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BAH, HAC, which are, each to each of this Book: And because AH is en equal to the three plane angles EDF, qual to DM, the point H coincides EDM,

MDF containing the solid an- with the point M: Wherefore HK, gle at D; the solid angles at A and D which is perpendicular to the plane are equal, and therefore coincide with BAC, coincides with MN, (13. 11.) one another; to wit, if the plane an- which is perpendicular to the plane gle BAC be applied to the plane angle EDF, because these planes coincide EDF, the straight line AH coincides with one another: Therefore HK is with DM, as was shewn in Prop. B. 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, the solid parallelopiped KO; And beviz. A to B, as B to C. The solid de- cause, as A is to B, so is B to C, and scribed from A, B, C is equal to the that A is equal to LK, and B to each equilateral solid described from B, e- of the straight lines DE, DF, and C quiangular to the other.

to KM; therefore LK is to ED, as Take a solid angle D contained by DF to KM; that is, the sides about three plane angles EDF, FDG, GDE; the equal angles are reciprocally proand make each of the straight lines portional; therefore the paraHelogram ED, DF, DG equal to B, and com- LM is equal (14. 6.) to EF: And be. plete the solid parallelopiped DH: cause EDF, LKM are two equal plane Make LK equal to A, and at the point angles, and the two equal straight K in the straight line LK make (26. lines DG, KN are drawn from their 11.) a solid angle contained by the vertices above their planes, and con. three plane angles LKM, MKN, NKL tain equal angles with their sides; equal to the angles EDF, FDG,GDE, therefore the perpendiculars from the

H

points G, N, to the planes EDF, LKM are equal (Cor.35. 11.) to one another:

Therefore the solids KO, DH are of N

G the same altitude; and they are upon

equal bases LM, EF, and therefore they are equal (31. 11.) to one ano ther; But the solid KO is described

from the three straight lines A, B, C, A

T and the solid DH from the straight each to each; and make KN equal to line B. If therefore three straight B, and KM equal to C; and complete lines, &c. Q. E. D.

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PROP. XXXVII. THEOREM.

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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, Take AB to CD, as EF to ST, and EF, GH be proportionals, viz. as AB from ST describe (27. 11.) a solid pato CD, 80 EF to GH; and let the si- rallelopiped SV similar and similarly milar parallelopipeds AK, CL, EM, situated to either of the solids EM, GN be similarly described from them. GN: And because AB is to CD, as AK is to CL, as EM to GN.

EF to ST, and that from AB, CD the Make (11. 6.) AB, CD, O, P conti- solid parallelopipeds AK, CL are sinual proportionals, as also EF, GH, milarly described: and in like manner Q, R: And because as AB is to CD, the solids EM, SV from the straight 80 EF to GH; and that CD is (11.5.) lines EF, ST; therefore AK is to ČL, to O, as GH to Q, and 0 to P, as Q to

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a R as EM to SV: But, by the hypothee

sis, AK is to CL, as EM to GN: R; therefore, ex æquali, (22. 5.) AB Therefore GN is equal (9. 6.) to SV: is to P, as EF to R: But as AB to P, But it is likewise similar and similarly 80 (Cor. 33. 11.) is the solid AK to the situated to SV; therefore the planes solid CL; and as EF to R, so (Cor. which contain the solids GN, SV are 33. 11.) is the solid EM to the solid similar and equal, and their homoloGN: Therefore (11.5.) as the solid AK gous sides GH, ST equal to one anoto the solid CL, so is the solid EM to ther : And because, as AB to CD, 802 the solid GN.

EF to ST, and that ST is equal to But let the solid AK be to the solid GH; AB'is to CD, as EF to GH. CL, as the solid EM to the solid GN: Therefore, if four straight lines, &r. The straight line AB is to CD, as EF Q. E. D. to GH.

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. “ Let the plane CD be perpendicu- “plane, meets it; therefore FGE is “ lar to the plane AB, and let AD be a right angle (3. def. 11): But EF “ their common section ; if any point “is also at right angles to the plane " E be taken in the plane CD, the per- AB; and therefore EFG is a right “pendicular drawn from E to the “ angle: Wherefore two of the angles “ plane AB shall fall on AD.

“ of the triangle EFG are equal toge" For, if it does not, let it, if possi- “ther to two right angles; which is "ble, fall elsewhere, as EF ; and let “ absurd: There" it meet the plane AB in the point "fore the perpenF; and from F draw, (12. 1.) in the « dicular from the

A

D “ plane AB a perpendicular FG to point E to the “ DA, which is also perpendicular (4. plane AB does

-В “ def. 11.) to the plane CD; and join “not fall else« EG: Then because FG is perpen- “ where than upon the straight line “ dicular to the plane CD, and the AD; it therefore, falls upon it. If

straight line EG, which is in that “ therefore a plane,” &c. Q.E.D.

