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 piane 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.

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.

Produce 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 figure 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 CM is equal (5. 2. Sup.) to the solid parallelopiped 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, MP; 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.

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 (7. 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 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 (29. 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 parallelopipeds 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,

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. Prob. 16. 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.

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 (4. 5.); that is, the prism BNM is to the parallelopiped CD as the triangle AME to the paral lelogram LG.

PROP. IX. THEOR.

Solid parallelopipeds 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 parallelopipeds, 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 parallelopiped (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..