Prove that a cyclic parallelogram is a rectangle.

(N/A) Let $ABCD$ be a cyclic parallelogram.
Since $ABCD$ is a cyclic quadrilateral,the sum of its opposite angles is $180^{\circ}$.
Therefore,$\angle A + \angle C = 180^{\circ}$ .... $(1)$
In a parallelogram,opposite angles are equal,so $\angle A = \angle C$ .... $(2)$
Substituting $(2)$ in $(1)$,we get:
$\angle A + \angle A = 180^{\circ}$
$2\angle A = 180^{\circ}$
$\angle A = 90^{\circ}$
Since $\angle A = \angle C$,we have $\angle C = 90^{\circ}$.
Similarly,for the other pair of opposite angles,$\angle B + \angle D = 180^{\circ}$ and $\angle B = \angle D$,which implies $\angle B = \angle D = 90^{\circ}$.
Since all angles of the parallelogram $ABCD$ are $90^{\circ}$,$ABCD$ is a rectangle.