex spherical polyg&nrisfUea than the circumference of a great circle. For, the sum of the face angles of the corresponding polyedral angle at the centre of the sphere is less than four right angles (VI. 70). ^, ^ V, ^ r PROPOSITION "XXIX:.— THEOREM. 87. The sum of the angles of a spherical triangle is greater than two, and less than six, right angles. For, denoting the angles of a spherical triangle by A, B, C, and the sides respectively opposite to them in its polar triangle by a',h',c', we have (70), ^ = 180° — o', jB = 180° — 6', C==180°— c', the sum of which is J. + jB + C= 540° — (a' + 6' -f c'). But a' + 6' + c' < 360° (86); therefore, A -^ B -\- C> 180°; that is, the sum of the three angles is greater than two right angles. Also, since each angle is less than two right angles, their sum is less than six right angles. 88. Corollary. A spherical triangle may have two or even three right angles ; also two or even three obtuse angles. 89. Definitions. If a spherical triangle ABC has two right angles, jB and C, it is called a bi-rectangular triangle ; and since the sides AB and A C must each pass through the pole of BC (58), the vertex A is that pole, and therefore AB and A C are quadrants. 264 GEOMETRY. If a triangle has three right angles it is called a tri-rectangular triangle; each of its sides is a quadrant, and each vertex is the pole of the opposite side. Three planes passed through the centre of the sphere, each perpendicular to the other two (VI. 48), divide the surface of the sphere into eight tri-rectangular triangles, ABC, A'BC, etc. RATIO OF THE SURFACES AND VOLUMES OF SPHERICAL FIGURES. 90. Definitions. A tune is a portion of the surface of a sphere included between two semi-circumferences of great circles ; as AMBNA. A. spherical ungula, or wedge, is a solid bounded by a lune and the two semicircles which intercept the lune on the surface of the sphere ; as the solid ABMANB. The common diameter AB, of the semi- circles, is called the edge of the ungula; the lune is called its base. 91. Definition. The excess of the sum of the angles of a spherical triangle over two right angles is callod the spherical excess. If the angles of a spherical triangle ABC are denoted by J., B and C, and its spherical excess by E, and if a right angle is the unit employed in expressing the angles, we shall have E=A-\^B-\-C—2. PROPOSITION XXX.— THEOREM. 92. Two symmetrical spherical triangles are equivalent. Let ABC and A'B'C be two symmetrical triangles with their homologous vertices diametrically opposite to each other on the sphere. Let P be the pole of the small circle which passes through the three points A, B and C. The great circle arcs PA, PB, PC, are equal (36). BOOK VIII, 265 Draw the diameter POP' and the great circle arcs F'A\ P'B\ P'C] these arcs being equal, respectively, to PJ., PB, P(7, are also equal to each other. The triangles PAB, P'A'B', are mu- tually equilateral, and also isosceles; therefore, they are superposable (65) and are equal in area. For the same reason the triangle PA C is equivalent to the tri- angle P'^'C, and PBC is equivalent to P'B'C. Therefore the triangle ABC, which is the sum of the triangles PAB, PAC and PBC, is equivalent to its symmetrical triangle A'B'C which is the sum of the triangles P'A'B', P'A'C and P'B'C. If the pole P should fall without the triangle ABC, the triangle would be equivalent to the sum of two of the isosceles triangles diminished by the third ; but as the same thing would occur for the symmetrical triangle, the conclusion would be the same. 93. Corollary I. If the arcs of two great circles, AC A', BCB', intersect on the sur- face of a hemisphere, the sum of the oppo- site triangles ACB, A' CB', is equivalent to a lune whose angle is the angle ACB, formed by the great circles. For, completing the great circle BCB' C\ the triangles A' CB', ACB, are symmetri- cal, and therefore equivalent. Hence, the sura of ACB and A' CB' is equivalent to the sum of ACB and ACB, that is, to the lune ACBCA, whose angle is the angle ACB. 94. Corollary II. The reasoning employed in the demonstration of the theorem may be applied also to the pyramids whose bases are two symmetrical triangles. Hence, two symmetrical spherical triangu- lar pyt amids are equivalent. Also by the reasoning in Corollary I. we infer that the sum of the volumes of two spherical triangular pyramids the sum of whose bases is equivalent to a lune, is equal to the volume of the ungula whose base is that lune. 23 266 GEOMETRY. PROPOSITION XXXI.— THEOREM. 95. A lune is to the surface of the sphere as the angle of the lune is four right angles. ^ Let ANBMA be a lune, and let MNP be the great circle whose poles are the ex- tremities of the diameter AB. Let the circumference of the circle MNP be divided into any number of equal parts J/a, ahy etc. ; and let planes be passed through the diameter AB and each of the points of division. The whole surface of the sphere will evidently be divided into equal lunes of which tbel given lune will contain the same number as there are parts in the] arc MN. Hence, whether the number of the parts in MN and the number of the parts in the whole circumference 3fNP, are commen- surable or incommensurable, the ratio of the lune ANBMA to th< surface of the sphere is the same as the ratio of the arc 3IN to th< circumference MNP; or, since MN is the measure of the angle of] the lune, and the circumference MNP is the measure of four right angles, the lune is to the surface of the sphere as the angle of the lune is to four right angles. 96. Corollary I. Two lunes, on the same or on equal spheres, are] to each other as their angles. 97. Corollary II. If we denote the surface of the tri-rectangularj triangle by T, the surface of the whole sphere will be ST (89) ;i therefore, denoting the surface of the lune by L and its angle by A,\ the unit of the angle being a right angle, we have — = ^, whence i = ^ X 2^. If, further, we take the tri-rectangular triangle as the unit of sur- face in comparing surfaces on the same sphere, we shall have L = 2A; that is, a right angle being the unit of angles y and the tri-rectangular ■"/ BOOK VIII. 267 triangle the unit of spherical surfaces, the area of a lune is expressed by twice its angle. 98. Corollary III. The tri-rectangular spherical pyramid (that whose base is the tri-rectangular triangle) being taken as the unit of volume, the same reasoning may be employed to prove that the volume of an ungula will be expressed by twice its angle. \ PKOPOSITION XXXII.— THEOKEM. 99. The area of a spherical triangle is equal to its spherical excess (the right angle being the unit of angles and the tri-rectangular triangle the unit of areas). For, let ABC be a spherical triangle. Complete the great circle ABA'B\ and produce the arcs AC and BG to meet this circle in A' and B\ 4- We have, by the figure, ABC+A'BC = lune A ABC-\- AB'C = lune A and by (93) ABC+A'B'C= lune a The sum of the first members of these equations is equal to twice the triangle ABC, plus the four triangles ABC, A'BC, AB'C, A'B'C, which compose the surface of the hemisphere. With the system of units -adopted, the surface of the hemisphere is expressed by 4; therefore, denoting the area of the triangle ABC by K, and the numerical measures of its angles by A, B and C, we have (97), 2^-1-4 = 2^ + 2^ + 20, whence K= A -\- B -{- C — 2 = spherical excess. 100. Corollary. The same reasoning, in connection with (94) and (98), may be employed to prove that, if Fis the volume of a spheri- cal triangular pyramid whose base is the spherical triangle ABC, and if the unit of volume is the volume of the tri-rectangular spheri* cal pyramid, we shall have V= A^ B ^ 0—2. 268 GEOMETRY. 101. Scholium. It must not be forgotten that the preceding results are merely the expression of the ratios of the figures considered to the adopted units. For example, suppose the angfes of a spherical triangle are given in degrees as follows : A = 80°, B = 100°, C = 120° ; then, reducing them to the right angle as the unit, IpJ^'"^' ^_80_ 100 120_2^4 -^ ^^ 90 ^ 90 ^ 90 3 therefore, the area of this triangle is f of the area of the tri-rectangu- lar triangle. Also, the volume of the spherical pyramid of which this triangle is the base is ^ of the volume of the tri-rectangular spherical pyramid. Hence, also, it follows that the volumes of two triangular spherical pyramids are to each other as the areas of their bases. PROPOSITION XXXIII.— THEOREM. 102. The area of a spherical polygon is measured by the sum of its angles minus the product of two right angles multiplied by the number of sides of the polygon less two. ^ Let ABODE be a spherical polygon. From /^^^'^^ any vertex, as J., draw the diagonals A C, AD ; l/i^—^../\ the polygon will be divided into as many tri- K y jj angles as there are sides less two. The surface \__ -^ of each triangle is measured by the sum of its angles minus two right angles ; and the sum of all the angles of the triangles is equal to the sum of the angles of the polygon ; therefore the surface of the polygon is measured by the sum of its angles minus two right angles multiplied by the number of triangles, that is, by the number of sides of the polygon less two. 103. Corollary I. Denoting the number of sides of the polygon by n, the sum of its angles by S, and its area by jBT, then, with the adopted system of units, we have K=S—2(n — 2) = /S— 2n + 4. 104. Corollary II. The tri-rectangular pyramid being taken as the unit of volume,- the volume of any spherical pyramid will have the BOOK VIII. 269 same numerical expression as the area of its base ; that is, the volume of a spherical pyramid is to the volume of the tri-rectangular pyramid as the base of the pyramid As to the tri-rectangular triangle. Now the volume of the tri-rectangular pyramid is one-eighth of the volume of the sphere, and the tri-rectangular triangle is one-eighth of the surface of the sphere ; therefore, the volume of a spherical pyramid is to the volume of the sphere as its base is to the surface of the sphere. SHORTEST LINE ON THE SURFACE OF A SPHERE BETWEEN TWO POINTS. PROPOSITION XXXIV.— THEOREM. "^ 1 05. The shortest line that can be drawn on the surface of a sphere between two points is the arc of a great circle, not greater tlian a semi- circumference, joining the two pohits. Let AB be an arc of a great circle, less than a semi-circumference, joining any two points A and B of the surface of a sphere ; and let C be any arbitrary point taken in that arc. Then we say that the shortest line from A to B, on the surface of the sphere, must pass through G. From A and B as poles, with the polar dis- tances J. C and BC, describe circumferences on the surface; these circumferences touch at G and lie wholly without each other. For, let M be any point in the circumference whose pole is A, and draw the arcs of great circles AM, BM, forming the spherical triangle AMB. We have, by (82), AM -f BM > AB, and subtracting from the two members of this inequality the equal arcs AM and A G, we have BM ^ BG; therefore, M lies without the circumference whose pole is B. Now let AFGB be any line from A to B, on the surface of the .sphere, which does not pass through the point C, and which therefore cuts the two circumferences in different points, one in F, the other in G. Whatever may be the nature of the line AF, an equal line can be drawn from A to G', for, li AG and AF be conceived to be drawn on two equal spheres having a common diameter passing through A, and therefore having their surfaces in coincidence, and if one of 23* 270 GEOMETRY. these spheres be turned upon the common diameter as an axis, thft point A will be fixed and the point F will come into coincidence with C; the surfaces of the two spheres continuing to coincide, the line AF will then lie on the common surface between A and C. For the same reason, a line can be drawn from B to C, equal to BG. There- forCj a line can be drawn from A to B, through (7, equal to the sum of AF siiid BG, and consequently less than any line AFGB that does not pass through C. The shortest line from ^ to -B therefore passes through C, that is, through any, or every, point in AB ; consequently it must be the arc AB itself. A BOOK IX. MEASUREMENT OF THE THREE ROUND BODIES. THE CYLINDER. 1. Definition. The area of the convex, or lateral, surface of a cylinder is called its lateral area. 2. Definition. A prism is inscribed in a cylinder when its bases are inscribed in the bases of the cylinder. If a polygon ABCDEF is inscribed in the base of a cylinder, planes passed through the sides of the polygon, parallel to the elements of the cylinder, intersect the cylinder in parallelograms, ABB'A'y etc. (VIII. 6), which evidently determine a prism inscribed in the cylinder. 3. Definition. A prism is circumscribed about a cylinder when its bases are circumscribed about the bases of the cylinder. If a polygon ABCD is circumscribed about the base of a cylinder, planes passed through the sides of the polygon, parallel to the elements of the cylinder, will evidently contain the elements, aa'y bb\ etc., drawn at the points of contact, and be tangent to the cylinder in these elements. The intersection of these planes with the plane of the upper base of the cylinder will therefore determine a polygon A'B'C'D\ equal to ABCD, circumscribed about the upper base, and a prism will be formed which is circumscribed about the cylinder. 271 272 GEOMETRY. 4. Definition. A right section of a cylin- der is a section made by a plane perpen- dicular to its elements ; as abcdef. The intei-section of the same plane with an 'inscribed or circumscribed prism is a right section of the prism. 5. Definition. Similar cylinders of revolution are those which are generated by similar rectangles revolving about homologous sides. PROPOSITION I.— THEOREM. 6. A cylinder is the limit of the inscribed and circumscribed prisms, the number of whose faces is indefinitely increased. Let any polygon abed be inscribed in the base of the cylinder ac' and at the vertices of this polygon let tangents be drawn to the base of the cylinder forming the circumscribed poly- gon ABCD. Upon these polygons as bases let prisms be formed, inscribed in, and circumscribed about, the cylinder. We shall assume, as evident, that the convex surface of the cylinder is greater than that of the inscribed prism and less than that of the circumscribed prism.* Suppose the arcs ab, be, etc., to be bisected and polygons to be formed having double the number of sides of the first; and upon these as bases suppose prisms to be constructed, inscribed and circum- scribed, as before; and let this process be repeated an indefinite number of times. The difference between the convex surface of the inscribed prism and that of the corresponding circumscribed prism will continually diminish and approach to zero as its limit. There * A proof, however, can be given analogous to that of (V. 32). BOOK IX. 273 fore these convex surfaces themselves approach to the convex surface of the cylinder as their common limit. At the same time, it is evident that the volumes of the inscribed and circumscribed prisms approach to the volume of the cylinder as their common limit. 7. Scholium. In the preceding demonstration, the base of the cylin- der is not required to be a circle, but may be any closed convex curve. We have, however, tacitly assumed that the curve is the limit of the perimeters of the inscribed and circumscribed polygons ; a principle which was rigorously proved in the case of regular poly- gons inscribed in a circle. PEOPOSITION II.— THEOKEM. 8. The lateral area of a cylinder is equal to the product of the perimeter of a right section of the cylinder by an element of the surface. Let ^^CD^i^bethe base and^^' any element of a cylinder, and let the curve abcdef be any right section of the surface. Denote the perimeter of the right section by P, the element A A ' by E, and the lat- eral area of the cylinder by 8. Inscribe in the cylinder a prism ABCDEF A' of any number of faces. The right section, abcdef of this prism will be a polygon inscribed in the right section of the cylinder formed by the same plane. Denote the lateral area of the prism by s, and the perimeter of its right section by p ; then, the lateral edge of the prism being equal to E, we have (VII. 16), 8=pXE. Let the number of lateral faces of the prism be indefinitely increased, as in the preceding proposition ; then s approaches indefinitely to /S as its limit, and p approaches to P; therefore, at the limit, we hav*- (V.31), S=PXE. 23** s 274 GEOMETRY. 9. Corollary I. The lateral area of a right cylinder is equal to the product of the perimeter of its base by its altitude. 10. Corollary II. Let a cylinder of revolution hh generated by the rectangle whose sides are R and H revolving about the side H. Then, R is the radius of the base, and H is the altitude of the cylinder. The perimeter of the base is ItzR (V. 40), and hence, for the lateral area S we have the expression S = 2t:R.H. The area of each base is t:R'^ (V. 43) ; hence the total area T of the cylinder of revolution, is expressed by T= 27:R.H-\- 27:R^ = 1tzR(^H-\- R). 11. Corollary III. Let S and s de- note the lateral areas of two similar cylinders of revolution (4) ; T and t their total areas; R and r the radii of their bases; H and h their alti- tudes. The generating rectangles be- ing similar, we have (III. 12) therefore, H R H+R^ h" r~ h-\-r ' 8 27: RH R H H^ R' 8 "" 27rrA ~ r ' h~ h' ~ r^ * T 27tR{H-\-R) R H-\-R H' R' t. 2;rr(A + r) r h 4- r h' r' That is, the lateral areas, or the total areas, of similar cylinders of revo- lution are to each other as the squares of their altitudes, or as the squares of the radii of their bases. BOOK IX 275 PKOPOSITION III.— PROBLEM. 12. The volume of a cylinder is equal to the product of its hose by iU altitude. Let the volume of the cylinder be denoted by F, its base by B, and its altitude by If. Let the volume of an inscribed prism be de- noted by V'i and its base by B' \ its altitude will also be H, and we shall have (VII. 38) V' = B' XS. Let the number of faces of the prism be indefinitely increased, as in (8) ; then the limit of V is F, and the limit of B' iaB; therefore (V. 31), V=BXH. 13. Corollary I. Let V be the volume of a cylinder of revolution, B the radius of its base, and H its altitude ; then the area of its base is tzB^ (V. 43) ; and therefore V=r:Bm, 14. Corollary II. Let V and v be the volumes of two similar cyl- inders of revolution ; B and r the radii of their bases ; H and h their altitudes; then, the generating rectangles being similar, we have JB[_B h r and F_ TtBm B' H H' B' V Tcr'h \r' ' h ~ h' ~ r' that is, the volumes of similar cylinders of revolution are to each other as the cubes of their altitudes, or as the cubes of their radii. 