Question

Prove that a cyclic parallelogram is rectangle.

Answer 1

Given : $ABCD$ be a cyclic parallelogram.

To prove :$ABCD$ is a rectangle.

A rectangle is a parallelogram with one angle ${90}^0$

So, we have to prove angle ${90}^0$

Since $ABCD$ is a paralleogram

$\angle A=\angle C$ (opposite angle of parallelogram are equal)......................(i)

In cyclic parallelogram $ABCD$

$\angle A+\angle C={180}^0$ (Sum of opposite angles of a cyclic quadrilateral is ${180}^0$)

$\angle A+\angle A={180}^0$ from(i)

$2\angle A={180}^0$

$\angle A=\frac{{180}^0}{2}={90}^0$

Hence,

$\angle A=\angle C={90}^o$

Similarly we can prove other two angles are also equal to ${90}^o$

Thus, a cyclic parallelogram is rectangle.