magnitudes be continual proportionals, the firft is faid to have to Book XI. the fourth the triplicate ratio of that which it has to the second; therefore the folid AB has to the folid CD, the triplicate ratio e11.Def.5. of that which AB has to EX: But as AB is to EX, fo is AE to EC; therefore, the folid AB has to the folid CD, the triplicate ratio of that which the fide AE has to the homologous fide EC. Wherefore, fimilar folids, &c. Q. E. D. COR. From this it is manifeft, that, if four ftraight lines be continual proportionals, as the firft is to the fourth, fo is the folid parallelopiped defcribed from the first to the fimilar folid fimilarly defcribed from the fecond; because the first straight line has to the fourth the triplicate ratio of that which it has to the second. SOL 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. 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 folid CD: The ratio which the folid AB has to the folid CD is the fame with that which is compounded of the ratios of the fides AE to EC, EF to EG, and EH to EK. F L a A. 11. Let the folids AB, CD be placed on different fides of the fame plane, fo that the bafes AF, CG be in the fame plane, and AE, EC in the fame ftraight line; and because the planes FH, GK have the fame a inclination to the cominon plane, they coincide with one another; therefore, because the angles AEF, AEH are equal to CEG, CEK, the ftraight line EG is in the fame ftraight line with EF, and KE with EH: Complete the parallelogram GC and the folid parallelopiped GL, of which the bafe is GC and HE one of its infifting lines. Take any straight line a, and as AE to EC, fo & abc d E G K D b 12. 6. make a to b; and as FE to EG, fo 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 to e; but as AF to CG, 23. 6. Cca fo e 22. 5. d; therefore, by equality, the straight line a is to d. Book XI. fod is the folid AB, to the folid GL, because they are of the fame altitude; therefore the folid AB is to the folid GL, as a to d 32. 11. c: and the folid 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 the folid AB is to the folid CD, as But the ratio of a to d is faid to be of a to b, b to c, and c to d, which are the fame 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. f Def.A.5. compounded f of the ratios COR. I. Hence it is manifeft, 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 infifting line HE to the infifting line EK. COR. 2. Hence alfo, 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 infifting lines are the fame with their altitudes: and if they be contained by any other parallelograms, they have the fame ratio with thofe contained by rectangles, which are upon equal bafes, and of the fame altitudes with them. TH PROP. XXXIV. THEOR. See N. HE bafes and altitudes of equal folid parallelopipeds, are reciprocally proportional; and if the bafes and altitudes be reciprocally proportional, the folid parallelopipeds are equal. Let AB, CD be equal parallelopipeds; their bases are reciprocally proportional to their altitudes; that is, as the base EH is to the base NP, fo is the altitude of the folid CD to the alti tude of the folid AB. If the altitudes be equal, the folids are to one another as their bafes and the folids are equal; therefore the bafes are b A. 5. equal. a 32. II. 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 c. the point M draw MO perpendicular to the plane PN, and let M D e z. 6. gDef.4.11. it meet that plane in O; then MO is the altitude of the folid Book XI. CD: make OQ equal to the altitude of AB; and join OP, and draw QT parallel to OP: Therefore MP is to PT, as MO to d 31. 1. OQ; that is, as the altitude of CD to the altitude AB. Draw TV parallel to PC and let the plane QTV cut the planes of the folid CD in the straight lines TW, WX, VX: and because QT, TV are parallel to OP, PC, the plane TX is parallel to the plane PN; therefore CW is a parallelopiped §: and it is of f 15. 11. the fame altitude with the folid AB; therefore, as the folid AB to the folid CW, fo is the base EH to the base PN a: and because the folid CD is cut by the plane TX, which is parallel to its oppofite planes, as the folid DT is to the folid CW, fo is the base MV to the base VP, and the ftraight line MT to a B T W X P A E C h 25. 11. the ftraight line TP : and, by compofition', as the folid CD k 1.6. to the folid CW, fo is MP to PT, that is, the altitude of CD1 18. 5. to the altitude of AB: But the folid AB is to the folid CW, as the folid CD to the folid CW, because the solids AB, CD are equal: Wherefore, as the base HE to the base PN. fo " is the n 11. 5. altitude of the folid CD to the altitude of the folid AB. 12 Next, Let the bafes of the folids AB, CD be reciprocally proportional to their altitudes; that is, as the bafe EH to the bafe NP, fo is the altitude of the folid CD to the altitude of the folid AB: the folid AB is equal to the folid CD. If the altitudes be equal, the bafes are alfo equal; and there- b A. 5. fore the folids are equal TM. But, if the altitudes be not equal, m 31. 11. let CD have the greater altitude: and the fame construction being made, it may be demonftrated, as before, that the folid AB is to the folid CW, as the base HE to the base PN; and that the folid CD is to the folid CW, as the altitude of the folid CD to the altitude of the folid AB: But as the base HE to the base PN, fo is the altitude of the folid CD to the altitude of the folid AB; therefore, as the folid AB to the folid CW, fo is the folid CD to the folid CW. Wherefore the folid AB n 11. is equal to the folid CD. Therefore, &c. Q. E. D. 5. P 9. S. PROP. BOOK XI. See N. a 12. 1. PROP. XXXV. THEOR. F from the vertices of two equal plane angles, equal ftraight lines be elevated above their planes, making equal angles with their fides, each to each; the perpendiculars drawn from the extremities of the equal lines to the planes of the firft angles, shall be equal to one another. Let BAC, EDF be equal plane angles, and AG, DH equal ftraight 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 perpendicular to the plane BAC; it makes right angles with AK, b3.Def.11. KB; and the plane GKB paffing through it is perpendicular c 18. 11. to the plane BAC ; therefore AB, which is at right angles to their common fection BK, is at right angles to the plane BKG d; d4.Def.11. and ABG is therefore a right angle d. C For the fame reafons, the angles 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 e to DE, EH: e 26. 1. BG are equal A, D are contained by three and because the folid angles at K D F H plane angles equal to one another, each to each; the planes in f A. 5. which they are have the fame inclination to one another f : g6.Def.11. and the angle GBK is the inclination of the planes BAG, BAC, because BG, BK are at right angles to their common fection AB: and, for the fame reafon, 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 fides BG, EH oppofite to equal angles, are equal; therefore the other fides are equal, each to each: Wherefore the perpendicular GK is equal to the perpendicular HL. Therefore, &c. QE. D. COR. COR. Likewife, if from the vertices of two equal plane angles, Book XI. ftraight lines be elevated, making equal angles with their fides; they fhall have the fame inclination to the planes of the first angles. Let the ftraight 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 ftraight lines AG, DH to the planes BAC, m s. Def. EDF: The angle GAK is equal to the angle HDL. II. a 12. 1. Draw a KB, LE perpendiculars to AB, DE; and join BG, EH: and it may be proved, as in the propofition, that AB, BK are equal to DE, EL: and they contain right angles; therefore the base AK is equal to DL: and AG is also equal to DH; h 4. 1. therefore the two fides GA, AK are equal to the two HD, DL: and the base GK is alfo equal to the bafe HL; therefore the angle GAK is equal to the angle HDL. Q. E. D. PROP. XXXVI. THEOR. [F three ftraight lines be proportionals, the folid defcribed fides, is equal to the equilateral parallelopiped defcribed 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 folid defcribed from A, B, C is equal to the equilateral folid described from B, contained by parallelograms equiangular to those of the other figure. k 8. I. Take a folid angle D contained by three plane angles EDF, FDG, GDE; and make each of the ftraight lines ED, DF, DG |