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plane angles. Therefore, the solid AB is cut into two equal parts by the plane CDEF. Q. E. D.

PROPOSITION XXIX.
Taror.Solid parallelopipeds upon the same base, and of See N.

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 CD, CE, BH, BK be terminated in the same straight line DK: the solid ah shall be 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 op

D IT K posite planes ALGF, CBHD, AH is cut into

F a

N two equal parts * by the plane AGHC;

• 28, 11. therefore the solid Al 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 t to the solid AK. † 6 Ax.

: Next, let the parallelograms DM, EN, opposite to the base, have no common side. Then, because CH, CK are parallelo

• 84. 1. grams, CB is equal * to each of the oppo

H E
K D E

H K

M G site sides DH, EK ;

F
N

N wherefore DH is equal

BI to EK: add, or take away the common part HE; then DE is equal + to HK: wherefore also the triangle +2 or 3 Ax. CDE is equal * to the triangle BHK, and the parallelogram DG .38.1.

B В

A

MA

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

36. 1.

.24. 11.

• C. 11.

is equal to the parallelogram HN: for the same reason, the triangle AFG is equal to the triangle LMN: and the parallelogram CF is equal * to the parallelogram BM, and 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 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 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 Al is equal + to the remaining parallelopiped AK. Therefore, solid parallelopipeds, &c. Q. E. D.

† 3 Ax.

PROPOSITION XXX.

See N.

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 shall be 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,

Nķ and that the plane LBHM is that in

E which are the parallels LB, MHPQ,

M. H in which also is the figure BLPQ ;

R and the plane ACDF is that in which F are the parallels AC, FDOR, in which also is the figure CAOR ; therefore

od 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 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, ORPQ are + parallel ; therefore the solid CP + Hyp. is a parallelopiped: but the solid cm is equal * to the solid • 29. 11. CP, because they are upon the same base ACBL, 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, • 29. 11. 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.

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Theor.- Solid parallelopipeds, which are upon equal bases, See N.

and of the same altitude, are equal to one another.

F
N

M

D G

K

B

Let the solid 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

р

R may be in a straight line; therefore the straight line LM, which is at

X right angles to the plane in which the bases are, in the point L, is com

AS mon * to the two solids AE, CF: let

HT

• 13. 11. 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, . 14. 1. & HB, and let them meet in Q, and complete the solid parallelo- N. 14. 6. piped 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 base LQ. And • 7.5. 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 AE to the 25. 11. solid LR: for the same reason, because the solid parallelopiped

+11.5.

* 9.5.

29. 11.

35. 1.

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 is the solid CF to the solid LR: and therefore the solid AE is equal * 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

P

R
bases SB, CD in the same plane, so
that CL, LB may be in a straight D G

Q
line ; and let the angles SLB, CLD
be unequal: the solid se shall be

AS HT
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 is equal * to the solid
SE, because 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 parallelogram AB is equal
they are upon the same base LB, and between the same pa-
rallels 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

M E
equal to the solid CF.

К.
From the points G, L
K, E, M; N, S, F, P,

R

A HQ T draw the straight lines GH, KT, EV, MX; NY, SZ, FI, PU, perpendicular to the planes in which are the bases AB, CD; and let them mext them in the points Q, T, V, X; Y, 2, 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

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to SB, for

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• 15. 11.

another : and MG, EK are parallels; 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 same reason, the planes MV, GT are parallel to one another : therefore the solid QE is a parallelopi ped. 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 solid AE, and the solid FY to the solid CF, because they are upon the same bases and of the same altitude; therefore the solid AE is equal to the solid CF. Wherefore, solid parallelopipeds, &c.

29, or so. 11.

Q. E. D.

PROPOSITION XXXII.

THEOR.-Solid parallelopipeds which have the same altitude, See N.

are to one another as their bases.

Let AB, CD be solid parallelopipeds of the same altitude: they shall be 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 . Cor. 45. 1. to AE, so that the angle FGH may be equal to the angle LCG; and upon the base FH, complete the solid parallelopiped GK, one of B

D K whose insisting lines is FD, where

OP

Q. by the solids CD, GK must be of

L F the same altitude: therefore the

A M solid AB is equal * to the solid GK,

• 31. ll. 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 * to the base FC, as the solid HD to the solid DC : but • 25. 11. 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.

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