Book XI. PROP. XXVIII. THEOR. IF a solid parallelopiped be cut by a plane passing See Note. through the diagonals of two of the opposite planes; it shall be cut in two equal parts. Let AB be a solid parallelopiped, 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 same plane with C it, CD, FE are parallel a; wherefore the diagonals CF, DE are a 9. 11. in the plane in which the parallels are, and are themselves parallels b; and the plane CDEF shall cut the solid AB into two equal parts. G Because the triangle CGF is equal e to the triangle CBF, and the triangle DAE to DHE; and that the parallelogram CA is equal d and similar to the opposite one BE; and the paral- A lelogram GE to CH: therefore the prism contained by the two triangles B b 16. 11. F c 34. 1. H d 24. 11. E CGF, DAE, and the three parallelograms CA, GE, EC, is equal e C. 11. to the prism contained by the two triangles CBF, DHE, and the three parallelograms BE, CH, EC; because they are contained by the same number of equal and similar planes, alike situated, and none of their solid angles are contained by more than three plane angles. Therefore the solid AB is cut into two equal parts by the plane CDEF. Q. E. D. N. B. The insisting straight lines of a parallelopiped, men⚫tioned in the next and some following propositions, are the sides ' of the parallelograms betwixt the base and the opposite plane 'parallel to it.' PROP. XXIX. THEOR. SOLID parallelopipeds upon the same base, and of See Note. the same altitude, the insisting straight lines of which are terminated in the same straight lines in the plane opposite to the base, are equal to one another. Book XI. See the figures below. Let the solid parallelopipeds AH, AK be upon the same base AB, and of the same altitude, and let their insisting straight lines AF, AG, LM, LN be terminated in the same straight line FN, and CD, CE, BH, BK be terminated in the same straight line DK; the solid Af is equal to the solid AK. First, Let the parallelograms DG, HN, which are opposite to the base AB, have a common side HG: then, because the solid AH is cut by the plane AGHC passing through the diagonals AG, CH of the opposite planes ALGF, CBHD, AH is cut into a 28. 11. two equal parts a by the plane AGHC: therefore the solid AH is double of the prism which is contained betwixt the triangles ALG, CBH: for the same reason, because the solid AK is D cut by the plane LGHB through the diagonals LG, BH of the op posite planes ALNG, CBKH, H K F G N But, Let the parallelograms DM, EN opposite to the base, have no common side: then, because CH, CK are parallelob 34. 1. grams, CB is equal to each of the opposite sides 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 DC is equal to the parallelogram HN: for the same reason, the triangle AFG is equal to the triangle LMN, and e 24. 11. the parallelogram CF is equal to the parallelogram BM, and c 38. 1. d 36. 1. : CG to BN; for they are opposite. Therefore the prism which is contained by the two triangles AFG, CDE, and the three pafC. 11. rallelograms AD, DG, GC is equal f to the prism contained by the two triangles LMN, BHK, and the three parallelagrams BM, MK, KL. If therefore the prism LMNBHK be taken from the solid of which the base is the parallelogram AB, Book XI. 記 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 parallelopiped AH, is equal to the remaining parallelopiped AK. Therefore solid parallelopipeds, &c. Q. E. D. PROP. XXX. THEOR. SOLID parallelopipeds upon the same base, and See Note. of the 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 parallelopipeds CM, CN be upon the same base AB, and of the same 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, OPGN, in which also is the Book XI. figure ALPO; and the plane CBKE is that in which are the pa rallels 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 parallelopiped; but the solid CM, of which the base is ACBL, to which a 29. 11. FDHM is the opposite parallelogram, is equal to the solid CP, of which the base is the parallelogram ACBL, to which ORQP is the one opposite; because they are upon the same base, and their insisting straight lines AF, AO, CD, CR; LM, LP, BH, BQ are in the same straight lines FR, MQ: and the solid CP is equal to the solid CN; for they are upon the same 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 solid CN. Wherefore solid parallelopipeds, &c. Q. E. D. See Note, PROP. XXXI. THEOR. SOLID parallelopipeds which are upon equal bases, and of the same altitude, are equal to one another. Let the solid parallelopipeds AE, CF be upon equal bases AB, CD, and be of the same altitude; the solid AE is equal to the solid CF. First, Let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be placed in the same plane, and so as that the sides CL, LB be in a straight line; therefore Book XI. the straight line LM, which is at right angles to the plane in r which the bases are, in the point L, is common a to the two, so- a 13. 11. lids 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. Pro- b 14. 1. duce OD, HB, and let them meet in Q, and complete the solid parallelopiped LR, the base of which is the parallelogram LQ, and of which LM is one of its insisting straight lines: therefore, because the parallelogram AB is equal to CD, as the base AB is to the base LQ, so is the base CD to the same LQ: and be- c7. 5. cause the solid parallelopiped 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 d the solid AE to the solid LR :d 25. 11. for the same reason, because the solid parallelopiped CR is cut by the plane LMFD, which is parallel to the opposite planes is the solid CF to the solid LR; and therefore the solid AE is equal to the solid CF. e 9.5: But let the solid parallelopipeds SE, CF be upon equal bases SB, CD, and be of the same altitude, and let their insisting straight lines be at right angles to the bases; and place the bases SB, CD in the same plane, so that CL, LB be in a straight line; and let the angles SLB, CLD be unequal; the solid SE is also in this case equal to the solid CF: produce DL, TS until they meet in A, and from B draw BH parallel to DA; and 'et HB, OD produced meet in Q, and complete the solids AE, LR: therefore the solid AE, of which the base is the parallelogram LE, and AK the one opposite to it, is equal to the solid SE, off 29. 11: which the base is LE, and to which SX is opposite : for they are upon the same base LE, and of the same altitude, and their insisting straight lines, viz. LA, LS, BH, BT; MG, MV, EK, EX, are in the same straight lines AT, GX: and because the |