is the figure BLPQ; and because the plane ACDF is Book III. 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, because 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; LM, LP, BH, BQ are terminated 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 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. b Supplement b 14. 1. 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 be 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, and so as that the sides CL, LB be in a straight line; therefore the straight line LM, which is at right angles to the plane in which the bases are, in a 11.2.Sup. the point L, is common to the two solids 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". Produce OD, HB, and let them 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, so is the base CD to the same 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 e 7. 5. .3.3. Sup. to the base LQ, so is the solid AE to the solid LR: for the same reason, because the solid parallelepiped CR is cut by the plane LMFD, which is parallel to the oppo- Book III. site planes CP, BR; as the base CD to the base LQ; so is the solid CF to the solid LR: but as the base AB to the base LQ, so the base CD to the base LQ, as has been proved therefore, as the solid AE to the solid LR, so is the solid CF to the solid LR; and therefore the solid AE is equal to the solid CF. e f € 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 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 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 plane opposite to it, is equal to f 5. 3. Sup. the solid 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, viz. LA, LS, BH, BT; MG, MU, EK, EX are in the same straight lines AT, GX: and because the parallelogram AB is equal to SB, for they are upon the same base g 35 1. LB, and between the same parallels LB, AT; and because 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, by the first case, the solid AE is equal to the solid CF; but the solid AE is equal to the solid SE, as was demonstrated; therefore the solid SE is equal to the solid CF. g CASE 2. If the insisting straight lines AG, HK, BE, Supplement LM; 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. Because solid parallelepipeds on the b 6. 3. Sup. same base, and of the same altitude, are equal h, if two solid parallelepipeds be constituted on the bases AB and CD of the same altitude with the solids AE and CF, and with their insisting lines perpendicular to their bases, they will be equal to the solids AE and CF; and, by the first case of this proposition, they will be equal to one another; wherefore the solids AE and CF are also equal. Wherefore, solid parallelepipeds, &c. Q. E. D. 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: they are to one another as their bases; that is, as the base AE to the base CF, so is the solid AB to the solid CD. To the straight line FG apply the parallelogram FH a Cor. 45. 1. equal a to 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 solids CD, GK must be of the same altitude. b 7. 3. Sup. Therefore the solid AB is equal to the solid GK, because they are upon equal bases AE, FH, and are of the same altitude and because the solid parallelepiped CK is cut by the plane DG which is parallel to its opposite c c 3. 3. Sup. planes, the base HF is to the base FC, as the solid HD Book III. to the solid 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 solid AB 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. Let the prisms BNM, DPG, the bases of which are the triangles AEM, CFG, have the same altitude; complete the parallelograms AE, CF, and the solid parallelopipeds AB, CD, in the first of which let AN, and in the other let CP be one of the insisting lines. And because the solid parallelepipeds AB, CĎ have the same altitude, they are to one another as the base AE is to the base CF; wherefore the prisms, which are their halves d are to one another, as the base AE to d 4. 3. Sup. the base CF; that is, as the triangle AEM to the triangle CFG. COR. 2. Also a prism and a parallelepiped, which have the same altitude, are to one another as their bases; that is, the prism BNM is to the parallelepiped CD as the triangle AEM to the parallelogram LG. For by the last Cor. the prism BNM is to the prism DPG as the triangle AME to the triangle CGF, and therefore the prism BNM is to twice the prism DPG as the triangle AME to twice the triangle CGF; that is the prism BNM is e 4. 5. to the parallelepiped CD as the triangle AME to the parallelogram LG. |