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BAH, HAC, which are, each to each, equal to the three plane angles EDF, EDM, MDF containing the solid angle at D; the solid angles at A and D are equal, and therefore coincide with one another; to wit, if the plane angle BAC be applied to the plane angle EDF, the straight line AH coincides with DM, as was shewn in Prop. B.

of this Book: And because AH is equal to DM, the point H coincides with the point M: Wherefore HK, which is perpendicular to the plane BAC, coincides with MN, (13. 11.) which is perpendicular to the plane EDF, because these planes coincide with one another: Therefore HK is equal to MN. Q. E. D.

PROP. XXXVI. THEOREM.

If three straight lines be proportionals, the solid parallelopiped described from all three as its sides, is equal to the equilateral parallelopiped described from the mean proportional, one of the solid angles of which is contained by three plane angles equal, each to each, to the three plane angles containing one of the solid angles of the other figure.

Let A, B, C be three proportionals, viz. A to B, as B to C. The solid described from A, B, C is equal to the equilateral solid described from B, equiangular to the other.

Take a solid angle D contained by three plane angles EDF, FDG, GDE; and make each of the straight lines ED, DF, DG equal to B, and complete the solid parallelopiped DH: Make LK equal to A, and at the point K in the straight line LK make (26. 11.) a solid angle contained by the three plane angles LKM, MKN, NKL equal to the angles EDF, FDG, GDE,

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the solid parallelopiped KO; And because, as A is to B, so is B to C, and that A is equal to LK, and B to each of the straight lines DE, DF, and C to KM; therefore LK is to ED, as DF to KM; that is, the sides about the equal angles are reciprocally proportional; therefore the paralelogram LM is equal (14. 6.) to EF: And be cause EDF, LKM are two equal plane angles, and the two equal straight lines DG, KN are drawn from their vertices above their planes, and con tain equal angles with their sides; therefore the perpendiculars from the points G, N, to the planes EDF, LKM are equal (Cor.35. 11.) to one another: Therefore the solids KO, DH are of the same altitude; and they are upon equal bases LM, EF, and therefore they are equal (31. 11.) to one another; But the solid KÓ is described from the three straight lines A, B, C, and the solid DH from the straight line B. If therefore three straight lines, &c. Q. E. D.

PROP. XXXVII. THEOREM.

If four straight lines be proportionals, the similar solid parallelopipeds similarly described from them shall also be proportionals. And if the similar parallelopipeds similarly described from four straight lines be proportionals, the straight lines shall be proportionals.

Let the four straight lines AB, CD, EF, GH be proportionals, viz. as AB to CD, so EF to GH; and let the similar parallelopipeds AK, CL, EM, GN be similarly described from them. AK is to CL, as EM to GN.

Make (11. 6.) AB, CD, O, P continual proportionals, as also EF, GH, Q, R And because as AB is to CD, so EF to GH; and that CD is (11.5.) to O, as GH to Q, and O to P, as Q to

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Take AB to CD, as EF to ST, and from ST describe (27. 11.) a solid parallelopiped SV similar and similarly situated to either of the solids EM, GN: And because AB is to CD, as EF to ST, and that from AB, CD the solid parallelopipeds AK, CL are similarly described: and in like manner the solids EM, SV from the straight lines EF, ST; therefore AK is to CL,

K

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R;

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therefore, ex æquali, (22. 5.) AB is to P, as EF to R: But as AB to P, so (Cor. 33. 11.) is the solid AK to the solid CL; and as EF to R, so (Cor. 33. 11.) is the solid EM to the solid GN: Therefore (11.5.) as the solid AK to the solid CL, so is the solid EM to the solid GN.

But let the solid AK be to the solid CL, as the solid EM to the solid GN: The straight line AB is to CD, as EF to GH.

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as EM to SV: But, by the hypothesis, AK is to CL, as EM to GN: Therefore GN is equal (9.5.) to SV: But it is likewise similar and similarly situated to SV; therefore the planes which contain the solids GN, SV are similar and equal, and their homologous sides GH, ST equal to one ano◄ ther: And because, as AB to CD, se 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, THEOREM.

If a plane be perpendicular to another plane, and a straight line be "drawn from a point in one of the planes perpendicular to the other plane, this straight line shall fall on the common section of the planes.

