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Book XI DG equal to B, and complete the parallelopiped DH: Make ~ LK equal to A, and at K, in LK, make a a folid angle contained a 26. 11. by three plane angles LKM, MKN, NKL equal to the angles EDF, FDG, GDE, each to each; and make KN equal to B, and KM equal to C, and complete the parallelopiped KO: and because, as A is to B, fo is B to C, and that A is equal to KL, and B to each of the ftraight lines DE, DF, and C to KM; therefore LK is to DE, as DF to KM; that is, the fides about the equal angles are reciprocally proportional; therefore the b 14. 6. parallelogram LM is equal to EF: and because EDF, LKM are two equal plane angles, and the equal lines DG, KN are drawn from their vertices above their planes, and containing equal angles with their fides; therefore the perpendiculars from 35. 11. G, N to the planes EDF, LKM are equal; and the folids KO, DH are therefore of the fame altitude; and they are upon equal 31. 11. bases LM, EF; therefore they are equal to one another: But the folid KO is described from the three ftraight lines A, B, C, and the folid DH from the ftraight line B. If, therefore, three ftraight lines, &c. Q. E. D.

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PROP. XXXVII. THEOR.

F four ftraight lines be proportionals, the fimilar folid parallelopipeds fimilarly described from them hall alfo be proportionals. And if the fimilar parallelopipeds fimilarly defcribed from four straight lines be proportionals, the ftraight lines fhall be proportionals.

Let the four ftraight lines AB, CD, EF, GH be proportionals, viz. AB to CD, as EF to GH; and let the fimilar parallelopipeds AK, CL, EM, GN be fimilarly described from them; AK fhall be to CL, as EM to GN.

Make a AB, CD, O, P continual proportionals, as alfo EF, GH, QR and because AB is to CD, as EF to GH; and that CD is to O, as GH to Q and O to P, as Q to R; therefore, 22. 5 by equality c, AB is to P, as EF to R: But as AB to P, fod is d Cor. 33. the folid AK to the folid CL; and as EF to R, fod is the folid 11. EM to the folid GN; therefore b as the folid AK to the folid CL. fo is the folid EM to the folid GN.

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But let the folid AK be to the folid CL, as the folid EM to the folid GN; the ftraight line AB is to CD, as EF to GH.

Take AB to CD, as e EF to ST, and from ST describe f the € 27. 11. parallelopiped SV fimilar and fimilarly fituated to EM or GN:

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and because AB is to CD, as FF to ST, and that from AB, CD Book XI. the fimilar parallelopipeds AK, CL are fimilarly described; and in like manner, the folids EM, V from EF, ST; therefore AK is to CL, as EM to SV: But, by the hypothefis, AK is to CL, as EM to GN; therefore GN is equal to SV: But it is like- g 9. 5.

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wife fimilar and fimilarly fituated to SV; therefore the planes which contain the folids are fimilar and equal, and their homologous fides GH, ST are equal to one another: and because, as AB to CD, fo is EF to ST, and that ST is equal to GH; AB is to CD, as EF to GH. Therefore, if four straight lines, &c, Q. E. D.

PROP. XXXVIII. THEOR.

F a plane be perpendicular to another plane, a ftraight line drawn from a point in one of them perpendicular to the other, fhall fall on the common fection of the planes.

For a ftraight line drawn from that point perpendicular to the common section of the planes is perpendicular to the other plane; and only one ftraight line can be drawn from that point a4.Def.11. perpendicular to that other plane.

PROP. XXXIX. THEOR.

N a folid parallelopiped, if the fides of two of the oppofite planes be bifected, the common fection of the planes paffing through the points of division, and the diameter of the parallelopiped bifect each other.

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BOOK XI.

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Let the fides of the oppofite planes CF, AH of the parallelopiped AF be bifected in the points K, L, M, N; X, O, P, R, and join KL, MN, XO, PR: and because DK, CL are equal a 33. 1. and parallel, KL is parallel to DC: For the fame reason, MN is parallel to BA: But BA is parallel to CD; therefore MN is b 9. 11. parallel to CD; and therefore alfo to KL. In like manner, it may be proved, that XO is parallel to PR. Let the planes paffing through the parallels cut one another in the straight lines YS; and let DG be the diameter of the parallelopiped AF: YS and DG do meet and bifect one another.

Jon DE, BG: and because DF, DC are doubles of DK, DX, the parallelograms CF, XK are fimilar and fimilarly fituated, and having the

angle at D common
to both, they are a-
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c 26. 6. meter ; therefore

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C

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rallel to CA, and CA to EG; therefore DB is equal and parallel b to EG and DE, BG join their extremities; therefore DE is equal and parallel to BG: and DG, YS are in the fame plane with them; therefore they must meet one another; let them meet in T: and because DE is parallel to BG, the alternate angles EDT, BGT are equal; and the angle DTY is equal to GTS: Therefore the triangles DTY, GTS have two angles of the one equal to two of the other, and the fide DY equal to GS, which are oppofite to equal angles; therefore the g 26. 1. other fides are equal, each to each. Wherefore DT is equal to TG, and YT equal to TS. Therefore, &c. Q. E. D.

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f 15. I.

PROP.

PROP. XL. THEOR.

F there be two triangular prisms of the fame altitude, the bafe of one of which is a parallelogram, and the bafe of the other a triangle; if the parallelogram be double of the triangle, the prifms fhall be equal to one another.

Let the prifm AEB-CFD standing upon the parallelogram AF, and contained by the planes AEB, CFD, AD, DE, be of the fame altitude with the prifm GHK-LMN ftanding upon the triangle GHK, and contained by the planes GM, MK, KL, MLN: and let the parallelogram AF be double of the triangle GHK, the prifm AEB. CFD is equal to the prifm GHK-LMÑ.

BOOK XI.

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Complete the folids AX, GO; and because the parallelogram AF is double of the triangle GHK; and the parallelogram HK is double of the fame triangle; therefore the parallelogram AF a 34. 1. is equal to HK. But parallelopipeds upon equal bafes, and of

the fame altitude, are equal to one another; therefore the folid b 31. 11. AX is equal to the folid GO: and the prifm AEB-CFD is the

half of the folid AX, and the prifm GHK-LMN the half of c 28. 11. the folid GO; therefore the prifm AEB-CFD is equal to the prifm GHK-LMN. Wherefore, &c. Q. E. D.

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