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Bisect AB in F, (1. 10.) and from the center F, at the distance FB, describe the semicircle AHB, and draw AH, BH to any point H in the circumference.

Therefore the angle AHB in a semicircle is equal to the right angle C. (111. 31.).

But if the angle C be not a right angle:

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at the point A, in the straight line AB, make the angle BAD equal to the angle C, (1. 23.) and from the point A draw À E at right angles to AD; (1. 11.)

bisect AB in F, (1. 10.) and from F draw FG at right angles to AB, (1. 11.) and join GB.

Because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, FG, each to each, and the angle AFG is equal to the angle BFG; (1. def. 10.)

therefore the base AG is equal to the base GB; (1. 4.) and the circle described from the center G, at the distance GA, shall pass through the point B:

let this be the circle AHB. The segment AHIB shall contain an angle equal to the given rectilineal angle C.

Because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE,

therefore AD touches the circle: (111. 16. Cor.) and because AB, drawn from the point of contact A, cuts the circle, the angle DAB is equal to the angle in the alternate segment

AHB: (111. 32.)

but the angle DAB is equal to the angle C; (constr.) therefore the angle C is equal to the angle in the segment AIIB.

Wherefore, upon the given straight line AB, the segment AHB of a circle is described, which contains an angle equal to the given angle C. Q.E.F.

PROPOSITION XXXIV. PROBLEM. . From a given circle to cut off a segment, 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 from the circle 4 BC a segment that shall contain an angle equal to the given angle D.

Draw the straight line EF touching the circle ABC in any point B, (111. 17.)

and at the point B, in the straight line BF, make the angle FBC equal to the angle D. (1. 23.) Then the segment BAC shall contain an angle equal to the given angle D.

Because the straight line EF touches the circle ABC,

and BC is drawn from the point of contact B, therefore the angle FBC is equal to the angle in the alternate

segment BAC of the circle: (111. 32.)

but the angle FBC is equal to the angle D; (constr.) therefore the angle in the segment BAČ is equal to the angle

D. (1. ax. 1.) Wherefore from the given circle ABC, the segment BĄC is cut off, containing an angle equal to the given angle D. Q.E.F.

PROPOSITION XXXV. THEOREM. If two straight lines cut one another within a circle, 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, cut one another in the point E, within the circle ABCD.

Then the rectangle contained by AE, EC shall be equal to the rectangle contained by BE, ED.

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First, if AC, BD pass each of them througła the center, so that E is the center; it is evident that since AE, EC, BE, ED, being all equal, (1. def. 15.) therefore the rectangle AE, EC is equal to the rectangle BE, ED.

Secondly, let one of them BD pass through the center, and cut the other AC, which does not pass through the center, at right angles, in the point E.

Then, if BD be bisected in F,
Fis the center of the circle ABCD.

Join AF. Because BD which passes through the center, cuts the straight line AC, which does not pass through the center, at right angles in E,

therefore AE is equal to EC: (111. 3.) and because the straight line BD is cut into two equal parts in the

point F, and into two unequal parts in the point E, therefore the rectangle BE, ED, together with the square on EF, is equal to the square on FB; (11. 5.)

that is, to the square on FA: but the squares on AE, EF, are equal to the square on FA: (1. 47.) therefore the rectangle BE, ED, together with the square on EF, is equal to the squares on AE, EF: (1. ax. 1.)

take away the common square on EF, and the remaining rectangle BE, ED is equal to the remaining square on AE; (I. ax. 3.)

that is, to the rectangle AE, EC. Thirdly, let BD, which passes through the center, cut the other AC, which does not pass through the center, in E, but not at right angles.

Then, as before, if BD be bisected in F,

Fis the center of the circle.
Join AF, and from F draw FG perpendicular to AC; (1. 12.)

therefore AG is equal to GC; (111. 3.)
wherefore the rectangle AE, EC, together with the square on EG,
is equal to the square on AG: (11. 5.)

to each of these equals add the square on GF; therefore the rectangle A E, EC, together with the squares on EG,

GF, is equal to the squares on AG, GF; (1. ax. 2.) but the squares on EG, GF, are equal to the square on EF; (1. 47.)

and the squares on AG, GF are equal to the square on AF: therefore the rectangle AE, EC, together with the square on EF, is equal to the square on AF';

that is, to the square on FB: but the square on FB is equal to the rectangle BE, ED, together

with the square on EF; (11. 5.) therefore the rectangle A E, EC, together with the square on EF, is equal to the rectangle BE, ED, together with the square on EF; (I. ax. 1.)

take away the common square on EF, and the remaining rectangle AE, EC, is therefore equal to the re

maining rectangle BE, ED. (ax. 3.) Lastly, let neither of the straight lines AC, BD pass through the center.

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Take the center F, (111. 1.)
and ihrough E the intersection of the straight lines AC, DB,

draw the diameter GEFH. And because the rectangle AE, EC' is equal, as has been shewn, to the rectangle GE, EH; and for the same reason, the rectangle BE, ED is equal to the

same rectangle GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED. (I. ax. 1.)

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

PROPOSITION XXXVI. THEOREM. If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; thc rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square on the line which touches it.

Let D be any point without the circle ABC,
and let DCA, DB be two straight lines drawn from it,

of which DCA cuts the circle, and DB touches the same. Then the rectangle AD, DC shall be equal to the square on DB. Either DCA passes through the center, or it does not:

first, let it pass through the center E.

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Join EB, therefore the angle EBD is a right angle. (111. 18.) And because the straight line A C is bisected in E, and produced

to the point D, therefore the rectangle AD, DC, together with the square on EC, is equal to the square on ED: (11. 6.)

but CE is equal to EB; therefore the rectangle AD, DC, together with the square on EB,

is equal to the square on ED: but the square on ED is equal to the squares on EB, BD, (1. 47.)

because ÈBD is a right angle: therefore the rectangle AD, DC, together with the square on EB,

is equal to the squares on EB, BD: (ax. 1.)

take away the common square on EB; therefore the remaining rectangle AD, DC is equal to the square

on the tangent DB. (ax. 3.) Next, if DCA does not pass through the center of the circle ABC.

Take E the center of the circle, (111. 1.) draw EF perpendicular to AC, (1. 12.) and join EB, EC, ED.

Because the straight line EF, which passes through the center, cuts the straight line AC, which does not pass through the center, at right angles; it also bisects AC, (111. 3.)

therefore AFis 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 on FC,

is equal to the square on FD: (11. 6.)

to each of these equals add the square on FE; therefore the rectangle AD, DC, together with the squares on CF, FE,

is equal to the squares on DF, FE: (1. ax. 2.) but the square on ED is equal to the squares on DF, FE, (I. 47.)

because EFD is a right angle;

and for the same reason, the square on EC' is equal to the squares on CF, FE; therefore the rectangle AD, DC, together with the square on EC,

is equal to the square on ED: (ax. 1.)

but CE is equal to EB; therefore the rectangle AD, DC, together with the square on EB,

is equal to the square on ED: but the squares on EB, BD, are equal to the square on ED, (1. 47.)

because EBD is a right angle: therefore the rectangle Al), DC, together with the square on EB,

is equal to the squares on EB, BD;

take away the common square on EB; and the remaining rectangle AD, DC is equal to the square on DB. (1. ax. 3.)

Wherefore, if from any point, &c. Q.E.D. COR. If from any point without a circle, there be drawn two straight

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