276 GEOMETKY I THE CONE. 15. Definition. The area of the convex, or lateral, surface of a cone is called its lateral area. 1§. Definition. A pyramid is inscribed in a cone when its base is inscribed in the base of the cone, and its .vertex coincides with the vertex of the cone. If a polygon ABCD is inscribed in the base of a cone and planes are passed through its sides and the vertex 8 of the cone, these planes intersect the convex surface of the cone in right lines (VIII. 18) and determine a pyramid inscribed in the cone. 17. Definition. A pyramid is circum- scribed about a cone when its base is cir- cumscribed about the base of the cone, and its vertex coincides with the vertex of the cone. If a polygon ABCD is circumscribed about the base of a cone, its points of con- tact with the base being a, b, c, d, and planes are passed through its sides and the vertex S of the cone, these planes will be tangent to the cone m the elements Sa, Sb, etc. (VIII. 21), and will determine a pyramid circumscribed about the cone. 18. Definition. A truncated cone is the portion of a cone included between its base and a plane cutting its convex surface. When the cutting plane is parallel to the base, the truncated cone is called a frustum of a cone; as ABCD-abcd. The altitude of a frustum is the perpendicular distance Tt between its bases. If a pyramid is inscribed in the cone, the cutting plane determines a truncated pyramid inscribed in the truncated cone; and if a pyramid is circumscribed about BOOK IX. 277 the cone, the cutting plane determines a truncated pyramid circum- scribed about the truncated cone. 19. Definition. In a cone of revolution, as JS-ABC, generated by the revolution of the right triangle SAO about the axis SO, all the elements, SA, SB, etc., are equal ; and any ele- ment is called the slant height of the cone. In a cone of revolution, the portion of an ele- ment included between the parallel bases of a frustum, as Aa, or Bb, is called the slant height of the frustum. 20. Definition. Similar cones of revolution are those which are generated by similar right triangles revolving about homologous sides. PROPOSITION IV.— THEOEEM. 21. A cone is the limit of the inscribed and circumscribed pyramids, the number of whose faces is indefinitely increased. The demonstration is precisely the same as that of Proposition I., substituting a cone for a cylinder, and pyramids for prisms. 22. Corollary. A frustum of a cone is the limit of the frustums of the inscribed and circumscribed frustums of pyramids, the number of whose faces is indefinitely increased. V PROPOSITION v.— THEOREM. 23. The lateral area of a cone of revolution is equal to the product of the circumference of its base by half its slant height. Let S-MNPQ be a cone generated by the revolution of the right triangle SOM about the axis SO. Denote its lateral area by S, the circumference of its base by C, and its slant height SM hy L. Circumscribe about the base any regular polygon ABCD, and upon this polygon as a base construct a regular pyramid S-ABCD circumscribed about the cone. Denote the 24 278 GEOMETRY. lateral area of the pyramid by s, and the perimeter of its base oy p ; its slant height is the same as that of the cone, since it is an ele- ment of contact, as SMot SN; therefore, we have (YII. 47), L s=pXj Th^ number of lateral faces of the pyramid being indefinitely in- creased, 8 approaches indefinitely to S, and p approaches indefinitely to O; therefore, at the limit, we have (V. 31), .= oxf 24. Corollary I. If jB is the radius of the base, we have C = 2^jB (V. 40) ; hence S = 27rB X - == ^BL. The area of the base being 7ri2^, the total area T of the cone is 25. CoToUary II. Hence, by the same process as was employed in (11), we can prove that the lateral areas, or the total areas, of similar cones of revolution are to each other as the squares of their slant heights, or as tike squares of their altitudes, or as the squares of the radii of their bases. PEOPOSITION VI.— THEOREM. 26. The lateral area of a frustum of a cone of revolution is equal to the half sum of the circumferences of its bases multiplied by its slant height. The plane which cuts off" the frustum MNPm, from the cone S-MNP, also cuts off from any circumscribed pyramid a frustum, as ABCDa, the lateral area of which is equal to the half sum of the pe- rimeters of its bases multiplied by its slant height Mm (VII. 48). When the number of faces of the frustum of the pyramid is indefinitely increased, its lateral area ap- proaches indefinitely to that of the frustum BOOK IX 279 of the cone, and the perimeters of its bases approach indefinitely to the circumferences of the bases of the frustum of the cone ; and the slant height Mm is common. Hence, if we express by area Mm, the area of the surface generated by the revolution of Mm about the axis, which is the lateral area of the frustum of the cone ; and by drc. OM, and circ. om, the circumferences of the bases whose radii are OM and om; we shall have, at the limit, area Mm = \ (circ. OM -\- drc. om) X Mm. 27. Corollary. Let IK be the radius of a ^^ section of the frustum equidistant from its bases ; then, IK=i{OM+ om), (1. 124), and ik ^.V^K since circumferences are proportional to their /,-- j- radii, ctrc. IK =^ ^ {circ. OM -\- drc. om); ^^^' '^ therefore, area Mm = drc. IK X Mm ; that is, the lateral area of a frustum of a cone of revolution is equal to the circumference of a section equidistant from its bases multiplied by its slant height, PROPOSITION VII.— THEOREM. 28. The volume of any cone is equal to one-third of the product of its base by its altitude. Let the volume of the cone be demoted by V, its base by B, and its altitude by jET. Let the volume of an inscribed pyramid be denoted by V\ and its base by B'; its alti- tude will also be jBT, and we shall have (VII 54), V' = iB'XH. When the number of lateral faces of the pyramid is indefinitely increased, V approaches indefinitely to F, and -B ' to jB ; therefore, at the limit, we have V=iBxJI^ 280 GEOMETRY. 29. Corollai-y I. If the cone is a cone of revolution, let R be the radius of the base, then B = izR ^ and we have V=\7:.R\H. 30. Corollary II. Let R and r be the radii of the bases of two syiiilar cones of revolution ; H and h their altitudes, V and v their volumes ; then, the generating triangles being similar, we have H_R h r and hence V _ijrJR\H_R^ S_H^R^ that is, similar cones of revolution are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases. PEOPOSITION VIII.— THEOKEM. 31. A frustum of any cone is equivalent to the sum of three cones whose common altitude is the altitude of the frustum, and whose ba^ea are the lower base, the upper base, and a mean proportional between the bases of the frustum. Let V denote the volume of the frustum, /(^^^^^ B its lower base, b its upper base, and h its / ///I altitude. /-'''' ''r^-"^J I Let V denote the volume of an inscribed \'^-.^ / ^--jy"'' frustum of a pyramid, B' its lower base, and b' its upper base; its altitude will also be h, and we shall havb (VIL 59), V = \h{B' -i- b' ^ ]/WF). When the number of lateral faces of the frustum of a pyramid is indefinitely increased, V, B' and b', approach indefinitely to V, B and b, respectively ; therefore, at the limit, we have V=ih{B-]-b-\-l/M), which is the algebraic expression of the theorem. BOOK IX. 281 32. Corollary. If the frustum is that of a cone of revolution, and the radii of its bases are B and r, we shall have B = 7:.R\ 6 = 7r.r^ \/Bb = 7t,Rr, and consequently, V= i7r./i(i2'-|-r' + Rr). THE SPHERE. 33. Definition. A spherical segment is a portion of a sphere in- cluded between two parallel planes. The sections of the sphere made by the parallel planes are the bases of the segment ; the distance of the planes is the altitude of the segment. Let the sphere be generated by the revolution of the semicircle EBF about the axis EF; and let Aa and Bb be two parallels, perpendicular to the axis. The solid generated by the figure ABba is a spheri- cal segment ; the circles generated by Aa and Bb are its bases ; and a6 is its altitude. If two parallels Aa and TE are taken, one of which is a tangent at E, the solid generated by the figure EAa is a spherical segment having but one base, which is the section generated by Aa. The segment is still in- cluded between two parallel planes, one of which is the tangent plane at E, generated by the line ET. 34. Definition. A zone is a portion of the surface of a sphere in- cluded between two parallel planes. The circumferences of the sections of the sphere made by the parallel planes are the bases of the zone ; the distance of the planes is its altitude. A zone is the curved surface of a spherical segment. In the revolution of the semicircle EBF about EF, an arc AB generates a zone ; the points A and B generate the bases of the zone ; and the altitude of the zone is ab. An arc, EA, one extremity of which is in the axis, generates a 24* 282 GEOMETRY. zone of one base, which is the circumference described by the ex- tremity A. 35. Definition. When a semicircle revolves about its diameter, the solid generated by any sector of the semicircle is called a spherical sector. Thus, when the semicircle EBF revolves about EFy the circular sector COD generates a spherical sector. The spherical sector is bounded by three curved surfaces ; namely, the two conical surfaces generated by the radii OC and OD^ and the zone generated by the arc CD. This zone is called the base of the spherical sector. PROPOSITION IX.— LEMMA. ^ 36. The area of the surface generated by a straight line revolving about an axis in its plane, is equal to the projection of the line on the axis multiplied by the circumference of the circle whose radius is the perpf.ndlcular erected at the middle of the line and terminated by the axis Let AB be the straight line revolving about the axis XY; ab its projection on the axis ; 01 the perpendicular to it, at its middle point 7, ^a terminating in the axis ; then, U^\ area AB = ab X circ. 01. ^ ~~"h""^\ For, draw IK perpendicular, and AH par- allel to the axis. The area generated by AB is that of a frustum of a cone ; hence (27), area AB = AB X drc IK. Now the triangles ABH and I OK, having their sides perpendicular each to each, are similar (III. 33), hence AHorab:AB = IK:Ol or, since circumferences are proportional to their radii, a6 : A^ = circ. IK: drc. 01, wheiice BOOK IX 283 therefore, AB X circ. IK=ab X cire, 01, area AB := ab X circ. 01. If AB is taken parallel to the axis, the result is the same, and in fact has already been proved, since in this case the surface generated is that of a cylin- der whose radius is 01 and whose altitude is ab (9). PROPOSITION X.— THEOREM. 37. The area of a zone is equal to the product of its altitude by the circumference of a great circle. Let the sphere be generated by the revolution of the semicircle ^J5i^ about the axis EF; and let the arc AD generate the zone whose area is required. Let the arc AD be divided into any number of equal parts, AB, BC, CD. The- chords AB, BC, CD, form a regular broken line, which differs from a portion of a regular polygon only in this, that the arc subtended by one of its sides, as AB, is not necessarily an aliquot part of the whole circumfer- ence. The sides being equidistant from the centre, a circle described with the perpendicular 01, let fall from the centre upon any side, would touch all the sides and be inscribed in the regular broken line. Drawing the perpendiculars Aa, Bb, Cc, Dd, we have by the preced- ing Lemma, area AB = ab X circ. 01, area BC = be X circ. 01, area CD = cd X circ. 01, the sum of which is or area ABCD = (ab -{- be -}- cd) X circ. 01, area ABCD = ad X circ. OL This being true whatever the number of sides of the regular broken line, let that number be indefinitely increased ; then area ABCD, 284 GEOMETRY. generated by the broken line, approaches indefinitely to the area of the zone generated by the arc AD, and circ. 01 approaches indefi- nitely to cire. OE, or the circumference of a great circle ; hence, at the limit, we have area of zone AD = ad y, cire. OE, which establishes the theorem. 38. Corollary I. Let S denote the surface of the zone whose alti- tude is J3", the radius of the sphere being M ; then, 39. Corollary II. Zones on the same sphere, or on equal spheres, are to each other as their altitudes. 40. Corollary III. Let the arc AD generate a zone ^^- of a single base ; its area is d/^... 1 tO Ad X 1it.0A = -K,Ad X AB = Tz , AD'' Qll. ^1)\ / that is, a zone of one base is'' equivalent to the circle \ whose radius is the chord b/ the generating arc of the \ zone. ■B PEOPOSITION XI.— THEOREM. 41. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle. This follows directly from the preceding proposition, since the sur- face of the whole sphere may be regarded as a zone whose altitude is the diameter of the sphere. 42. Corollary I. Let S denote the area of the surface of a sphere whose radius is i2 ; then S=2^R X 2B = 47rE^; that is, the surface of a sphere is equivalent to four great circles. 43. Corollary II. Let S and 8' be the surfaces of two spheres whose radii are R and R' ; then, S JTtR' (2Ry _R^, S'~ Ar.R'^~ (2Ry'~ R"' hence, the surfaces of two spheres are to each other as the squares of their diameters, or as the squares of their radii. BOOK IX. 285 PKOPOSITION XII.— LEMMA. y 44. If a triangle revolves about an axis situated in its plane and passing through the vertex without crossing its surface, the volume generated is equal to the area generated by the base multiplied by one- third of the altitude. Let ABC be the triangle revolving about an axis XY passing through the vertex A ; then, the volume generated is equal to the area generated by the base BG multiplied by one-third of the alti- tude AD. We shall distinguish three cases : 1st. When one of the sides of the triangle, as ABy lies in the axis. (Figs. 1 and 2.) Fig. 1. Fig. 2. D Draw CE perpendicular to the axis. According as this perpen- dicular falls within the triangle (Fig. 1) or without it (Fig. 2), the volume generated is the sum or the difference of the cones generated by the right triangles JlCiKand BCE. The volumes of these cones are (29), voI.ACE=\t.. CE'x AE, vol. BCE =^7:.CE'X BE; if we take their sum, we have in Fig. 1, AE -f BE = AB; if we take their difference, we have in Fig. 2, AE — BE = AB ; there- fore, in either case, vol.ABC=i7c.CE'xAB = i^. CEX CEXAB; or, since CE X AB and BCX AD are each double the area of the triangle, (IV. 13), vol. ABC = i7t .CEX BCX AD. But re . CE X BC is the measure of the surface generated hj BC C24) ; therefore, vol. ABC=area BCXi AD. 286 GEOMETRY. 1 2d. When the triangle has only the vertex A in the axis, and the base BC when produced meets the axis in F (Fig. 3). ^- The volume generated is then the difference of the volumes generated by the triangles J. (ZP and ABF^ and, by the first case, these volumes are vol ACF= area FC X i AD, vol ABF=areaFB X ^AD, the difference of which is vol ABC= (area FC— area FB) X \AD = area BC X \AD, 3d. When the triangle has only the vertex A in the axis, and the base J5(7is parallel to the axis (Figs. 4 and 5). The volume generated is the sum (Fig. 4), or the difference (Fig. 5), of the volumes generated by the right triangles ABD and ACD, Draw ^jffand C^ perpendicular to the axis. The volume gener- ated by the triangle ABD is the difference of the volumes of the cylinder generated by the rectangle ADBH and the cone generated by the triangle ABH; therefore, vol ABD = n.AD'x BD — \7: . AD' X BD= liz . AD^ X BD = 27t.ADX BDX \AD, or, since 2t: . AD X BD is the lateral .area of the cylinder gener- ated by the rectangle AHBD (9), vol ABD = area BD X\AD; and in the same manner we have vol ACD = area CD X ^AD. BOOK IX. 287 Taking the sura of these (Fig. 4), or their difference (Fig. 5), we have vol. ABG= area BGX iAD. Therefore, in all cases, the volume generated by the triangle is equal to the area generated by its base multiplied by one-third of its altitude. PROPOSITION XIII.— THEOREM. 45. The volume of a spherical sector is equal to the area of the zone which forms its base multiplied by one-third the radius of the sphere. Let the sphere be generated by the revolution of the semicircle EBF ahout the axis EF; and let the circular sector AOD generate a spherical sector whose volume is required. Inscribe in the arc AD a regular broken line ABCDy as in Proposition X., forming with the radii OA and OD a regular polygonal sector/ OABCD. Decompose this polygonal sector into triangles A OB, BOC, COD, by drawing radii to its vertices. Taking the sides AB, BC, CD, as bases, the perpen- dicular 01 from the centre upon any side is the common altitude of these triangles. The volume generated by the polygonal sector is the sum of the volumes generated by the triangles, and the volume generated by any triangle is equal to the area generated by its base multiplied by one-third of its altitude 01 (44) ; therefore, vol. OABCD == area ABCD X — • o When the number of sides of the regular polygonal sector is in- definitely increased, vol. OABCD approaches indefinitely to the volume of the spherical sector OAD, area ABCD to the area of the zone AD, and 01 to the radius OA of the sphere ; therefore, at the limit, we have vol. spherical sector OAD = zone ADy^\ OA ; which establishes the theorem. ^ '»^t-^i^jl^ 288 GEOMETRY. PROPOSITION XIV.— THEOREM. 46. The volume of a sphere is equal to the area of its surface multi- plied by one-third of its radius. This follows directly from the preceding proposition ; for. if a cir- cula^r sector is increased until it becomes the semicircle which gener- ates the sphere, the spherical sector which it generates becomes the sphere itself, ^fcud'lis surface 1)ecomes the surface of the sphere. 47. Corollary I. If V denotes the volume of a sphere whose radius is R, we have (42) F=4;r.i22 X \R=-i^.R\ Or, if D is the diameter of the sphere, whence D'^ = (2i?)' = SR\ 48. Corollary II. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters. PROPOSITION XV.— THEOREM. 49. The solid generated by a circular segment revolving about a diameter exterior to it, is equivalent to one-sixth of the cylinder whose radius is the chord of the segment and whose altitude is the projection of that chord on the axis. Let ANBIA be a circular segment revolving about the diameter EF, and ab the projection of the chord AB on the axis. The volume generated is the difference of the volumes generated by the circular sector A OB and the triangle A OB. Draw- ing 01 perpendicular to AB, we have (45), (44), (38) and (36), vol. sph. sector A OB ^= zone AB X i OA = | tt . OA^ . ab, vol. triangle A OB = area AB X ^ 01 = -f ^^ . 7)1^ . ab, the difference of which gives vol. segment ANB = f 7r(02' — 07') X ah. BOOK IX. 289 But 0A^—0I' = AI' = ilB'; hence vol. segment ANB = ^t: . AB^ .ab, which establishes the theorem, since tt . AB^ . ah is the volu.xie of the cylinder whose radius is AB and whose altitude is ab (13). PROPOSITION XVI.— THEOREM. 50. The volume of a spherical segment is equal to the half sum of its baxes multiplied by its altitude plus the volume of a sphere of ivhich that altitude is the diameter. E Lot Aa and Bb be the radii of the bases of a a^^^ spherical segment, and ab its altitude, so that the /T segment is generated by the revolution of the figure y \ ANBba about the axis EF. \~" The segment is the sum of the solid generated by \ the circular segment ANB and the frustum of a cone \ generated by the trapezoid ABba ; hence, denoting ^^^ the volume of the spherical segment by V, we have (49) and (32), V=ir:.AB\ab + i7z.{Bb' + J^' + Bb.Aa).ab. Drawing AH parallel to EF^ we have BH = Bb — Aa, and hence BH' = Bb' -^ A^' — 2Bb . Aa, and AB' = AH' + BH' = '^' J^ Bb'-i-A^' — 2Bb . Aa. Substituting this value of AB', we have, after reduction, V=i{r:.Bb' -i- ^.'Aa').ab J^ ^tz .'^\ which establishes the theorem, since tt . Bb' and tt . Aa' represent the bases of the segment, and I t: . ab^ is the volume of the sphere whose diameter is ab (47). 51. Corollary. Denoting the radii of the**bases of the spherical segment by E and r, and its altitude by h, we have, for its volume, V=i7r(R--^ryi-{-in.h\ 25 T ft . ,^ 'I )^s U 290 ^ GEOMETRY. If the point A coincides with Ey the upper base becomes zero, and the solid generated becomes a segment of one base. Therefore, making r == in the above expression, the volume of a spherical segment of one base is PROPOSITION XVII.