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PROP. XXXIX., THEOREM. 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.

Let the sides of the opposite planes therefore, because KL, BA are each CF, AH of the solid parallelopiped of them parallel to DC, and not in the D к I

same plane with it, KL is parallel (9.

11.) to BA: And because KL.MN E

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, B

PR are in one plane. Let YS be the

common section of the planes KN, N

XR; and DG the diameter of the so

lid parallelopiped AF: YS and DG AF, be divided each into two equal do meet, and cut one another into two parts in the points K, L, M, N, X, equal parts. O, P, R; and join KL, MN, XO, PR: Join DY, YE, BS,SG. Because DX And because DK, CL are equal and is parallel to OE, the alternate angles parallel, KL is parallel (33. 1.) to DC: DŠY, YOE are equal (29.1.) to one For the same reason, ÀN is parallel another: And because DX is equal to to BA; And BA is parallel to DC; OE, and XY to Yo, and contain e.

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qual angles, the base DY is equal (4. (33. 1.) to EG: And DE, BG join 1.) to the base YE, and the other an- their extremities; therefore DE is e

qual and parallel (33. 1.) to BG: And F

DG, YS are drawn from points in the 0

one to points in the other; and are E

therefore in one plane : Whence it is

manifest, that DG, YS must meet one IT

another; let them meet in T: And because DE is parallel to BG, the al

ternate angles EDT, BGT are equal ; B

(29. 1.) and the angle DTY, is equal

(15. 1.) to the angle GTS: Therefore А 1 N

in the triangles DTY, GTS there are G

two angles in the one equal to two gles are equal; therefore the angle angles in the other, and one side equal XYD is equal to the angle OYE, and to one side, opposite to two of the DYE is a straight (14. 1.) line: For equal angles, viz. DY to GS; for they the same reason BSG is a straight are the halves of DE, BG: Therefore line, and BS equal to SG: And be the remaining sides are equal, (26. 1.) cause CA is equal and parallel to DB, each to each. Wherefore DT is equal and also equal and parallel to EG; to TG, and YT equal to TS. Where. therefore DB is equal and parallel fore, if in a solid, &c. Q. E. D.

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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.

Let the prisms ABCDEF, GHKL- the triangle GHK, the prism ABCDEF MN be of the same altitude, the first is equal to the prism GHKLMN. whereof is contained by the two tri- Complete the solids AX, GO; and angles ABE, CDF, and the three pa. because the parallelogram AF is rallelograms AD, DE, EC; and the double of the triangle GHK; and the other by the two triangles GHK, parallelogram HK double (34. 1.) of LMN, and the three parallelograms the came triangle; therefore the paLH, HN, NG; and let one of them rallelogram AF is equal to HK. But have a parallelogram AF, and the solid parallelopipeds upon equal bases, other a triangle GHK for its base; and of the same altitude, are equal if the parallelogram AF be double of (31. 11.) to one another. Therefore

the solid AX is equal to the solid B D

M

GO; and the prism ABCDEF is half X

(28. 11.) of the solid GO. Therefore N

the prism ABCDEF is equal to the HI

prism GHKLMN. Wherefore, if there be two, &c. Q. E. D.

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BOOK XII. .

LEMMA 1.

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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.

Let AB and C be two unequal are in DE: And let the divisions in magnitudes, of which AB is the great- AB be AK, KH, HB; and the divi

If from AB there be taken more sions in ED be EF, FG, GE: And than its half, and from

D

because DE is greater than AB, and the remainder

that EG taken from DE is not greater than its half, and so on; K+

than its half, but BH taken from AB there shall at length re

is greater than its half; therefore the main a magnitude less H

rernainder GD is greater than the rethan C.

mainder HA. Again, because GD is For C may be multi

greater than HA, and that GF is not plied so as at length to

B
C! H

greater than the half of GD, but HK become greater than

is greater than the half of HA; thereAB. Let it be so multiplied, and let fore the remainder FD is greater than DE its multiple be greater than AB, the remainder AK. And FD is equal and let DE be divided into DF, FG, to C, therefore C is greater than AK; GE, each equal to C. From AB take that is, AK less than C. Q. E. D. BH greater than its half, and from And if only the halves be taken the remainder AH take HK greater away, the same thing may in the same than its half, and so on, until there way be demonstrated. be as many divisions in AB as there

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PROP. I. THEOREM.

Similar polygons inscribed in circles are to one another as the squares

of their diamelcrs. Let ABCDE, FGHKL be two cir. polygons are divided into similar tricles, and in them the similar poly, angles; the triangle ABE, FGL, are gons ABCDE, FGHKL; and let BM, GN, be the diameters of the circies :

F
As the square of BM is to the square
of GN, so is the polygon ABCDE to

B
E G

L the polygon FGHKL. Join BE, AM, GL, FN: And be

N

M cause the polygon ABCDE is similar to the polygon FGHKL, and similar

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