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Book III. ference of the lefs fegment ADC falls within the right angle CAF, And this is all that is meant, when in the Greek text, and the tranflations from it, the angle of the greater fegment is faid to ⚫ be greater, and the angle of the less fegment is faid to be lefs than a right angle.'

II. I.

b. 19. 3.

C. 31. 3.

d. 31. x.

COR. From this it is manifeft, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the fame two; and when the adjacent angles are equal, they are right angles.

IF

PROP. XXXII. THEOR.

a ftraight line touches a circle, and from the point of contact a ftraight line be drawn cutting the circle, the angles made by this line with the line touching the circle, fhall be equal to the angles which are in the alternate fegments of the circle.

Let the ftraight line EF touch the circle ABCD in B, and from the point B let the traight line BD be drawn cutting the circle. the angles which BD makes with the touching line EF fhall be equal to the angles in the alternate fegments of the circle; that is, the angle FBD is equal to the angle which is in the fegment DAB, and the angle DBE to the angle in the fegment BCD.

A

From the point B draw BA at right angles to EF, and take any point in the circumference BD, and join AD, DC, CB; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line from the point of contact B, the center of the circle is in BA; therefore the angle ADB in a femicircle is a right angle, and confequently the other two angles BAD, ABD are equal to a right angle. but ABF is likewife a right angle; therefore the angle ABF is equal to the angles BAD, ABD. take from thefe equals the common angle

d

c

E B

F

ABD; therefore the remaining angle DBF is equal to the angle

BAD, which is in the alternate fegment of the circle; and because Book III. ABCD is a quadrilateral figure in a circle, the oppofite angles

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BAD, BCD are equal to two right angles; therefore the anglese. 22. 3. DBF, DBE, being likewife equal f to two right angles, are equalf. 13 1. to the angles BAD, BCD; and DBF has been proved equal to BAD; therefore the remaining angle DBE is equal to the angle BCD in the alternate fegment of the circle. Wherefore, if a ftraight line, &c. Q. E. D.

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UPON a given ftraight line to defcribe a fegment See N.

of a circle, containing an angle equal to a given

rectilineal angle.

Let AB be the given straight line, and the angle at C the given rectilineal angle; it is required to defcribe upon the given straight line AB a fegment of a circle, containing an angle equal to the angle C.

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But if the angle C be not a right angle, at the point A in the ftraight line AB, make the angle BAD equal to the angle C, c. 23. 1.

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to the bafe GB; and the circle defcribed from the center G, at

C. 4. 10

Book III. the distance GA, fhall pass through the point B; let this be the circle AHB. and becaufe from the point A the extremity of the f.Cor. 16. 3. diameter AE, AD is drawn at right angles to AE, therefore AD f touches the circle; and because

32. 3.

2. 17. 3.

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H

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wherefore upon the given ftraight line AB the fegment AHB of
a circle is defcribed which contains an angle equal to the given
angle at C.
Which was to be done.

PROP. XXXIV. PRO B.

Tcoutain an

O cut off a fegment from a given circle which shall 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 fegment from the circle ABC that fhall contain an angle equal to the angle D...

Draw the straight line EF touching the circle ABC in the b. 13. 1. point B, and at the point B, in the ftraight line BF, make the angle FBC equal to the an

C. 32. 3.

c

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gle D. therefore, because the
ftraight line EF touches the
circle ABC, and BC is drawn
from the point of contact B,
the angle FBC is equal to D
the angle in the alternate.
fegment BAC of the circle.
but the angle FBC is equal
to the angle D; therefore

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the angle in the fegment BAC is equal to the angle D. wherefore the fegment BAC is cut off from the given circle ABC containing an angle equal to the given angle D. Which was to be done.

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Book III.

IF

PROP. XXXV. THEOR.

F two straight 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 fegments 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

BE, ED.

A

E

If AC, BD pafs each of them through the center, fo that E is the center; it is evident, that AE, EC, BE, ED, being all e-. qual, the rectangle AE, EC is likewife c-B qual to the rectangle BE, ED.

e

But let one of them BD pass 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 becaufe BD, which paffes thro' the center, cuts the ftraight line AC, which does not pafs

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c

F

Cb. 5. 2.

E

B

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 fquare
of EF, and the remaining rectangle BE, ED is equal to the re-
maining square of AE; that is, to the rectangle AE, EC.

C. 47. I.

Next, Let BD which paffes thro' the center, cut the other AC, which does not pafs 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 1. 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 fquare 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 equal to the fquare of EF; and the 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 fquare of AF; that

C. 47. I.

c

F

E

C

G

B

b

is, to the fquare of FB. but the fquare of FB is equal to the rectangle BE, ED together with the fquare 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 pass thro' the center. take the center F, and thro'

E the interfection of the ftraight

lines AC, DB draw the diameter H

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

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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, fhall be equal to the fquare of the line which

touches it.

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

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