— THEOREM. 52. The volume of a spherical pyramid is equal to the area of its base multiplied by one-third of the radius of the sphere. For, let V denote the volume of a spherical pyramid, and s the area of the spherical polygon which forms its base. Let F, S and R denote the vol ime, surface and radius of the sphere ; then (VIII. 104), — = -, whence t; = « X ~- ir But- = J-R (46) ; therefore, S v=sX\R APPENDIX I. EXERCISES IN ELEMENTARY GEOMETRY. f:^i p A ^ =*tf • & ,4'D '^i^^^-^i 4^ EXERCISES IN ELEMENTARY GEOMETRY. In order to make these exercises progressive as to difficulty, and to bring them fairly within the grasp of the student at the successive stages of hia progress, many of them are accompanied by diagrams in which the necessary auxiliary lines are drawn, or by references to the articles in the Geometry on which the exercise immediately depends, or by both. These aids are less and less freely given in the later exercises, and the student is finally left wholly to his own resources. GEOMETRY.— BOOK I. THEOREMS. 1. The sum of the three straight lines drawn from any point within a triangle to the three vertices, is less than the sum and greater than the half sum of the three sides of the tri- angle (I. 33, 66). 1/ 2. The medial line to any side of a triangle is less than the half sum of the other two sides, and greater than the excess of that half sum above half the thu-d side (I. 66, 67, 112). 3. The sum of the three medial lines of a triangle is less than the perimeter (sum of the three sides), and greater than the semi-porimeter of the triangle. 4. If from two points, A and B^ on the same side of a straight line MN^ straight lines, AP^ BP^ are drawn to a point P in that line, making with it equal angles APM and BPN, the sum of the lines APand BPis less than the sum of any other two Hnes, AQ and BQ, drawn from A and B to any other point Q in MN{I. 38, 66). 25* 2t3 294 EXERCISES 5. If f]x»m C^o points, A and B^ on opposite sides of a straight line MN^ straight lines AP^ BP, are drawn to a point P in that line, making with it equal angles APN and BPN^ the difference of the lines AP and BP is greater than the difference of any other two straight lines AQ and BQ^ drawn from A and B to any other point Q in MN. V 6. The three straight lines joining the middle points of the angle divide the triangle into four equal triangles (I. 122). 7. The straight line AE which bisects the angle ex- terior to the vertical angle of an isosceles triangle ABd is parallel to the base BC. ' ^ 8. In 'any right triangle, the straight line drawn from the vertex of the right angle to the middle of the hy- potenuse is equal to one-half the hypotenuse (I. 121, 38, 46). 9. If one of the acute angles of a right triangle is double the other, the hypotenuse is double the shortest side (Ex. 8), (I. 69, 86, 90). 10. If ABC is any right triangle, and if from the acute angle A, AD is drawn cutting BCin E and a parallel to AC m D ^o that ED = 2 AB; then, the angle DA C is one-third the angle BA C, (Ex. 8), (I. 69, 86, 49). 11. If J5C is the base of an isosceles triangle ABC, and BD is drawn perpendicular to AC, the angle DBC is equal to one-half the angled. (I. 73). 12. If from a variable point in the base of an isosceles tri- angle parallels to the sides are drawn, a parallelogram is formed whose perimeter is constant. 1 3. If from a variable point P in the base of an isos- celes triangle ABC, perpendiculars, PM, PN, to the sides, are drawn, the sum of PM and PN is constant, and equal to the perpendicular from C upon AB (I. 104, 83). I What modification of this statement is required when Pis taken in produced? BC EX ERCISES 295 14. If from any point within an equilateral triangle, per- pendiculars to the three sides are drawn, the sum of these lines is constant, and equal to the perpendicular from any vertex upon the opposite side (Ex. 13). *W''hat modification of this statement is required when the point is taken without the triangle? 15. If ^4jBC is an equilateral triangle, and if BD and CD bisect the angles B and C, the lines DE^ DF^ paral- lel to AB^ AC, respectively, divide BC into three equal parts. 16. The locus of aU the points which are equally distant from two inter- secting straight lines consists of two perpendicular lines (I. 126, 25). What is the locus of all the points which are equally distant from two parallel Hnes? ' 17. Let the three medial lines of a triangle ABC meet in 0. Let one of them, AD^ be produced to G^ making DG= DO, and join CG. Then, the sides of the triangle OCG are, respectively, two-thirds of the medial lines of ^.SC (L 134). Also, if the three medial lines of the triangle OCG be drawn, they will be respectively equal to ^AB, iBC andMC. 18. In any triangle ABC, if AD is drawn perpen- dicular to BC, and AE bisecting the angle BAC, the angle DAE is equal to one-half the difierence of the b angles B and C (L 68). 19. If ii^ bisects the angle jB of a triangle ABC, and Cj& bisects the exterior angle A CD, the angle E is equal to one-half the angle A. b 20. If from the diagonal BD of a square ABCD, BE is cut off equal to BC, and EF is drawn perpendicular to BD^ then, DE=EF=FC. 296 EXERCISES. 21. If Eund Fare the middle points of the oppo- site sides, AD, BC, of a parallelogram AjBCI), the straight lines BE, DF, trisect the diagonal AC. '22. Tlie sum of the four lines drawn to the vertices of a quadrilateral from !in\' point except the intersection of the diagonals, is greater than tiie sum of the diagonals. • , r» 23. The straight lines joining the middle points of the adjacent sides of anj^ quadrilateral, form a paral- lelogram whose perimeter is equal to the sum of the diagonals of the quadrilateral (I. 122). 24. The intersection of the straight lines which join the middle points of opposite sides of any quadrilateral, is the middle point of the straight line which joins the middle points of the diagonals (I. 122, 108, 109). 25. The four bisectors of the angles of a quadrilateral form a second quadrilateral, the opposite angles of which are supplementary. If the first quadrilateral is a parallelogram, the second is a rectangle. If the first is a rectangle, the second is a square. A ED 26. A parallelogi-am is a sjTnmetrical figure with re- spect to its centre (intersection of the diagonals), (I. 140). 27. If in a parallelogram ABCD, E and G are any two symmetrical points in the sides AD, BO, and F and H any two sjnoimetrical points in the sides AB, DC, the figure EFGH ya a parallelo- gram concentnc with ABCD. / >^ t7 R p c A E D \ K, U '\ y. 28. If the diametf'rs (I. 140) EG, Flf, joining syuniu.'trical points in the opposite sides of a s, so that CD shall be equal to the radius of the circle, then, the angle C will be equal to one-third the angle A OB. Note. There is no method known of drawing BC, under these conditioiH. and with the use of straight lines and circles only, A OB being nni/ given angle : so that the trisection of an angle, in general, is a problem that cannot be solved by elementary geometry. 39. If ABC is an equilateral triangle inscribed in a circle, and P any point in the arc BC, then PA = PB -j PC. EXERCISES 299 40. If a triangle ABC is formed by the intersection of three tangents to a circumfer- ence, two of which, AM and AN^ are fixed, while the third BC touches the circumference at a variable point F^ prove that the perimeter of the triangle ABC'i^ constant, and equal to AM-\-AN, or 2 AN. Also, prove that the angle BOC is constant. 41. If ABC is a triangle inscribed in a circle, and from the middle point D of the arc BC sl perpen- dicular i>^ is drawn to AB; then, (11. 57), (I. 87), AE=H^^ + ^C\ BE=IUB — AC). If the perpendicular D^E^ is drawn from the middle point 2>^ of the arc BAC, then AE'=HAB—AC), BE' = h[AB + AC). Also join AD and draw the diameter Diy ; then, the angle ADJy is equal to one-half the difierence of the angles ACB and ABC. 42. If two straight lines are drawn through the point of contact of two circles, the chords of the intercepted arcs are parallel. 43. Two circles are tangent internally at P, and a chord AB of the circle touches the smaller at C\ prove that FC bisects the angle (II. 62). 44. If through P, one of the points of in- tersection of two circumferences, any two secants, AFB, CFB, are drawn, the straight lines, AC, DB, joining the extremities of the secants, make a constant angle E, equal to the angle MFN formed by the tangents at F. larger AFB 45. If through one of the points of intersection of two circumferences, a diameter of each circle is drawn, the straiglit line which joins the extremities of these diameters passes through the other point of intersection, is parallel to the line joining the centres, and is longer than any other line drawn through a point of intersection and terminated by the two circumferences. 300 !•: X K u 46. The feet, a, 6, c, of the perpendiculars let faL from any point P in a circumference on the sides of an inscribed triangle ABC, are in a straight line. Join fib, be, and prove that the angle abC =^ the angle Abe (11. 99). 47 . If equilateral triangles ABC/^ BCA\ CAB\ are constructed on the sides of any triangle ABC: 1st. The circumferences circumscrib- ed about the equilateral triangles in- tersect in the same point F\ 2d. The straight lines AA', BB% CC, are equal and intersect in P\ 3d. The centres of the three circumferences are the vertices of an equilateral tri- angle ocny. 48. The inscribed and the three escribed circles of a triangle ABC being drawn, as in the figure on p. 86, let />, JY, B^\ D'^\ be the four points of contact on the same side BC. Designate the sidi' BC, opposite to the angle A, by «, the side ^IC by h, and the side AB by c; and let « = ^ (a-f6 + c). Prove the following properties : BD'' ^Ciy''-=s, BD'''=CD'' =s—a, BD =Ciy =8-b, Bir =CD =s—c. BIT' =D'''D'-=h, BD"'' =D'D'' =c, BD" =--6 — 0, I Also, let a circumference be circumscribed about the triangle ABC. Prove that this circumference bisects each of the six lines 0^, 0^^, 0^^\ 0' C, 0-' 0''\ C' 0' ; and that the points of bisection are the centres of circumferences that may be circumscribed about the quadrilaterals BO CO', COAO'', AOBO''', ABO'O'', BCO'^O'"', CAO'^'O', re- spectively. Finally, designating the radius of the circumscribed circle by R ; the radius of the inscribed circle by r ; the radii of the escribed circles by r\ i/^ '/'' ; the perpendiculars from the centre of the circumscribed circle to the three sides by p\ p^'', p^'^ ; prove the following relations : ^4.^//+ r^^^ = 4i2 + r, E X E U C 1 S E S . ;u)i LOCI. 49. Find the locus of tlie centre of a circumference which passes through two given points. 50. Find the locus of the centre of a circumference which is tangent to two given straight lines. 51. Find the locus of the centre of a circumference which is tangent to a given straight line at a given point of that line, or to a given circumfer- ence at a given point of that circumference. 52. Find the locus of the centre of a circumference passing through a given point and having a given radius. 53. Find the locus of the centre of a circumference tangent to a given straight line and having a given radius. 54. Find the locus of the centre of a circumference of given radius, tap- gent externally or internally to a given circumference. c N 55. A straight line JiiV, of given length, is placed with its extremities on two given perpen- dicular Hues, AB, CD; find the locus of its middle point F (Ex. 8). M iV 56. A straight line of given length is inscribed in a given circle ; find the locus of its middle point. 57. A straight line is drawn through a given point J., intersecting a given circumference in B and C; find the locus of the middle point, P, of the inter- cepted chord BC. Note the special case in which the point A is on ihe given circumference. 58. From any point A in a given circumference, a straight hne AP o£ fixed length is drawn i)avallel to a given line 3IN; find the locus of the extremity P. 50. Upon a ^iven base BC^ a triangle ABC'ib constructed having a given vertical angle A ; find the locus of the intersection of the perpendiculars from the vertices of this triangle upon the opposite sides (II. 97). 2R '602 EXERCISES, n 60. The angle ACB is any inscribed angle in a given segment of a circle ; AC is produced to P, making CP equal to CB : find the locus of P. 61 . From one extremity J^ of a fixed diameter AB, any chord ^C is drawn, and at C a tangent CD. From B, a perpendicular BD to the tan- gent is drawn, meeting J. C in P. Find the locus of P. 62. A triangle ABC is given, right angled at A. Any perpendicular, EF, to BC, \s drawn, cutting AB in Z>, and CA in F. Find the locus of P, the inter- ^ section of BF and CD. 63. The base BC of Si triangle is given, and the me- dial line BE^ from B, is of a given length. Find the locus of the vertex A. Draw A parallel to EB. Since B0= BC, is a fixed point ; and since A0=2 BE, OA is a constant distance. 64, An angle BAC is given in position, and l)oints B and C are taken in its sides so that AB -{- AC shall be a given constant length. Find the locus of the centre of the circle circumscribed about the triangle ABC (Ex. 41). Also, if the points B and C are so taken that AB — AC IS B. given constant length, find the locus of the centre of the circle circumscribing ABC (Ex.41). Also find the locus of the middle point of BC, 65. The base BC of sl triangle ^^^is given in position, and the verti^-.-ii EXERCISES. 803 angle J is of a given magnitude ; find the loci of the centres of the inscribed and escribed circles- In the figure, p. 86, we have the angles BOC= 90° -\- ^ A, BO'G = \S0° — BOC=90° — iA, BO''C=BO'''C=hA. The loci are circumferences whose centres lie in the circumference of the circle circumscribed about ABC. 66. Find the locus of all the points the sum of the distances of each of which from two given straight lines is a given constant length (Ex. 13). Show that the locus consists of four straight lines forming a rectangle. 67. Find the locus of all the points the difierence of the distances of each of which from two given straight lines is a given constant length (Ex. 13). Show that the locus consists of parts of four straight lines whose intersec- tions form a rectangle. 68. If in Ex. 66 by sum is understood algebraic sum, and distances falling on opposite sides of the same line have opposite algebraic signs, show that Ex. 66 includes Ex. 67, and the locus consists of the whole of four indefinite lines whose intersections form a rectangle. PROBLEMS. The most useful general precept that can be given, to aid the student m his search for the solution of a problem, is the following. Suppose the problem solved, and construct a figure accordingly : study the properties of this figure, drawing auxiliary lines when necessary, and endeavor to discover the dependence of the problem upon previously solved problems. This is an analysis of the problem. The reverse process, or synthesis, then furnishes a constraction of the problem. In the analysis, the student's ingenuity will be exercised especially in drawing useful auxiliary lines ; in the sjnthesis, he will often find room for invention in combining in the most simple form the several steps suggested by the analysis. The analysis frequently leads to the solution of a problem by the intersec- tion of loci. The solution may turn upon the determination of the position of a particular point. By one condition of the problem it may appear that this required point is necessarily one of the points of a certain line ; this line is a locus of the point satisfying that condition. A second condition of the problem may furnish a second locus of the point ; and the point is then fully determined, being the intersection of the two loci. Some of the following problems are accompanied by an analysis to illus- trate the process. 69. A triangle ABC being given, to draw DE parallel to the base ^^ so that DE= DB + EC. Analysis. Suppose the problem solved, and that DE is the required parallel. Since DE=^ DB -\- EC, we may divide it into two portions, DP and PE, respectively equal to DB and EC. Join PB, PC. Then we have the angle DBP^ DPB =^PBC; therefore, the line PB bi- sects the angle ABC. In the same manner it is shown that CP bisects 304 XERCISES the angle A OB. The point P, then, Ues in each of the bisectors of the base angles of the triangle, and is therefore the intersection of these bisectors. Hence we derive the following construction. Const ruction. Bisect the angles B and C by straight lines. Through the intersection F of the bisectors, draw the line BPE parallel to BC. Thia line satisfies the conditions. For we have, by, the constniction, the angle DBP= PBC= BPD\ therefore, PD = DB\ and in the same manner, PE'= EG\ hence, DE = DB-{- EG. We have, however, tacitly assumed that DE is to be drawn so as to cut the sides of the triangle ABC between the vertex and the base. Suppose it drawn cutting AB and AC produced. Then the same analysis shows that the point P is found by bisecting the exterior angles CBD, BCE. Thus the problem has two solutions, if the position of DE is not limited to one side of the base BC. 70. To determine a point whose distances from two given inter- secting straight lines, AB, A^ B\ are given. Anahisis. The locus of all the points which are at a given dis- tance from AB consists of two parallels to AB, CE and DF, each at the given distance from AB. The locus of all the points at a given distance from A^B'' consists of two parallels, C^E^ and D^F\ each at the given distance from A^ B\ The required point must be in both loci, and therefore in their intersection. There are in this case four intersections of the loci, and the problem has four solutions. Construction. At any point of AB, as A, erect a perpendicular CD, and make AC "= AD = the given distance from AB ; through C and D draw parallels to AB. In the same manner, draw parallels to A^ B^ at the given distance A^C^ = A^D^. The intersection of the four parallels determines the four points Py, Pi, P^, P^, each of which satisfies the conditions. E c 71. Given two perpendiculars, AB and CD, inter- secting in 0. To construct a square, one of whose angles shall coincide with one oi' the right angles at 0, and the vertex of the opposite angle of the square shall lie on a given straight line EF. (Two solutions. ) B05 EXERCISES. 12. In a given rhombus, ABCD, to inscribe A sqasLie EFGII. (Ex. 71.) 73. In a given straight line, to find a point equally distant from two given points without the line. 74. To construct a square, given its diagonal. 75. In a given square, to inscribe a square of given magnitude. 76. From two given ]ioints on the same side of a given straight line, tc draw two straight lines meeting in the given straight line and making equal angles with it. (Ex. 4.) 77. Through a given point P within a given angle, to draw a straight line, terminated by the sides of the angle, which shall be bisected at P. 78. Given two straight lines which cannot be produced to their intersec- tion, to draw a third which would pass through their intersection and bisect their contained angle. 79. Through a given point, to draw a straight line making equal angles with two given straight lines. 80. Given the middle point of a chord in a given circle, to draw the chord. 81. To draw a tangent to a given circle which shall be parallel to a given straight line. 82. In the prolongation of any given chord ^J5 of a circle, to find a point jP, such that the tangent FT, drawn from it to the circle, shall be of a given length. 