taken from the folid of which the base is the parallelogram AB, Book XI. and in which FDKN is the one opposite to it; and if from this fame solid there be taken the prism AFG, CDE; the remaining folid, viz. the parallelepiped AH, is equal to the remaining parallelepiped AK, Therefore solid parallelepipeds, &c. Q. E. D. PROP. XXX. THEOR. OLID parallelepipeds upon the same base, and of See N. the same altitude, the infifting straight lines of which are not terminated in the fame 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 same altitude, but their insisting straight lines AF, AG, LM, LN, CD, CE, BH, BK not terminated in the same straight lines. the folids 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 Book XI. are the parallels AL, OPGN, in which also is the figure ALPO; and the plane CBKE is that in which are the parallels CB, ROEK, in which also is the figure CBOR; 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 opposite a. 29. II. parallelogram, is equal a to the solid CP of which the base is the parallelogram ACBL, to which OROP 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 a to the folid 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 fame straight lines ON, RK. therefore the solid CM is equal to the solid CN. Wherefore solid parallelepipeds, &c. Q. E. D. See N. OLID parallelepipeds which are upon equal bases, another. Let the folid parallelepipeds AE, CF, be upon equal bafes 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 bases AB, CD, and let the bases be placed in the same plane, and so as G that the sides CL, LB be in a straight line; therefore the straight Book XI. line LM, which is at right angles to the plane in which the bases are, in the point L, is common a to the two solids AE, CF; let the a. 13. 11. 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. produce OD, HB, and let them b. 14. It meet in Q, and complete the solid parallelepiped 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, fo is the base c. 7. 5. CD to the fame LQ. and because the solid parallelepiped AR is cut by the plané LMEB which is parallel to the opposite planes AK, DR; as the base AB is to the base LQ , fo is d the folid d. 25. 11. AE to the solid LR. for the same reason, because the folid parallelepiped CR is cut by the plane LMFD which is parallel to the opposite planes CP, BR; as the base CD P K R. to the base LQ , so N is the solid CF to the V X G) folid LR. but as the K D base AB to the base o Q LQ, so the base B CD to the base LQ, C I as before was prov AS HT ed. therefore as the folid AE to the folid LR, so is the folid CF to the folid LR; and therefore the folid AE is equal e to the folid CF. e. 9. 5 But let the solid parallelepipeds 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 folid 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 let 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 f to the solid SE, of which the base is LE, f. 29. 11. and to which SX is opposite; for they are upon the fame base LE, and of the same altitude, and their infifting straight lines, viz. LA, LS, BH, BT; MG, MV, EK, EX are in the same straight P g. 35. I. Book XI. lines AT, GX. and because the parallelogram AB is equal to SB, for they are upon the same base LB, and between the same paral- R V X is equal to the base K D e B angle CLD, ther- C L fore, by the first case, AS HT the solid AE is equal to the solid CF; but the folid AE is equal to the solid SE, as was demonstrated; therefore the solid SE is equal to the solid CF. But if the insisting straight lines AG, HK, BE, LM; CN, RS, DF, OP, be not at right angles to the bafes AB, CD; in this case likewise the folid AE is equal to the solid CF. from the points G, K, E, M; N, S, F, P, draw the straight lines GQ, KT, EV, MX; h. II. II. NY, SZ, FI, PU, perpendicular h to the plane in which are the bases AB, CD; and let them meet it in the points Q, T, V, X; i. 6. II. Y, Z, I, U, and join QT, TV, VX, XQ; YZ, ZI, IU, UY. then because GQ, KT, are at right angles to the same plane, they are parallel i to one another. and MG, EK are parallels; therefore the planes MQ, ET of which one passes through MG, GQ, and the other through EK, KT which are parallel to MG, GQ, k. 15. 11. and not in the same plane with them, are parallel k to one another. for the same reason, the planes MV, GT are parallel to one another. therefore the folid QE is a parallelepiped. in like manner, it may be proved, that the folid YF is a parallelepiped. but, from what has been demonstrated, the solid EQ is equal to the solid FY, because they are upon equal bases MK, PS, and of the same altitude, and have their insisting straight lines at right angles to the bases. and Book XI. the solid EQ_is equal to the solid AE; and the folid FY to 1. 29. or 30. the folid CF; because they are upon the fame bafes and of the fame altitude. therefore the solid AE is equal to the solid CF. Wherefore folid parallelepipeds, &c. Q. E. D. II. OLID parallelepipeds which have the same altitude, Sec N. are to one another as their bases. Let AB, CD be folid parallelepipeds of the same altitude. they are to one another as their bases; that is, as the base AE to the bafe CF, so the solid AB to the solid CD. To the straight line FG-apply the parallelogram FH equal a to a. Cor.45. In AE, so that the angle FGH be equal to the angle LCG; and complete the solid parallelepiped GK upon the base FH, one of whose insisting lines is FD, whereby the folids CD, GK must be of the fame altitude. therefore the folid AB is equal to the solid GK, b. 31. II. because they are upon equal bases B D K AE, FH, and are of the same alti. N 0 tude. and because L F the folid parallelepiped CK is cut by the plane DG A M C G H which is parallel to its opposite planes, the base HF is c to the base FC, as the folid C. 25. 11, HD to the folid DC. but the base HF is equal to the base AE, and the solid GK to the solid AB. therefore as the base AE to the base CF, so is the folid AB to the folid CD. Wherefore folid paralielepipeds, &c. Q. E. D. Cor. From this it is manifest that prisms upon triangular bases, of the same altitude, are to one another as their bases. Let the prisms the bases of which are the triangles AEM, CFG, and NBO, PDQ the triangles opposite to them, have the same altitude; and complete the parallelograms AE, CF, and the solid parallelepipeds AB, CD, in the first of which let MO, and in the other let GQ be one of the insisting lines, and because the folid parallelepipeds AB, CD have the same altitude, they are to one |