Question

Prove that a cyclic parallelogram is a rectangle.

Answer 1

Let ABCD be a cyclic parallelogram.

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

We know that opposite angles of a parallelogram are equal.

$\therefore \angle A=\angle Cand\angle B=\angle D$

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

Then $\angle A+\angle A={180}^{\circ }$

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

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

Similarly, $\angle B=\angle D={90}^{\circ }$

Parallelogram ABCD has all the angles as ${90}^{\circ }$. Therefore, it is a rectangle.