magnitudes be continual proportionals, the first is said to have to Book XI. the fourth the triplicate ratio of that which it has to the secondo; therefore the folid AB has to the solid CD, the triplicate ratio eu.Def.5. of that which AB has to EX: But as AB is to EX, so is AE to EC; therefore, the folid AB has to the folid CD, the triplicate ratio of that which the side AE has to the homologous fide EC. Wherefore, fiinilar folids, &c. Q. E. D. Cor. From this it is manifest, that, if four straight lines be continual proportionals, as the first is to the fourth, so is the solid parallelopiped described from the first to the similar folid fimilarly described from the second; because the first straight line has to the fourth the triplicate ratio of that which it has to the second. PROP. D. THEOR. OLID parallelopipeds contained by parallelograms equiangular to one another, each to each, have to one another the ratio which is compounded of the ratios of their fides. ܪ Let AB, CD be parallelopipeds, of which AB is contained by the parallelograms AF, FH, AH equiangular, each to each, to the parallelograms CG, GK, CK which contain the solid CD: The ratio which the folid AB has to the solid CD is the same with that which is compounded of the ratios of the fides AE to EC, EF to EG, and EH 10 EK. Let the folids AB, CD be placed on different fides of the same plane, so that the bases AF, CG be in the same plane, and AE, EC in the same straight line ; and because the planes FH, GK have B the same a inclination to the com. L a A, II. inon plane, they coincide with one another; therefore, because the angles AEF, AEH are equal to CEG, CEK, the straight line EG is in the same straight line with С EF, and KE with EH : Complete the parallelogram GC and the G folid parallelopiped GL, of which KA the base is GC and HE one of its insisting lines. Take any straight D a b c line a, and as AE to EC, so 6 b 12. 6. make a to b; and as FE to EG, so make b to c; and as HE to EK, fo make c to d: Then, because the parallelogram AF is equiangular to CG, AF is to CG, as a toc®; but as AF to CG, C 23.6. Cca To Book XI. so d is the solid AB, to the folid GL, because they are of the mfame altitude; therefore the folid AB is to the folid GL, as a to d 32. 11. c: and the solid GL is to the folid CD, as & the base GH is to the base GK; that is, as the straight line EH to EK, or as c to e 22. 5. d; therefore, by equality, the solid AB is to the solid CD, as the straight line a is to d. But the ratio of a to d is said to be f Def.A.5. compounded f of the ratios of a to b, b to c, and c to d, which are the same with the ratios of the fides AE to EC, FE to EG, and HE to EK: Therefore the folid AB has to the folid CD the ratio which is the fame with that which is compounded of the ratios of the fides AE to EC, FE to EG, and HE to EK. Wherefore, &c. Q. E. D. Cor. 1. Hence it is manifest, that equiangular parallelopipeds have to one another the ratio which is compounded of the ratios of their bases, and of their infifting lines. For the ratio of AB to CD is compounded of the ratios of AB to GL, and GL to CD; that is, of the ratios of the base AF to the base GC, and of the insisting line HE to the insisting line EK. Cor. 2. Hence also, any parallelopipeds have to one another the ratio which is compounded of the ratios of their bases and of their altitudes. For if they be contained by rectangles, they are equiangular, and their insisting lines are the same with their altitudes : and if they be contained by any other parallelograms, they have the same ratio with those contained by rectangles, which are upon equal bases, and of the same altitudes with them. PROP. XXXIV. THEOR. See N. HE bases and altitudes of equal folid parallelo pipeds, are reciprocally proportional; and if the bases and altitudes de reciprocally proportional, the solid parallelopípeds are equal. Let AB, CD be equal parallelopipeds; their bases are reciprocally proportional to their oltitudes ; that is, as the base EH is to the base NP, fo is the altitude of the solid CD to the alci. tude of the solid AB. If the altitudes be equal, the solids are to one another as their bases 8: and the solids are equal ; therefore the bases are b A. s. equal But let the altitudes not be equal, and let the altitude of CD be the greater: and if MP be at right angles to the base PN, it is the altitude of CD: make PT equal to the altitude of AB. But, if PM be not at right angles to the base PN, from C1n. the point M draw MO perpendicular to the plane PN, and let a 32. II it meet that plane in 0; then MO is the altitude of the solid Book XI. g Def.4.11. folid AB; therefore, as the M D B T W X a PS folid CW, fo is the base h 25. 