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plane LBHM is parallel to the opposite plane ACDF, and that LBHM is the plane in which are the parallels LB, MHPQ, in which also is the figure BLPQ; and ACDF is the plane in which are the parallels AC, FDOR, in which also is the figure CAOR; therefore the figures BLPQ, CAOR are in parallel planes. In like manner, because the plane ALNG is parallel to the opposite plane CBKE, and that ALNG is the plane in which are the parallels AL, OPGN, in which also is the figure ALPO; and CBKE is the plane in which are the parallels CB, RQEK, in which also is the figure CBQR; therefore the figures ALPO, CBQR are in parallel planes: and the planes A CBL,ORQP are parallel; therefore the solid CP is a parallelepiped. But the solid CM, of which the base is ACBL, to which FDHM is the opposite parallelogram, is equal (XI. 29.)

the base is the parallelogram ACBL, to which ORQP 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. For the same reason the solid CP is equal to 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 (I. ax. l.) CM is equal to CN. Wherefore parallelepipeds, &c.

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PROP. XXXI. THEOR. PARALLELEPIPEDS which are upon equal bases, and of the same altitude, are equal to one another.

Let the parallelepipeds AE, CF, be upon equal bases AB, CD, and be of the same altitude; AE is equal to CF.

First, let the insisting lines be at right angles to the bases AB, CD, and let the bases be placed in the same plane, and so that the sides CL, LB may be in a straight line; therefore (XI. 13.) the straight line LM, which is at right angles to the 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; DF, OP, CN: and, first, let the angle ALB be equal to CLD; then (I. 14. cor.) AL, LD are in a straight line. Produce OD, HB, and let them meet in Q; complete also the parallelepiped LR, the base of which is the parallelogram LQ, and of which LM is one of its insisting lines. Therefore (V. 7.) because the parallelogram

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AB is equal to CD, as AB: LQ :: CD : LQ: and because the 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 LA, so is (XI. 25.) the solid AE to the solid LR. For the same reason, because the 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 AB: LQ:: CD: LQ, as has been proved: therefore (V. 11.) as the solid AE to the solid LR, so is the solid CF to the solid LR; and therefore (V. 9.) the solid AE is equal to the solid CF.

But let the parallelepipeds SE, CF be upon equal bases SB, CD, and be of the same altitude, and let their insisting lines be at right angles to the bases; and 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 also in this case equal to the solid CF. Produce DL, TS, till 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 (XI. 29.) the solid AE, of which the base is the parallelogram LE, and AK the one opposite to it, is equal 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 lines, viz., LA, LS, BH, BT; MG, MV, EK, EX are in the same straight lines AT, GX. Also, because (I. 35.) the parallelogram AB is equal 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 CD; therefore the base AB is equal to the base CD, and the angle ALB to CLD. Therefore, by the first case, the solid AE is equal to the solid CF; but AE is equal to SE, as was demonstrated; therefore SE is equal to CF.

But, if the insisting straight lines AG, HK, BE; CN, &c., be not at right angles to the bases AB, CD; in this case likewise the solid AE is equal to the solid CF.

For if parallelepipeds be described on the bases AB, CD, of the same altitude as the solids AE, CF, and having their insisting lines perpendicular to their bases, they will be respectively equal (XI. 29. or 30.) to AE, CF; and by the first case of this proposition they are equal to one another. Therefore (I. ax. 1.) AE, CF are equal. Wherefore parallelepipeds, &c.

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PROP. XXXII. THEOR.

PARALLELEPIPEDS which have the same altitude, are to one another as their bases.

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Let AB, CD be 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 (I. 45. cor.) the parallelogram FH equal to AĚ, so that the angle FGH may be equal to the angle LCG; and complete the parallelepiped GK on the base FH, one of whose insisting lines is FD, the solids CD, GK being thus of the same altitude. Therefore (XI. 31.) the solids AB, GK are equal, because they are upon equal bases AE, FH, and are of the same altitude. Now, the parallelepiped CK is cut by the plane DG, which is parallel to its opposite planes ; therefore (XI. 25.) the base HF is to GL, as the solid HD to DC. But HF is equal to AE, and the solid GK to AB: therefore, as the base AE to the base GL, so is the solid AB to the solid CD. Wherefore parallelepipeds, &c.

