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9. 11.

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• 34. 1.

• 24. 11.

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 * parallel ; wherefore the diagonals CF, DE, are in the plane in which the parallels are, and are themselves * parallels : and the plane CDEF shall cut the solid AB into two equal parts.

ЭН 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 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 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.'

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6

PROP. XXIX. THEOR.

See N.

Solid parallelopipeds upon the same base, and of 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

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C С

the opposite planes ALGF, CBHD, AH is cut into two equal parts * by the plane AGHC; therefore the • 28. 11. solid AH is double of the prism which is contained betwixt the triangles

K ALG, CBH: for the same reason,

N because the solid AK is cut by the plane LGHB, through the diago

A nals 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 +6 Ax. AK.

Next, let the parallelograms DM, EN, opposite to the base, have no common side. Then, because CH, CK, are parallelograms, CB is equal to each of the opposite • 34. 1. sides DH, EK; wherefore DH is equal to EK: add, or take away the common part HE;

H E then DE is equal +

K D E

H K
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| 2 or 3 Ax. F to HK: wherefore

N also the triangle

BI CDE is equal * to

• 38. 1. the triangle BHK, and the parallelogram DG is equal * to the paral

• 56. 1. lelogram 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 * 24. 11. 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 C. 11. 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 AH, is equal + to the re- *3 Ax. maining parallelopiped AK. Therefore solid parallelopipeds, &c. Q. E. D.

PROP. XXX. THEOR. Solid parallelopipeds upon the same base, and of the same See N.

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.

E

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,

N. K Κ
BQ, CR. And because the
plane LBHM is parallel to the

M H
opposite plane ACDF, and that
theplane LBHM is thatin which F

R are the parallels LB, MHPQ, in which also is the figure BLPQ;

А с
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 planc
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 is a parallelopiped:
but the solid CM is equal to the solid CP, because they
are upon the same base ACBL,
and their insisting straight lines

N K
AF, AO, CD, CR; LM, LP,
BH, BQ, are in the same

^ H

R straight lines FR, MQ: and the solid CP is equal* to the solid F CN, for they are upon the same base ACBL, and their insisting straight lines AO, AG, LP,

ST 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.

+ Hyp.
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29. 11.

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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 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 may 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 in Q, * 14. 1. 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 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 * 25.11. AE to the solid LR: for the same reason, because the solid P 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 before was proved ; therefore, as the solid AE to the solid LRt, so is the solid CF to + 11. 5. the solid LR: and therefore the solid AE is equal* to * 9.5. 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 may 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 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

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29. 11.

* 35. 1.

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* 11. 11.

the same straight lines AT, GX: and because the
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

HT
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 planes in which are the bases AB, CD; and let them meet them 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* one another :

M E and MG, EK are parallels; therefore L 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 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 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. Q. E. D.

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