ABCD IS A RECTANGLE.PROVE TAT THE CENTER OF THE CIRCLE THROUGH A,B,C,D IS THE POINT OF INTERSECTION OF ITS DIAGNOLS

Here is the solution of your question-

Given, ABCD is a cyclic rectangle whose diagonals intersect at O.

To show : O is the centre of circle.

We know that ∠BCD = 90° (since it is a rectangle)

So, BD is the diameter of the circle (If angle made by the chord at the circle is right angle then the chord is the diameter).

Also, diagonals of a rectangle bisect each other and are equal.

∴ BO = OD

BD is the diameter therefore, BO and OD are the radius.

Thus, O is the centre of the circle.

Hence, centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.