Let equal chords AB and CD intersect at P. Since the chords are equal, their corresponding parts have equal sums. Using the intersecting-chords theorem and equality of the chords, we get AP = CP and BP = DP. Hence proved.
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Draw the perpendicular bisectors of any two sides of the cyclic quadrilateral. Their intersection gives the circumcentre. Then observe whether this point lies inside or outside the quadrilateral.
Opposite angles of a cyclic quadrilateral are supplementary. (2x + 10) + (3x – 20) = 180 5x – 10 = 180 x = 38 Thus, ∠P = 86° and ∠R = 94°.
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.