83. To draw a tangent to a given circle, such that its segment intercepted between the point of contact and a given straight line shall have a given length. In general, there are four solutions. Show when there will be but three solutions, and when but two ; also, when no solution is possible. 84. Through a given point within or without a given circle, to draw a straight line, intersecting the circumference, so that the intercepted chord shall have a given length. (Two solutions. ) 85. Through a given point, to draw a straight line, intersecting two given circumferences, so that the portion of it intercepted between the circumfer- ^'nces. shall have a given length. (Two solutions. ) 86 In a given circle, inscribe a chord of a given length which produced chall be tangent to another given circle. 87. Construct an angle of 60°, one of 120°, one of 30°, one of 150°, one of 45° and one of 1 35°. 88. To find a point within a given triangle, such that the three straight lines drawn from it to the vertices of the triangle shall make three equal angles with each other. When will the problem be impossible ? 89. Construct a i)arallelogram, given, 1st, two adjacent sides and one diago- 26 » U 30G EXERCISES, K # nal ; Hd, one side and the diagonals ; 3d, the diagonals and the angle they contain. 90. Construct a triangle, given the base, the angle opposite to the base, and the altitude. Analysis. Suppose BAG to be the required tri- x"' "^^^ angle. The side jSC being fixed in position and ~7^^'^7~^^^^\\~ ^ magnitude, the vertex A is to be determined. One j; ^^ -.^ n locus of A is an arc of a segment, described upon ^^— -^ AB. containing the given angle. Another locus of ^ is a straight line MN drawn parallel to ^C, at a distance from it equal to the given altitude. Hence the position of A will be found by the inter- section of these two loci, both of which are readily constructed. Limitation. If the given altitude were greater than the perpendicular distance from the middle of ^BC to the arc BAC^ the arc would not inter- sect the line MN^ and there would be no solution possible. The hmits of the data, within which the solution of any problem is pos- sible, should always be determined. 91. Construct a triangle, given the base, the medial Hne to the base, and the angle opposite to the base. 92. Construct a triangle, given the base, an angle at the base, and the sum or difference of the other two sides. Analysis. On the sides of the given angle, jB, take BG = given ba^e, and BD = given sum or difference of the sides. Join GD. The problem is reduced to drawing, from C, a line GA, which shall cut BD, or BD produced, in a point A, so that GA shall be equal to AD, which is obviously effected by making the angle DGA = the angle ADG. If, when the difference of AB and AG is given, AG is to be the greater side, BD = AG — AB is to be taken in AB produced through B. i/ 93. Construct a triangle, given the base, the angle opposite to the base, ^^ and the sum or difference of the other two sides. Analysis. Suppose ABG is the required triangle. First, when the sum of AB and AG is given, produce BA to D, making BD = AB + AG. Join GD. The angle D is one-half of BAG, and is therefore known. Hence the following construction. Make an angle BDG equal to one-half the given angle. Take DB = given sum of sides. From B as centre, and with radius equal to the given base, describe an arc cutting DG in G. Draw GA, making the angle DGA = the angle BDG. Secondly, when the difference of AB and ^C is given; take BD^ ^ AB — AG, and join GD^. The angle AD^G is one-half the supplement of BA G, and hence the construction can readily be found. This problem can also be solved by an application of Ex. 41. I EXERCISES. 807 94. Construct a triangle, given the base, the sum or difference of the other two sides, and the difference of the angles at the base. Anali/sis. In the preceding figure, the angle BCD^ = ^ [ACB — AJ^C)^ and BCD = 90° + BCD^ ; and hence a construction can readily be found. 95. Construct a triangle, given, 1st, two angles and the sum of two sides ; 2d, two angles and the perimeter. 96. Construct a triangle, given, 1st. Two sides and one medial line 2d. One side and two medial lines ; 3d. The three medial lines. See Exercise 17. 97. Construct a triangle, given an angle, the bisector of that angle, and the perpendicular from that angle to the opposite side. 98. Construct a triangle, given the middle points of its sides. 99. Construct a triangle, given the feet of the perpendiculars from the imgles on the opposite sides. (Ex. 36. ) 100. Construct a triangle, given the perimeter, one angle, and one altitude. 101. Construct a triangle, given an angle, together with the medial line and the perpendicular from that angle to the opposite side. 1 02. Construct a triangle, given the base, the sum or the difference of the other two sides, and the radius of the inscribed circle. (Ex. 48. ) 103. Construct a triangle, given the centres of the three ascribed circles. 104. Construct a triangle having its vertices on two given concentric cir- cumferences, its angles being given. 105. Divide a given arc into two parts such that the sum of their chords shall be a given length. (Ex. 41.) 106. Construct a. square, given the sum or the difference of its diagonal and side. (Ex. 20. ) 107. With a given radius, describe a circumference, 1st, tangent to two given straight lines ; 2d, tangent to a given straight line and to a given cir- cumference ; 3d, tangent to two given circumferences ; 4th, passing through a given point and tangent to a given straight line ; 5th, passing through a given point and tangent to a given circumference ; 6thj having its centre on a given straight line, or a given circumference, and tangent to a ^ven straight line, or to a given circumference. (Exs. 52, 53, 54.) 108. Describe a circumference, 1st, tangent to two given straight lines, and touching one of them at a given point (Exs. 50, 51) ; 2d, tangent to a given circumference at a given point and tangent to a given straight line ; 3d, tan- gent to a given straight line at a given point and tangent to a given cir- cumference (Ex. 51) ; 4th, passing through a given point and tangent to a given straight line at a given point ; 5th, passing through a given point and tangent to a given circumference at a given point. 109. Draw a straight line equally distant from three given points. When will there be but three solutions, and when an indefinite number of solutions ? 308 EXERCISES. 1 110. Describe a circumference equally distant from four given points; the distance from a point to the circumference being measured on a radius, or radius produced. In general there are four solutions. If three of the given "points are in a straight line, one of the four circumferences becomes a straight line. 111. An angle A is given in position, and a point P in its plane. It is required to draw a straight line through jP, intersecting the sides of the angle and forming a triangle ABC of a given perimeter. (Ex. 40.) 112. With a given point as a centre, describe a circle which shall be divided by a given straight hne into segments containing given angles. 113. Through a given point without a given circle, draw a secant, so that the portion of it without the circle shall be equal to the portion within. (Ex. 96.) 114. Inscribe a straight line MN, of given length, between two given straight lines AB, CD, and parallel to a given straight line EF. 115. Inscribe a straight line of given length between two given circumfcr ences, and parallel to a given straight line. 116. Through P, one of the points of intersection of two circumferences draw a straight line, terminated by the circumferences, which shall be bi sected in P. 117. Through one of the points of intersection of two circumferences, di*aw a straight line, terminated by the circumferences, which shall have a given length. 118. Given two parallels ^5, CD, and two other parallels ^^^^ C'D\ inclined to the first ; through a given point P, in their plane, draw a straight line such that the portion of it intercepted between the parallels ^Z^, CD, shall be equal to the portion of it intercepted between the parallels A^ B\ C'D\ (Ex. 77.) 119. From two given P9ints, J., B, on the same side of a given straight line il/iV, draw straight lines, meeting in a point P of MN, so that the angle APM shall be equal to double the angle BPN. Analyais. The solution of Exercise 76, suggests the possible advantage of EXERCISES. 309 emplojnng the symmetrical point of one of the given points. Let B^ be the symmetrical point of B with respect to MN[l. 135). Join ^^7^ and produce it toward E. Then, since APM = 2 BFN=^ 2 B'PN - 2 MPE, B'PE bi- sects the angle APM. Therefore, the problem is reduced to finding a point Pin MN such that B'PE shall bisect the angle APM. With B' as centre describe an arc through A^ cutting MN in C. The perpendicular B^E upon A C, evidently intersects MN in the required point. 120. With the vertices of a given triangle as centres, describe three cir- cumferences each of which shall be tangent to the other two. (Four solu- tions.) (Ex. 48.) 121. Construct a quadrilateral, given its four sides and the straight line joining the middle points of two opposite sides. (Ex. 24.) 1 22. Construct a pentagon, given the middle points of its sides. The middle points of all the diagonals can be determined by the principle of Ex. 23. 1 23. Find a point in a given straight line, such that tangents drawn from it to two given circumferences shall make equal angles with the line. (Four solutions. ) Compare Ex. 76. 124. If a figure is moved in a plane, it may be brought from one position to any other, by revolving it about a certain fixed point (that is, by causing each point of the figure to move in the circumference of a circle whose centre is the fixed point). Find that point, for two given positions of the figure. GEOMETRY.— BOOK IH. THEOREMS. 125. If three parallels AA\ BB^^ CC\ intercept on two straight lines J.C, A'C\ the segments AB and BC^ or A^ B^ and B^ C\ in a given ratio m : n, that is, if AB: BC^A'B': B'C' = m:n; then, (m + n). BB' = n.AA' -f m. CG\ (in. 25, 10.) 126. In any triangle ABC^ if from the vertex J., AE is drawn to the circumference of the circumscribed circle, and AD to the base BC^ making the angles CAE and BAD equal to each other, then (III. 25), ABXAC = ADXAE. 310 EXERCISES. 127. From the preceding theorem, deduce as a corol- lary the following: In any triangle ABC, if from the vertex A, AE is drawn bisecting the angle J., meeting the circumference of the circumscribed circle in E and the base BC m D^ then ABXAC = ADXAE, Also deduce (III. 65). 128. If ABCD is a given parallelogram, and AN a variable straight Kne drawn through A cutting EG m M and CD in N\ then, the product 5J/i)JV is constant. (III. 25.) 129. If ABCD is any parallelogram, and from any point P in the diago- nal AC (or in AC produced) PM is drawn cutting BA in J/, EC m iV, AD in M' and DC in N'; then, PM.P]Sr= PM'.PN\ (III. 25.) 130. If a square DEFG is inscribed in a righc triangle ABC, so that a side DE coincides with the hypotenuse EC (the vertices F and G being in the sides AC and AE) ; then, the side DE is a mean proportional between the segments ED and EC of the hypotenuse. (III. 25.) 131. If a straight Hne AE is divided at C and D so that AE.AD = t^, and if from A any straight line AEis drawn equal to J. (7; then, JE'C 3ts the angle DEB. (III. 10, 32, 23.) 132. If «, &, c, denote the three perpendiculars from the three vertices of a triangle upon any straight line MN in its plane, and p the perpendicular from the intersection of the three medial lines of the triangle upon the same straight line MN\ then, ^Ex. 125,) a + 64- c ^ 3 133. If ABC and A^BC are two triangles having a common base EG and their vertices in a line AA^ parallel to the base, and if any parallel to the base cuts the sides AE and AC in D and E^ and the sides A^E and A'C in D" and E' ; then DE = D'E'. 134. If two sides of a triangle are divided proportionally, the straight lines drawn from corresponding points of section to the opposite angles inter- sect on the line joining the vertex of the third angle and the middle of the third side. 135. The difference of the squares of two sides of any triangle is equal to the difference of the squares of the projections of these sides on the third side. (III. 48.) 136. If from any point in the plane of a polygon, perpendiculars are drawn to all the sides, the two sums of the squares of the alternate segments of the sides are equal. (Ex. 135.) 137. If is the centre of the circle circumscribed about a triangle ABCy and P is the intersection of the perpendiculars from the angles upon the opposite sides ; the perpendicular from upon the side EG ia equal to one- half the,distance AP (I. 132), (III. 25, 30.) I A EXERCISES. 311 138. In any triangle, the centre of the circumscribed circle, the intersec- tion Df the medial lines, and the intersection of the perpendiculars from the angles upon the opposite sides, are in the same straight line ; and the dis- ince of the first point from the second is one-half the distance of the second t)m the third. 139. If d denotes the distance of a point P from the centre of a circle, 'od r the radius ; and if any straight line drawn through P cuts the cir- "umfcrence in the points A and B\ then, the product PA.PB is equal to ^ — d^ if Pis within the circle, and to d^ — r^ if Pis without the circle. 140. In any quadrilateral, the sum of the squares of the diagonals is equal MO twice the sum of the squares of the straight lines joining the middle points of the opposite sides. (III. 64) and (Ex. 23.) 141. In any quadrilateral ABCD inscribed in a cir- cle, the product of the diagonals is equal to the sum of the products of the opposite sides ; that is, Df AC.DB = AD.BC + AB.DC. (Make the angle DAE = BAG, and prove that AD.BC=AG.DE, ^nd AB.DC = AC. BK) 142. In an inscribed quadrilateral ABCD^ if P is the intersection of th; then AD.AB _AF ,.^^ „. . crcd-fU ^^'^-"^ 143. In an inscribed quadrilateral ABCD., AD.AB-j-CB.CD ^ AC BA.BC+DA.DC BD 144. In an inscribed quadrilateral, the product of the perpendiculars let fall from any point of the circumference upon two opposite sides is equal to the product of the perpendiculars let fall from the same point upon the other two sides. (III. 65.) 145. If from a point P in a circumference, any chords P4, PjB, PC, are drawn, and any straight line MN parallel to the tangent at P, cutting the chords (or chords produced) in a, A, c; then, the products PA. Pa., PB.Pb, PC.Pc, are equal. 146. If two tangents are drawn to a circle at the extremities of a diameter, the portion of any third tangent intercepted between them is divided at its point of contact into segments whose product is equal to the square of the radius. 147. If on a diameter of a circle two points are taken equally distant from the centre, the sum of the squares of the distances of any point cf the cir cumference from these two points is constant. 312 EXERCISES. 148. If a point P on the circumference of a circle is taken as the centre of a second circle, and any tangent is drawn to the second circle cutting the first in M and N; then, the product FM.FN is constant. 149. The perpendicular from any point of a circumference upon a chord in a mean proportional between the perpendiculars from the same point upon the tangents drawn at the extremities of the chord. 150. If AB is the chord of a quadrant of a circle whose centre is 0, anl anj'' chord MN parallel to AB cuts the radii OA, OB in P and Q ; then MP^ + PN^ = AB\ (III. 48) and (Ex. 139.) 151. If ABCD is any parallelogram, and any circumference is described passing through A and cutting AB, A C, AD, in the points F, G, II, re- spectively; then AFAB + AII^D=:AG.Aa 152. In any isosceles triangle, the square of one of the equal sides is equal to the square of any straight line drawn from the vertex to the base plus the product of the segments of the base. 153. If ^ and B are the centres of two circles which touch at (7, and Pis a point at which the angles AFC and BFC are equal, and if from P tan- gents PI) and FE are drawn to the two circles ; then, PD.PE = PC\ (III. 21 and 66. ) 154. If two circles touch each other, secants drawn through their point of contact and terminating in the two circumferences are divided proportionally at the point of contact. 155. If two circles are tangent externally, the portions of their common tangent included between the points of contact is a mean proportional be- tween the diameters of the circles. 156. Two circles are tangent internally at A, and from any point P in the circumference of the exterior circle a tangent FM is drawn to the interior circle ; prove that the ratio FA : FM is constant. 157. If two circles intersect in the points A and B, and through A any secant CAD is drawn terminated by the ciicumferences at C and i), the straight lines BC and BD are to each other as the diameters of the circles. 158. If a fixed circumference is cut by any circumference which passes through two fixed points, the common chord passes through a fixed point. 159. Two chords AB and CD, perpendicular to each other, intersect in a point P either within or without the circle, and the line OF is drawn from the centre 0. Prove that if J9 is the diameter of the circle, PA^ + PB^ + W^ + FD^ = D\ and AB^ + CD' + aUP^ - 2 DK EXERCISES 313 CI 160. If any number of circumferences pass through the same two points, and if through one of these points any two straight Unes are drawn, the cor- respon iing segments of these Unes intercepted between the circumferences are proportional. 161. If a triangle ABC is inscribed in a circle, and from the vertex A^ AD and AE are drawn parallel to the tangents at B and G respectively and cutting the base BC in D and E\ then BD.DE=AD' = AE\ BB:nE=AB':AC\ 162. Let AB be a given straight line. At A erect the perpendicular Al) =AB; in BA produced take AO = h AB ; with centre and radius OD describe a circumference, cutting AB and AB produced in C and C^ ; prove that AB is divided in extreme and mean ratio, internally at C, and externally at C\ 163. If a rhombus ABCD is circumscribed about a circle, any tangent MN determines on two adjacent sides AB^ AD, two segments BM, DN, whose product is constant. 164. If in a semicircle whose diameter is AB^ any two chords AC and BD are drawn intersecting in P, then AB^ = AC.AP+ BD.BP. 165. If is the intersection of the three medial lines of a triangle ABC, prove the relations . AB'+AC' + BC' = S(OA' +OB'-i-OC'), BG^ -{-SOA^^AC' +SOB' =AB^+3 0C\ 166. If is the intersection of the three medial lines of a triangle ABC, and F any point in the plane of the triangle ; then, -p2.^ +PB^+FC' =1)1' + OB' +0C' -\-SFO\ 167. If R, r, r', r // „./// , are respectively the radii of the circumscribed, the inscribed, and the three escribed circles in any triangle, and if J, d', df\ d^^\ are respectively the distances from the centre of the circumscribed circle to the centres of the inscribed and escribed circles, prove the relations -2Rr' = d''^ —2Rr'' d^ -\-d'^+d''^^d '''- 12 2Rr'' 314 EXERCISES, LOCI. ■ 168. From a fixed point 0, a straight line OA is drawn ID any point in a given straight Hne MN^ and divided at P in a given ratio m : n [i. e., so that OF : PA = m : n) ; find the locus of P. N 169. From a fixed point 0, a straight line OA is drawn to any point in a given circumference, and divided at P in a given ratio ; find tho locus of P. 170. Find the locus of a point whose distances from two given straight fines are in a given ratio. (The locus consists of two straight lines.) 