Let O be the centre. Join CO, and produce it to meet the circle again in E. Join DE. Now since O is the centre, the segment ACE is greater than a semicircle; hence, by the first case, fig. (α), the angle ACE is equal to ADE. In like manner the angle ECB is equal to EDB. Hence the whole angle ACB is equal to the whole angle ADB. Cor. 1.—If two triangles ACB, ADB on the same base AB, and on the same side of it, have equal vertical angles, the four points A, C, D, B are concyclic. Cor. 2.—If A, B be two fixed points, and if C varies its position in such a way that the angle ACB retains the same value throughout, the locus of C is a circle. In other words—Given the base of a triangle and the vertical angle, the locus of the vertex is a circle. Exercises. 1. Given the base of a triangle and the vertical angle, find the locus— (1) of the intersection of its perpendiculars; (2) of the intersection of the internal bisectors of its base angles; (3) of the intersection of the external bisectors of the base angles; (4) of the intersection of the external bisector of one base angle and the internal bisector of the other. 2. If the sum of the squares of two lines be given, their sum is a maximum when the lines are equal. 3. Of all triangles having the same base and vertical angle, the sum of the sides of an isosceles triangle is a maximum. 4. Of all triangles inscribed in a circle, the equilateral triangle has the maximum perimeter. 5. Of all concyclic figures having a given number of sides, the area is a maximum when the sides are equal. PROP. XXII.—Theorem. The sum of the opposite angles of a quadrilateral (ABCD) inscribed in a circle is two right angles. Dem.—Join AC, BD. The angle ABD is equal to ACD, being in the same segment ABCD [xxi.]; and the angle DBC is equal to DAC, because they are in the same segment DABC. Hence the whole angle ABC is equal to the sum of the two angles ACD, DAC. To each add the angle CDA, and we have the sum of the two angles ABC, CDA equal to the sum of the three angles ACD, DAC, CDA of the triangle ACD; but the sum of the three angles of a triangle is equal to two right angles [I. xxxii.]. Therefore the sum of ABC, CDA is two right angles. Or thus: Let O be the centre of the circle. Join OA, OC (see fig. 2). Now the angle AOC is double of CDA [xx.], and the angle COA is double of ABC. Hence the sum of the angles [I. Def. ix., note] AOC, COA is double of the sum of the angles CDA, ABC; but the sum of two angles AOC, COA is four right angles. Therefore the sum of the angles CDA, ABC is two right angles.