lid EQ is equal to the solid AE; and the solid FY, to the solid Book XI. CF: because they are upon the same bases and of the same altitude: therefore the solid AE is equal to the solid CF. Where- 1 29. or fore, solid parallelepipeds, &c. Q. E. D. 30. 11. PROP. XXXII. THEOR. SOLID parallelepipeds which have the same alti. See N. tude, 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 a a Cor. 45, to AE, so that the angle FGH be equal to the angle LCG, and 1. 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 to the b 13.11. solid GK, because D K c to its opposite planes, the base HF is to the base FC, as the so- c 25. 11. lid 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. 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 ÑBO, 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 solid parallelepipeds AB, CD have the same altitude, they are to one another as the base AE is to the base CF; where Book XI. fore the prisms, which are their halvesd, are to one another as the base AE to the base CF; that is, as the triangle AEM to the d 28. 11. triangle CFG. PROP. XXXIII. THEOR. SIMILAR solid parallelepipeds are one to another in the triplicate ratio of their homologous sides. Let AB, CD be similar solid parallelepipeds, and the side AE homologous to the side CF: the solid AB has to the solid CD the triplicate ratio of that which AE has to CF. Produce AE, GE, HE, and in these produced take EK equal to CF, EL equal to FN, and EM equal to FR; and complete the parallelogram KL, and the solid KO: because KE, EL are equal to CF, FN, and the angle KEL equal to the angle CFN, because it is equal to the angle AEG, which is equal to CFN, by reason that the solids AB, CD are similar; therefore the parallelogram KL is similar and equal to the parallelogram CN: for the same reason, the parallelogram MK is similar and equal to CR, and also OE to FD. Therefore three parallelograms of the solid KO are equal and similar to three parallelograms of the solid CD and the three opposite ones in each solid are a 24. 11. equal a and similar to these: therefore the solid KO b C. 11. is equalb and si c 1.6. B X D H P milar to the solid CD: complete the parallelogram GK, and complete the solids EX, LP upon the bases GK, KL, so that EH be an insisting straight line in each of them, whereby they must be of the same altitude with the solid AB: and because the solids AB, CD are similar, and, by permutation, as AE is to CF, so is EG to FN, and so is EH to FR and FC is equal to EK, and FN to EL, and FR to EM; therefore as AE to EK, so is EG to EL, and so is HE to EM: but, as AE to EK, so is the parallelogram AG to the parallelogram GK; and as GE to EL, so is GK to KL; and as HE to EM, so is PE to KM: therefore as Book XI. the parallelogram AG to the parallelogram GK, so is GK to KL, and PE to KM: but as AG to GK, sod is the solid AB to the c 1. 6. solid EX; and as GK to KL, sod is the solid EX to the solid d 25. 11. PL; and as PE to KM, sod is the solid PL to the solid KO; and therefore as the solid AB to the solid EX, so is EX to PL, and PL to KO: but if four magnitudes be continual proportionals, the first is said to have to the fourth the triplicate ratio of that which it has to the second: therefore the solid AB has to the solid KO the triplicate ratio of that which AB has to EX: but as AB is to EX, so is the parallelogram AG to the parallelogram GK, and the straight line AE to the straight line EK. Wherefore the solid AB has to the solid KO, the triplicate ratio of that which AE has to EK. And the solid KO is equal to the solid CD, and the straight line EK is equal to the straight line CF. Therefore the solid AB has to the solid CD, the triplicate ratio of that which the side AE has to the homologous side CF, &c. Q. E. D. COR. From this it is manifest, that, if four straight lines be continual proportionals, as the first is to the fourth, so is the solid parallelepiped described from the first to the similar solid similarly described from the second; because the first straight line has to the fourth the triplicate ratio of that which it has to the second. PROP. D. THEOR. SOLID parallelepipeds contained by parallelograms See N equiangular to one another, each to each, that is, of which the solid angles are equal, each to each, have to one another the ratio which is the same with the ratio compounded of the ratios of their sides. Let AB, CD be solid parallelepipeds, of which AB is contain ed by the parallelograms AE, AF, AG, equiangular, each to each, to the parallelograms CH, CK, CL, which contain the solid CD. The ratio which the solid AB has to the solid CD is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH. Book XI. Produce MA, NA, OA to P, Q, R, so that AP be equal to DL, AQ to DK, and AR to DH; and complete the solid parallelepiped AX contained by the parallelograms AS, AT, AV similar and equal to CH, CK, CL, each to each. Therefore the a C. 11. solid AX is equal to the solid CD. Complete likewise the solid AY, the base of which is AS, and of which AO is one of its insisting straight lines. Take any straight line a, and as MA to AP, so make a to b; and as NA to AQ, so make b to c; and as AO to AR, so c to d: then, because the parallelogram AE is equiangular to AS, AE is to AS, as the straight line a to c, as is demonstrated in the 23d prop. book 6, and the solids AB, AY, being betwixt the parallel planes BOY, EAS, are of the same b 32. 11. altitude. Therefore the solid AB is to the solid AY, as b the base AE to the base AS; that is, as the straight line a is to c. the base OQ is to the c 25. 11. And the solid AY is to the solid AX, as base QR; that is, as the straight line OA to AR; that is, as the straight line c to the straight line d. And because the solid AB is to the solid AY, as a is to c, and the solid AY to the solid AX as c is to d; ex æquali, the solid AB is to the solid Ax, or CD which is equal to it, as the straight line a is to d. But d def. A. the ratio of a to d is said to be compounded dof the ratios of a to b, b to c, and c to d, which are the same with the ratios of the sides MA to AP, NA to AQ, and OA to AR, each to each. And the sides AP, AQ, AR are equal to the sides DL, DK, DH, each to each. Therefore the solid AB has to the solid CD the ratio which is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH. Q. E. D. 5. PROP. XXXIV. THEOR. Book XI. THE bases and altitudes of equal solid parallelepi. See N. peds are reciprocally proportional; and, if the bases and altitudes be reciprocally proportional, the solid parallelepipeds are equal. Let AB, CD be equal solid parallelepipeds; their bases are reciprocally proportional to their altitudes; that is, as the base EH is to the base NP, so is the altitude of the solid CD to the altitude of the solid AB. X First, Let the insisting straight lines AG, EF, LB, HK; CM, NX, OD, PR be at right angles to the bases. As the base EH to the base NP, so is CM to AG. If the base EH be equal to the base NP, then because the solid AB is likewise equal to the solid CD, CM shall be equal to AG. Because if the bases EH, NP be equal, but the altitudes AG, CM be not equal, neither shall the solid AB H A E N be equal to the solid CD. But the solids are equal, by the hypothesis. Therefore the altitude CM is not unequal to the altitude AG; that is, they are equal. Wherefore, as the base EH to the base NP, so is CM to AG. Next, Let the bases EH, NP not be equal, but EH greater than the other: since then the solid AB is equal to the solid CD, CM is therefore greater than K B AG; for, if it be not, neither also in this case |