point on the circumference of a circle; it is required to prove that the angle APB is half of the angle AOB standing on the same arc.

Proof. Join PO, and produce it to Q.

Then because OA is equal to OP;

therefore the angle OAP is equal to the angle OPA: (1. 6.) but the exterior angle AOQ is equal to the two interior and opposite angles OAP and OPA;

(1. 25.) therefore the angle AOQ is double of the angle OPA. Similarly the angle QOB is double of the angle OPB.

Hence in figs. (1 and 2) the sum, or (in fig. 3) the difference of the angles AOQ, QOB is double of the sum or difference of OPA and OPB,

that is, the angle AOB is double of the angle APB;

and therefore the angle APB is half of the angle AOB on the same arc*.

THEOREM 14.

Angles in the same segment of a circle are equal to one another.

Part. En. Let APB, AQB be angles in the same segment APQB; it is required to prove that the angle APB is equal to the angle AQB.

Proof. Take O the centre; and join AO, BO.

Then because the angles APB,

AQB at the circumference are each of them half the angle

AOB at the centre on the same arc,

(III. 13.)

therefore the angle APB is equal to the angle AQBt.

COR. The angle subtended by the chord of a segment at a point within it is greater than, and at a point outside its segment on the same side of the chord as the segment, is less than, the angle in the segment.

Part. Ex. Let APB be a segment of a circle, and C a point on the same side of AB as the segment; it is required to prove that the angle

D B

ACB is greater or less than the angle in the segment APB according as C is within or without the segment.

Proof. Take any point D in AB and join CD, and let CD (produced if necessary) meet the curved boundary of the segment in P. Join PA, PB.

Then if C is within the segment APB it is evidently within the triangle APB, and therefore the angle ACB is greater than the angle APB. (I. 13.)

Again if C is without the segment APB, P is evidently within the triangle ACB, and therefore the angle ACB is less than the angle (I. 13.)

APB.

Remark. From this theorem and its corollary we learn that the locus of a point on one side of a given straight line at which that straight line subtends a constant angle is an arc of a circle of which that line is the chord.

THEOREM 15.

The angle in a segment is greater than, equal to, or less than a right angle, according as the segment is less than, equal to, or greater than a semicircle*.

* Euclid, III. 31.

Part. En. Let AB be a diameter of a circle, cutting off the semicircle ADB; and let any other chord BC divide the circle into the segment BDC less than a semicircle, and BEC greater than a semicircle;

B

it is required to prove that the angle in the segment BDC less than a semicircle is greater than a right angle ;

and that the angle in the semicircle BDA is equal to a right angle; and that the angle in the segment BEC greater than a semicircle is less than a right angle.

Proof. Let O be the centre; join CO.

Then the angle in the segment CDB is half the angle COB subtended at the centre by the same arc BEC.

(III. 13.)

But this is a reflex angle, and is greater than two right angles;

therefore the angle in the segment CDB is greater than one right angle.

Again, the angle in the semicircle ADB is half the angle AOB upon the same arc AEB.

(III. 13.)

But the angle AOB is equal to two right angles; therefore the angle in the semicircle is equal to a right angle.

Lastly, the angle in the segment CEB is half the angle COB. (III. 13.)

But the angle COB is less than two right angles;

therefore the angle in the segment CEB is less than a right angle.

THEOREM 16.

A segment is less than, equal to, or greater than a semicircle according as the angle in it is greater than, equal to, or less than a right angle.

Proof. According as the angle in the segment, that is at the circumference, is greater than, equal to, or less than a right angle; the angle at the centre will be greater than, equal to, or less than two right angles;

that is, the segment is less than, equal to, or greater than a semicircle.

Alternative Proof. For of the hypotheses that a segment is either greater than, equal to, or less than a semicircle, one must be true; and of the conclusions proved in Th. 15 that the angle in that segment is either less than, equal to, or greater than a right angle, no two can be true at the same time;

therefore the converses of the theorems in Th. 15 are true, that is, according as the angle in a segment is less than, equal to, or greater than a right angle, that segment is greater than, equal to, or less than a semicircle.

THEOREM 17.

The opposite angles of a quadrilateral inscribed in a circle are supplementary*.

* Euclid, III. 22.

Part. En. Let ABCD be a quadrilateral inscribed in

a circle; it is required to prove

that its opposite angles ABC, CDA are supplementary.

Proof. Take O the centre, and join AO, CO.

Then the angles ABC, CDA are respectively the halves of the angles made by AO, OC at the centre O.

(III. 13.)

......

But the sum of the angles at the centre O is four right angles : (I. 4, COR.) therefore the sum of the angles ABC, CDA is two right angles;

that is, the angles ABC, CDA are supplementary.

COR. I. The exterior angle of a quadrilateral inscribed in a circle is equal to the interior opposite angle.

For if CD is produced to E, the exterior angle ADE is supplementary to ADC, and is therefore equal to ABC.

COR. 2. If the opposite angles of a quadrilateral are supplementary, the quadrilateral can be inscribed in a circle.

EXERCISES ON SECTION III.

I. Prove that the lines which join the extremities of equal arcs in a circle are either equal or parallel.

2. If two opposite sides of a quadrilateral inscribed in a circle are equal, prove that the other two are parallel.