The relation between these angles may be reasoned out as follows:

CAD is an isosceles triangle; and therefore the angles CAD, CDA are equal. Hence the exterior angle ACE, which is equal to their sum (Ch. V., 1), must be twice ADC. Similarly BCE is twice BDC. Therefore the sum

(or difference, see second figure) ACB is twice, ADB.

That is, the angle at the centre of a circle is double the angle at the circumference which stands upon the same arc (here AB).

E

B

B

The truth of this should be tested by describing a number of circles, constructing, in each case, an angle at the centre and another at the circumference on the same arc, and using the protractor to determine the magnitudes of these angles.

the angles at the circumference is half the same angle, ACB, at the centre.

Hence angles described in the same segment of a circle, i.e., angles standing on the same arc of a circle, being on the circumference, are equal to one another.

Using the bevel, construct a number of angles as in the annexed figure, all of the same magnitude, and with the sides of each passing

through the points A and B. Then

taking any three of the angular
points, and, by the method of
Ch. X., 6, constructing for the
circle through these three points,
show, by describing the circle, that A

it passes through the other angu

lar points, and also through the points A and B.

3. Take any four points, A, B, C, D, on the circumference of a circle, and join them as in the figure, so constructing a quadrilateral in the circle. Adjust the bevel to the opposite angles B and D, and construct angles equal to them adjacent to one another. What do you observe with reference to the sum of the angles B and D? What with reference to the sum of the angles A and C ?

Repeat this measurement with

B

respect to the opposite angles of other quadrilaterals in circles.

The annexed figure suggests what conclusion should be reached with respect to the sum of the angles at B and D and at A and C. For the angle marked at O is double of the angle ADC (Ch. XII., B 1); and the other angle at O is double the angle ABC. Therefore the angles at O are together double

the sum of the angles ADC and ABC. But the angles at O make up four right angles.

Hence the angles ABC, ADC are together equal to two right angles.

Hence the opposite angles of a quadrilateral inscribed in a circle are together equal to two right angles.

Using the protractor, construct a quadrilateral with two of its opposite angles together equal to two right angles. Taking any three of the angular points, and, by the method

A

D

of Ch. X., 6, constructing for the circle through these three points, and describing the circle, note the position of the quadrilateral with respect to the circle.

Repeat the construction for several such quadrilaterals.

The result of such observations is that if the opposite angles of a quadrilateral are together equal to two right angles, a circle can be described about it.

Since a quadrilateral can be divided into two triangles by joining its opposite angles, the sum of all the angles of any quadrilateral is four right angles. Hence if the sum of a pair of opposite angles be two right angles, the sum of the other pair is two right angles also.

in a semicircle may be proved thus: The straight angle ACB at the centre is (Ch. XII., 1) double any of the angles at the circumference. But the straight angle ACB is 180°. Hence the angle in a semicircle is 90°.

ADB being a right-angled triangle, find the centre of the circle through A, D and B, by bisecting AD, BD and drawing the perpendiculars A EC, FC. Note that these

F

B

perpendiculars intersect in AB; and note also that C, being the centre of the circle through A, D and B, the centre of the hypotenuse of a right-angled triangle is equidistant from the three angles of the triangle.

5. A chord, such as AB, which does not pass through the centre, divides the circle into two segments, one of which, ADB, is greater, and the other, ACB, less than a semicircle. Evidently the marked angle AOB is greater than two right angles, and therefore the angle ACB, which is

B

half of the marked angle AOB, is greater than one right angle. Similarly the angle ADB, being half the other angle at O, is less than a right angle.

Hence the angle in a segment of a circle. less than a semicircle is greater than a right angle; and the angle in a segment of a circle greater than a semicircle is less than a right angle.

AC, CB contain an angle which has, in succession, the magnitudes 80°, 85°, 89°, 91, 95°. Construct in the different cases the circles through A, C and B, and note the positions of the centre with respect to the side AB.

6. Draw with accuracy the tangent CAB at any point A on the circumference of a circle. From A draw any chord AD, and construct the angles AED, AFD in the segments into which AD divides the circle. Then, using the bevel, discover the relation in

B

size between the angle CAD and the angle AFD in the alternate segment; and the relation between the angle BAD and the angle AED in the alternate segment.