Book III. the distance GA, fhall pass through the point B; let this be the circle AHB. and because from the point A the extremity of the

f. Cor. 16.3. diameter AE, AD is drawn at right angles to AE, therefore ADʼ touches the circle; and because

g. 32. 3.

AB drawn from the point of
contact A cuts the circle, the
angle DAB is equal to the an-
gle in the alternate fegment
AHB . but the angle DAB is
equal to the angle C, therefore
also the angle C is equal to the
angle in the fegment AHB.

wherefore upon

C

H

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

a. 17. 3.

T

O cut off a fegment from a given circle which fhall contain an angle equal to a given rectilineal angle.

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

Draw a the ftraight line EF touching the circle ABC in the 23. 1. point B, and at the point B, in the straight line BF, make the

the angle in the segment BAC is equal to the angle D. wherefore

the fegment BAC is cut off from the given
ing an angle equal to the given angle D.
done,

circle ABC contain

Which was to be

PROP. XXXV.

THEOR.

Book III.

IF

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

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 rectangle contained by

But let one of them BD pafs thro' the center, and cut the other AC, which does not pafs thro' the center, at right angles, in the point E. then, if BD be bifected in F, F is the center of the circle ABCD; and join AF. and because BD, which passes thro' the center, cuts the ftraight line AC, which does not pass

thro' the center, at right angles in E,

AE, EC are equal 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, the rectangle BE, ED together with
the fquare of EF, is equal to the A
fquare of FB; that is, to the square of
FA; but the fquares of AE, EF are

equal to the fquare of FA; there

fore the rectangle BE, ED, together with the fquare of EF, is equal to the fquares of AE, EF. take away the common square of EF, and the remaining rectangle BE, ED is equal to the remaining square of AE; that is, to the rectangle AE, EC.

Next, Let BD which passes thro' the center, cut the other AC, which does not pass thro' the center, in E, but not at right angles. then, as before, if BD be bifected in F, F is the center of the circle. Join AF, and from F draw FG perpendicular to AC; therefore d. 12. 1.

a. 3. 3.

b. 5. 2.

b

Book III. AG is equal to GC; wherefore the rectangle AE, EC, together with the fquare of EG, is equal to the fquare of AG. to each of these equals add the square of GF, therefore the rectangle AE, EC, together with the fquares of EG, GF is equal to the fquares of AG, GF. but the fquares of EG, GF are

c

c. 47. I. equal to the fquare of EF; and the

D

F

E

G

B

fquares of AG, GF are equal to the
square of AF. therefore the rectangle A
AE, EC, together with the fquare of

EF is equal to the square of AF; that

is, to the fquare of FB. but the fquare of FB is equal to the rectangle BE, ED together with the square of EF; therefore the rectangle AE, EC, together with the fquare of EF, is equal to the rectangle BE, ED together with the fquare of EF. take away the common fquare of EF, and the remaining rectangle AE, EC is therefore equal to the remaining rectangle BE, ED.

Laftly, Let neither of the straight lines AC, BD pafs thro' the center. take the center F, and thro'

E the interfection of the ftraight

lines AC, DB draw the diameter H
GEFH. and because the rectangle
AE, EC is equal, as has been fhewn,
to the rectangle GE, EH; and for
the same reason, the rectangle BE,
ED is equal to the fame rectangle
GE, EH; therefore the rectangle

D

F

E

A

B

AE, EC is equal to the rectangle BE, ED. Wherefore, if two ftraight lines, &c. Q. E. D.

PROP. XXXVI. THEOR.

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, fhall be equal to the square 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 circle,

and DB touches the fame. the rectangle AD, DC is equal to Book III,

the fquare of DB.

Either DCA paffes thro' the center, or it does not; first, let it pass thro' the center E, and join EB;

therefore the angle EBD is a right

D

A

ther with the fquare of EB, is equal to the fquares of EB, BD. take away the common fquare of EB; therefore the remaining rectangle AD, DC is equal to the fquare of the tangent DB.

a. 18. 3.

b. 6. 2.

c. 47. L.

But if DCA does not pass thro' the center of the circle ABC, take the center E. and draw EF perpendicular to AC, and d. 1. 3. join EB, EC, ED; then EFD is a right angle. and because the e. 12. I. ftraight line EF, which paffes thro' the center, cuts the straight

line AC, which does not pass thro' the
center, at right angles, it shall likewise
bifect fit; therefore AF is equal to FC,
and because the straight line AC is bi-
fected in F, and produced to D, the rec-
tangle AD, DC together with the fquare
of FC, is equal to the fquare of
FD. to each of thefe equals add the B
fquare of FE, therefore the rectangle
AD, DC together with the squares of
CF, FE, is equal to the fquares of DF, A
FE. but the fquare of ED is equal to
the fquares of DF, FE, because EFD is
a right angle; and the fquare of EC is

c

D

f. 3.3

E

equal to the squares of CF, FE; therefore the rectangle AD, DC together with the fquare of EC, is equal to the fquare of ED. and CE is equal to EB, therefore the rectangle AD, DC toge

C. 47. I.

C

Book III. ther with the fquare of EB, is equal to the square of ED. but the fquares of EB, BD are equal to the fquare of ED, because EBD is a right angle; therefore the rectangle AD, DC together with the fquare of EB, is equal to the fquares of EB, BD. take away the common fquare of EB, therefore the remaining 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 circle, there be drawn two ftraight 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. the rectangle BA, AE to the rectangle CA, AF. for each of them is equal to the square of the straight line AD which touches the circle.

D

A

E

C

B

a. 17. 3.

b. 18. 3.

PROP. XXXVII. THEOR.

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 fquare 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.

Draw a the straight line DE touching the circle ABC, find its center F, and join FE, FB, FD; then FED is a right angle. and because DE touches the circle ABC, and DCA cuts it, the recf. 36. 3. tangle 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 square of DE is equal to the square of DB, and the straight line DE equal to the straight line DB. and FE is equal to FB,