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Bol XI.

а

C 22.

PRO P. XXVII. PRO B.
T
O describe from a given straight line a solid paralle-

lepiped similar, and similarly situated to one given. Let AB be the given straight line, and CD the given solid ра• rallelepiped. It is required from AB to describe a solid paralle.

lepiped similar, and fimilarly situated to CD. in 26. 11. At the point A of the given straight line AB, make a solid

angle equal to the solid angle at C, and let BAK, KAH, HAB be the three plane angles which contain it, so that BAK be e

qual to the angle ECG, and KAH to GCF, and HAB to b 12. 6. FCE: And as EC to CG, so make b BA to AK; and as GC to

CF, so make b KA to AH; wherefore, ex aequali“, as EC to
CF, so is BA to AH : Complete the parallelogram BH, and
the solid AL: And
because, as EC to

L
CG, so BA to AK, H
the sides about the

M м

D
equal angles ECG,

F
BAK are proporti-
onals; therefore the

K
parallelogram BK

G
is similar to EG.
For the same rea- A B

13
fon, the parallelo-
gram KH is similar to GF, and HB to FE, Wherefore three

parallelograms of the solid AL are similar to three of the solid d 24. 11.

CD; and the three opposite ones in each folid are equal d and similar to these, each to each. Also, because the plane angles which contain the solid angles of the figures are equal, each to

each, and fituated in the same order, the folid angles are e. e B. II.

qual o, each to each. Therefore the folid AL is similar f to the 115. def. II. folid CD. Wherefore from a given straight line AB a solid pa

rallelepiped AL has been described similar, and fimilarly situated to the given one CD. Which was to be done.

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PROP.

Book XI.

PRO P. XXVII.

THEOR.

I'

a folid parallelepiped be cut by a plane palling thro’ See N.

the diagonals of two of the opposite planes; it shall be cut in two equal parts.

F

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, and not in the same plane with it, CD, FF are parallel"; wherefore the diagonals CF, a 9.11. DE are in the plane in which the parallels are, and are themselves paral

С

B lelsb: And the plane CDEF shall cut

b 16. 11 the folid AB into two equal parts.

G Because the triangle CGF is equal to the triangle CBF, and the triangle

C 34. I. DAE to DHE; and that the paral

D lelogram CA is equal and fimilar to

H d 24.11 the opposite one BE ; and the parallelogram GE to CH : Therefore the A

E 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, C. sI. DHE, and the three parallelograms BE, CH, EC; because they are contained by the same number of equal and similar planes, alike fituated, and none of their solid angles are contained by more than three plane angles. Therefore the folid AB is cut into two equal parts by the plane CDEF. Q. E. D.

· N. B. The insisting straight lines of a parallelepiped, mentioned in the next and some following propositions, are the * Gdes of the parallelograms betwixt the base and the opposite

plane parallel to it.'

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SOLID parallelepipeds upon the same base, and of the Sec N.

same altitude, the insisting straight lines of which are terminated in the fame straight lines in the plane opposite to the base, are equal to one another.

28. II.

Book XI. Let the foli! parallelepipeds AH, AK be upon the same ta fe WAB, and of the fame altitude, and let their infifting straight See the fi- lines AF, AG, LM, LN, be terminated in the same straight line gures below. 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 folid AH is cut by the plane AGHC passing through the diagonals AG, CH of the opposite planes ALGF, CBHD, AH is cut into two equal parts * by the plane AGHC: Therefore the folid AH is double of the

D H K
prism which is contained be-
twixt the triangles ALG, F

N
CBH: For the same realon,
because the folid AK is cut
by the plane LGHB through

C

B the diagonals LG, BH of the opposite planes ALNG, A L CBKH, the solid AK is double of the same prism which is contained betwixt the triangles ALG, CBH. Therefore the folid AH is equal to the folid AK.

But, let the parallelograms DM, EN oppofite to the base, have no common side : Then, because CH, CK are parallelograms, CB is equal to each of the opposite fides 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 tric 38. 1. angle CDE is equal to the triangle BHK : And the parallelod 36. S.

gram DG is equal d to the parallelogram HN: For the same

reason, the triangle AFG is equal to the triangle LMN, and C24.11. the parallelogram CF is equal to the parallelogram BM, and D H E

K

) E H K
M
G

G
F

M
N

N

b 34. S.

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A
L
A

L
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 priim, contain-
ed by the two triangles LMN, BHK, and the three parallelo-
grams BM, MK, KL. If therefore the prism LMNBHK be

fC. In:

taken

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

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Same

OLID parallelepipeds upon the same base, and of the

See Ni fame altitude, the insisting straight lines of which are not terminated in the fame straight lines in the plane opposite to the base, are equal to one another.

Let the parallelepipeds CM, CN be upon the fame base AB, and of the fame 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

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A

С 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 BLPO, 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

Book XI. 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, OROP are parallel ; therefore the solid CP is a parallelepiped : But the so

lid CM, of which the base is ACBL, to which FDHM is the a 29. 11. opposite parallelogram, is equal to the solid CP of which, the

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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 folid CP is equal to the fold CN; for they are upon the same base ACBL, and their insisting straight lines AO, AG, LP, LN; CR, CE, BO, BK are in the same straight lines ON, RK: Therefore the folia CM is equal to the folid CN. Wherefore solid parallelepipeds, &c. Q. E. D.

See N.

PRO P. XXXI. THE O R. OLID 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

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