Prove that a cyclic parallelogram is a rectangle.

Solution:



Let ABCD be a cyclic parallelogram.

$\angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$ (Opposite angles of a cyclic quadrilateral) ... (1)

We know that opposite angles of a parallelogram are equal.

$\therefore \angle A=\angle C$ and $\angle B=\angle D$

From equation (1),

$\angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$

$\Rightarrow \angle \mathrm{A}+\angle \mathrm{A}=180^{\circ}$

$\Rightarrow 2 \angle A=180^{\circ}$

$\Rightarrow \angle \mathrm{A}=90^{\circ}$

Parallelogram ABCD has one of its interior angles as 90°. Therefore, it is a rectangle.