Cor. From this it is manifest that triangular prisms 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 parallelepipeds AB, CD, in the first of which let MO, and in the other let GQ be one of the insisting lines. Aud because the parallelepipeds AB, CD have the same altitude, they are to one another as the base AE is to GL; wherefore the prisms, which (XI. 28.) are their halves, are to one another, as the base AE to GL; that is, (V. 15.) as the triangle AEM to CFG.

PROP. XXXIII. THEOR.

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

Let AB, CD be similar parallelepipeds, and the side AE homologous to CF: AB has to CD the triplicate ratio of that which AE has to CF.

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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, since it is equal to the angle AEG, which is equal to CFN, because the solids AB, CD are similar; therefore the parallelogram KL is similar and equal to 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 (XI. 24.) the three opposite ones in each solid are equal and similar to these. Therefore (XI. C.) the solid KO is equal and similar to CD. Complete the parallelogram GK, and the solids EX, LP upon the bases GK, KL, so that EH may be an insisting line in each of them ; and thus they are of the same altitude with the solid AB. Then, because the solids AB, CD are similar, (XI. def. 8. and alternately,) 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 (VI. 1.) as AE to EK, so is the parallelogram AG to GK; and as GE to EL, so is GK to KL; and as HE to EM, so is BE to KM: therefore (V. 11.) as the parallelogram AG to GK, so is GK to KL, and PE to KM. But (XI. 25.) as AG to GK, so is the solid AB to EX; and as GK to KL, só is the solid EX to PL: and as PE to KM, so is the solid PL to KO: and therefore (V. 11.) as the solid AB to 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 KO the triplicate ratio of that which AB has to EX. But as AB is to EX, so is the parallelogram AG to GK, and the straight line AE to EK. Wherefore the solid AB has to the solid KO, the triplicate ratio of that which A E has to EK ; and the solid KO is equal to CD, and the straight line EK to CF. Therefore the solid AB has to CD the triplicate ratio of that which the side A E has to the homologous side CF. Wherefore similar parallelepipeds, &c.

Cor. From this it is manifest, that, if four straight lines be continual proportionals, as the first is to the fourth, so is the 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.

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PROP. D. THEOR.

PARALLELEPIPEDs 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.

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Let AB, CD be parallelepipeds, of which AB is contained by the parallelograms AE, AF, AG, equiangular, each to each, to CH, CK, CL, which contain CD. The ratio which AB has to 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, making AP equal to DL, AQ to DK, and AR to DH; and complete the parallelepiped AX contained by the parallelograms AS, AT, AV, similar and equal to CH, CK, CL, each to each. Therefore (XI. C.) AX is equal to CD. Complete likewise the solid AY, the base of which is AS, and of which AO is one of its insisting lines. Take any straight

b line a, and (VI. 12.) find b, c, d, such that MA: AP:: a;b; NA:AQ :: b:c, and AO: AR::c:d. Then, (VI. 23.) because the parallelogram AE is equiangular to AS, AE is to AS, as a to c: and the solids AB, AY, being between the parallel planes BOY, EAS, are of the same altitude : therefore (XI. 32.) the solid AB is to the solid ÀY, as AE to AS; that is, as a to c. Also (XI. 25.) the solid AY is to AX as the base OQ is to QR; that is, (VI. 1.) as OA to AR; that is, as c to d. And because the solid AB is to A Y as a is to c, and AY to AX, as c is to d; ex æquo, AB is to AX or CD, which is equal to it, as a is to d. But (V. def. 10.) the ratio of a to d is said to be compounded of 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 AB has to 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. Parallelepipeds, therefore, &c.

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