straight lines from their centres are equal. Therefore B K, K C are equal to E L, L F, and they contain equal angles: Therefore the base B C is equal (1. 4.) to the base EF: Therefore, in equal circles, &c. Q. E. D.

To bisect a given circumference, that is, to divide it into two equal parts.

Let A D B be the given circumference; it is required to bisect it.

D

Join A B, and bisect (1. 10.) it in c; from the point c draw CD at right angles to AB, and join A D, D B: The circumference A D B is bisected in the point D. Because a C is equal to C B, and C D common to the triangles ACD, BC D, the two sides a C, CD are equal to the two BC, CD; and the angle ACD is equal to the angle B C D, because each of them is a right A

B

angle; therefore the base AD is equal (1. 4.) to the base B D. But equal straight lines cut off equal (III. 28.) circumferences, the greater equal to the greater, and the less to the less, and A D, DB are each of them less than a semicircle; because DC passes through the centre (II. 1. Cor.). Wherefore the circumference A D is equal to the circumference DB: Therefore the given circumference is bisected in D. Which was to be done.

In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

K

Let ABCD be a circle, of which the diameter is B C, and centre E; and draw CA dividing the circle into the segments A B C, A D C, and join B A, AD, DC; the angle in the semicircle B A C is a right angle; and the angle in the segment A B C, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle.

Join A E, and produce BA to F; and because BE is equal to EA, the angle E A B is equal (1. 5.) E BA; also, because A E is equal to

E C the angle E A C is equal to ECA; wherefore the whole angle BAC is equal to the two angles A B C, A CB; But FAC, the exterior angle of the triangle ABC, is equal (1. 32.) to в the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a

F

C

E

right (1. Def. 10.) angle: Wherefore the angle BAC in a semicircle is a right angle.

And because the two angles A B C, B A C of the triangle ABC are together less (1. 17.) than two right angles, and that B A C is a right angle, A B C must be less than a right angle; and therefore the angle in a segment ABC greater than a semicircle, is less than a right angle.

And because A B C D is a quadrilateral figure in a circle, any two of its opposite angles are equal (III. 22.) to two right angles; therefore the angles A B C, A D C are equal to two right angles; and A B C is less than a right angle; wherefore the other ADC is greater than a right angle.

Besides, it is manifest, that the circumference of

the greater segment A B C falls without the right angle CA B, but the circumference of the less segment ADC falls within the right angle CA F. "And this is all that is meant, when in the Greek text, and the translations from it, the angle of the greater segment is said to be greater, and the angle of the less segment is said to be less, than a right angle."

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 same two; and when the adjacent angles are equal, they are right angles.

PROP. XXXII. THEOR.

If a straight line touches 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 touching the circle, shall be equal to the angles which are in the alternate segments of the circle.

Let the straight line EF touch the circle A B C D in B, and from the point в let the straight line B D be drawn cutting the circle: The angles which B D makes with the touching line EF shall be equal to the angles in the alternate segments of the circle: that is, the angle F BD is equal to the angle which is in the segment DA B, and the angle DBE to the angle in the segment B C D.

A

From the point в draw (I. 11.) BA at right angles to EF, and take any point c in the circumference B D, and join AD, DC, CB; and because the straight line EF touches the circle A B C D in the point B, and BA is drawn at right angles to the touching line from the point of contact B, the centre of the E

B

F

circle is (m. 19.) in BA; therefore the angle A D B in a semicircle is a right (III. 31.) angle, and consequently the other two angles BA D, A B D are equal (1. 32.) to a right angle: But A B F is likewise a right angle; therefore the angle ABF is equal to the angles BAD, ABD: Take from these equals the common angle A B D; therefore the remaining angle D B F is equal to the angle B A D, which is in the alternate segment of the circle; and because A B C D is a quadrilateral figure in a circle, the opposite angles B A D, BCD are equal (III. 22.) to two right angles; therefore the angles D B F, DBE, being likewise equal (1. 13.) to two right angles, are equal to the angles BA D, BCD; and DBF has been proved equal to B AD: Therefore the remaining angle D BE is equal to the angle B CD in the alternate segment of the circle. Wherefore, if a straight line, &c. Q. E. D.

PROP. XXXIII. PROB.

Upon 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, and bisect (1. 10.)
AB in F, and from the centre F,
at the distance FB, describe the
semicircle AHB; therefore the
angle AHB in a semicircle is
(II. 31.) equal to the right angle at c.

A

H

But, if the angle c be not a right angle, at the point A, in the straight line AB, make (1. 23.) the angle BAD equal to the angle c, and from the point

A draw (1. 11.) AE at right angles to AD; bisect (1. 10.) AB in F, and from F draw (1. 11.) FG at right angles to AB, and join GB: And 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; and the angle AFG is equal to the angle BFG; therefore the base AG is equal (1. 4.) to

the base GB; and the circle described from the centre G, at the distance 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 (III. 16. Cor.) touches the circle; and because A B drawn from the point of contact A cuts the circle, the angle DAB is equal to the angle in the alternate segment AHB (III. 32.). But the angle DAB is equal to the angle c, therefore also the angle c is equal to the angle in the segment AHB: Wherefore, upon the given straight line A B 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.