PROP. XXXI. THEOR.

Parallelopipeds which are upon equal bases, and of the same altitude, are equal to one another.

Let the parallelopipeds AE, CF, be upon equal bases AB, CD, and be of the same altitude; the solid AE shall be 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 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 to the two solids AE, CF; * 13. 11. 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.* Produce OD, HB, and let them meet * 14. 1. in Q, and complete the 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 the base CD to the same LQ. 7. 5. And because the 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 the * 25. 11. solid AE to the solid LR: for the same reason, because the parallelopiped CR is cut by the plane LMFD,

z 2

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 was before proved; there11. 5. fore as the solid AE to the solid LR, so is the solid CF to the solid LR; and therefore the solid AE is equal to the solid CF.

9.5

But let the 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 shall be 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 29. 11. solids AE, LR: therefore the solid AE is equal to the solid SE; 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 35. 1. 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 the base CD; therefore the base AB is equal to the base CD, and 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.

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 shall be 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 to the plane * 11. 11. 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 to one another: and MG, EK are parallels; * 6. 11. therefore the planes 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 to one another: for the 15. 11. 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 to the * 29 or solid AE, because they are upon the same base and of 30, 11. the same altitude: and for the same reason, the solid FY is equal to the solid CF: therefore the solid AE is equal to the solid CF: wherefore parallelopipeds, &c. Q. E. D.

* Cor.

45. 1.

PROP. XXXII. THEOR.

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

Let AB, CD be parallelopipeds of the same altitude; they are to one another as their bases; that is, as the base AE, to the base CF, so shall the solid AB be to the solid CD.

To the straight line FG apply the parallelogram FH equal to AE, so that the angle FGH be equal to the angle LCG; and complete the parallelopiped GK upon the base FH, one of whose insisting lines is FD, whereby the solids CD, GK must be of the same alti31. 11. tude: therefore the solid AB is equal to the solid GK; because they are upon equal bases AE, FH, and

are of the same altitude; and because the parallelopiped CK is cut by the plane DG which is parallel to 25. 11. its opposite planes, the base HF is 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 parallelopipeds, &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 NBO, PDQ the triangles opposite to them, have the same altitude; they shall be to one

another as their bases. Complete the parallelograms AE, CF, and the 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 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 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 parallelopipeds are one to another in the triplicate ratio of their homologous sides.

Let AB, CD be similar parallelopipeds, and the side AE homologous to the side CF: the solid AB shall have 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, each to each, 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 pa

* 28. 11.

rallelogram CN. For the same reason, the parallelogram MK is similar and equal to CR, and also OE to