Book III. AD is equal to the arch DB: and therefore the given arch ADB is bisected in D. Which was to be done.

Igles

Na circle, the angle in a semicircle is a right anbut the angle in a segment greater than a femicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

Let ABCD be a circle, of which the diameter is BC, and centre E; draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC; the angle in the femicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a femicircle, is greater than a right angle.

Join AE, and produce BA to F; and because BE is equal a 5. r. to EA, the angle EAB is equal a to EBA; also, because

AE is equal to EC the angle
EAC is equal to ECA; where-

F

fore the whole angle BAC is e

A

qual to the two angles ABC,

D

ACB. But FAC, the exterior angle of the triangle ABC is b 32. 1. alfo equal to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is c 7. def. 1. therefore a right angle: wherefore the angle BAČ in a femicircle is a right angle.

And because the two angles ABC, BAC of the triangle d 17. 1. ABC are together lessd than two right angles, and BAC is a right angle, ABC must be less than a right angle; and

therefore

therefore the angle in a segment ABC, greater than a semi- Book III. circle, is less than a right angle. Alfo because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal e to two right angles; therefore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle; wherefore the other ADC is greater than a right angle. Therefore, in a circle, &c. QE. D.

€ 22.3.

COR. From this it is manifest, 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.

F

a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line which touches the circle, shall be equal to the angles in the alternate segments of the circle.

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Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn cutting the circle: The angles which BD makes with the touching line EF shall be equal to the angles in the alternate segments of the circle: that is, the angle FDB is equal to the angle which is in the segment DAB, and the angle DBE to the angle in the segment BCD.

From the point B draw a BA at right angles to EF, and a 11. t. take any point Cin the arch 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

H2

E

B

D

C

F

Book III. touching line, from the point of contact B, the centre of b 19.3. the circle is in BA; therefore the angle ABD, in a femic 31. 3. circle, is a right angle, and consequently the other two d 32. 1. angles BAD, ABD are equal d to a right angle: but ABF is likewise a right angle; therefore the angle ABF is equal to the angles BAD, ABD: take from these equals the common angle ABD; and there will remain the angle DBF equal to the angle BAD, which is in the alternate segment of the circle. And because ABCD is a quadrilateral figure in a circle, the oppofite e 22. 3. angles BAD, BCD are equale to two right angles; therefore f 13. 1. the angles DBF, DBE, being likewise equalf to two right angles, are equal 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 segment of the circle.

Wherefore, if a straight line, &c.

E. D.

PROP. XXXIII. PROB.

A 10. I.

U

PON a given straight line to describe a segment 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 describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C.

First, let the angle at C be a right angle; bifecta AB in F, and from the centre F, at the distance FB, describe

the

a right angle, at the point A,

A

F

B

in the straight line AB, make the angle BAD equal to the c 23. 1.

angle C, and from the

point A draw d AE at

H

angle BFG; therefore the base AG is equale to the base e 4.1. GB, and the circle described from the centre G, at the di

stance GA, shall pass through the point B; let this be the circle AHB: And because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD f touches the circle; and because AB, drawn f Cor.16.3.

from the point of contact

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.

Book III.

a 17.3.

PROP. XXXIV. PROB.

10 cut off a segment from a given circle which shall contain an angle equal to a given rectili

T

neal 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 point B, and at the

€ 32. 3. equal to the angle in

the alternate fegment BAC: but the angle FBC is equal to the angle D; therefore 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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PROP.