Book XI DG equal to B, and complete the parallelopiped DH: Make EDF, FDG, GDE, each to each ; and make KN equal to B, the equal angles are reciprocally proportional; therefore the b 14. 6. parallelogram LM is equal to EF : and because EDF, LKM are two equal plane angles, and the equal lines DG, KN are equal angles with their sides; therefore the perpendiculars from C 35. 11. G, N to the planes EDF, LKM are equal €; and the solids KO, DH are therefore of the same altitude; and they are upon equal 31.11. bases LM, EF; therefore they are equal d to one another : But the solid KO is described from the three straight lines A, B, C, PROP. XXXVII. THEOR. IS solid parallelopipeds fimilarly described from them Let the four straight lines AB, CD, EF, GH be propora tionals, viz. AB to CD, as EF to GH; and let the fimilar parallelopipeds AK, CL, EM, GN be similarly described from them; AK shall be to CL, as EM to GN. Make a AB, CD, O, P continual proportionals, as also EF, GH, Q. R: and because AB is to CD, as EF to GH; and that EM to the folid GN; therefore 6, as the solid AK to the folid But let the solid AK be to the folid CL, as the solid EM to Take AB to CD, as į EF to ST, and from ST describe f the f 27.11. parallelopiped SV similar and similarly situated to EM or GN: and a 11, 6. II. e 12. 6. and because AB is to CD, as FF to ST, and that from AB, CD Book XI. the fimilar parallelopipeds AK, CL are fimilarly described; and in like manner, the folids EM, V from EF, ST, therefore AK is to CL, as EM to SV: But, by the hypothesis, AK is to CL, as EM to GN; therefore GN is equal to SV: But it is like. 8 9. S. wise fimilar and fimilarly situated to SV; therefore the planes which contain the solids are similar and equal, and their homologous fides GH, ST are equal to one another: and because, as AB to CD, fo is EF to ST, and that ST is equal to GH; AB is to CD, as EF to GH. Therefore, if four straight lines, &c, Q. E. D. PROP. XXXVIII. THEOR. See N. I perpendicular to the other, shall fall on the common section of the planes. For a straight line drawn from that point perpendicular to the common section of the planes is perpendicular to the other plane * ; and only one straight line can be drawn from that point a 4.Def.11. perpendicular to that other plane, b 13. 11. PROP. XXXIX. THEOR. a solid parallelopiped, if the sides of two of the opposite planes be bisected, the common section of the planes passing through the points of division, and the diameter of the parallelopiped bisect each other. Dd Let Воок XI. . Let the sides of the opposite planes CF, AH of the parallelo piped AF be bisected in the points K, L, M, N; X, O, P, R, and join KL, MN, XO, PR: and because DK, CL are equal a 33. 1. and parallel, KL is parallel to DC: For the same reason, MN is parallel to BA: But BA is parallel to CD; therefore MN is b 9. 11. parallel to CD; and therefore also to KL. In like manner, it may be proved, that XO is parallel to PR. Let the planes passing through the parallels cut one another in the straight lines YS; and let DG be the diameter of the parallelopiped AF: : YS and DG do meet and bifect one another. Jon DE, BG: and because DF, DC are doubles of DK, DX, the parallelograms CF, XK are similar and similarly fituated, and having the angle ar D common c to both, they are about the same dia X the diameter DE DK is equal to KF, d 2. 6. & DY is equal d to YE. 9. 5. For the same reason, P M to EG: and DE, BG join their extremities; therefore DE is equal and parallel a to BG: and DG, YS are in the same plane with them; therefore they must meet one another ; let them meet in T: and because DE is parallel to BG, the altere 29. 1. nate angles EDT, BGT are equal e; and the angle DTY is f 15. i. equal i to GTS: Therefore the triangles DTY, GTS have two angles of the one equal to two of the other, and the side DY equal to GS, which are opposite to equal angles; therefore the 8 26. 1. other fides are equal , each to each. Wherefore DT is equal to TG, and YT equal to TS. Therefore, &c. Q. E. D. b PROP, I PROP. XL. THEOR. Book XI. F there be two triangular prisms of the same alti tude, the base of one of which is a parallelogram, and the base of the other a triangle; if the parallelogram be double of the triangle, the prisms shall be equal to one another. Let the prism AEB-CFD standing upon the parallelograin AF, and contained by the planes AEB, CFD, AD, DE, be of the same altitude with the prism GHK L M N ftanding upon the triangle GHK, and contained by the planes GM, MK, KL, MLN: and let the parallelogram AF be double of the triangle GHK, the prism AEB.CFD is equal to the prism GHK-LMN. Complete the folids AX, GO; and because the parallelogram AF is double of the triangle GHK ; and the parallelogram HK is double a of the same triangle ; therefore the parallelogram AF a 34. I. is equal to HK. But parallelopipeds upon equal bases, and of the same altitude, are equal to one another; therefore the solid b 31. 11. AX is equal to the solid GO: and the prism AEB-CFD is the half c of the solid AX, and the prism GHK-LMN the half c of c 28. 11. the folid GO; therefore the prism AEB-CFD is equal to the prism GAK-LMN. Wherefore, &c. Q. E. D. |