$AC$ and $BD$ are chords of a circle that bisect each other. Prove that $ABCD$ is a rectangle.

(A) Let the circle have its centre at $O'$. Let the chords $AC$ and $BD$ intersect at point $O$.
Since the chords bisect each other,$AO = OC$ and $BO = OD$.
In $\Delta AOB$ and $\Delta COD$:
$AO = OC$ (given),
$BO = OD$ (given),
$\angle AOB = \angle COD$ (vertically opposite angles).
Therefore,$\Delta AOB \cong \Delta COD$ by the $SAS$ congruence rule.
This implies $AB = CD$ and $\angle OAB = \angle OCD$. Since these are alternate interior angles,$AB \parallel CD$.
$A$ quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. Thus,$ABCD$ is a parallelogram.
Since $ABCD$ is a cyclic quadrilateral (as its vertices lie on the circle),the sum of its opposite angles is $180^{\circ}$.
In a parallelogram,opposite angles are equal,so $\angle A = \angle C$ and $\angle B = \angle D$.
Since $\angle A + \angle C = 180^{\circ}$,we have $2\angle A = 180^{\circ}$,which means $\angle A = 90^{\circ}$.
$A$ parallelogram with one angle equal to $90^{\circ}$ is a rectangle. Therefore,$ABCD$ is a rectangle.