In a cyclic quadrilateral, opposite angles are supplementary. Therefore, ∠C = 180° – 75° = 105°. Similarly, ∠D = 180° – 110° = 70°.
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Since AB is the diameter of the circle, the angle subtended by the diameter at any point on the circumference is 90°. Therefore, ∠ACB = 90°.
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.
The perpendicular from the centre to the chord bisects the chord. With radius 15 cm and distance 9 cm, half the chord is 12 cm. Therefore, the chord length is 24 cm.
The radius is half the diameter, so it is 13 cm. The perpendicular from the centre bisects the chord, giving half chord 12 cm. Hence, the distance is 5 cm.