171. Find the locus of the points which divide the various chords of a given circle into segments (external or internal) whose product is equal to a given constant, h^. (III. 56, 59.) 172. Find the locus of a point the sum of whose distances from twx> given straight lines is equal to a given constant h. 1 73. Find the locus of a point the difference of whose distances from two given straight lines is equal to a given constant k. 174. Find the locus of a point such that the sum of the squares of its dis- tances from two given points is equal to a given constant, h^. (III. 62.) 1 75. Find the locus of a point such that the difference of the squares of its distances from two given points is equal to a given constant k^. (III. 62. ) 176. If J., B and C are three given points in the same straight Hne, find the locus of a point P at which the angles APB and BPC are equal. (Ex. 131.) 177. Through Jl, one of the points of intersection of two given circles, any secant is drawn cutting the two circumferences in the points B and C ; find the locus of the middle point of BC 178. Through A, one of the points of intersection of two given circles, any secant is drawn cutting the two circumferences in the points B and C, a ad on this secant AP is laid off equal to the sum of AB and J. (7; find the locus of P. 179. From a given point 0, any straight line OA is drawn to a given straight line ifA^ and divided at P (internally or externally) so that the product OA. OPia equal to a given constant ; find the locus of P. (P]x. 145.) 180. From a given point in the circumference of a given circle, any chord I J A is drawn and divided at P (internally or externally) so that the produof OA. OP is et^ual to a given constant ; find the locus of P. EXERCISES. 315 181. From a given point 0, any straight line OA s drawn to a given straight line MN, and OP is drawn making a given angle with OA, and such that OP is to OA in a given ratio ; find the locus of P. With the same construction, if OP is so taken that the product OP. OA is equal to a given constant ; find the locus of P. N 182. From a given point 0, any straight line OA is drawn to a given circumference, and OP is drawn making a given angle with OA, and such that OP is to 0^ in a given ratio; find the locus of P. With the same construction, if OP is so taken that the product OP OA is equal to a given constant ; find the l"»cus of P. 183. One vertex of a triangle whose angles are given is fixed, while the second vertex moves on the circumference of a given circle ; what is the locus of the third vertex ? 184. Given a circle and a point A ; find the locus of the point P such that the distance PA is equal to the tangent from P to the circle 0. 1 85. Find the locus of a point from which two given circles are seen under equal angles. Mote. The angle under which a circle is seen from a point is the angle contained by the two tangents from that point. 186. Find the locus of a point, such that the sum of the squares of its dis- tances from the vertices of a given triangle is equal to the square of a given Hne. (Ex. 166.) 187. From any point A within a given circle, two straight lines AAf and -47V are drawn perpendicular to each other, intersecting the circumference in M and JV; find the locus of the middle point of the chord MN^. PROBLEMS. 188. To divide a given straight Hne into three segments. A, Bsmd C, such that A and B shall be in the ratio of two given straight lines M and iV, and B and C shall be in the ratio of two other given straight lines P and Q. 1 89. Through a given point, to draw a straight line so that the portion of it intercepted between two given straight lines shall be divided at the point in a given ratio. 190. Through a given point, to draw a straight line so that the distances from two other given points to this line shall be in a given ratio. (Two solu- tions. ) 191. Through a given point P, to draw a straight line cutting two given parallels in 31 and K, so that the distances A3I and BN of the points of intersection from two giver, loints A and B on these parallels shall be in a given ratio. 316 EXELCISBS. 1 192. To d termine a point whose distances from tliree given i oints snail be proportional to three given straight lines. (III. 79. ) 193. To determine a point whose distances from three given indefinite straight lines shall be proportional to three given straight lines. (Ex. 170.) 194. Given two straight lines which cannot be produced to their intersec- tion, to draw a straight line through 9 g.ven point which would, if sufficiently produced, pass through the unknown point of intersection of the given Imes. (HI. 35.) 195. In a given triangle ABC to draw a parallel EF to the base BQ intersecting the sides AB and AC in U and F respectively, so that BE + CF= BC\ or so that BE— CF= BC. (in. 19, 21.) 196. In a given triangle ABC^ to inscribe a square DEFG. (Exs. 71 and 133.) 197. In a given triangle ABC^ to inscribe a paral- lelogram DEFG^ such that the adjacent sides DE and DG shall be in a given ratio and contain a given angle. (Remark, that the solution of this problem includes that of the preceding. ) 198. Construct a triangle, given its base, the ratio of the other two sides, and one angle. (III. 79. ) 199. To determine a point in a given arc of a circle, such that the chorda drawn from it to the extremities of the arc shall have a given ratio. 200. To find a point P in the prolongation of a given . chord CD of a given circle, such that the sum of the two tangents PA and PB^ drawn from it to the circle, shall be equal to the entire secant PC 201. To divide a given straight line into two segments, such that the sum of their squares shall be equal to the square of a given straight .ine. 202. Given an obtuse-angled triangle ; it is required to draw a straight line from the vertex of the obtuse angle to the opposite side, the square of wnich shall be equal to the product of the segments into which it divides that side. 203. Through a given point P to draw a straight line intersecting a given circumference in two points A and B^ so that PA shall be to PB in a given ratio. 204. Given two circumferences intersecting in J. ; to draw through A a secant, BA C, such that AB s lall be to J. 6^ in a given ratio. EXERCISES. 317 205. Given two circumferences intersecting in A ; to draw through A a secant ABC, such that the product AB.AC shall be equal to the square of a given line. Construction. Produce the common chord AD, and take E so that AD.AE = the square of the given line (III. 71). Make the angle AEC equal to the angle inscribed in the segment ABD, and let EC cut the circumference in C and C^ Join A C and A C\ Either of these hues satisfies the conditions of the problem. 206. To describe a circumference passing through two given points A and B, and tan- gent to a given circumference 0. Analysis. Suppose A TB is the required cir- cumference tangent to the given circumference at T, and ACDB any circumference passing « through A and B and cutting the given cir- cumference in C and D. The common chords AB and CD, and the common tangent at T, all pass through a common point P (Ex. 15i) ; from which a simple construction may be infeiTed. There are two solutions, given by the two tangents that can be drawn from P. 207. To describe a circumference passing through two given points and tangent to a given straight line. (Two solutions. ) 208. To describe a circumference passing through a given point and tan- gent to two given straight lines. c 209. To describe a circumference passing through a given point P, and tangent to a given straight line MN and to a given circumference 0. Analysis. Suppose EPD is one of the circumferences which satisfy the conditions, passing through P, touch- ing 3IN at E and the circumference lit D. Through the centre 0, draw COBA perpendicular to MI^', join CP meeting the circumference EPD in Q ; also join CE. It can be proved ^ ^^jb""' that CE passes through D, and that EP CRCQ= CECD^ CA.CB; the point Q is therefore determined, and the problem is reduced to that of Ex. 206 or 207. The point Q may be taken either in PC or in PC pro- duced through C, and thus *\ere will be obtained, in all, four solutions. 27 * 318 EXERCISES 210. To describe a circumfer- ence passing through a given point A and tangent to two given circumferences, and 0^. Analysis. If ACDB is one of the circumferences satisfying the conditions, we can show that the Ym^CD^ joining the points of con- tact with the given circles, passes through F^ the intersection of the line of centres, 0\ with a com- mon tangent TT^^ and that FC.PD = PT.PT'. Hence, joining PA, we have PA.PB = PT.PT\ and PB is known ; or ^ is a known point on the required circumference. The problem is thus reduced to Ex. 20G. By employing also the internal common tangent, we find, in all, four solutions. 211. To describe a circle tangent to two given straight lines and to a given circle. This is reducible to Ex. 208. If both the given straight lines intersect the given circle, there may be eight solutions. 212. To describe a circle tangfet to two given circles, and to a given straight line. This is reducible to Ex. 209. There may be eight solutions. 21 3. To describe a circle tangent to three given circles. This is reducible to Ex. 210. There may be eight solutions. *214. To describe a circumference which shall bisect three given circum- ferences. *215. To construct a triangle, given its base in position and magnitude, and the sum (or the difference) of the other two sides, the locus of the vertex opposite the base being a given straight line. *216. To construct a triangle, given the product of two sides, the medial line to the third side, and the difference of the angles adjacent to the third side. *217. To construct a triangle, similar to a given triangle, and having its three vertices on the circumferences of three given concentric circles. The same problem, substituting three parallel straight lines for the three circumferences. *218. In a given circle, to inscribe a triangle, such that 1 St. Its base shall be parallel to a given straight line, and its other two sides shall pass through two given points in that line ; or, 2d. Its base shall be parallel to a given straight line, and its other two sides shall pass through two given i)oints not in that line ; or, 3d. Its three sides shall pass through three given points. I Exercises 214 to 218 are intended only for the most advanced stidents. EXERCISES. GEOMETRY.— BOOK THEOKEMS. 219. Two triangles which have an angle of the one equal to the supple- ment of an angle of the other are to each other as the products of the siden including the supplementary angles. (IV. 22.) 220. Prove, geometrically, that the square described upon the sum of two straight lines is equivalent to the sum of the squares described on the two lines plus twice their rectangle. Note. By the "rectangle of two lines" is here meant the rectangle of which the two lines are the adjacent sides. 221. Prove, geometrically, that the square described upon the difference of two straight lines is equivalent to the sum of the squares described on the two lines minus twice their rectangle. 222. Prove, geometrically, that the rectangle of the sum and the differ- ence of two straight lines is equivalent to the difference of the scjuares of those lines. 223. Prove, geometrically, that the sum of the squares on two lines is equivalent to twice the square on half th-'ir sum plus twice the square on half their difference. Or, the sum of the S(iuares on the two segments of a line is equivalent to twice the square on half the line plus twice the square on the distance ox the point of section from the middle of the line. 224. The area of a triangle is equal to the product of half its perimet<;r by the radius of the inscribed circle ; that is, if a, b and c denote the sidt^ op- posite the angles^, i? and (7- respectively, r the radius of the ins-. -^ \ / I \ \l/^' ^^^^'**»l/ .' ^ V p o ; \ \ 236. If three straight lines Aa, Bh, Cc, drawn from the vertices of a triangle ABC to the opposite sides, pass through a common point within the triangle, then b Oa ,0h ,0c^. Aa'^ Bb^ Cc What modification of this statement is necessaiy if the point is with- out the triangle ? 237. If from any point within a triangle ABC, any three straight lines, Oa, Ob, Oc, are drawn to the three sides, and through the vertices of the triangle three straight lines Aa'', Bb\ Cc'', are drawn parallel respectively to Oa, Ob, Oc, then Oa^ ,0b , Oc _, Aa' "^ Bb' "^ C(/ What modification of this statement is necessaiy if the point is taken without the triangle ? Deduce the preceding theorem from this. 238. If from the vertices of a triangle ABC, three straight lines, AA\ BB% CC\ are drawn to the opposite sides (or these sides produced), each equal to a given line L, and from any point within the triangle, Oa^ Ob- 27** V 322 EXEKCISES. Oc, are drawn parallel respectively to AA^, BB\ CC\ and terminating in the same sides, then, the sum of Oa, Ob and Oc is equal to the given length L 239. If flf, />, c ana d denote the four sides of any quadrilateral, m and n it° diagonals, and S its area, then S = il/(2 /?m + a^ — lj^-hc^ — d-') {2m)i — a^ -f />2 _ ^2 + cP"). If ths quadrilateral is inscribed in a circle, this formula becomes S = i/{p-a){p-h){p-c){p-d), in which p ^= ^ {a -\- h -\- c -{- d). If th« quadrilateral is such that it can be circumscribed about a circle and also inscribed in a circle, then the formula becomes S = \/ahcd. PROBLEMS. 240. To construct a triangle, given one angle, the side opposite to that angle, and the area (equal to that of a given square). 241. To construct a triangle, given its angles and its area. 242. To construct a triangle, given one angle, the medial Hne from one of the other angles, and the area. 243. To construct a triangle, given its area, the radius of the inscribed circle, and the radius of one of the escribed circles ; or, given its area and the radii of two escribed circles. (Exercises 48 and 224. ) 244. Griven any triangle, to construct an isosceles triangle of the same area, whose vertical angle is an angle of the given triangle. 245. Given any triangle, to construct an equilateral triangle of the same area. "~24f). Given the three straight lines EF^ GH and DK^ in the figure of Exercise 235, to construct the triangle ABC. 247. Bisect a given triangle by a parallel to one of its sides. Or, more generally, divide a given triangle into two or more parts propor- tional to given lines, by parallels to one of its sides. EXERCISES. 323 248. Bisect a triangle by a straight line drawn through a given point in one of its sides. 249. Through a given point, draw a straight line which shall form with two given intersecting straight lines a triangle of a given area. llemark that the area and an angle being known, the product of the sides uicluding tliat angle is known. (lY. 22. ) 250. Bisect a trapezoid bj'' a parallel to its bases. 251. Inscribe a rectangle of a given area in a given circle. 252. Inscribe a trapezoid in a given circle, knowing its area and the common length of its inclined sides. (See Ex. 229. ) 253. Given three points, A, B and C, to find a fourth point -P, such that the areas of the triangles AFB^ AFC, J3F(y\ shall be equal. (Four solu- tions. ) 254. Given three points. A, B and C, to find a fourth point jP, such that the areas of the triangles AFB, AFC, BFC, shall be proportional to three given lines L, M, N. (Four solutions. ) See Exercise 1 70. GEOMETRY.— BOOK Y. THEOREMS. 255. An equilateral polygon inscribed in a circle is regular. 256. An equilateral polygon circumscribed about a circle is regular if the number of its sides is odd. 257. An equiangular polygon inscribed in a circle is regular if the number of its sides is odd. 258. An equiangular polj^gon circumscribed about a circle is regular. 259. The area of the regular inscribed hexagon is three-fourths of that of the regular circumscribed hexagon. 260. The area of the regular inscnibed hexagon is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles. 261. A plane surface may be entirely covered (as in the construction ;:f a pavement) by equal regular polygons of either three, four, or six sides. 262. A plane surface may be entirely covered by a combination of squares and regular octagons having the same side, or by dodecagons and equilateral triangles having the same side. 263. The area of a regular inscribed octagon is equal to that of a rectangle whose adjacent sides are equal to the sides of the inscribed and circumscribed squares. 264. The area of a circle is a mean proportional between the areas of any two similar polygons, one of which is circumscribed about the circle and the other isoperimetrical with the circle. ( Galileo's Theorem. ) 324 E X E K (^ I S E S . 265. Two diagonals of a regular pentagon, not drawn from a common vertex, divide each other in extreme and mean ratio. 266. If a = the side of a regular pentagon inscribed in % circle whose radius is R, then, R a = -^V 10 — 2i/5. 26*k, If a = the side of a regular octagon inscribed in a circle whose radius is R, then, a = rV'I — i/2. 268. If a = the side of a regular dodecagon inscribed in a circle whose radius is R^ then, a = RV2 — j/3. 269. If a = the side of a regular pentedecagon inscribed in a circle whose radius is /?, then, « = ^ (l^lO + 2i/5 + i/3 — 1/15). 270. K c? = the diagonal of a regular pentagon inscribed in a circle whose radius is i?, then, ^= 2 V^IO + 2v'5. 271. IS a = the side of a regular polygon inscribed in a circle whose radius is i?, and A = the side of the similar circumscribed polygon, then, A = 2aR 2AR V'{iR-'—a^) i/(4jK2 4-^2) 272. K a = the side of a regular polygon inscribed in a circle whose radius is jff, and a^ = the side of the regular inscribed polygon of double the number of sides, then. a'^ = r{2R — \/4R^ — a^)'. 3 _ a'-^{iR^ — a'^] R^ 273. K AB and CD are two perpendicular di- ameters in a circle, and E the middle point of the radius OC, and if EF is taken equal to EA, then OF is e(iual to the side of the regular inscribed decagon, and ^i^^ is equal to the side of the regular inscribed pentagon. Corollary. If a = the side of a regular pentagon and a^ = the side of a regular decagon, inscribed «n a circle whose radius is R, then, ^2_,^/a /,>2. EXERCISES. 325 274. In two circles of diifercnt radii, angles at the centres subtended by arcs of equal length are to each other inversely as the radii. 275. From any point within a regular pol^'gon of n sides, perpendiculars are drawn to the several sides ; prove that the sum of these perpendiculars is equal to n times the apothem. (V. 22. ) What modification of this statement is required if the point is taken with- out the polygon ? 276. If perpendiculars are dropped from the vertices of a regular polygon upon any diameter of the circumscribed circle, the sum of the perpendicu- lars which fall on one side of this diameter is equal to the sum of those which fall on the opposite side. 277. If n is the number of sides of a regular polygon inscribed in a circle whose radius is R, and a point P is taken such that the sum of the squares of its distances from the vertices of the polygon is equal to a given quantity k^, the locus of F is the circumference of a circle, concentric with the given circle, whose radius r is^determined by the relation n (III. 52 and 53), (Ex. 276.) PROBLEMS. 278. Divide a given circle into a given number of equivalent parts, by con- centric circumferences. Also, divide it into a given number of parts proportional to given Imes, by concentric circumferences. 279. A circle being given, to find a given number of circles whose radii shall be proportional to given lines, and the sum of whose areas shall be equal to the area of the given circle. 