Suppl. a 34. 1. b 38. 1. c 36. 1. Because CH, CK are parallelograms, CB is equala 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 the triangle CDE is equal to the triangle BHK, and the parallelogram DG is equal to the parallelogram HN. For the same reason the triangle AFG is equal to the triangle LMN. The parallelogram CF is d 2.3. Sup. equald to the parallelogram BM, and CG to BN, because they are opposite. Therefore the planes which contain the prism DAG are similar and equal to the corresponding planes which contain the prism HLN, each to each; and the contie15.2. Sup. guous planes are also equally inclined to one another, because the parallel planes AD, LH, and AE, LK are cut by the same plane DN; therefore the prisms DAG, HLN are f 1. 3. Sup. equalf. Therefore, if the prism LNH be taken from the solid, of which the base is the parallelogram AB, and FDKN the plane opposite to the base; and if the prism AGD be taken from the same solid, the remaining parallelepiped AH is equal to the remaining parallelepiped AK. Therefore, solid parallelepipeds, &c. Q. E. D. PROP. VI. THEOR. 1 SOLID parallelepipeds upon the same base and 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 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 Book III. the same straight lines; the solids CM, CN are equal to each other. 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. The planes LBHM and ACDF are parallela, and the plane LBHM is that in which are the parallels LB, MHPQa, a Def. 5. and in which also is the figure BLPQ; and the plane ACDF 3. Sup. is that in which are the parallels AC, FDOR, and in which also is the figure CAOR; therefore the figures BLPQ, CAOR are in parallel planes. In like manner the planes ALNG and CBKE are parallel, and the plane ALNG is that in which are the parallels AL, OPGN, and in which also is the figure ALPO; and the plane CBKE is that in which are the parallels CB, RQEK, and in which also is the figure CBQR; therefore the figures ALPO, CBQR are in parallel planes. But the planes ACBL, ORQP are also parallel; therefore the solid CP is a parallelepiped. Now the solid parallelepiped CM is equal to the solid parallelepiped CP, b 5. 3. Sup. because they are upon the same base, and their insisting straight lines AF, AO, CD, CR, and LM, LP, BH, BQ are terminated in the same straight lines FR, MQ. But 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, and CR, CE, BQ, BK are terminated in the same straight lines ON, RK. Therefore the solid CM is equal to the solid CN. Wherefore, solid parallelepipeds, &c. Q. E. D. Suppl. PROP. VII. THEOR. SOLID parallelepipeds, which are upon equal bases and of the same altitude, are equal to one another. Let the solid parallelepipeds AE, CF be upon equal bases AB, CD, and of the same altitude; the solid AE is equal to the solid CF. Case 1. Let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be placed in the same plane, so that the sides CL, LB may be in a straight line; therefore the straight line LM, which is at right angles to the a 11.2. Sup. plane in which the bases are, in the point L, is common to the two solids AE, CF. Let the other insisting lines of the solids be AG, HK, BE, and DF, OP, CN; and, first, let the angle ALB be equal to the angle CLD; then AL, LD are in b 14. 1. c 7.5. a straight lineb. Produce OD, HB, and let them meet in Q, and complete the solid parallelepiped LR, of which the base is the parallelogram LQ, and LM one of its insisting straight lines. Because the parallelogram AB is equal to CD, the base AB: base LQ:: base CD: base LQ. Because the solid parallelepiped AR is cut by the plane LMEB, which is parallel to the opposite planes AK, DR, the base AB : d 3.3. Sup. base LQ :: solid AE: solid LRd; and because the solid parallelepiped CR is cut by the plane LMFD, which is parallel to the opposite planes CP, BR, the base CD: base LQ:: solid CF: solid LR. Therefore the solid AE : solid Book III. LR:: solid CF: LR; therefore the solid AE is equal to the solid CFf. e 11. 5. f 9.5. Let the solid parallelepipeds SE, CF be upon upon equal bases SB, CD, and of the same altitude, and let their insisting straight lines be at right angles to the bases; place the bases SB, CD in the same plane, so that CL, LB may be in a straight line; and let the angles SLB, CLD be unequal; the solid SE is equal to the solid CF. Produce DL, TS until they meet in A, and from B draw BH parallel to DA; let HB, OD produced meet in Q, and complete the solids AE, LR. The solid AE, of which the base is the parallelogram LE, and AK the plane opposite to it, is equals to the solid g 5.3. Sup. SE, of which the base is LE, and SX the plane opposite; for they are upon the same base LE, and of the same altitude, and their insisting straight lines LA, LS, BH, BT, and MG, MU, EK, EX are in the same straight lines AT, GX. The parallelogram AB is equal to SB, for they are upon the same h 35. 1. base LB, and between the same parallels LB, AT; and the base SB is equal to the base CD; therefore the base AB is equal to the base CD. But the angle ALB is equal to the angle CLD; therefore, as before, the solid AE is equal to the solid CF. But the solid AE has been proved to be equal to the solid SE; therefore the solid SE is equal to the solid CF. Case 2. If the insisting straight lines AG, HK, BE, LM, and CN, RS, DF, OP be not at right angles to the bases AB, CD, in this case likewise the solid AE is equal to the solid : CF. If two solid parallelepipeds be constituted on the bases with their insisting lines perpendicular to their bases, they Suppl. will be equali to the solids AE, CF; and by the first case of this proposition, they will be equal to each other; wherefore i 6. 3. Sup. the solids AE, CF are also equal. Wherefore, solid parallelepipeds, &c. Q. E. D./ 1 PROP. VIII. THEOR. / SOLID parallelepipeds, which have the same altitude, are to one another as their bases. Let AB, CD be solid parallelepipeds of the same altitude; the base AE: base CF :: solid AB: solid CD. a Cor. 45. 1. To the straight line FG apply the parallelogram FH equala to AE, so that the angle FGH may 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 so lids CD, GK must be of the same altitude. Therefore the b7.3. Sup. solid AB is equal to the solid GK, because they are upon equal bases AE, FH, and are of the same altitude. Because the solid parallelepiped CK is cut by the plane DG, c 3.3. Sup. which is parallel to its opposite planes, the base HF is to the base FC as the solid HD to the solid DC, But the base HF is equal to the base AE, and the solid GK to the solid AB; therefore the base AE is to the base CF as the solid AB is to the solid CD. Wherefore, solid parallelepipeds, &c. Q. E. D. / COR. 1. From this it is manifest that prisms upon triangular bases, and of the same altitude, are to one another as their bases, |