# 14. 6.


DF, and C to KM; therefore
LK is to ED as DF to KM;

that is, the sides about the
equal angles are reciprocally


M proportional; therefore the parallelogram LM is equal to

E EF: and because EDF, LKM

B C are two equal plane angles, and 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.


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 ihe similar parallelopipeds AK, CL, EM, GN be similarly described from them, AK shall be to CL, as EM to GN.

Make* AB, CD, O, P continual proportionals, as • 11. 6. also EF, GH, Q, R: and because as AB is to CD so EF


I to GH; and that CD is* to

11. 5. O, as GH to Q, and 0 to P, as Q to R; therefore, ex

ор. æquali*, AB is to P, as EF

* 22. 5. to R: but as AB to P, so*

Cor. 33. is the solid AK, to the solid

11. CL; and as EF to R, so* is


G H Q R * Cor. 33. the solid EM to the solid GN; therefore* as the solid AK to the solid CL, so is * 11.5. the solid EM to the solid GN.


* 27. 11.


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 simi-
larly situated to either of the
solids EM, GN. And be-

cause 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

T lines EF, ST; therefore AK

H Q R is to CL, as EM to SV; but 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.



* 9. 5.


any point

See N.

If a plane be perpendicular to another plane, and a 6 straight line be drawn from a point in one of the planes perpendicular to the other plane, this straight sline shall fall on the common section of the planes.

“ Let the plane CD be perpendicular to the plane 66 AB, and let AD be the common section : if “ 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; 6 and from F draw* in the plane AB, a perpendicular

“ FG to DA, which is also perpendicu* 4 Def. 11. 6 lar* to the plane CD; and join EG.

“ Then because FG is perpendicular to
“ the plane CD, and the straight line

6 EG, which is in that plane, meets
* 3 Def. 11. - it; therefore FGE is a right angle*:

6 but EF is also at right angles to the

* 12. 1.

C с


“ 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.


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

site 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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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 ss. 1. 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

D K parallel* to BA: and because

• 9. 11. 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

P 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 shall 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* to 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 ihe other angles * 4. 1.


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29. 1,

* 14. 1.




9. 11. * 33. 1.

are equal; therefore the angle
XYD is equal to the angle

OYE, and DYE is a straight*

E 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 parallel* 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 ineet 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, BĞ: 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. R. E. D.

* 29. 1. * 15. 1.

* 26. 1.


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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34. 1.


and the parallelogram HK
double* of the same triangle; B D
therefore the parallelogram


N AF is equal to HK: but solid A

H н parallelopipeds upon equal bases, and of the same altitude, are equal* to one another; therefore the solid * 31. 11. AX is equal to the solid GO: and the prism ABCDEF is half* of the solid AX; and the prism GHKLMN * 28. 11. half* of the solid GO: therefore the prism ABCDEF * 28. 11. is equal to the prism GHKLMN. Wherefore, if there be two, &c. Q. E. D.

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