280. In a given equilateral triangle, inscribe three equal circles tangent to each other and to the sides of the triangle. Determine the radius of these circles in terms of the side of the triangle. 281. In a given circle, inscribe three equal circles tangent- to each other and to the given circle, y^ Detei-mine the radius of these circles in terms of the radius of the given circle. GEOMETRY.— BOOK VI. THEOREMS. 282. If a straight line AB is parallel to a plane MN^ any plane perpen- dicular to the line AB is perpendicular to the plane MN. 283. If a plane is passed through one of the diagonals of a parallelogram, :he perpendiculars to this plane from the extremities of the other diagonal •1.0 equal. 326 EXERCISES. 284. If the intersections of a number of planes are parallel, all the per- pendiculars to these planes, drawn from a common point in space, lie in one plane. 285. If the projections of a number of points on a plane are* in a straight line, these points are in one plane. 286. If each of the projections of a line AB upon two intersecting planes is a straight line, the line ^^ is a straight line. 28Ti Let A and B be two points, and M and N two planes. If the sum of the two perpendiculars from the point A upon the planes M and N is equal to the sum x)f those from B upon these planes, this sum is the same for every other point in the straight line AB. (Ex. 125. ) Extend the theorem to any number of planes. 288. Let J., B and C be three points, and M and N two planes. If the sum of the two perpendiculars from each of the points J., B and C, upon the planes M and -A^, is the same for the three points, it will be the same for every point in the plane ABC. (Ex. 287.) Extend the theorem to any number of planes. 289. A plane passed through the middle point of the common perpen- dicular to two straight lines in space (VI. 63), and parallel to both these lines, bisects every straight line joining a point of one of these lines to a point of the other. 290. In any triedral angle, the three planes bisecting the three diedral angles, intersect in the same straight line. 291. In any triedral angle, the three planes passed through the edges, per- pendicular to the opposite faces respectively, intersect in the same straight line. ^ 292. In any triedral angle, the three planes passed through the edges and the bisectors of the opposite face angles respectively, intersect in the same straight line. 293. In any triedral angle, the three planes passed through the bisectors of the face angles, and perpendicular to these faces respectively, intersect in the same straight line. 294. If through the vertex of a triedral angle, three straight lines are drawn, one in the plane of each face and perpendicular to the opposite edge, these three straight lines are in one plane. LOCI. 295. Find the locus of the points in space which are equally distant froTu two given points. 296. Locus of the points which are equally distant from two given planes ; or whose distances from two given planes are in a given ratio. (Compare Ex. 170.) 297. Locus of the points which are equally distant from two given straight lines in the same plane. 298. Locus of the points which are equally distant from three given points. EXERCISES. 327 299. Locus of the points which are equally distant from three given planes. ooO. Locus of the points which are equally distant from three given straight lines in the same plane. 301. Locus of the points which are equally distant from the three edges of a triedral angle. (Ex. 293. ) 302. Lovus of the points in a given plane which are equally distant from two given points out of the plane. 303. Locus of the points which are equally distant from two given planes, and at the samo time e(iually distant from two given points. ^Exs. 295 and 296.) 304. Locus ot 1 point in a given plane such that the straight lines drawn from it to two gi i-on points out of the plane make equal angles with the plane. (IIL 79.) 305. Locus of a point such that the sum of its distances from two given planes is equal to a given straight line. 306. Locus of a point such that the difference of the squares of its dis- tances from two given points is equal to a given constant. 307. Locus of a point in a given plane such that the difi'erence of the squares of its distances from two given points is equal to a given constant. 308. A straight line of a given length moves so that its extremities are constantly upon two given perpendicular but non-intersecting straight lines ; what is the locus of the middle point of the moving line? 309. Two given non-intei>ecting straight lines in space are cut by an indefinite number of parallel planes, the two intersections of each plane with the given lines are joined by a straight line, and each of these joining lines is divided in a given ratio in : n; what is the locus of the points of division ? \/ PEOBLEMS. In the solution of problems in space, we assume — 1st, that a plane can be drawn passing through three given points (or two intersecting straight lines) and its intersections with given straight Hues or planes determined — and 2d, that a perjiendicular to a given jilane can be drawn at a given point in the plane, or from a given point without it ; and the solution of a problem will consist, not in giving a graphic constraction, but in determining the con- ditious under which the proposed problem is solved by the application of these elementary problems. The graphic solution of problems belongs to Descriptive Geometry. 310. Through a given straight line, to pass a plane perpendicular to a given plane. 311. Through a given point, to pass a plane perpendicular to a given straight line. 312. Through a given point, to pass a plane parallel to a given plane. 313. To determine that point in a given straight line which is equidistant from two given points not in the same plane with the given line. o28 EXERCISES. 3J4. To find a point in a plane which shall be equidistant from three given points in space. 315. Through a given point in space, to draw a straight Kne which shall cut two given straight lines not in the same plane. 316. Given a straight line AB parallel to a plane M', from any point A in AB, to draw a straight line AP, of a given length, to the plane M, nialfing the angle BAP equal to a given angle. 317. Through a given point ^ in a plane, to draw a straight line AT m that plane, which shall be at a given distance PT from a given point P without the plane. 318. A given straight line AB meets a given plane at the point A ; to draw through A a straight line AP in the given plane, making the angle BAP equal to a given angle. 319. Through a given point A, to draw to a given plane M a straight Une which shall be parallel to a given plane JV and of a given length. 320. Through a given point A, to draw to a given plane M a straight line which shall be parallel to a given plane iV^ and make a given angle with the plane M. 321. Given two straight lines, CD and EF, not in the same plane, and AB any third straight line in space ; to draw a straight line PQ from AB to EF which shall be parallel to CD. 322. Given two straight lines AB and CD, not in the same plane ; to draw a straight line PQ from AB to CD which shall make a given angle with AB. 323. Given two straight lines, AB and CD, not in the same plane, to find a point in J. 5 at a given perpendicular distance from CD. 324. Through a given point , to draw a straight line which shall meet a given straight line and the circumference of a given circle not in the same plane. 325. In a given plane and through a given point of the plane, to draw a straight line which shall be perpendicular to a given line in space. (VI. 62.) 326. In a given plane, to determine a point such that the sum of its dis- tances from two given points on the same side of the plane shall be a minimum. 327. In a given plane, to determine a point such that the difference of its distances from two given points on opposite sides of the plane shall be a maximum. 328. To cut a given polyedral angle of four faces by a plane so that tlin section shall be a parallelogram. ^"Z?"*'- — .....aexfeRcisEs. A p' 829 GEOMETRY.— BOOK VIL THEOREMS. ^. 329. The volume of a triangular prism is equal to ine ^luduct of the area of a lateral face by one-half the perpendicular distance of that face from the opposite edge. 330. In any quadrangular prism, the sum of the squares of the twelve edges is equal to the sum of the squares of its four diagonals phis eight times the square of the line joining the common middle points of the diago- nals taken two and two. Deduce (VII. 20) from this. 331. Of all quadrangular prisms having equivalent surfaces, the cube has the greatest volume. 332. The lateral surface of a pyramid is greater than the base. 333. At any point in the base of a regular pyramid a perpendicular to the base is erected which intersects the several lateral faces of the pyramid, or these faces produced. Prove that the sum of the distances of the points of intersection from the base is constant. (See Ex. 275.) _, 334. In a tetraedron, the planes passed through the three lateral edges and the middle points of the edges of the base intersect in a straight line. The four straight lines so determined, by taking each face as a base, meet in a point which divides each line in the ratio 1 : 4. Note. This point is the centre of graviti/ of the tetraedron. 335. The perpendicular from the centre of gravity of a tetraedron to any plane is equal to the arithmetical mean of the four perpendiculars from the vertices of the tetraedron to the same plane. (Ex. 125.) 336. In any tetraedron, the straight lines joining the middle pomts of the pposite edges meet in a point and bisect each other in that point. 337. The plane which bisects a diedral angle of a tetraedron divides tho opposite edge into segments which are proportional to the areas of the adja- cent faces. 338. Any plane passing through the middle points of two opposite edges of a tetraedron divides the tetraedron into two equivalent solids. 339. If one of the triedral angles of a tetraedron is tri-rectangnlar [i. <•., composed of three right angles), the square of the area of the face opposite to it is equal to the sum of the squares of the areas of the three other faces. 340. If a, h, c, d, are the perpendiculars from the vertices of a tetraedron upon the opposite faces, and r/, y, t. pL ^j^ 360. The volumes of polyedrons circumscribed about the same sphere aro .^ proportional to their surfaces. 832 EXERCISES. LOCI. 361. Locus of the points in space whicli are at a given distance from a given straight line. 362. Locus of the points which are at the distjftice a from a point J., and at the distance h from a point B. 363. #Locus of the centres of the spheres which are tangent to three given planes. 364. Locus of a point in space the ratio of whose distances from two fixed points is a given constant. 365. Locus of the centres of the sections of a given sphere made by planes passing through a given straight Hne. 366. Locus of the centres of the sections of a given sphere made by planes passing through a given point 367. Locus of a point in space the sum of the squares of whose distances from two fixed points is a given constant. (Ex. 174.) 368. Locus of a point in space the diiference of the squares of whose dis- tances from two fixed points is a given constant. (Ex. 175.) PROBLEMS. 369. To cut a given sphere by a plane passing through a given straight line so that the section shall have a given radius. 370. To construct a spherical surface with a given radius, 1st, passing througli three given points ; 2d, passing through two given points and tan- gent to a given plane, or to a given sphere ; 3d, passing through a given point and tangent to two given planes, or to two given spheres, or to a given plane and a given sphere ; 4th, tangent to three given planes, or to three given spheres, or to two given planes and a given sphere, or to a given plane and two given spheres. 371. Through a given point on the surface of a sphere, to draw a great circle tangent to a given small circle. 372. To draw a great circle tangent to two given small circles. 373. At a given point in a great circle, to draw an arc of a great circle which shall make a given angle with the first. 374. To find the ratio of the volumes of two cylinders whose convex areas are equal. 375. To find the ratio of the convex areas of two cylinders whose volumes are equal. 376. To find the ratio of the volumes generated by a rectangle revolving successively about its two adjacent sides. APPENDIX II. mTRODUCTION TO MODERN GEOMETRY. Vi o- V INTRODUCTION TO MODERN GEOMETRY. TRANSVERSALS. 1. DEFimnoN. Any straight line cutting a system of lines is called a transversal. PROPOSITION I.— THEOREM. 2. If a tmii/wersal cuts the sides of a triarif/le {produced if necessanj), the product of three non-adjacent segments of the sides is equal to the product of the other three segments. Let ABC be the given triangle, a and ale the transversal. When the transversal cuts a side produced, as the side ^C at a, the segments are the distances, aB, aC, of the point of section from the extremities of the line (III. 22). The segments aB, bC, cA^ having no extremity in common, are non-adjacent. Draw CN parallel to AB. By similar triangles, we have aB cB bC NO aC JV^ bA cA' and multiplying, aBXbO_cB whence, aCXbA cA' aBXbCXcA = aCX bA X cB. 3. CoroUary. Conversely, if three points are taken on the sides of a tri- angle {one of the points, or all three^ lying in the sides produced), so that the product of three non-adjncent segments of the sides is equal to the product of the other three, then the three points lie in the same straight line. Let Of, 6, c, be so taken on the sides of the triangle ABC, that the rela- tion [I] is satisfied. Join ah, and let ab produced be supposed to cut 335 836 MODERN GEOMETRY. AB izi ^ point whicli we shall call c\ Then by the above theorem, we hftve aBXhGXc'A = aCXbAX c'B, 1 ad since by hypothesis we also have aBXhCXcA = aOXhAX cB, lere follows, by division, cA cB' which can evidently be true only when c^ coincides with c; that is, the three points a, h and c are in the same straight line, 4. Scholium. The principle in the corollary often serves to determine, in a vrery simple manner, whether three points lie in the same straight line. For example, take the following theorem : The middle points of the three diagonals of a complete quadrilateral are in a straight line. A complete quadrilateral is the figure formed by four straight lines inter- secting in six points, as ABCDEF. The line EF is called the third diagonal. •' Let X, Jf, JV, be the middle points of the three diagonals. Let (?, H, K, be the middle points of the sides of "the triangle FDC. The sides of the triangle GIIK pass through the points LMN, respectively (I. 121 and 122). The line ABE^ considered as a transversal of the triangle CDF^ gives AD.BF.EG = AF.BG.ED. Dividing each factor of this equation by 2, and observing that we have Mi> = LK, h BF= MG, etc. (L 121), we find LKMG.NH=LRMKNG] therefore, the points L, if, iV, lying in the sides of the triangle GITK, satisfy the condition of the preceding corollaiy, and are in a straight line. MODEKN GEOMETRY. 337 ■ PROPOSITION II.— THEOREM. 5. Three straight lines, drawn throvgh the vertices of a triangle and any point in its plane, divide the sides into segments such that the product of three non-adjacent segments is equal to the product of the other three. Let ABC be the triangle, and any point in its plane, through which Aa, Bb, Cc, are drawn. The triangle A Ca is cut by the transversal Bh] hence, by (2), aB.ba'A^d^ BC.hA.aC^, and the triangle ABa, cut by the transversal Cc, gives ^ BCdO.cA^aCA^cB. Multiplying these equations together, and omitting the common factors, we obtain aB.hQ.cA = aChA.cB. 6. CoroUanj. Conversely, if three points are taken on the sideji of a tri- angle {all the points being on the sides themselves or two on the sides pro- duced), so that the product of three non-adjacent segments of the sides « equal to the product of the other three, the straight lines joining these poinU with the opposite vertices of the triangle meet in one point. . The proof is similar to that of (3). 7. Scholium. The principle of this corollary often serves to determine whether three straight lines meet in a ])oint. For example, if Aa, Bb, Cc, are the bisectors of the angles of the triangle ABC, we have by (111. 21 ), aB aC bC bA cA cB AB A& BC AB' AC BC and the product of these equalities, omitting the common denominator ABXBCX AC, is aB.bCcA = aCbA.cB; therefore, the three bisectors of the angles of a triangle pass through the same point. With the same facility, it can be shown that the straight lines joining the points of contact of the i)iscribed circle with the opposite va^tices of the tn- angle meet in a point ; that the three perpendiculars from the vertices of a triangle to the opposite sides meet in a point ; and that the three medial liaeb of a triangh meet in a point. 29 ' W 338 MODERN GEOMETKY ANHARMONIC RATIO. 8. Definition. If foui* points are taken in a straight line, the quotient obtained by dividing the ratio of the distances of the first two from the third by the ratio of the distances of the first two from the fourth, is called the (nihqrmovic ratio of the four points. Thus the anharmonic ratio of the four \ — -— — 7-\ '■ |.oints A, B, C\ A is ^ bo d CA , DA CB ' DB' which for brevity is denoted by [ABCD]. In applying the definition the points may be taken in any order we please, but the adopted order is always to be indicated in the notation. Thus, the same points, considered in the order J., C, B, Z), give the anharmonic ratio 9. The anharmonic ratio of four points iS'not changed in value when tico of the points are interchanged^ provided the other two are interchanged at the same time. Thus [ABCD] [BADC] CA . DA ^ CA.DB CB ' DB CB.DA DB , CB^ CA.DB DA ' CA CB.DA ^^^^^^-Jd- BD~BCAD rnnj^Ai-^^ AD _ ACBD ^^^^^^-'bc'-ac'bcad' Tlierefore, [^i?CZ)] = [BADC] = [CDAB] = [DCBA]. There are then four different ways in which the same anharmonic ratio can be ex- pressed. There are, in all, twenty-four ways in which the four letters may be written, and therefore four points give rise to six anharmonic ratios differing in value. Three of these six are the reciprocals of the other three. 10. Definition. A system of straight Hnes diverging from a point is called a pencil; each diverging line is called a ray; and the point from which they diverge is called the v^-ffx of the pencil. MODERN GEOMETRY. 331) PROPOSITION III.— THEOREM. 11. If a pencil of four rays is cut by a transversal, any anJiarmonic ratio of the four points of intersection is constant for all positions of the trans- vei'sal. Fig. 1. Fig. 2. Let 0-MNPQU the pencil; and let ABCD, A'B'C'D', be any two positions of a transversal ; then [ABCD] = [A'B'C'iy]. For, drawing Bed parallel to 03/, we have by similar triangles, CA OA DA^OA DB dB CB cB Dividing the first of these equations by the second, we have dB [ABUU\ cB Drawing B^cfd^ parallel to J/, we have in the same manner, d'B' [A'B'G'I)'] c'B' The second members of these two equations being equal (III. 35), we have [ABCD] = [A'B'C'D'l It is important to observe that the preceding demonstration applies when the transversals cat one or more of the rays on opposite sides of the vertex, as in Fig. 2. 