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.
Or again: Let O be the centre (fig. 2). Join OA, OB OC, OD. Then the four triangles AOB, BOC, COD, DOA are each isosceles. Hence the angle OAB is equal to the angle OBA, and the angle OAD equal to the angle ODA; therefore the angle BAD is equal to the sum of the angles OBA, ODA. In like manner the angle BCD is equal to the sum of the angles OBC, ODC. Hence the sum of the two angles BAD, BCD is equal to the sum of the two angles ABC, ADC, and hence each sum is two right angles.
Cor.—If a parallelogram be inscribed in a circle it is a rectangle.
Exercises.