Question

Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals. 

Answer 1

Let O be the circle circumscribing the cycle rectangle ABCD. Since ∠ABC= 90^o and AC is a chord of the circle, so AC is a diameter of a circle. Similarly BD is a diameter 

Hence, point of intersection of AC and BD is the center of the circle.