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

Question

Philoid · Accepted Answer

Let O be the circle circumscribing the cyclic rectangle ABCD.

Since, ∠ABC = 90^o and AC is the chord of the circle. Similarly, BD is a diameter

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