Question
Prove that a cyclic parallelogram is a rectangle.
✓
Answer
Given: $\text{ABCD}$ is a cyclic parallelogram.
To prove: $\text{ABCD}$ 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$]$
$\therefore ABCD$ is a parallelogram
$\therefore \angle 1 = \angle 2$
From $(1)$ and $(2),$
$\angle 1 = \angle 2 = 90^\circ$
$\therefore ||\ gm\ \text{ABCD}$ is a rectangle.
To prove: $\text{ABCD}$ 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$]$
$\therefore ABCD$ is a parallelogram
$\therefore \angle 1 = \angle 2$
From $(1)$ and $(2),$
$\angle 1 = \angle 2 = 90^\circ$
$\therefore ||\ gm\ \text{ABCD}$ is a rectangle.
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