in which the parallels are, and are themselves parallels (14. 2. Sup.): the plane CDEF cuts the solid AB into two equal parts.

G

Because the triangle CGF is equal (34. 1.) to the triangle CBF, and the triangle DAE to DHE; and since the parallelogram CA is equal (2. 3. Sup.) and similar to the opposite one BE; and the parallelogram GE to CH: therefore the planes which contain the prisms CAE, CBE, are equal and similar, each to each; and they are also equally in- A clined to one another, because the planes

AC, EB are parallel, as also AF and BD, and they are cut by the plane CE (15. 2. Sup.). Therefore the prism CAE is equal to the prism CBE (1. 3. Sup.), and the solid AB is cut into two equal prisms by the plane CDEF. Q. E. D.

N. B. The insisting straight lines of a parallelopiped, mentioned in the following propositions, are the sides of the parallelograms betwixt the base and the plane parallel to it.

PROP. V. THEOR.

Solid parallelopipeds upon the same base, and of the same altitude, the insisting straight lines of which are terminated in the same straight lines in the plane opposite to the base, are equal to one another.

Let the solid parallelopipeds AH, AK be upon the same base AB, and of the same altitude, and let their insisting straight lines AF, AG, LM, LN, be terminated in the same straight line FN, and let the insisting lines CD, CE, BH, BK be terminated in the same straight line DK; the solid AH is equal to the solid AK.

Because CH, CK are parallelograms, CB is equal (34. 1.) to each of the opposite sides DH, EK; wherefore DH is equal to EK: add, or take away the common part HE; then DE is equal to HK: Wherefore also the triangle CDE is equal (38. 1.) to the triangle BHK: and the parallelogram DG is equal (36.1.) to the parallelogram HN. For the same reason, the triangle AFG is equal to the triangle LMN, and the parallelogram CF is equal (2. 3. Sup.) to the parallelogram BM, and CG to BN; for they are opposite. Therefore the planes which

contain the prism DAG are similar and equal to those which contain the prism HLN, each to each; and the contiguous planes are also equally inclined to one another (15. 2. Sup.), because that the parallel planes AD and LH, as also AE and LK, are cut by the same plane DN: therefore the prisms DAG, HLN are equal (1. 3. Sup.). If therefore the prism LNH be taken from the solid, of which the base is the parallelogram AB, and FDKN the plane opposite to the base; and if from this same solid there be taken the prism AGD, the remaining solid, viz. the parallelopiped AH is equal to the remaining parallelopiped AK. Therefore solid parallelopipeds, &c. Q. E. D.

PROP. VI. THEOR.

Solid parallelopipeds upon the same base, and of the same altitude, the insisting straight lines of which are not terminated in the same straight lines in the plane opposite to the base, are equal to one another.

Let the parallelopipeds, CM, CN, be upon the same base AB, and of the same altitude, but their insisting straight lines AF, AG, LM, LN, CD, CE, BH, BK, not terminated in the same straight lines; the solids CM, CN are equal to one another.

Prodnce FD, MH, and NG, KE, and let them meet one another in the points O, P, Q, R; and join AO, LP, BQ, CR. Because the planes (def. 5. 3. Sup.) LBHM and ACDF are parallel, and because the

plane LBHM is that in which are the parallels LB, MHPQ (def. 5.3. Sup.), and in which also is the figure BLPQ; and because the plane ACDF is that in which are the parallels AC, FDOR, and in which also is the figures 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 parallelopiped. Now the solid parallelopiped (Mis equal (5. 2. Sup.) to the solid parallelopiped CP; because they are upon the same base, and their insisting straight lines AF; AC, 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 parallelopipeds, &c. Q. E. D.

PROP. VII. 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. 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 parallelopiped 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 parallelopiped 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 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 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 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 sanie 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 paralJelopipeds on the same base, and of the same altitude, are equal (6. 3. Sup.), if two solid parallelopipeds 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 parallelopipeds, &c. Q. E. D.

PROP. VIII. THEOR.

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

Let AB, CD be solid parallelopipeds 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 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 (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 parallelopiped 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 parallelopipeds, &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 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 (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 parallelopiped, which have the same altitude, are to one another as their bases; that is, the prism BNM is to the parallelopiped 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

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