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Supplement

PROP. III. THEOR.

If a solid parallelopiped be cut by a plane parallel to two of its opposite planes, it will be divided into two solids, which will be to one another as their bases.

Let the solid parallelopiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, ·HD, and divides the whole into the solids ABFV, EGCD; as the base AEFY to the base EHCF, so is the solid ABFV to the solid EGCD.

Produce AH both ways, and take any number of straight lines HM, MN, each equal to EH, and any number AK, KL each equal to EA, and complete the parallelograms LO, KY, HQ, MS, and the solids LP, KR,

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HU, MT; then, because the straight lines LK, KA, AE are all equal, and also the straight lines KO, AY, EF, which make equal angles with LK, KA, AE, the parala 36. 1. & lelograms LO, KY, AF are equal and similara; and likewise the parallelograms KX, KB, AG; as also the b 2. 3. Sup- parallelograms LZ, KP, AR, because they are opposite

def. 1. 6.

b

planes. For the same reason, the parallelograms EC, HQ, MS are equal; and the parallelograms HG, HI, IN, as also b HD, MU, NT; therefore three planes of the solid LP, are equal and similar to three planes of the solid KR, as also to three planes of the solid AV: but the three planes opposite to these three are equal and similar to them in the several solids; therefore the solids

c 15. 2. Sup.

LP, KR, AV are contained by equal and similar planes Book III. And because the planes LZ, KP, AR are parallel, and are cut by the plane XV, the inclination of LZ to XP is equal to that of KP to PB; or of AR to BV : and the same is true of the other contiguous planes, therefore the solids LP, KR, and AV, are equal to one another. Ford 1.3. Sup. the same reason, the three solids ED, HU, MT, are equal to one another; therefore what multiple soever the base LF is of the base AF, the same multiple is the solid LV of the solid AV; for the same reason, whatever multiple the base NF is of the base HF, the same multiple is the solid NV of the solid ED: And if the base LF be equal to the base NF, the solid LV is equal to the solid NV; and if the base LF be greater than the base NF, the solid LV is greater than the solid NV; and if less, less. Since then there are four magnitudes, viz. the two bases AF, FH, and the two solids AV, ED, and of the base AF and solid AV, the base LF and solid LV are any equimultiples whatever; and of the base FH and solid ED, the base FN and solid NV are any equimultiples whatever; and it has been proved, that if the base LF is greater than the base FN, the solid LV is greater than the solid NV; and if equal, equal; and if less, less: Therefore, as the base AF is to the base FH, so is the e def, 5. 5. solid AV to the solid ED. Wherefore, if a solid, &c. Q. E. D.

COR. Because the parallelogram AF is to the parallelogram FH as YF to FC, therefore the solid AV is f 1. 6. to the solid ED as YF to FC.

Supplement

PROP. IV. THEOR.

If a solid parallelepiped be cut by a plane passing through the diagonals of two of the opposite planes, it will be cut into two equal prisms.

Let AB be a solid parallelepiped, 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, though not in the same plane with it, CD, FE are parala 8. 2. Sup. lela; wherefore the diagonals CF,

b 14. 2. Sup.

DE are in the plane in which the
parallels are, and are themselves
parallels b: the plane CDEF cuts
the solid AB into two equal parts.
Because the triangle CGF is

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c 34. 1. equal to the triangle CBF, and the triangle DAE to DHE; and

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since the parallelogram CA is

d 2. 3. Sup. equal and similar to the opposite one BE; and the parallelogram

G

B

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GE to CH: therefore the planes which contain the prisms CAE, CBE, are equal and similar, each to each; and they are also equally inclined to one another, because the planes AC, EB are parallel, as also AF and e 15.2, Sup. BD, and they are cut by the plane CE. Therefore the £1. 3. Sup. prism CAE is equal to the prism CBE, and the solid AB is cut into two equal prisms by the plane CDEF. Q. E. D.

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

Book III.

PROP. V. THEOR.

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

a

Because CH, CK are parallelograms, CB is equal to a 34. F. 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 parallelo- 6 38. 1. gram DG is equal to the parallelogram HN. For the c 36. 1. same reason, the triangle AFG is equal to the triangle LMN, and the parallelogram CF is equal to the paral- d 2. 3. Sup. lelogram BM, and CG to BN; for they are opposite.

c

d

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Therefore the planes which contain the prism DAG are similar and equal to those which contain the prism HLN, each to each; and the contiguous planes are also equally inclined to one another, because that the parallel planes e 15. 2. AD and LH, as also AE and LK, are cut by the same plane DN; therefore the prisms DAG, HLN are equal £. f 1. 3. Sup.

Sup.

Supplement If therefore the prism LNH be taken from the solid, of which the base is the parallelogram AB, and FDKN the plane opposite to the base; and if from this same solid there be taken the prism AGD, the remaining solid, viz. the parallelepiped AH, is equal to the remaining parallelepiped AK., Therefore solid parallelepipeds, &c. Q. E. D.

PROP. VI. THEOR.

Solid parallelepipeds 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 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, * def. 5. 3. LP, BQ, CR. Because the planes LBHM and ACDF

3

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are parallel, and because the plane LBHM is that in which are the parallels LB, MHPQ, and in which also

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