I PROP. XXVIII. THEOR. F a folid parallelepiped be cut by a plane paffing through See N. the diagonals of two of the opposite planes ; it shall be cut in two equal parts. Let AB be a folid parallelepiped, and DE, CF, the diagonals of the opposite parallelograms AH, GB, viz. those which are drawn betwixt the equal angles in each: And because CD, FE are each of them parallel to GA, and not in the fame plane with it, CD, FF are parallel; wherefore the diagonals CF, 49. 11 DE are in the plane in which the pa rallels are, and are themselves paral lels: And the plane CDEF shall cut Because the triangle CGF is equal C 34. 1. D H d 24. II. E prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is N. B The insisting straight lines of a parallelepiped, men- S PROP. ΧΧΙΧ. ΤHEOR. OLID parallelepipeds upon the fame base, and of the se: N. fame altitude, the insisting straight lines of which are terminated in the same straight lines in the plane oppofite to the base, are equal to one another. P4 Let Book XI. Let the folid parallelepipeds AH, AK be upon the fame base ~ AB, and of the same altitude, and let their insisting straight See the fi- lines AF, AG, LM, LN, be terminated in the fame straight gures below. line FN, and CD, CE, BH, BK be terminated in the fame straight line DK; the solid AH is equal to the folid AK. a 28. 11. First, Let the parallelograms DG, HN, which are oppofite to the base AB, have a common fide HG: Then, because the folid AH is cut by the plane AGHC passing through the diagonals AG, CH of the opposite planes ALGF, CBHD, AH is cut into two equal parts by the plane AGHC: Therefore the folid AH is double of the prism which is contained be twixt the triangles ALG, CBH: For the fame reason, because the folid AK is cut by the plane LGHB through C B the diagonals LG, BH of the opposite planes ALNG, b 34. I c 38. I. d 36. 1. с 24. II. D H E. CBKH, the folid AK is double of the same prism which is contained betwixt the triangles ALG, CBH. Therefore the folid AH is equal to the folid AK. But, let the parallelograms DM, EN opposite to the base, have no common fide: Then, because CH, CK are parallelograms, CB is equal to each of the oppofise fides DH, EK; wherefore DH is equal to EK: Add, or take away the common part HE; then DE is equal to HK: Wherefore also the triangle CDE is equal to the triangle BHK: And the parallelogram DG is equald to the parallelogram HN: For the fame reason, the triangle AFG is equal to the triangle LMN, and the parallelogram CF is equal to the parallelogram BM, and C KDEHK CG to BN; for they are opposite. Therefore the prifm which is contained by the two triangles AFG, CDE, and the three parallelograms AD, DG, GC is equalf to the prifm, contain ed by the two triangles LMN, BHK, and the three parallelograins BM, MK, KL. If therefore the prism LMNBHK be taken from the solid of which the base is the parallelogram Book XI. AB, and in which FDKN is the one opposite to it, and if from this same solid there be taken the prism AFGCDE; the remaining solid, viz. the parallelepiped AH, is equal to the remaining parallelepiped AK. Therefore solid parallelepipeds, &c. Q. E. D. f C. II. taken SOLID parallelepipeds upon the fame base, and of the see N. same altitude, the insisting straight lines of which are not terminated in the same straight lines in the plane opposite to the base, are equal to one another. Let the parallelepipeds CM, CN be upon the same base AB, and of the fame altitude, but their insisting straight lines AF, AG, LM, LN, CD, CE, BH, BK not terminated in the same straight lines: The solids CM, CN are equal to one another. Produce FD, MH, and NG, KE, and let them meet one another in the points O, P, Q, R; and join AO, LP, BQ, CR: And because the plane LBHM is parallel to the opposite plane ACDF, and that the plane LBHM is that in which are the parallels LB, MHPQ, in which also is the figure BLPQ and the plane ACDF is that in which are the parallels AC, FDOR, in which also is the figure CAOR; therefore the figures BLPQ, CAOR are in parallel planes: In like manner, because the plane ALNG is parallel to the opposite plane CBKE, and that the plane ALNG is that in which are the parallels AL P3 Book XI. AL, OPGN, in which also is the figure ALPO; and the plane ☑ CBKE is that in which are the parallels CB, RQEK, in which also is the figure CBQR; therefore the figures ALPO, CBQR are in parallel planes: And the planes ACBL, ORQP are parallel; therefore the solid CP is a parallelepiped: But the solid CM, of which the base is ACBL, to which FDHM is the a 29. 11. opposite parallelogram, is equal to the folid CP, of which the See N. base is the parallelogram ACBL, to which ORQP is the one opposite; because they are upon the fame base, and their infifting straight lines AF, AO, CD, CR; LM, LP, BH, BQ are in the same straight lines FR, MQ: And the folid CP is equal to the solid CN; for they are upon the fame base ACBL, and their insisting straight lines AO, AG, LP, LN; CR, CE, BQ, BK are in the same straight lines ON, RK: Therefore the solid CM is equal to the folid CN. Wherefore solid parallele pipeds, &c. Q. E. D. PROP. XXXI. THEOR. SOLID parallelepipeds which are upon equal bafes, and of the fame altitude, are equal to one another. Let the folid parallelepipeds AE, CF, be upon equal bases AB, CD, and be of the fame altitude; the solid AE is equal to the folid CF. First, Let the insisting straight lines be at right angles to the bafes AB, CD, and let the bases be placed in the fame plane, and & 13. 11. and so as that the fides CL, LB be in a straight line; there- Book XI. fore the straight line LM, which is at right angles to the plane { in which the bases are, in the point L, is common to the two folids AE, CF; let the other insisting lines of the solids be AG, HK, BE; DF, OP, CN: And first, let the angle ALB be equal to the angle CLD; then AL, LD are in a straight line b. b 14. 1. Produce OD, HB, and let them meet in Q, and complete the folid parallelepiped LR, the base of which is the parallelogram LQ, and of which LM is one of its infisting straight lines : Therefore, because the parallelogram AB is equal to CD, as the bafe AB is to the base LQ, so is the base CD to the same c 7.5. LQ: And because the solid parallelepiped AR is cut by the plane LMEB, which is parallel to the opposite planes AK, DR; as the base AB is to the base LQ, so is the solid AE to the d 25. 11. folid LR: For the fame reason, because the folid parallelepiped CR is cut by the plane LMFD, which is parallel to the oppofite ( planes CP, BR; as R the bafe CD to the bafe AB to the base LQ, so the bafe CD to the base LQ, as before was proved : Therefore as the solid AE to the folid LR, so is the solid CF to the solid LR; and therefore the solid -AE is equal to the folid CF. € 9.5 But let the folid parallelepipeds SE, CF be upon equal bases SB, CD, and be of the fame altitude, and let their insisting ftraight lines be at right angles to the bases; and place the bafes SB, CD in the same plane, so that CL, LB be in a straight line; and let the angles SLB, CLD be unequal; the folid SE is also in this cafe equal to the solid CF: Produce DL, Ts until they meet in A, and from B draw BH parallel to DA; and let HB, OD produced meet in Q, and complete the solids AE, LR: Therefore the folid AE, of which the base is the parallelogram LE, and AK the one opposite to it, is equalf to the fo- f 29. 11. lid SE, of which the base is LE, and to which SX is opposite; for they are upon the fame base LE, and of the fame altitude, and their infisting straight lines, viz. LA, LS, BH, BT; MG, MV, EK, EX are in the same straight lines AT, GX: And becaufe |