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"For if it does not, let it, if poffible, fall elsewhere, as EF; and Book XI. "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 alfo per- a. 12. 1. દ pendicular to the plane CD; and join EG. then becaufe FG is b.4.Def.11. "perpendicular to the plane CD,

" and the ftraight line EG, which "is in that plane, meets it; there"fore FGE is a right angle. but "EF is alfo at right angles to the "plane AB; and therefore EFG "is a right angle. wherefore two "of the angles of the triangle

&

E

B

"EFG are equal together to two right angles; which is abfurd. "therefore the perpendicular from the point E to the plane AB "does not fall elfewhere than upon the ftraight line AD. it there"fore falls upon it. If therefore a plane, &c. Q. E. D."

IN

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6.3.Def.11.

'N a folid parallelepiped, if the fides of two of the op- See N. pofite planes be divided each into two equal parts, the common fection of the planes pafling through the points of division, and the diameter of the folid parallelepiped cut each other into two equal parts."

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b. 9. 11.

Book XI. BA. and BA is parallel to DC; therefore becaufe KL, BA are each of them parallel to DC, and not in the fame plane with it, KL is parallel to BA. and because KL, MN are each of them parallel to BA, and not in the fame plane with it, KL is parallel b 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 fection of the planes KN, XR; and DG the diameter of the folid parallelepiped AF. YS and DG do meet, and cut one another into two equal parts.

c. 29. 1.

d. 4. 1.

D

K

X

C

F

Join DY, YE, BS, SG. because DX is parallel to OE, the alternate angles DXY, YOE are equal to one another. and becaufe DX is equal to OE, and XY to YO, and contain equal angles, the bafe DY is equal to the bafe YE, and the other angles are equal; therefore the angle XYD is e

E

T

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CA is equal and parallel to DB, and alfo equal and parallel to EG; therefore DB is equal and parallel to EG. and DE, BG 2. 33. 1. 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 manifeft that DG, YS must meet one another; let them meet in T. and becaufe DE is parallel to BG, the alternate angles EDT, BGT are equal; and the angle DTY is equal f 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 fide equal to one fide, oppofite to two of the equal angles, viz. DY to GS; for they are the halves of DE, BG. thesefore the remaining fides are equal , each to each. wherefore DT is equal to TG, and YT equal to TS. Therefore if in a folid, &c. Q, E. d.

f. 15. 1.

S: 26. 1.

Book XI.

PROP. XL.

THEOR.

IF there be two triangular prifmus of the fame altitude, the bafe of one of which is a parallelogram, and the base of the other a triangle; if the parallelogram be double of the triangle, the prifins fhall be equal to one another.

Let the prifms ABCDEF, GHKLMN be of the fame 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 bafe; if the parallelogram AF be double of the triangle GHK, the prifm ABCDEF is equal to the prifm GHKLMN.

Complete the folids AX, GO; and because the parallelogram AF is double of the triangle GHK; and the parallelogram HK double of the fame triangle; therefore the parallelogram AF is a. 34. f.

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equal to HK. but folid parallelepipeds upon equal bafes, and of

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the fame altitude are equal to one another. therefore the folid b. 31. 11. AX is equal to the folid GO. and the prifm ABCDEF is half. c. 23.11. of the folid AX; and the prifin GHKLMN half of the folid GO. therefore the prifm ABCDEF is equal to the prifm GHKLMN. Wherefore if there be two, &c. Q. E. D.'

Q 2

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Which is the first Propofition of the tenth Book, and is neceffary to fome of the Propofitions of this Book.

F from the greater of two unequal magnitudes there

IF

be taken more than its half, and from the remainder more than its half; and fo on. there fhall at length remain a magnitude lefs than the leaft of the propofed magnitudes.

Let AB and C be two unequal magnitudes, of which AB is the greater. if from AB there be taken more

than its half, and from the remainder more A

than its half, and fo on; there fhall at length
remain a magnitude less than C.

K

F

G

D

For C may be multiplied fo as at length to become greater than AB. let it be fo multiplied, and let DE its multiple be greater than H AB, and let DE be divided into DF, FG, GE, each equal to C. from AB take BH greater than its half, and from the remainder AH take HK greater than its half, and fo on until there be as many divifions in AB as there are in DE. and let the divifions AK, KH, HB be as many as the divifions DF, FG, GE. and becaufe DE is greater than AB, and that EG taken from DE is lefs than its

BC

E

half, but BH taken from AB is greater than its half, therefore the Book XII. remainder GD is greater than the remainder HA. again, because GD is greater than HA, and that GF is 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 is less than C. Q. E. D. And if only the halves be taken away, the fame thing may in the fame way be demonftrated,

S

PROP. I. THEO R.

IMILAR polygons infcribed in circles, are to one
another as the fquares of their diameters.

Let ABCDE, FGHKL be two circles, and in them the fimilar polygons ABCDE, FGHKL; and let BM, GN be the diameters. of the circles. as the fquare of BM is to the fquare of GN, fo is the polygon ABCDE to the polygon FGHKL.

Join BE, AM, GL, FN. and because the polygon ABCDE is fimilar to the polygon FGHKL, the angle BAE is equal to the angle a. 1. Def.6, GFL, and as BA to AE, fo is GF to FL. wherefore the two triangles BAE, GFL having one angle in one equal to one angle in the

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other, and the fides about the equal angles proportionals, are equi

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angular; and therefore the angle AEB is equal to the angle FLG. b. 6. 6, but AEB is equal to the angle AMB, because they stand upon the c. 21. 3, fame circumference; and the angle FLG is, for the fame reafon,

equal to the angle FNG. therefore alfo the angle AMB is equal to FNG. and the right angle BAM is equal to the right angle GFN; d. 31 3, wherefore the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one another, therefore as BM to

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