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"Let the plane CD be perpendicu"lar to the plane AB, and let AD be "their common section; if any point "E be taken in the plane CD, the per"pendicular drawn from E to the "plane AB shall fall on AD.

66 For, if it does not, let it, if possi"ble, fall elsewhere, as EF; and let "it meet the plane AB in the point "F; and from F draw, (12.1.) in the "plane AB a perpendicular FG to "DA, which is also perpendicular (4. "def. 11.) to the plane CD; and join "EG: Then because FG is perpen"dicular to the plane CD, and the straight line EG, which is in that

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"plane, meets it; therefore FGE is

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E

a right angle (3. def. 11): But EF "is also at right angles to the plane "AB; and therefore EFG is a right "angle: Wherefore two of the angles "of the triangle EFG are equal toge"ther to two right angles; which is "absurd: There"fore the perpen"dicular from the "point. E to the "plane AB does "not fall else"where than upon the straight line "AD; it therefore, falls upon it. If "therefore a plane," &c. Q. E. D.

PROP. XXXIX.” THEOREM.

A

In a solid parallelopiped, if the sides of two of the opposite planes be divided each into two equal parts, the common section of the planes passing through the points of division, and the diameter of the solid parallelopiped cut each other into two equal parts.

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therefore, because KL, BA are each of them parallel to DC, and not in the same plane with it, KL is parallel (9. 11.) to BA: And because KL,MN are each of them parallel to BA, and not in the same plane with it, KL is parallel (9. 11.) to MN; wherefore KL, MN are in one plane. In like manner, it may be proved, that XO, PR are in one plane. Let YS be the common section of the planes KN, XR; and DG the diameter of the solid parallelopiped AF: YS and DG do meet, and cut one another into two equal parts.

Join DY,YE, BS, SG. Because DX is parallel to OE, the alternate angles DXY, YOE are equal (29.1.) to one another: And because DX is equal to OE, and XY to YO, and contain e

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(33.1.) to EG: And DE, BG join their extremities; therefore DE is equal and parallel (33. 1.) to BG: And DG, YS are drawn from points in the one to points in the other; and are therefore in one plane: Whence it is manifest, that DG, YS must meet one another; let them meet in T: And because DE is parallel to BG, the alternate angles EDT, BGT are equal; (29. 1.) and the angle DTY, is equal (15. 1.) to the angle GTS: Therefore in the triangles DTY, GTS there are two angles in the one equal to two angles in the other, and one side equal to one side, opposite to two of the equal angles, viz. DY to GS; for they are the halves of DE, BG: Therefore the remaining sides are equal, (26. 1.) each to each. Wherefore DT is equal to TG, and YT equal to TS. Wherefore, if in a solid, &c. Q. E. D.

PROP. XL. THEOREM.

If there be two triangular prisms of the same altitude, the base of one of which is a parallelogram, and the base of the other a triangle; if the parallelogram be double of the triangle, the prisms shall be equal to one another.

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the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN.

Complete the solids AX, GO; and because the parallelogram AF is double of the triangle GHK; and the parallelogram HK double (34. 1.) of the same triangle; therefore the parallelogram AF is equal to HK. But solid parallelopipeds upon equal bases, and of the same altitude, are equal (31. 11.) to one another. Therefore the solid AX is equal to the solid GO; and the prism ABCDEF is half (28. 11.) of the solid GO. Therefore the prism ABCDEF is equal to the prism GHKLMN. Wherefore, if there be two, &c. - Q. E. D.

BOOK XII.

LEMMA I.

Which is the first proposition of the tenth book, and is necessary to some of the propositions of this book.

If from the greater of two unequal magnitudes, there be taken more than its half, and from the remainder more than its half, and so on: There shall at length remain a magnitude less than the least of the proposed magnitudes.

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are in DE: And let the divisions in AB be AK, KH, HB; and the divisions in ED be EF, FG, GE: And because DE is greater than AB, and that EG taken from DE is not greater than its half, but BH taken from AB is greater than its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA; therefore the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK less than C. Q. E. D.

And if only the halves be taken away, the same thing may in the same way be demonstrated.

PROP. I. THEOREM.

Similar polygons inscribed in circles are to one another as the squares of their diameters.

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