Let $ABCD$ be a rectangle inscribed in a circle. Let its diagonals $AC$ and $BD$ intersect at point $O$. We need to show that $O$ is the centre of the circle.

Since $ABCD$ is a rectangle, all its interior angles are $90°$. Consider triangle $ABC$ inside the rectangle. It is a right-angled triangle with the right angle at $B$, and its hypotenuse is the diagonal $AC$.

A key property states that for a right-angled triangle, the circumcentre lies at the midpoint of the hypotenuse. The circle passing through the vertices of the rectangle also passes through $A$, $B$, and $C$, making it the circumcircle of triangle $ABC$.

Since the diagonals of a rectangle bisect each other, their point of intersection $O$ is the midpoint of the diagonal $AC$. Because $AC$ is the hypotenuse of right-angled triangle $ABC$, its midpoint $O$ is the circumcentre of the triangle.

As the circumcentre of triangle $ABC$ is the centre of its circumcircle, the point $O$ must be the centre of the circle in which the rectangle is inscribed. Similarly, $O$ is the midpoint of the hypotenuse of triangle $ADC$. Hence, the point of intersection of the diagonals lies at the centre of the circle.