Book XI.

See the

figures

below.

Let the solid parallelepipeds 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 CD, CE, BH, BK be terminated in the same straight line DK; the solid AH is equal to the solid AK.

First, let the parallelograms DG, HN, which are opposite to the base AB, have a common side HG: then, because the solid AH is cut by the plane AGHC passing through the diagonals AG, CH of the opposite planes ALGF, CBHD, AH is cut into a 28. 11. two equal parts a by the plane AGHC: therefore the solid AH is double of the prism which is contained betwixt the triangles ALG, CBH: for the same reason, because the solid AK is cut by the plane LGHB through the diagonals LG, BH of the opposite planes ALNG, CBKH, the solid AK is double of the same prism which is contained betwixt the triangles ALG, CBH. Therefore the solid AH is equal to the solid AK.

Ꭰ

H

K

F

G

N

C

B

A

L

But, let the parallelograms DM, EN opposite to the base, have no common side: then, because CH, CK are parallelob 34. 1. grams, CB is equal 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 to the triangle BHK and the parallelogram DG is equal to the parallelogram HN: for the same reason, the triangle AFG is equal to the triangle LMN, and the parallelogram CF is equale to the parallelogram BM, and K D E

c 38. 1.

d 36. 1.

e 24. 11.

D

HE

c

CG to BN; for they are opposite. Therefore the prism which is contained by the two triangles AFG, CDE, and the three parallelograms AD, DG, GC, is equalf to the prism containFC. 11. ed by the two triangles LMN, BHK, and the three parallelograms BM, MK, KL. If therefore the prism LMN BHK be

taken from the solid of which the base is the parallelogram AB, Book XI. and in which FDKN is the one opposite to it; and if from this same solid there be taken the prism AFGCDE, the remaining solid, viz. the parallelepiped AH, is equal to the remaining parallelepiped AK. Therefore, solid parallelepipeds, &c. Q. E. D.

PROP. XXX. THEOR.

SOLID parallelepipeds upon the same base, and See N. 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 parallelepipeds 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: and because the plane LBHM is parallel to the opposite plane ACDF,

and that the plane LBHM is that in which are the parallels LB, MHPQ, in which also is the figure BLPQ; and the plane ACDF is that 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 the plane ALNG is that in which are the parallels AL, OPGN, in which also is the

Book XI. figure ALPO; and the plane CBKE is that 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 ACBL, ORQP are parallel; therefore the solid CP is a parallelepiped: but the solid CM, of which the base is ACBL, to which a 29. 11. FDHM is the opposite parallelogram, is equal a to the solid CP, of which 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: and the solid CP is equal a 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 OL, RK; therefore the solid CM is equal to the solid CN. Wherefore, solid parallelepipeds, &c. Q. E. D.

See N.

PROP. XXXI. THEOR.

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

Let the solid parallelepipeds 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 Book XI. the straight line LM, which is at right angles to the plane in which the bases are, in the point L, is common a to the two so- a 13. 11. lids 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. Pro- b 14.1. duce OD, HB, and let them meet in Q, and complete the solid parallelepiped 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: and be- c 7. 5. cause the solid 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 LQ, so isd the solid AE to the solid LR: d 25. 11. for the same reason, because the solid 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 as the base AB to the base LQ, so

the base CD to the base LQ, as before

was proved: therefore as the solid AE

to the solid LR, so

A S

X

K

B

H T

is the solid CF to the solid LR; and therefore the solid AE is equal to the solid CF.

e 9. 5.

But let the solid parallelepipeds 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 equalf to the solid SE, off 29. 11 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

P

Book XI. parallelogram AB is equals to SB, for they are upon the same base LB, and between the same parallels LB, AT; and that the

g 35. 1. base SB is equal to the base CD; there

lid AE is equal to the

HT

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 is equal to the solid CF : from the points G, K, E, M; N, S, F, P draw the straight lines GQ, 11. 11. KT, EV, MX; NY, SZ, FI, PU, perpendicular 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

T

C

R

Y

Ꮓ

i 6. 11.

HQ

A the same plane, they are paralleli 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, k 15. 11. are parallel to one another: for the same reason, the planes MV, GT are parallel to one another: therefore the solid QE is a parallelepiped: in like manner, it may be proved, that the solid YF is a parallelepiped: but, from what has been domonstrated, 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 so