12. Defimtion. The anharmonic ratio of a pencil of four rays is the anharmonic ratio of the four points on these rays determined by any trans- versal. Thus, [ABCI)]^ [ACBD]^ etc., are the anharmonic ratios of the pencil formed by the rays OJ/, OiV, OP^ (9 ^, in the preceding figure. To 840 MODERN GEOMETRY distinguish the pencil in which the ratio is considered, the letter at thfi vertex is prefixed to the ratio ; thus, [ O.ABCD]^ [ O.ACBD]^ etc. 13. The angles of a pencil are the six angles which the rays, ^ taken two and two, form with each other. It follows from the preceding proposition that the values of the anharmonic ratios in two pencils will be equal, if the angles of the pencils are equal, each to each. 14. Definition. The anhhrmonic ratio of four fixed points A, jB, C, D, on the circumference of a circle, is the anharmonio ratio of the pencil formed by join- ing the four points to any variable point on the circumference. PROPOSITION IV.— THEOREM. 15. The anharmonic ratio of four fixed points on the circumference of a circle is constant. For, the angles of the pencil remain the same for all positions, 0, 0^, etc., of its vertex, on the circumference. (II. 58.) 16. Definition. If four fixed tangents to a circle are cut by a fifth (vari- able) tangent, the anharmonic ratio of the four points of intersection is called the anharmonic ratio of the four tangents. PROPOSITION V. 17. The anharmonic ratio of four fixed tangents to a circle is constant. For, let four tangents, touch- ing the circle at the points A, B, C, D, be intersected by any fifth tangent in J/, N, F, Q. The pencil formed by the rays Oi¥, ON, OF, 0<2, will have constant angles for all positions of the variable tangent, since (as the reader can readily prove) the angle MON will be meas- ured by one-half of the fixed arc AB, the angle NOPhy one-half of the arc BC, and the angle POQ by one-half of the arc CD. The angles of the pencil being constant, the anharmonic ratio [ 0. MNFQ] is constant. 18. Corollary. The anharmonic ratio of four tangents to a circle ?> equal to the anharmonic ratio of the four points of contact. For, if any point 0^ in the circumference be joined to A, B, C, D, the pencil formed will have the same angles as the pencil formed by the rays OM, ON, OP, 0^,*since these angles will also be measured by one-half the arcs AB, BC, CD, respectively. MODERN GEOMETRY 341 1 9. The properties of anharmonic ratios can be applied to the demonstra- tion of two classes of theorems, in one of which certain points are to be shown to lie in the same straight line, and in the other certain straight lines are to be shown to meet in the same point. Corresponding theorems in these two classes are placed side by side, in the following propositions, in order to exhibit their analogy. PKOPOSITION VI. Theorem,. 2C. When two pencils 0-ABCD, 0^-A^B^C^D^^ have the same anhar- monic ratio and a homologous ray OA common^ the intersections 6, c, c?, of the other three pairs of homx)logous rays, are in a straight line. Theorem. \V- B B^ C C For, let the straight line joining h and c meet OA in a, and let the points in which it meets OD and O^D^ be called d and d\ respectively. By hypothesis we have (11), [abed] = [abcd^], which can be true only when 8 and d^ coincide ; but 8 and d^ being on the different Hues OD and O^D'' can co- incide only when they are identical with their intersection d. Therefore, rt, 6, c, dj are in a straight Una. 22. Corollary. If one of the anhar- monic ratios of a pencil is equal to one of those of a second pencil, the 21. When two right-lined figures of four points A, B, C, D, and A, B^^ C^, ly^ have the same an- harmonic ratio and. a homologous point A common J the straight lines joining the other three pairs of ho- mologous points meet in the same point 0. For, let be the point of meeting of BB' and CC ; draw OA and 0D\ and let the point in which 01/ meets AD be called d. Then we have (11), [AB'C'D^ = [ABC81 and, by hypothesis, [AB'C'J)']^[ABCD]; hence, [ABCd] = [ABCDl Therefore S coincides with Z>, and the straight line DD^ also passes through the point 0. 23. Corollani. If one of the an- hai^monic ratios of a system of four points is equal to one of those of a remaining anharmonic ratios of the \ second system, the remaining anhar- 29* 1342 MODERN GEOMETRY, first pencil are equal to those of the iecond^ each to each. For, let the pencils be placed so as to have a common homologous ray. Since one of the ratios has the same value in both pencils, the intersec- tions of the other three pairs of ho- mologous rays lie in a straight line, which is a common transversal ; and then any two corresponding anhar- monic ratios in the two pencils will be equal to that of the four points on the common transversal (11), and there- lore equal to each other. monic ratios of the two systems art equal, each to each. For, let the two systems, be placed so as to have a common homologous point. . Since one of the anharmonic ratios has the same value in both sys- tems, the straight lines joining the other three pairs of homologous points meet in a point; and then any two corresponding anharmonic ratios in the two systems are equal, being de- termined in the same pencil (11). PROPOSITION VII. Theorem. 24. Iftico triangles, ABC, A'B'C, are so situated that the three straight lines, AA\ BB\ CC\ joiinng their corresponding vertice.'i, meet in a point, 0, the three intei'sections, a, h, c, of their corresponding sides, are in a straight line. Theorem. 25. If two triangles, ABC, A'B'C, are so situated that the three intersec- tions, a, b, c, of their corre., the lines LN, PC, EQ, joining the other three pairs of homologous points, meet in a common point M (21). Therefore L, M, J^, are in the same straight line. This theorem is due to Pascal. 29. Corollai-y 1. If the vertex D is brought nearer and nearer to the vertex C, the side CD will approach to the tangent at C', therefore, when the point D is finally made to coin- cide with C, the theorem will still a]>ply to the resulting pentagon if we substitute the tangent at C for the side CD. Tlie theorem then takes the following fonu. hi any pentagon AB CEF inscribed diagonals, AD, BE, CF, joining the opposite vertices, intersect in the same point. For, regarding AB, BC,' CD and EF, as fixed tangents cut by the tan- gent ED in P N, D, E, and by the tangent FA m A, L, M, F, we have (IT), [PiVDE] = [ALMF], or, considering these anharmonic ra- tios as belonging to pencils whose ver- tices are B and O, respectively, [B.PNDE] = [CAL3JFI These two pencils, having a common anharmonic ratio and a common ho- mologous ray XiV, the intersections, A, D, 0, of the other three pairs of homologous rays are in a straight hne (20). Therefore AD, BE and CF, meet in the same point 0. This theorem is due to Brianchon. 30. Corollary 1. If a vertex C is brought into the circumference, the sides i^C and CD will become a single line touching the circle at C. The theorem will still apply to the result- ing jientagon if we regard the point of contact of this side as the vertex of a circumscribed hexagon. The theorem then takes the following form. In any pentagon ABDEF circnm- scrihed about a circle, the line joining a vertex and the point of contact of the opposite side, and the diagonals joining the other non-conse^viive ver- tices meet in a point. MODERN GEOMETKY 345 in a circle, tht intersection N of a side with the tangent drawn at the oppo- s^ite vertex, and the intersections L, M, of the other non-consecutive side^ are three points in a straight line. By the same process we can reduce the hexagon to a quadrilateral and finally to a triangle ; whence the following corollaries. 31. Corollary II. In any quadri- 32. Corollary II. In any quadri- lateral inscribed in a circle, if tan- lateral circumscribed about a circle, if we take the points of contact of two gents are drawn at two consecutive vertices, the point of intersection of eavh of them with the side passing through the point of contact of the other, and the intersection of the other two sides, are three points in a straight line. 33. Corollary III. In any quadri- 29** adjacent sides, and join the point of contact of each side with the vertex on the other side, and if the remaining two vertices are joined, the three straight lines so drawn meet in a point 34. Corollary III. In any quadri- lato'al circumscrihed about a circle^ the straight lines joining the points of contact of opposite sides, and. the di- agonals, are four straight lines which meet in a point. 846 MODERN GEOMETRY lateral inscrihed in a circle, the inter- sections of the tangents drawn at op- posite vertices and the intersections of the opposite sides are four points iti a straight line. 35. Corollanj TV. In any triangle inscribed in a circle, the intersections 36. Corollary IV. In any triangU circumscribed about a circle, the straight lines joining the point of con- tact of each side with the opposite ver- tex meet in a point. of each side icith the tangent drawn at the opposite vertex are in a straight line. 37. ScJioJ/'um.. Pascal's Theorem (27) may be applied to the figure ABODE FA, formed by joining any six points of the circumference by consecutive straight Hnes in any oi'der whatever, a figure which may still be called a hexagon (non-convex), but which for distinction has been called a hexrigram. The demonstration (27) applies to this figure, word for word. Brianchon's Theorem (28) may also be extended to a circumscribed hexagram, formed by six tangents at any six points taken in any assumed order of succession. MODERN GEOMETRY. 347 HARMONIC PROPORTION. 38. Definition. Four points A, B, C, D, ! 1 \ are called four harmonic points when their ^ c B D anharmonic ratio [ABCD] is equal to unity ; that is, when CA_DA^ CA_DA CB ' DB ' ""^ CB D^ which agrees with the definition of harmonic points in (III. 76). 39. Definition. A harmonic pencil is a pencil of four rays whose anhar- monic ratio is equal to unity ; that is, a pencil 0, which determines upon any transversal a system of four harmonic points A, B, C, D. From (11) it follows that if one transversal of a pencil is divided harmonically, all other transversals of the pencil are also divided harmonically. The points A and B are called conjugate points with respect to C and 7>, that is, they divide the distance CD harmonically; and C and D < are called conjugate points with respect to A and /?, that is, they divide the distance AB harmonically (III. 76). In like manner, the rays OA and OB are called conjugate rays with respect to the rays OC and OD., and are said to divide the angle COD harmonicaUy ; and the rays OC and OD are con- jugate rays with respect to OA and Oi?, and divide tJie angle A OB har- monically. PROPOSITION IX.— THEOREM. 40. If a straight line AB is divided harmonically at the points C and D, the half of AB is a mean proportional between the distances of its middle point from the conjugate points C and D; that is, OB^ = OC. OD. For, the harmonic proportion, CA: CB = DA: DB, ^ > j, ^ j, gives, by composition and division ( III. 10) and (III. 9), CA — CB CA + CB DA — DB DA + DB or, OC:OB=OB: OD. [IJ 41. Corollary. Conversely, if we have given AB and its middle point 0, and if C and D are so taken that WB^ = OC. OD, then, A, B, C and D are fmir harmonic points. 848 MODERN GEOMETRY. For, the proportion [1] gives OB + OC .OB—OC=OD+OB: OB— OB; *hiit ii CA: CB-=DA: DB. [2] 42. Scholium. The three straight lines AC, AB, AD, are in liarmonic progression. For, the harmonic proportion [2] may be wiitten thus, AC : AD = AB — AC : AD — AB, or, AC, AB, AD, are such that the fir^t is to the third as the difference between the first and second is to the difference between the second and third ; that is, they are in harmonic progression, according to the definition commonly given in algebra. Of three straight lines ^C, AB, AD, in harmonic progression, the second AB is called a harmonic mean between the extremes AC and AD. PROPOSITION X.— THEOREM. 43. In a complete quadrilateral, each diagonal iorems. PROPOSITION XIV.— THEOREM. 60. The angle contained by two straight lines is equal to the angle con- tained by the straight lines joining their poles to the centre of the auxilicmj circle. For, the poles Pand Q of two straight lines AB and CD are situated respectively on the perpendiculars let fall from the centre of the auxiliary circle upon the lines AB and CJD (45). PROPOSITION XV.— THEOREM. 61. The ratio of the distances of any two points from the centre of the auxiUarij circle is equal to the ratio of the distances of each point from the polar of the other. (Salmon's Theorem. ) Let F and Q be the points, AB and CD their polars, PF the distance of P from CD, and QE the distance of Q from AB, and the centre of the auxiliary circle. Draw PM and OQJSf, which will be parallel to ^£'and /^/'^respectively; draw /*£r perpen- dicular to OiV^and ^^ perpendicular to OM. IP ^ is the radius of the circle, we have R' = OF. 0M= OQ. Oi\r(48), whence OP OQ ON om' The similar triangles POH and Q OK give 30 OP OQ OH OK 354 MODERN GEOMETRY. tberefbre (III. 12), OP^ 0N+ OH ^HN ^ PF OQ 0M-{- OK KM QE PROPOSITION XVI.— THEOREM. 62. The anharmonic ratio of four jioints in a straight line is equal to thai of the pencil formed hy the four polars of these points. For, the pencil formed by joining the four points to the centre of the circle has the same angles as the pencil formed by their polars (60), and these pencils have equal anharmonic ratios (13). PROPOSITION XVII.— PROBLEM. 63. It is a Imown theorem that the three perpendiculars from the vertices of a triangle to the opposite sides meet in a point ; it is required to determine its reciprocal theorem hy the method of reciprocal polars. Let the perpendiculars from the angles upon the opposite sides of the triangle ABC meet in P Let A'B' 6" be the reciprocal polar triangle oi'ABO, A^ being the pole of J5C, B^ the pole of AC, and C the pole of AB. The pole of the perpendicular J.P is a point L on the line B' C\ since B' C^ is the polar of ^ (51) ; the pole of BP is a point Jf on A^ C\ and the pole of CP is a point N on A'B\ The direct theorem being that the three lines AP^ BP, CP, meet in a point, the recipro- \ cal theorem will be that their poles X, M, N, are in a straight line, the polar of P] but we must express the reciprocal theorem in relation to the triangle A^B'C\ Now joining OL, OM, ON, and OA', OB', 0C\ the angle A' OL is aright angle, by (6U) ; and 80 also B' 031 and C' ON are right angles. Hence, the reciprocal theorem may be expressed as follows : If from any point in the plane of a triangle A'B'C, straight lines OA', OB', OC, are draivn to its vertices, the lines OL, OM^ ON, drawn at perpendicular respectively to the lines OA', OB', OC, meet the sides respectively opposite to the corresponding vertices in three points, L, M, N, which are in a straight line. MODERN GEOMETRY. 35/; RADICAL AXIS OF TWO CIRCLES. 64. Definition. If through a point /*, in the plane of a circle, a straight line is drawn cutting the circumference in the points A and B, the product of the segments into which the chord AB is divided at /*, namely, the product FA.PB, is constant; that is, independent of the direction of (he secant (III. 61 ). This constant product, depending upon the position of the point ^ P with respect to the circle, is called the power of the point with respect to the circle. If we consider especially the secant PCD, drawn through P and the centre of the circle, and designate the dis- tance PO by d and the radius of the circle by ?•, we have, when the point P is without the circle, PC = d — r, PD = d -{- r, and hence the power of the point is expressed by the product {d — r) [d + r) or d^ — r^. If the point P is within the circle, the absolute values of PC and PD are r — d and r -{- d; but the segments PC and 2*D lying in opposite directions with respect to P, are conceived to have opi)Osite algebraic signs, so that the product PC. PD must be negative ; hence the power of the point P is expressed by the product — (r — d) [r -{- d) = — (r^ — d^) '=d''^ — r^. Thus, in all cases, whether the point is without or within the circle., its power is expressed by the square of its distance from the centre diminished by the square of the radius. 65. When the point P is without the circle, its power is equal to the square of the tangent to the circle drawn from that point. When the point is on the circumference, its power is zero. PROPOSITION XVIIL— THEOREM. 66. The locus of all the points whose powers with respect to two given circles are equal, is a straight line perpendicular to the line joining the centra of the circles, and dividing this line so that the difference of the squares of the two segments is equal to the difference of the squares of the radii. liCt and 0^ be the centres of the two circles whose radii are r and r^ ; let P be any point whose distances from and 0^ are d and d'', then the powers of P with respect to the two circles are d^ — r^and d^^ — r^^, and these being equal, by hypothesis, we have d^ — r^ = <^^2 — ^2^ whence d'^ — d'^ ^r^ —r^^. Now, drawing PX perpendicular to 0\ we have from the right triangles POX, PO' X, OT' ~WX'' PO' PO'^ =d^ —d'^^r^—r"' 356 MODERN GEOMETRY. therefore, the quantity OX^ — O^X^, being equal Xor^ -~r^^, is constant, a»id X is a fixed point. Hence the point P is always in the perpendicular Fig. 1. io 0(y erected at the fixed point X; that is, this perpendicular is the locua of P. 67. Definition. The locus, PX^ of the points whose powers with respect to two given circles and 0^ are equal, is called the radical axis of the two circles. 68. Corollary I. If the two circles have no point in common, the radical axis does not intersect either of them. Fig. 1. If the circles intersect, the power of each of the points of intersection is equal to zero ; therefore, each of these points is a point in the radical axis ; hence, in the case of two intersecting circles^ their common chord is their radical axis. Fig. 2. If the circles touch each other, either externally or internally, their common tangent at the point of contact is their radical axis. 69. Corollary II. From (65) and (67) it follows that the tangents PT, PT", drawn to the two circles from any point of the radical axis witliout the circles^ are equal. Hence, if SS^ is a common tangent to the two circles, intersecting the radical axis in N, we have NS = NS\ Therefore, the radical axis can be constructed by joining the middle points of any two common tangents. PROPOSITION XIX.— THEOREM. 70. The radical axes of « system of three circles, taJcen two and two^ meet in a point. Let 0, 0^, 0^^, be the given circles. Designate the radical axis of 0^ and 0'' by X, that of and 0'' by X'\ and that of and 0' by X'\ The three centres not being in the same straight line, the axes X and X^, perpendicular to the intersecting Knes 00^^ and 0^^ 0\ will meet in a MODERN GEOMETRY. 357 certain point V. This point will have equal powers with respect to 0^ and 0'^, and with respect to and 0'^, consequently it will also have equal ij-// JT" powers with respect to and (9^, and is therefore a point in their radical axis A'^^''. - 71. Definition. The point in which the radical axes of a system of three circles meet is called the radical centre of the system. If the three centres of the circles are in a straight hne, the three axes are parallel, and the radical centre is at an infinite distance. 72. Definition. Two circles and 0^ intersect orthogonally^ that is, at right angles^ when their tangents at the point of intersection are at right angles, or, which is the same thing, when their radii, OT^ 0^2] drawn to the common point, are at right angles. Denoting 0^ by d, and the radii by r and r^, we have in the right triangle 0T0\ d^ — r^ = r^^ ; hence, when two cir- cles info'sect orthogonally, the square of the radius of either is equal to the power of its centre with respect to the other circle. PROPOSITION XX.