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 (5. 2. Sup.) to the solid parallelepiped CP; 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 (5. 2. Sup) 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. 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 the point L, is common (11.2. Sup.) 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 (14. 1.). 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 (7. 5.) 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 to the base LQ, so is (3.3. Sup.) 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 opposite 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 (9.5.) to the solid CF. 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 (5. 3. Sup.) to 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 (35. 1.) to SB, for they are upon the same base 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. Case 2. If the insisting straight lines AG, HK, BE, 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 same base, and of the same altitude, are equal (6. 3. Sup.), 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 equal (Cor. 45. 1.) 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. Therefore the solid AB is equal (7.3. Sup.) 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 planes, the base HF is (3. 3. Sup.) 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, 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 parallelepipeds 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, CD 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 (4. 3. Sup.) are to one another, as the base AE to 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 (4. 5.); that is, the prism BNM is to the parallelepiped CD as the triangle AME to the parallelogram LG. PROP. IX. THEOR. Solid parallelepipeds are to one another in the ratio that is compounded of the ratios of the areas of their bases, and of their altitudes. Let AF and GO be two solid parallelepipeds, of which the bases are the parallelograms AC and GK, and the altitudes, the perpendiculars let fall on the planes of these bases from any point in the opposite planes EF and MO; the solid AF is to the solid GO in a ratio compounded of the ratios of the base AC to the base GK, and of the perpendicular on AC, to the perpendicular on GK. Case 1. When the insisting lines are perpendicular to the bases AC and GK, or when the solids are upright. In GM, one of the insisting lines of the solid GO, take GQ equal to AE, one of the insisting lines of the solid AF, and through Q let a plane pass parallel to the plane GK, meeting the other insisting lines of the solid GO in the points R, S and T. It is evident that GS is a solid parallelepiped (def. 5. 3. Sup.), and that it has the same altitude with AF, viz. GQ or AE. Now the solid AF is to the solid GO in a ratio compounded of the ratios of the solid AF to the solid GS (def. 10. 5.), and of the solid GS to the solid GO; but the ratio of the solid AF to the solid GS, is the same with that of the base AC to the base GK (8. 3. Sup.), because their altitudes AE and GQ are equal; and the ratio of the solid GS to the solid GO, is the same with that of GQ to GM (3. 2. Sup.); therefore, the ratio which is compounded of the ratios of the solid AF to the solid GS, and of the solid GS to the solid GO, is the same with the ratio which is compounded of the ratios of the base AC to the base GK, and of the altitude AE to the altitude GM (F. 5.). But the ratio of the solid AF to the solid GO, is that which is compounded of the ratios of AF to GS, and of GS to GO; therefore, the ratio of the solid AF to the solid GO is compounded of the ratios of the base AC to the base GK, and of the altitude AE to the altitude GM. Case 2. When the insisting lines are not perpendicular to the bases. Let the parallelograms AC and GK be the bases as before, and let AE and GM be the altitudes of two parallelepipeds Y and Z on these bases. Then, if the upright parallelepipeds AF and GO be constitut-. ed on the bases AC and GK, with the altitudes AE and GM, they will be equal to the parallelepipeds Y and Z (7. 3. Sup.). Now, the solids AF and GO, by the first case, are in the ratio compounded of the ratios of the bases AC and GK, and of the altitudes AE and GM; therefore also the solids Y and Z have to one another a ratio that is compounded of the same ratios. Therefore, &c. Q. E. D. COR. 1. Hence, two straight lines may be found having the same ratio with the two parallelepipeds AF and GO. To AB, one of the sides of the parallelogram AC, apply the parallelogram BV equal to GK, having an angle equal to the angle BAD (44. 1.); and as AE to GM, so let AV be to AX (12. 6.), then AD is to AX as the solid AF to the solid GO. For the ratio of AD to AX is compounded of the ratios (def. 10. 5.) of AD to AV, and of AV to AX; but the ratio of AD to AV is the same with that of the parallelogram AC to the parallelogram BV (1. 6.) or GK; and the ratio of AV to AX is the same with that of AE to GM; therefore the ratio of AD to AX is compounded of the ratios of AC to GK, and of AE to GM (E. 5.). But the ratio of the solid AF to the solid GO is compounded of the same ratios; therefore, as AD to AX, so is the solid AF to the solid GO. COR. 2. If AF and GO are two parallelepipeds, and if to AB, to the perpendicular from A upon DC, and to the altitude of the parallellepiped AF, the numbers L, M, N be proportional: and if to AB, to GH, to the perpendicular from G on LK, and to the altitude of the parallelepided GO, the numbers L, l, m, n be proportional; the solid AF is to the solid GO as LXMXN to lXmXn. For it may be proved, as in the 7th of the 1st of the Sup. that LX MXN is to lXmXn in the ratio compounded of the ratio of LXM to lXm, and of the ratio of N to n. Now the ratio of LXM to 1+m is that of the area of the parallelogram AC to that of the parallelogram GK; and the ratio of N to n is the ratio of the altitudes of the parallelepipeds, by hypothesis, therefore, the ratio of LXMXN to 1Xmxn is compounded of the ratio of the areas of the bases, and of the ratio of the altitudes of the parallelepipeds AF and GO; and the ratio of the parallelepipeds themselves is shewn, in this proposition, to be compounded of the same ratios; therefore it is the same with that of the product LX MXN to the product lxmXn. COR. 3. Hence all prisms are to one another in the ratio compounded of the ratios of their bases, and of their altitudes. For every prism is equal to a parallelepiped of the same altitude with it, and of an equal base (2. Cor. 8. 3. Sup.). |