Question
Prove that a cyclic parallelogram is a rectangle.
✓
Answer
Given: $A B C D$ is a cyclic parallelogram.
To prove: $A B C D$ is a rectangle.
Proof: $\because ABCD$ is a cyclic quadrilateral
$\therefore \angle 1+\angle 2=180^{\circ}---- (1)$
[ $\because$ Opposite angles of a cyclic quadrilateral are supplementary]
$\therefore A B C D$ is a parallelogram
$\therefore \angle 1=\angle 2$
From $(1)$ and $(2),$
$\angle 1=\angle 2=90^{\circ}$
$\therefore \| \mathrm{~gm} ~A B C D$ is a rectangle.
To prove: $A B C D$ is a rectangle.
Proof: $\because ABCD$ is a cyclic quadrilateral
$\therefore \angle 1+\angle 2=180^{\circ}---- (1)$
[ $\because$ Opposite angles of a cyclic quadrilateral are supplementary]
$\therefore A B C D$ is a parallelogram
$\therefore \angle 1=\angle 2$
From $(1)$ and $(2),$
$\angle 1=\angle 2=90^{\circ}$
$\therefore \| \mathrm{~gm} ~A B C D$ is a rectangle.
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