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

Let ADB be the given circumference; it is required to bisect it.

Join AB, and bisect it in C; from C draw CD at right angles to AB, and join AD, DB: the circumference ADB shall be bisected in the point D.

A. C B

Because AC is equal to CB, and CD common to the two triangles, ACD, BCD, the two sides AC, CD are equal to the two BC, CD, each to each, and the angle ACD is equal to the angle BCD, because each of them is a right angle therefore the base AD is equal to the base BD: But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less (3. 28); and AD, DB are each of them less than a semicircle, because DC passes through the centre (3. 1. Cor.); therefore the circumference AD is equal to the circumference DB: Wherefore the given circumference is bisected in D. Q.E. F.

In any circle, the angle in a semicircle is a right angle, and 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.

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

Then, because EA equal to the angle

angle in the segment ABC, which is greater than a semicircle, shall be less than a right angle; and the angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle. Join AE, and produce BA to F: is equal to EB, the angle EAB is EBA, and, because EA is equal to EC, the angle EAC is equal to ECA; therefore the whole angle BAC is equal to the two angles ABC, ACB: But FAC, the exterior angle of the triangle ABC, is equal to the two angles ABC, ACB (1. 32); therefore the angle BAC is equal to the angle FAC, and each of them is there- B fore a right angle: Therefore the angle BAC in a semicircle is a right angle.

F

E

୯

Again, because the two angles ABC, BAC of the triangle ABC are together less than two right angles (1. 17), and that BAC is a right angle, therefore ABC 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.

Lastly, because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are together equal to two right angles (3. 22); therefore the angles ABC, ADC are together equal to two right angles: But ABC is less than a right angle; therefore ADC is greater than a right angle, and therefore the angle in a segment ADC, less than a semicircle, is greater than a right angle.

Wherefore, The angle &c. Q. E.D.

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, each of them is a right angle.

L

PROP. XXXII. THEOR.

If 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 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 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 FBD shall be equal to the angle in the segment BAD, and the angle EBD to the angle in the segment BCD.

From the point B draw BA at right angles to EF, and take any point C in the circumference

E

B

F

BD, and join AD, DC, CB: Then, 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 centre of the circle is in BA (3. 19): Therefore the angle ADB in a semicircle is a right angle (3. 31); and, consequently, the other two angles DAB, ABD, are together equal to a right angle: But ABF is also a right angle; therefore the angle ABF is equal to the two angles DAB, ABD: From each of these equals take away the common angle ABD; then the remaining angle FBD is equal to the angle DAB, which is in the alternate segment of the circle: Again, because ABCD is a quadrilateral figure in a circle, the opposite angles DAB, DCB are together equal to two right angles (3. 22): But the angles FBD, EBD, are together

equal to two right angles; therefore the angles FBD, EBD, are equal to the angles DAB, DCB: and the angle FBD has been proved equal to the angle DAB; therefore the remaining angle EBD is equal to the angle DCB in the alternate segment of the circle.

Wherefore, If a straight line &c.

PROP. XXXIII.

Q.E. D.

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 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.

H

A F B

First, let the angle at C be a right angle: bisect AB in F, and from the centre F, at the distance FB, describe the semicircle AHB: then the angle AHB in a semicircle (3. 31) is equal to the right angle at C.

But if the angle C be not a right angle, at the point A,

in the straight line AB, make the angle BAD equal to the angle C, and from the point A draw AE at right angles to AD; bisect AB in F, and from F draw FG at right angles to AB, and join GB.

A

H

E

F

B

Ꭰ

Then, because AF is equal to FB, and FG common to the two 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-therefore the base AG is equal to the base BG, and the circle described from the centre G, at the distance GA, shall pass through the point B: Let this be the circle AEB: And because

from the point A, the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD touches the circle (3. 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 (3. 32): But the angle DAB is equal to the angle C; therefore also the angle in the segment AHB is equal to the angle C: Wherefore, upon the given straight line

C

H

F

B

E

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

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 ABC a segment that shall contain an angle equal to the given angle D.

A

Draw (3. 17) the straight line EF touching the circle ABC in the point B ; and at the point B, in the straight line BF, make the angle FBC equal to the angle D. Then, because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal to the angle in the alternate segment, BAC, of the circle (3. 32): But the angle FBC is equal to the angle D; therefore the angle in the segment BAC is equal to the angle D: Wherefore from the given circle ABC has been cut off the segment BAC, containing an angle equal to the given angle D. Q. E. F.

E

B