— THEOREM. 73. The radical axis of two given circles is the locus of the centres jf a system of circles which intersect both the given cirdes orthogonally; and the line joining the centres of the given circles is the common radical axis of all *he circles of that system. Let P be the centre of any circle which cuts the two given circles and 0^ orthogonally; then, by (72), the powers of the point P with respect to the two circles are each equal to the square of the radius of the circle P, that is, equal to each other ; therefore, the centre P is in the radical axis of the two given circles. . Again, let P and Q be the centres of any two of the circles which cut both 358 MODERN GEOMETRY and 0' orthogonally. Since the circle cuts the two circles P and Q orthogonally, its centre lies in the radical axis of P and ^; and for the Fig. L Fig. 2. same reas( n the centre 0^ lies in the radical axis of P and Q\ therefore, the Hne 0^ is that radical axis, and is consequently the common radical axis of all the circles which cut both and 0^ orthogonally. 74. Scholium. When the given circles intersect, Fig. 1, the radius of any ^ne of the circles P, Q, etc. , is evidently less than the distance of its centre from 0\ and therefore no one of these circles cuts 0\ But when the circles have no point in common, Fig. 2 (whether one circle is wholly without the other, as in Fig. 2, or wholly within the other), all the circles, P, Q^ etc., cut the line 00^\ and since 00^ is their common radical axis, it is their common chord ; therefore, these circles all pass through two fixed points L and U in the line 00^. Also, since 03" is a tangent to the circle P, we have OL. OU =^ OT^ — ~UW ; therefore, the diameter ^jB is divided harmonically at L and U (41). For a like reason. A' B^ is divided harmonically at L and U. CENTRES OF SIMILITUDE OF TWO CIRCLES. 75. Definition, If the straight line joining the centres of two circles is divided externally and internally in the ratio of the corresponding radii, the points of section are called, respectively, the external and tJie internalcentres of similitude of the two circles. PROPOSITION XXI.— THEOREM. 76. Jf in two circles two parallel radii are drawn, one in each circle, the straight line joining their extremities intersects the line of centres in the exter- nal centre of similitude if the parallel radii are in the same direction, and in the internal centre of similitude if these radii are in opposite directions. MODERN GEOMETRY 359 For, OA and 0^ A^ being any two parallel radii in the same direction, and the line A^A intersecting the line of centres in /S^, the similar triangles SOA, SO'A', give SO : SO' = OA : O'A', and therefore, by the definition (75), S is the external centre of similitude. Also, OA and O'^Ai being parallel radii in opposite directions, and the line AAi intersecting the line of centres in T, the similar triangles TOA^ TO' A I, give TO : T0'= OA: O'A,, and therefore T is the internal centre of similitude. 77. Corollary I. It is easily shown that, conversely, if any transversal is drawn through a centre of similitude^ the radii drawn to the points in which it cuts the circumferences will he parallel^ two and two. Of the four points in which the transversal cuts the circumferences, two points at the extremities of parallel radii, as A and A\ or B and B\ are called homologous points; and two points at the extremities of non-parallel radii, as A and B', or B and A', are called anti-homologous points. 78. Corollary 11. Hence, if a transversal drawn through a centre of similitude is a tangent to one of the circles it is also a tangent to the other ; so that when one circle is wholly without the other., the centres of similitude are the intersections of the pairs of external and internal common tangents^ re^ectively. Fig. 1. If the circles touch each other externally (Fig. 1), the point of contact is 360 MODERN GEOMETRY. their internal centre of similitude. If they touch internally (Fig. 2), point of contact is their external centre of similitude. Fig. 8. If one circle is wholly within the other (Fig. 3), both (fehlres of similitude lie within both circles. 79. Corollary III. The distances (as SA and SA\ or TA and TA^, etxj.) of a centre of similitude from two homologous points are to each other as the radii of the circles. 80. Corollary IV. Since we have SO : SO' := TO : TO', the line 00^ is divided harmonically at S and T; that is, the centres o/ two circles and their two centres of similitude are four harmonic points. PltOPOSITION XXII.— THEOREM. 81. The product of the distances of a centre of similitude of two circles from two anti-homologous points is constant Let a transversal through the centre of similitude >S intersect the circum- ferences and 0' in the homologous points A, A\ and i^, B\ The line of centres intersects the circumferences in the homologous points M, M\ and N, N\ respectively. Hence, by (79), ON O'N' SA SA' SB SB' SM SM' SN . SN'' MODERN GEOMETRY. ^^^ irom which equations we deduce SA. SB' - SA\ SB = ^,X SA\ SB\ But >S^^'. 6B' = SM\ SN' (lU. 58) ; therefore we have SA. SB' = SA'. SB = SM. SN' = SM'. SN. The products SM. SN', SM'. SN, are constant ; therefore, the products SA.SB', SA'.SB, are constant. 82. Corollary I. Hence, if A and B' are anti-homologous points of one secant drawn through S, and b and a' are anti-homologous points of a second secant, we have SA.SB'= Sb.Sa'; therefore, the fcmr points A, B', a', b, lie on the circumference of a cirde 0". 83. Corollary II. The chords Ab, a'B', joining pairs of anti-homologous points in the two given circles, may be called anti-homologous chords. The chord Ab is the radical axis of the circles and 0" ; the chord a'B' is the radical axis of the circles 0' and 0" (68) ; and these intersect the radical axis PX o? the circles and 0' in the same point P (70). Hence, pairs of anti-homologous chords in two circles intersect on the radical axis of the circles. 84. Corollary III. K the secant Sa' approaches indefinitely to SA', the anti-homologous chords a' A', bB, approach indefinitely to the tangents at A' and B. Hence, at the limit, we infer that the tangents at two anti- homologous points in two circles intersect on the radical axis. PROPOSITION XXIII.— THEOREM. 85. Three circles being given, and considered when taken two and two cw f-orming three pairs of circles; then, 1st. The straight lines joining the centre of each circle and the internal centre of similitude of the other two meet in a point; 2d. The external centres of similitude of the three pairs of circles are in a straight line ; 3d. The external centre of similitude of any pair and the internal centres of similitude of the other two pairs are in a straight line. Let 0, 0', 0", be the given circles ; S and T the external and internal centres of simUitude of 0' and 0" ; S' and T' those of and 0" ; S" and T" those of and 0'. Let R, R' and R" denote the radii of the three circles. 31 362 MODERN GEOMETRY. •^s 1st By the definition (75) we haye T''0 R TO' R' T''(y R' TO'r R'' tihe product of wMcli equations gives T'' 0. TO'. T' 0'' R.R'.R'' T'O R ' 1, or T'' 0^. T0'\ T' R\R'\R T'' 0. T0\ T' O'' = T'' a. TO''. T' ; therefore, in the triangle OO'O" the three straight lines OT, 0'T\ 0"T", meet in a point (6). 2d. By the definition we also have S"0 ^R Sa_^R^ S'O" ^R'\ S"0' R'' SO" R"' S'O i? ' whence, by mnltiplying these equations together, S" 0. SO'. S' 0" = S" 0'. SO". S'O; MODERN GEOMETRY. 363 therefore, the points S, S% S^^, being in the ades (produced) of the tri- angle 00' 0^^, are in a straight line (3). 3d. The product of the equations T''0 R TO' R' S'O'' R'' T''0' R'' TO'' R"' S'O R gives T'' 0. TO'. S' a' = T'' 0'. TO". S' ; therefore, the points T^ T"^ /S^^, are in a straight line (3). In the same manner it is shown that jT, T'^ S", are in a straight line ; and T'^ T'\ S^ are in a straight hne. 86. Be/imtion. The straight line SS'S", on which the three external centres of similitude lie, is called the external axis of similitude of the three circles ; and the hues ST"T', S'T"T, S"T'T, are called the three inter- nal axes of similitude. PROPOSITION XXIV.— THEOREM. 87. If a variable circle touch two fixed circles^ the chord of contact passes through their external centre of similitude when the contax^ts are of the same kind [hoth external or both internal)^ and through their internal centre of similitude when the contacts are of different kinds. Let the circle C touch the circles and 0' in A and B' ; and let the chord of contact AB' cut the two circles again in B and A'. The lines OC, O'Cf pass through Uie points of contact Drawing the radii OB, O'A'^ Fig. 1. ihe isosceles triangles GAB\ OAB, 0'A'B\ are similar ; consequently the radii OA and O'A' are parallel. Therefore (76), the chord AB' passes through the ext<»rnal centre of similitude iS', when the contacts are of the 364 MODERN GEOMETRY. same kind (Fig. 1), and tlirough the internal centre of similitude T, when the contacts are of a different kind (Fig. 2). Fig. 2. • ^ 88. Corollary. If two variable circles C and c touch two Jixed circles O and 0^, their radical axis passes through the external centre of similitude of the fixed circles when the contacts of each of the two circles are of the same kind, and through their internal centre of similitude when these contacts are of different kinds. For, the four points of contact A, B% V, a, (Fig. 1), lie on the circumfer- ence of a circle (82) which may be designated as the circle Q. The chord AB^ is the radical axis of the circle Q and the circle C ; the chord ab^ is the radical axis of the circle Q and the circle c ; and these two axes meet the radical axis of the circles C and c in the same point (70), that is, in the point S (87). The proof is similar when the contacts are of different, kinds. PROPOSITION XXV.— THEOREM. 89. The radical axis of two circles which touch three given circles is an axis of similitude of the three given circles. Let the circles 31 and N (figure on next page) touch the three given circles 0, 0^, 0^^, the contacts of each of the two cu'cles being all of the same kind, that is, all internal in the case of the circle 31, and all external in the case of the circle H. Let *S', *S^^, S^^, be the three external centres of similitude of the given circles taken in pairs, so that SS^S'''' is their external axis of similitude (86). Since the circles 31 and N touch the two given circles 0^ and 0^^, and, the contacts are of the same kind in each case, the radical axis of 31 and A'' passes through >S' (88). For the same reason, it passes through S^ and through S^\ Therefore SS^S'^ is the radical axis of the circles 31 and N. In the same manner it may be shown that if each of the two circles 31 and N has like contacts with the pair of circles 0^ and 0^^, but unlike contacts with the other two pairs (that is, if 31 touches both 0^ and 0^^ internally and externally, and N touches both 0^ and 0^^ externally and internally), the radical axis of 31 and iV is the internal axis of similitude which passes through the external centre of similitude, S^ of the circles (/ and 0''. MODERN GEOMETRY. 365 ys .AS' 90. Scholium. There are in general eight different circles which can be drawn tangent to three given circles, and these eight circles exist in pairs the four radical axes of which are the four axes of similitude of the three given circles. 91. Corollary. I. When two circles M and N touch three given circles 0, 0^, 0^^, the three chords of contact mn, mfr/, m/^nf^ meet in a point V^ which is a centre of similitude of the two circles M and Kand the radical centre of the three given circles 0, 0\ 0^\ For, since the circle touches the circles M and iV", and the contacts are of different kinds, the chord of contact mn passes through the internal centre .31 » 366 MODERN GEOMETRY. of similitude V of M and N (87) ; and for the same reason the chords m^?/ and m^^'n^^ pass through V. Also, since the two circles and 0^ touch the two circles M and iV, and the contacts are of diiferent kinds, the radical axis of and 0' passes through the internal centre of similitude, F, of M and i\^ (88). For the same reason, the radical axis of 0^ and 0^^, and the radical axis of 0^^ and (y, pa^ through V. Therefore, V is the radical centre of the three circles (9, 0^ 0'\ 92. Corollary II. 7'Ae pole of the radical axis of the circles M and N, with reference to any one of the three given circles, lies in the chord of contact of that circle. Thus, in the case represented in the figure, the pole of SS^S^' with respect to the circle is a point F lying in the chord of contact run. For, let R hi the point of meeting of the tangents to the circle drawn at m and n. These tangents are equal and touch the circles M and N\ therefore, the point R is on the radical axis of M and iV, that is, upon the line SS^S^^. But mn is the polar of the point R with respect to the chcle (49), and therefore the pole of SS^S^^ with respect to the cii'cle is a point P on the chord of contact mn (51). 1 ROPOSITION XXVL— PROBLEM. 93. To desaihe a circle tangent to three given circles. As remarked in (90), there are in general eight solutions of this problem. The solutions may all be brought under two cases: viz. — 1st. A pair of circles can be found one of which will touch all the given circles internally, and the other will touch all the given circles externally. 2d. A pair of circles can be found one of which will touch the first of the given circles internally and the other two externally, and the other will touch the first externally and the other two internally. By taking each of the given circles successively as the "first," this second case gives six circles, thus making, in all, the eight solutions. The principles developed in the preceding proposition furnish the follow- ing simple and elegant solution of the problem, first given by Gergonne.* Let 0, 0^ 0^' (preceding figure) be the three given circles. Let SS^S^^ be their external axis of similitude and V their radical centre. Find the poles /*, P\ P'\ of SS' S^' with, respect to each of the given circles, and draw VF, VF\ VF", intersecting the three circles in the points m and «, m' and n\ m" and n" , respectively. The circumference described through the three points m, m^ ?>i ', will touch the three given circles internally; and the circumference described through the three points w, ?t^, n''^ will touch the tlnee given circles externally. By substituting successively each internal axis of similitude for SS^S^ vve obtain the other three i)airs of circles. y/ Annahs de MaifUmatiques, t. IV MODERN GEOMETRY. 367 94. Scholium. This general solution embraces the solution of ten distinct problems, special cases of the general problem, in which one or more of the given circles may be reduced to points (that is, circles of infinitely small radius) or to straight lines (that is, circles of infinitely great radius). EXERCISES, 1. If X and U are two fixed straight lines and a fixed point, and if through any two straight lines OAA\ OBB'^ are drawn cutting iy in ^ and B and U in A^ and i?^, find the locus of the intersection of the fines AB' and A'B (43). 2. If the three sides of a triangle pass through three fixed points which are in a straight line, and two vertices of the triangle move on two fixed straight lines, the third vertex moves on a straight line which passes through the intersection of the two fixed lines (25). 3. If the three vertices of a triangle move on three fixed straight lines which meet in a point, and two sides of the triangle pass through two fixed points, the third side passes through a fixed point which is in a straight line with the other two (24). 4. If Q is any point in the polar of a point P with respect to a given circle, the circle described upon FQ as a diameter cuts the given circle orthogonally (48). 5. Let the polars of any point P, with respect to two given circles and 0^, intersect in Q. Then, the circle described upon PQ as a diameter cuts both the given circles orthogonally, and its centre is on the radical axis of the given circles. 6. Describe a circumference which shall pass through a given point and cut two given circles orthogonally. 7. The polars of any point in the radical axis of two circles intersect on that axis. 8. The poles of the radical axis of two circles taken with respect to each circle, and the two centres of similitude of the circles, are four harmonic points. 9. The radical axis of two circles is equally distant from the two polars of either centre of similitude. 10. If the sides AB^ BC, CD^ DA, of a quadrilateral circumscribed about a circle whose centre is touch the circumference at the points F, F, G, II, respectively, and if the chords HE and GF meet in P, the line PO is perpendicular to the diagonal A C. 11. If a quadrilateral is divided into two other quadrilaterals by any secant, the intersections of the diagonals in the three quadrilaterals are in a straight line. 368 MODERN GEOMETRY. ] 2. The anharmonic ratio of four points on the circumference of a circle is equal to the ratio of the products of the opposite sides of the quadrilateral determined by these points. 13. If a series of circles ha\ing their centres in a given straight line cut a given circle orthogonally, they have a common radical axis, which is the perpendicular let fall from the centre of the given circle upon the given straight line. 14. The three circles described upon the diagonals of a complete quadri- lateral as diameters have a common radical axis and cut orthogonally the circle described about the triangle formed by the three diagonals. 15. Three circles Oi, O2, O3, being given, any fourth circle Q is described and the radical axes of Q and each of the given circles are drawn forming a triangle ABC. Another circle Q^ being drawn, a second triangle A'B'C^ is formed in the same manner. Prove that the triangles ABC 2i\A A'B'C are homological (24), (70). 16. If two triangles are reciprocal polars with respect to a circle, they are homological (51), (62), (20)! 17. K from the vertices of a triangle ABC perpendiculars J.a, ^6, Cc, are let fall upon the opposite sides, the three pairs of sides BC and fee, A C and ac, AB and afe, intersect on the radical axis of the circles circum- scribed about the triangles ABC and ahe (64, 67). 18. Any common tangent to two circles is divided harmonically by any circle which has a common radical axis with the two given circles (41). 19. If the sides of a quadrilateral ABCD inscribed in a circle are pro- duced to meet in E and F^ forming a complete quadrilateral, the square of the third diagonal EF is equal to the sum of the squares of the tangents from E and F\ and the tangent from the middle point of EF\& equal to one-half of EF. 20. Given the three diagonals of a complete quadrilateral inscribed in a given circle, it is required to construct the quadrilateral (4, 48, 49, 57). 21. Given three circles, it is required to describe a fourth such that the three radical axes of this circle, combined successively with each of the given ones, shall pass through three given points (Exercises 3 and 15). ■^'.^^^■ i-w; THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETTURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. FEB 201935 FEB 211935 TTCT m^ JUN 1 VM MQ 11 t045 LIBftAHV USE ONLY MAY ; a 19 84 CIRCULATION 0£PT. /> LD 21-100m-8,'34 ^\x\\^ ' *i*'^r. -^r*^;i ,I^:J"H V\\ ^ W \ i UNIVERSITY OF CALIFORNIA LlBRAR'^ I \\ X 1 \ '-n^ ^y\J ;\r--