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Book III. right angles to AE, therefore AD f touches the circle; and be. cause AB drawn from the point C

H f Cor.

of contact A cuts the circle, the 16. 3. angle DAB is equal to the

F

A angle in the alternate segment

B g 32. 3. AHB 8: but the angle DAB is equal to the angle C, therefore

G also the angle C is equal to the

E angle in the segment AHB :

D wherefore, upon

the given straight line AB the segment AHB of a circle is described, which contains an angle equal to the given angle at C. Which was to be done.

PROP. XXXIV. PROB.

TO cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle.

Let ABC be the given circle, and the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall

contain an angle equal to the given angle D. a 17. 3. Draw a the straight line EF touching the circle ABC in the point B, and at the point B,

A
in the straight line BF,
b 23. 1. make b the angle FBC equal

to the angle D; therefore,
because the straight line

с
EF touches the circle ABC,
and BC is drawn from the D

point of contact B, the angle ( 32. 3. FBC is equal e to the angle in the alternate segment

E

B

F BAC of the circle : but the angle FBC is equal to the angle D; therefore the angle in the segment BAC is equal to the angle D: wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D. Which was to be done.

Book III.

PROP. XXXV. THEOR.

If two straight lines within a circle cut one ano. See N, ther, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

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

A

E D If AC, BD pass each of them through the centre, so that E is the centre; it is evident, that AE, EC, BE, ED, being all

B equal, the rectangle AE, EC is likewise

C equal to the rectangle BE, ED.

But let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E: then, if BD be bisected in F, F is the centre of the circle ABCD; join AF: and because BD, which passes 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 ano

D

a 3. 3. ther: and because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, the rectangle BE, ED, together with

F the square of EF, is equal to the square

b 5.2. of FB, that is, to the square of FA; but A the squares of AE, EF are equal to the

Cc 47. 1,

E
square of FA; therefore the rectangle
BE, ED, together with the square of
EF, is equal to the squares of AE, EF:

B
take away the common square of EF, and the remaining rectan-
gle BE, ED is equal to the remaining square of AE; that is, to
the rectangle AE, EC.

Next, Let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles: then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F drawd FG d 12.1,

N

Book III. perpendicular to AC; therefore AG is equal a to GC; where

fore the rectangle AE, EC, together with the square of EG, is a 3. 3. equal b to the square of AG: to each of these equals add the b 5. 2. square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF, is equal to

D the

squares of AG, GF: but the c 47. 1. squares of EG,GF are equal c to the square of EF; and the squares of

F
AG, GF are equal to the square of
AF: therefore the rectangle AE,

I
A

C
EC, together with the square of EF,

G is equal to the square of AF; that

B is, to the square of FB: but the square of FB is equal to the rectangle BE, ED, together with the square of EF: therefore the rectangle AE, EC, together with the square of EF, is equal to the rectangle BE, ED, toge. ther with the square of EF : take away the common square of EF, and the remaining rectangle AE, ÉC is therefore equal to the remaining rectangle BE, ED.

Lastly, Let neither of the straight lines AC, BD pass through the centre: take the centre F, and

H
through E, the intersection of the
straight lines AC, DB, draw the dia-
meter GEFH: and because the rect-

D

.F
angle AE, EC is equal, as has been
shown, to the rectangle GE, EH; and,

E

А
for the saine reason, the rectangle BE,
ED is equal to the same rectangle GE,
EH; therefore the rectangle AE, EC

G is equal to the rectangle BE, ED.

B
Wherefore, if two straight lines, &c. Q.E.D.

PROP. XXXVI. THEOR.

IF from any point without a circle two straight 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, shall be equal to the square of the line which touches it.

Let D be any point without the circle ABC, and DCA, DB Book III. two straight lines drawn from it, of which DCA cuts the circle, and DB touches the same : the rectangle AD, DC is equal to the square of DB.

Either DCA passes 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 a angle : and be

a 18. 3. cause the straight line AC is bisected

D in E, and produced to the point D, the rectangle AD, DC, together with the square of EC, is equal b to the square

C

b 6.2. of ED, and CE is equal to EB: therefore the rectangle AD, DC, together

B with the square of EB, is equal to the square of ED : but the square of ED is equal< to the squares of EB, BD, be

E

c 47. 1. cause EBD is a right angle : therefore the rectangle AD, DC, together with the

square of EB, is equal to the squares of EB, BD: take away the common square of EB; therefore the remain

А ing rectangle AD, DC is equal to the square of the tangent DB.

But if DCA does not pass through the centre of the circle ABC, take d the centre E, and draw EF perpendiculare to AC, d 1. 3. and join EB, EC, ED: and because the straight line EF, which e 12. 1, passes through the centre, cuts the straight line AC, which does not pass through the centre, at right

D angles, it shall likewise bisect fit;

f 3. 3. therefore AF is equal to FC: and because the straight line AC is bisected in F, and produced to D, the rectangle AD, DC, together with the square of FC, is equal to the square of FD: to B each of these equals add the square of FE; therefore the rectangle AD, DC, I

E together with the squares of CF, FE, is equal to the squares of DF, FE ; but the square of ED is equal é to the squares

A А of DF, FE, because EFD is a right angle ; and the square of EC is equal to the squares of CF, FE; therefore the rectangle AD, DC, together with the square of EC, is equal to the square of ED: and CE is equal to EB; therefore the rectangle AD, DC, together with the square of EB, is equal to the square of ED: but the squares of EB, BD are equal to the square o of ED, be

Book III. cause EBD is a right angle; therefore the rectangle AD, DC,

together with the square of EB, is equal to the squares of EB,
BD : take away the common square of EB; therefore the re-
maining rectangle AD, DC is equal to the square of DB.
Wherefore, if from any point, &c. Q. E. D.
Cor. If from any point without a

A
circle, there be drawn two straight
lines cutting it, as AB, AC, the rect-
angles contained by the whole lines
and the parts of them without the

E

F
circle, are equal to one another, viz. D
the rectangle BA, AE to the rectan-
gle CA, AF: for each of them is
equal to the square of the straight
line AD which touches the circle.

с

B

PROP. XXXVII. THEOR.

See N.

IF from a point without a circle there be drawntwo straight 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 shall touch the circle.

Let any point D be taken without the circle ABC, and from it let two straight 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 square of DB, DB touches the circle. a 17. 3. Draw a the straight line DE touching the circle ABC, find b 18. 3. its centre F, and join FE, FB, FD ; then FED is a right ban

gle : and because DE touches the circle ABC, and DCA cuts c 36. 3. it, the rectangle AD, DC is equal c to the square of DE: but

the rectangle AD, DC is, by hypothesis, equal to the square of DB : therefore the square of DE is equal to the square of DB; and the straight line DE equal to the straight line DB; and

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