BOOK XI.

PROP. XXVIII. THEOR.

Ir a solid parallelopiped be cut by a plane passing See N. through the diagonals of two of the opposite planes; it shall be cut in two equal parts.

Let AB be a solid parallelopiped, and DE, CF, the diagonals of the opposite parallelograms AH, GB, viz. those which are drawn betwixt the equal angles in each: And because CD, FE, are each of them parallel to GA, and not in the same plane with it, CD, !FE, are parallela; where- 9. 11. fore the diagonals CF, DE, are in the plane in which the parallels are, and are themselves parallels: AndG the plane CDEF shall cut the solid AB into two equal parts.

c

Because the triangle CGF is equal to the triangle CBF, and the triangle DAE to DHE; and that the parallelogram CA is equal and similar to the opposite one BE; and

C

B

b 16. 11.

IF

the parallelogram GE to CH: therefore the prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal to the prism contained⚫ C. 11. by the two triangles CBF, DHE, and the three parallelograms BE, CH, EC; because they are contained by the same number of equal and similar planes, alike situated, and none of their solid angles are contained by more than three plane angles. Therefore the solid AB is cut into two equal parts by the plane CDEF. Q. E. D.

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

PROP. XXIX. THEOR.

SOLID parallelopipeds upon the same base, and of See N. the same altitude, the insisting straight lines of which are terininated in the same straight lines in the plane opposite to the base, are equal to one another.

BOOK XI.

See the figures be

low.

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 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 *28. 11. ALGF, CBHD, AH is cut into two equal parts a by the plane AGHC: Therefore the solid AH is double of the prism which is contained betwixt the triangles ALG,

K

F

CBH: For the same rea

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.

But let the parallelograms DM, EN, opposite to the base, have no common side: Then, because CH, CK, are 34. 1. parallelograms, CB is equalb to each of the opposite sides

C

DH, EK; wherefore DH is equal to EK; Add, or take away, the common part HE; then DE is equal to HK: 38. 1. Wherefore also the triangle CDE is equal to the triangle $36. 1. 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 e 24. 11. equale to the parallelogram BM, and CG to BÑ; for they K D E

D

H K

are opposite. Therefore the prism which is contained by the two triangles AFG, CDE, and the three parallelograms f C. 11. AD, DG, GC, is equal to the prism contained by the two triangles LMN, BHK, and the three parallelograms BM,

MK, KL. If therefore the prism LMN, BHK, be taken Book XI. from the solid of which the base is the parallelogram AB, and in which FDKN is the one opposite to it; and if from this same solid there be taken the prism AFG, CDE, the remaining solid, viz. the parallelopiped AH, is equal to the remaining parallelopiped AK. Therefore solid parallelopipeds, &c. Q. E. D.

PROP. XXX. THEOR.

SOLID parallelopipeds upon the same base, and of See N. 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: And because the plane LBHM is parallel to the op

posite 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:

Book XI. 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 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 parallelopiped: But the solid CM, of which the base is ACBL, to which FDHM is the opposite 29. 11. parallelogram, is equal 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 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 ON, RK: Therefore the solid CM is equal to the solid CN. Wherefore solid parallelopipeds, &c. Q. E.D.

PROP. XXXI. THEOR.

See N. 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.

First, let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be placed in the same

a

plane, and so as that the sides CL, LB be in a straight line; BOOK XI. therefore the straight line LM, which is at right angles to the plane in which the bases are, in the point L, is commona to the two solids AE, CF; let the other insisting * 13.11. 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 14. 1. 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 the base CD to the same LQ: 7. 5. 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 the solid 425. 11. 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

e

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 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 one opposite to it, is equalf to the solid SE, of which 29. 11. 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 be

Q