lelopiped; but the solid CM, of which the base is ACBL, to which FDHM is the opposite parallelogram, is equal (29. 11.) to the solid CP, of which the base is the parallelogram ACBL, to which ORQP N K

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 (29. 11.) 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 in the same straight lines ON, RK: therefore the solid CM is equal to the solid CN. Wherefore, solid parallelopipeds, &c. Q. E. D.

PROP. XXXI. THEOR.

SOLID parallelopipeds which are upon equal bases and of the same altitude, are equal to one another.*

Let the solid parallelopipeds AE, CF be upon equal bases AB, CD, and be of the same altitude; the solid AE is equal to the solid 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 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 (13. 11.) 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 parallelopiped LR, the base of which is the paralelogram 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 parallelopiped AR is cut by the plane LMEB, which

* See Note.

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is parallel to the opposite planes AK, DR; as the base AB is to the base LQ, so is (25. 11.) the solid AE to the solid LR: for the same reason, because the solid parallelopiped 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 to the base LQ, as before was 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 parallelopipeds 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 one opposite to it, is equal (29. 11.) to the solid SE, of which the base is LE, and to which SX is 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, MV, 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 that the base SB is equal to the base CD; therefore the base AB is equal to the base CD, and P

But 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: from the points G, K, E, M; N, S, F, P, draw the straight lines GQ, KT, EV, MX; NY, SZ, FI, PU, perpendicular (11. 11.) to the plane in which are the bases AB, CD; and let them meet it in the points Q, T, V, X; 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 (6.

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A H Q 11.) to one another: and MG, EK are parallels; therefore the plane MQ, ET, of which one passes through MG, GQ, and the other through EK, KT, which are parallel to MG, GQ, and not in the same plane with them, are parallel (15. 11.) to one another. For the same reason the planes MV, GT are parallel to one another: therefore the solid QE is a parallelopiped: in like manner it may be proved, that the solid YF is a parallelopiped: 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 the solid EQ is equal (29. or 30. 11.) to the solid AE; and the solid FY to the solid CF; because they are upon the same bases and of the same altitude: therefore the solid AE is equal to the solid CF. Wherefore solid parallelopipeds, &c. Q. E. D.

PROP. XXXII. THEOR.

SOLID parallelopipeds which have the same altitude, are to one another as their bases.*

Let AB, CD be solid parallelopipeds of the same altidude; 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 parallelopiped 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 (31. 11.) to the solid GK, because

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which is parallel to

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its opposite planes, the base HF is (25. 11.) 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. lelopipeds, &c. Q. E. D.

Wherefore solid paral

* See Note.

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 parallelopipeds 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 parallelopipeds 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 (28. 11.) are to one another as the base AE to the base CF; that is, as the triangle AEM to the triangle CFG.

PROP. XXXIII. THEOR.

SIMILAR Solid parallelopipeds are one to another in the triplicate ratio of their homologous sides.

Let AB, CD be similar solid parallelopipeds, 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 equal (24. 11.) and similar to these therefore the solid KO is equal (C. 11.) and similar to the solid CD: complete the parallelogram GK and complete the solids EX, LP upon the bases GK, KL, so that EH be an

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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 (1. 6.) is the parallelogram AG to the parallelogram GK; and as GE to EL, so is (1. 6.) GK to KL, and as HE to EM, so (1. 6.) is PE to KM: therefore as the parallelogram AG to the parallelogram GK, so is GK to KL, and PE to KM: but as AG to GK, so (25. 11.) is the solid AB to the solid EX; and as GK to KL, so (25. 11.) is the solid EX to the solid PL; and as PE

to KM, so (25. 11.) 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 he 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 parallelopiped 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 parallelopipeds contained by parallelograms 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 parallelopipeds, of which AB is contained by the parallelograms AE, AF, AG, equiangular, each to each, to the parallelograms CH, CK, CL, which contains 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.

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 parallelopiped AX contained by the parallelograms AS, AF, AV, similar and equal to CH, CK, CL, each to each. Therefore the solid AX is equal (C. 11.) 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 altitude. Therefore the solid AB is to the solid AY, as (32. 11.) the base AE to the base AS; that is as the straight line a is to c. And the solid AY, is to the solid AX, as (25. 11.) the base OQ is to the base QR; that is, as the straight

* See Note.