11. MV to the base VP, and A. C N the straight line MT to the straight line TP*: and, by composition', as the solid CD k 1.6. to the folid CW, so is MP to PT, that is, the altitude of CD to the altitude of AB: But the folid AB is to the solid CW, as the folid CD to the solid CW, because the solids AB, CD are equal : Wherefore, as the base HE to the base PN, 1o ? is the n 11. 5. altitude of the solid CD to the altitude of the solid AB. Next, Let the bases of the solids AB, CD be reciprocally proportional to their altitudes; that is, as the base EH to the base NP, fo is the altitude of the solid CD to the altitude of the folid AB: the solid AB is equal to the folid CD. If the altitudes be equal, the bases are also equal b; and there. b A. s. fore the folids are equal m. But, if the altitudes be not equal, m 31. 15. let CD have the greater altitude: and the same construction being made, it may be demonstrated, as before, that the solid AB is to the solid CW, as the base HE to the base PN ; and that the solid CD is to the solid CW, as the altitude of the folid CD to the aliitude of the folid AB: But as the base HE to the base PN, so is the altitude of the folid CD to the altitude of the folid AB; therefore, as the folid AB to the folid CW, so n is the folid CD to the folid CW. Wherefore the solid AB is equal P to the folid CD. Therefore, &c. Q. E. D. P 9. So 11. 5. PROP. Book XI. PROP. XXXV. THEOR. See N. TF from the vertices of two equal plane angles, ; a 12. 1. C 18. Il. making equal angles with their fides, each to each ; the perpendiculars drawn from the extremities of the equal lines to the planes of the first angles, shall be equal to one another. Let BAC, EDF be equal plane angles, and AG, DH equal straight lines elevated above the planes BAC, EDF, making the angle BAG equal to EDH, and GAC to HDF; and from G, H let GK, HL be drawn perpendicular to the planes BAC, EDF, meeting them in K, L: the perpendicular GK is equal to HL. From the points K, L draw a KB, LE perpendiculars to AB, DE ; and join BG, AK; EH, DL: and because GK is per pendicular to the plane BAC; it makes right angles with AK, b 3.Def.us. KB b; and the plane GKB paffing through it is perpendicular to the plane BAC °; therefore AB, which is at right angles to their coinmon section BK, is at right angles to the plane BKG d; d4.Def.il. and ABG is therefore a right angle a. For the same reasons, the apgles HLD, HLE, HED are right angles : and because the angle BAG is equal to EDH, and ABG, DEH are right angles ; the two triangles ABG, DEH have two angles equal to two; and the sides AG, DH opposite to equal angles, are also equal, B therefore the other fides AB, e 26.1. BG are equal to DE, EH: F. and because the solid angles at K H plane angles equal to one another, each to each ; the planes in f A. s. which they are have the same inclination to one another gó.Defolt. and the angle GBK is the inclination of the planes BAG, BAC, because BG, BK are at right angles to their common section AB: and, for the same reason, the angle HEL is the inclination of the planes EDH, EDF; therefore the angle GBK is equal to HEL: and BKG, ELH are right angles ; therefore, in the triangles GBK, HEL, there are two angles equal to two: and the sides BG, EH opposite to equal angles, are equal; therefore the other sides are equal o, each to each : Wherefore the perpendicular GK is equal to the perpendicular HL. Therefore, &c. Q. E. D. Cor. Cor. Likewise, if from the vertices of two equal plane angles, Book XI. straight lines be elevated, making equal angles with their fides; they shall have the same inclination to the planes of the first angles. Let the straight lines AG, DH be elevated above the planes of the equal angles BAC, EDF, and making the angle BAG equal to EDH, and the angle GAC to HDF. And take AG equal to DH, and draw GK, HL perpendiculars to the planes BAC, EDF, and join AK, DL: the angles GAK, HDL are the inclination of the straight lines AG, DH to the planes BAC, m s. Del. EDF: The angle GAK is equal to the angle HDL. Draw a KB, LE perpendiculars to AB, DE; and join BG, a 12. . EH: and it may be proved, as in the proposition, that AB, BK are equal to DE, EL: and they contain right angles ; therefore the base AK is equal to to DL: and AG is also equal to DH; 54. 1. therefore the two sides GA, AK are equal to the two HD, DL: and the base GK is also equal to the base HL; therefore the angle GAK is equal k to the angle HDL. Q. E. D. k 8. I. PROP. XXXVI. THEOR. F three straight lines be proportionals, the solid I fides, is equal to the equilateral parallelopiped described from the mean proportional, equiangular to the other, Let A, B, C be three proportionals; that is, A to B, as B to C: The solid described from A, B, C is equal to the equilateral solid described from B, contained by parallelograms equiangular to those of the other figure. Take a solid angle D contained by three plane angles EDF, FDG, GDE ; and make each of the straight lines ED, DF, DG |