Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.

Let ABCD be a rectangle inscribed in a circle. Let the diagonals AC and BD intersect at P.

In a rectangle, the diagonals are equal and bisect each other.

So, AP = PC, BP = PD, and AC = BD.

Also, each angle of the rectangle is $90^\circ$, so $\angle ABC = 90^\circ$.
By the converse of the corollary (angle in semicircle theorem), AC is a diameter. Similarly, BD is also a diameter.

Two diameters of a circle intersect at the centre of the circle. Therefore P (the point where AC and BD intersect) is the centre of the circle.