Take a right triangle with one side 3 cm and another side 3 cm. Its hypotenuse is the required 6 cm chord. Draw its circumcircle. The distance from the centre to the chord is 3 cm.
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The perpendicular from the centre to the chord bisects it. Half the chord is 8 cm. Using the right triangle, radius² = 8² + 6² = 100. Therefore, radius = 10 cm.
Opposite angles of a cyclic quadrilateral are supplementary. (2x + 10) + (3x – 20) = 180 5x – 10 = 180 x = 38 Thus, ∠P = 86° and ∠R = 94°.
Let AB be a chord and O be the centre. Since OA = OB, O is equidistant from A and B. Therefore, O lies on the perpendicular bisector of AB. Hence proved.
Yes, provided X and Y lie on the same side of AB. Angles subtended by the same chord at points on the same segment of a circle are equal.