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PROP. XXXV. THEOR.

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two ftraight lines within a circle cut one another, the rectangle contained by the fegments of one of them is equal to the rectangle contained by the fegments of the other.

Book III.

Let the two ftraight lines AC, BD, within the circle ABCD, cut one another in the point E: the rectangle contained by AE, EC is equal to the rect

angle contained by BE, ED.

If AC, BD pafs each of them through

the centre, fo that E is the centre; it
is evident, that AE, EC, BE, ED, be- B
ing all equal, the rectangle AE.EC
is like wife equal to the rectangle

BE.ED.

E

But let one of them BD pass through the centre, and cut the other AC, which does not pafs through the centre, at right angles in the point E: then, if BD be bifected in F, F is the centre of the circle ABCD; join AF: and because BD, which paffes through the centre, cuts the straight line AC, which does not pass through the centre at right angles in E, AE, EC are equal a to one another: and because the straight line BD is cut into two equal parts, in the point F, and into two unequal, in the point E, BE.EDь+ EF2 = FB2 = AF2. But AF AE2+ EF2, therefore A BE.ED+EFAE+EF, and taking EF2 from each, BE.ED—A£2= AE.EC.

a 3. 3.

F

E

C

b 5. 2.

C 47. I

B

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Book II.

d 12. 1.

a 3. 3. b 5. 2.

D

Next, Let BD, which paffes through the centre, "cut the other AC, which does not pafs through the centre, in E, but not at right angles: then, as before, if BD be bifected in F, F is the centre of the circle. Join AF, and from F draw d FG perpendicular to AC; therefore AG is equala to GC; wherefore AE.EC+ bEGAG2, and adding GF2 to both, AE.EC+EG2+GF2=AG2+

E

C

GF2. Now EG2+GF EF2, and AG2+GF-AF2; therefore AE.EC+EF AF FB2. But FB2-BE.ED+ b EF2, therefore AE.EC+EF BE.ED+EF2, and taking EF from both, AE.EC BE.ED.

H

Lastly, Let neither of the ftraight lines AC, BD pass through the centre: take the centre F, and through E, the interfection of the ftraight lines AC, DB, draw the diameter GEFH : and because, as has been shown, AE.EC GE.EH, and BE.ED GE.EH; therefore AE.EC = BE.ED. Wherefore, if two

ftraight lines, &c. Q. E. D.

T

E

A

B G

I'

PROP. XXXVI. THE OR.

F from any point without a circle two ftraight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, is equal to the fquare of the line which touches it.

Let D be any point without the circle ABC, and DCA, DB two ftraight lines drawn from it, of which DCA cuts

the

the circle, and DB touches it: the rectangle AD.DC is equal Book III. to the fquare of DB.

Either DCA paffes through the centre, or it does not; first, Let it pass through the centre E, and join EB; therefore the angle EBD is a right angle: and because the straight line AC is bifected in E, and produced B to the point D, AD.DC+EC2 ED2.b But EC EB, therefore AD.DC+ EB ED2. Now ED EB2+BD2,

because EBD is a right angle; therefore AD.DC+EBEB2+BD2, and taking EB2 from each, AD.DC= BD2.

D

a 18. 3.

b 6.2.

E

C 47. I.

A

D

But, if DCA does not pass through the centre of the circle ABC, take d the centre E, and draw EF perpendicular e to AC, and join EB, EC, ED: and because the ftraight line EF, which paffes through the centre, cuts the ftraight line AC, which does not pass through the centre, at right angles, it likewife bifects fit; therefore AF is equal to FC and because the straight line AC is bisected in F, and produced to D, b AD.DC + FC2FD2; add FE to both, then AD.DC+FC2+ FE FD2+FE2. But e EC2=FC2+

B

F

FE2, and ED2FD2+FE2, becaufe DFE is a right angle;
therefore AD.DC+EC2=ED2. Now, because EBD is a
right angle, EDEB2+ BDEC2+BD2, and therefore,
AD.DC+ECEC2+BD2, and AD.DC=BD.
fore, if from any point, &c. Q. E. D.

Where

d 1. 3.

e 12. I,

f 3.3.

COR.

Book III.

COR. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another, viz. BA.AE CA.AF; for each of thefe rectangles is equal to the fquare of the ftraight line AD, which touches the circle.

D

A

B

a 17.3.

b 18. 3.

c 36. 3.

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F from a point without a circle there be drawn two ftraight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line, which cuts the circle, and the part of it without the circle, be equal to the square of the line which meets it, the line which meets fhall touch the circle.

Let any point D be taken without the circle ABC, and from it let two ftraight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it; if the rectangle AD.DC, be equal to the fquare of DB, DB touches the circle.

a

Draw the ftraight line DE touching the circle ABC; find the centre F, and join FE, FB, FD; then FED is a right b angle: and because DE touches the circle ABC, and DCA cuts it, the rectangle AD.DC is equal to the fquare of DE; but the rectangle AD.DC is, by hypothefis, equal to the fquare of DB: therefore the fquare of DE is equal to the fquare of DB; and the ftraight line DE equal to the

ftraight

D

d S. I.

E

straight line DB: but FE is equal to FB, wherefore DE, Book III. EF are equal to DB, BF; and the bafe FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal to the angle DBF; and DEF is a right angle, therefore alfo DBF is a right angle: but FB, if produced, is a diameter, and the ftraight line which is drawn at right angles to a diameter, from the extremity of it, touches e the circle: therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D.

B

F

e 16. 3.

ELE.

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