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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 2, in "the plane AB, a perpendicular FG to DA, which is alfo perpen- a. 12. 1. "dicular to the plane CD; and join EG. then becaufe FG is per-b.4. Def.11. pendicular to the plane CD, and

the ftraight line EG, which is "in that plane, meets it; there

C

"fore FGE is a right angle. but

"EF is alfo at right angles to the A

c

plane AB; and therefore EFC

" is a right angle. wherefore two "of the angles of the triangle

“EFG are equal together to two right angles; which is abfurd. "therefore the perpendicular from the point E to the plane AB does 4 not fall elsewhere than upon the ftraight line AD. it therefore falls upon it. If therefore a plane, &c. Q. E. D."

.

IN

PROP. XXXIX. THEOR.

Na folid parallelepiped, if the fides of two of the pofite planes be divided each into two equal parts, the common fection of the planes paffing thro' the points of divifion, and the diameter of the folid parallelepiped cur each other into two equal parts.

Book XI. BA. and BA is parallel to DC; therefore because KL, BA are each of them parallel to DC, and not in the fame plane with it, KL 5.9. 1. is parallel b 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.

Join DY, YE, ES, SG. because DX is parallel to OE, the alé. 29. 1. ternate angles DXY, YOE are equal to one another. and becaufe

DX is equal to OE,

c

and XY to YO,

D

and contain equal

angles, the bafe DY

d. 4. 1. is equal to the bafe

YE, and the other

angles are equal;

therefore the angle
XYD is equal to
the angle OYE, and

. 14. 1. DYE is a ftraight

line. for the fame

a

and parallel to DB, and alfo equal and parallel to EG; therefore DB is equal and parallel to EG. and DE, BG join their extremi33.. tics; 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 muft 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 15.1. 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 fide equal to one fide, oppofite to two of the equal angles, viz. DY to GS; for they are the halves of DE, BG. there15.1.fore the remaining fides are equals, each to each, wherefore DT is equal to TG, and YT equal to TS. Therefore if in a folid, &c.

2 E. D

PROP.

IF

PROP. XL. THEOR.

F there be two triangular prifms of the fame 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 prifms 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

Book XI..

b. 31. 11.

is double of the triangle GHK; and the parallelogram HK double 2. 34. I. of the fame triangle; therefore the parallelogram AF is equal to HK. but folid parallelepipeds upon equal bafes, and of the fame altitude are equal b to one another. therefore the folid AX is equal to the fold GO. and the prifm ABCDEF is half of the folid AX; and the prism 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.

c. 28. 11.

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

IF

F from the greater of two unequal magnitudes there be taken more than its half, more than its half; and fo on.

and from the remainder

there fhall at length re

main 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 A B there 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 C.

D

A

F

G

For C may be multiplied fo as at length to be- K+ come greater than AB. let it be fo multiplied, and let DE its multiple be greater than AB, and let DE H be divided into DF, FG, GE, each equal to C. from AB take BH greater than its half, and from tile 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 Becaufs DE is greater than AB, and that EG taken from DE is lefs

BCE

than

than its half, but BH taken from AB is greater than its half, therefore Book XII. the remainder GD is greater than the remainder HA. again, becaufe 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 lefs than C. Q. E. D.

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

SIMI

PROP. I. THEOR.

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 square of BM is to the square of GN, so is the polygon ABCDE to the polygon FGHKL.

a

Join BE, AM, GL, FN. and because the polygon ABCDE is fimilar to the polygon FGHKL, the angle BAE is equal to the angle" 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

A

2. 1. Def. $.

B

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

angular b; 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