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* 13. 11.

See N.

* 26. 11.

HAB equal to MDE, and HAC equal to the angle
MDF; and from H, M, let HK, MN be perpendicu-
lars to the planes BAC, EDF: HK shall be equal to
MN.

Because the solid angle at A is contained by the three plane angles BAC, 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* which is perpendicular to the plane EDF, because these planes coincide with one another. Therefore HK is equal to MN. Q. E. D.

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PROP. XXXVI. THEOR.

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. Arto B, as B to C: the solid described from A, B, C shall be 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* a solid angle contained by the three plane angles LKM, MKN, NKL, equal to the angles EDF, FÖG, GDE, each to each; and make KN equal to B, and KM equal to C; and complete 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,

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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 parallelogram LM is equal to EF: and because EDF, LKM are two equal plane angles, and

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the two equal straight lines DG, KN are drawn from their vertices above their planes; and contain equal angles with their sides; therefore the perpendiculars from the points G, N, to the planes EDF, LKM are equal to one another: therefore the solids KO, DH * Cor. 35. are of the same altitude: and they are upon equal bases 11.

*

*

LM, EF; and therefore they are equal to one another: * 31. 11. but the solid KO is described from the three straight

lines A, B, C, and the solid DH from the straight line B: therefore if three straight lines, &c. Q. E. D.

PROP. XXXVII. THEOR.

If four straight lines be proportionals, the similar solid See N. 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 shall be to CL, as EM to GN.

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Make AB, CD, O, P continual proportionals, as 11.6. also EF, GH, Q, R and be

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* 11. 5.

GN; therefore * as the solid AK to the solid CL, so

is the solid EM to the solid GN.

* 27. 11.

*9.5.

See N.

* 12. 1.

Next let the solid AK be to the solid CL, as the solid EM to the solid GN: the straight line AB shall be to CD, as EF to GH.

*

Take as AB to CD, so EF to ST, and from ST describe 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, as EM to SV; but

O P

M

HQ R

by the hypothesis, AK is to CL, as EM to GN; therefore GN is equal* 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 another: and because as AB to CD, so EF to ST, and that ST is equal to GH, therefore AB is to CD, as EF to GH. Therefore, if four straight lines, &c.

Q. E. D.

PROP. XXXVIII. THEOR.

"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."

"Let the plane CD be perpendicular to the plane "AB, and let AD be the common section: if any point "E be taken in the plane CD, the perpendicular drawn "from E to the plane AB shall fall on AD.

"For, if it does not, let it, if possible, fall elsewhere, "as EF; and let it meet the plane AB in the point F; "and from F draw * in the plane AB, a perpendicular "FG to DA, which is also perpendicu4 Def. 11. "lar* to the plane CD; and join EG. "Then because FG is perpendicular to "the plane CD, and the straight line "EG, which is in that plane, meets *3 Def. 11. it; therefore FGE is a right angle*:

"but EF is also at right angles to the

E

B

"plane AB; and therefore EFG is a right angle: "wherefore two of the angles of the triangle EFG are "equal together to two right angles; which ist absurd: † 17.1. "therefore the perpendicular from the point E to the "plane AB, does not fall elsewhere than upon the "straight line AD; it therefore falls upon it. If there"fore a plane, &c. Q. E. D."

PROP. XXXIX. THEOR.

In a solid parallelopiped, if the sides of two of the opposite See N. 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.

D

Let the sides of the opposite planes CF, AH, of the solid parallelopiped AF, be divided each into two equal parts in the points K, L, M, N; X, O, P, R; and join KL, MN, XO, PR: and because DK, CL are equal and parallel, KL is parallel* to DC: for the same reason, MN is parallel to BA: and BA is parallel to DC; therefore, because KL, BA are each of them parallel to DC, and not in the same plane with it, KL is parallel to BA: and because KL, MN are each of them parallel to BA, and not in the same plane with it, KL is parallel 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

*

B

A

the common section of the planes KN, XR; and DG the diameter of the solid parallelopiped AF: YS and DG shall meet, and cut one another into two equal parts.

33. 1.

*9.11.

* 9.11.

Join DY, YE, BS, SG. Because DX is parallel to OE, the alternate angles DXY, YOE are equal * to * 29. 1. one another and because DX is equal to OE, and XY to YO, and that they contain equal angles, the

base DY is equal to the base YE, and the other angles * 4. 1.

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* 14. 1.

9.11.

* 33. 1.

* 29. 1.

* 15.1.

*26. 1.

are equal; therefore the angle
XYD is equal to the angle
OYE, and DYE is a straight*
line: for the same reason BSG
is a straight line, and BS equal
to SG. And because CA is
equal and parallel to DB, and B
also equal and parallel to EG;
therefore DB is equal and paral-

*

lel* to EG: and DE, BG join their extremities; therefore DE is equal and parallel* 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: and the angle DTY is equal 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 *, 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. THEOR.

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.

Let the prisms ABCDEF, GHKLMN be of the same altitude, the first whereof is contained by the two triangles ABE, CDF, and the three parallelograms AD, DE, EC; and the other by the two triangles GHK, LMN, and the three parallelograms LH, HN, NG; and let one of them have a parallelogram AF, and the other a triangle GHK, for its base: if the parallelogram AF be double of the triangle GHK, the prism ABCDEF shall be equal to the prism GHKLMN.

Complete the solids AX, GO: and because the parallelogram AF is double of the triangle GHK;

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