Question

Prove that a cyclic parallelogram is a rectangle.
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Answer 1

Lets solve it !!

Prove that a cyclic parallelogram is a rectangle.

Given;

Let ABCD be a cyclic parallelogram.

To Prove;

ABCD is a rectangle.

Proof;

A rectangle is a parallelogram with one angle 90°, so we have to prove angle 90°.

Since ABCD is a parallelogram.

$\sf{\angle A = \angle C} - - \boxed{ \sf{ \orange{(opposite \: angles \: of \: parallelogram \: are \: equal)}}} ......(1)$

In cyclic parallelogram ABCD

$\sf{\angle A + \angle C = 180°} \\ \sf{\angle A + \angle A = 180°} \\ \sf{2 \angle A = 180°}$

$\sf{\angle A = \cancel\frac{180}{2} = 90} - - - \boxed{ \sf{ \orange{(sum \: of \: opposite \: angles \: of \: a \: cyclic \: \: quadrilateral \: is \: 180 \degree)}}} ......from \: 1$

So, ABCD is a parallelogram with one angle 90°.

Hence, ABCD is a rectangle.

________________

And we are done! :D

Answer 2

$\\ \huge\mathfrak {\underline \purple{ \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}$

Here , as the given query asked to prove that a cyclic parallelogram is a rectangle . Let us consider a cyclic parallelogram whose angles are ∠A , ∠B , ∠C , ∠D . The diagram is given in the attachment which is created by me & not from any sources . Let us proved it .

$\\ \huge\mathfrak {\underline \purple{ \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}$

$\\ \huge\mathfrak {\underline \purple{ \: \: \: \: \: \: \: \: solution\mapsto \: \: \: \: \: \: \: \: }}$

We know that,

• Opposite angles of parallelogram are equal.

This means ➲

$\large \tt\blue➦ ∠A \red\longrightarrow ∠C \\ \\ \large \tt\blue➦ ∠B \red\longrightarrow ∠D$

• Sum of the opposite angles if a cyclic quadrilateral is 180°

This means ➲

$\large \tt\blue➦ ∠A \: + ∠C = 180° .........(equation \: 1)$

$\large \tt \: from \: equation \: (1), \\ \\ \large \tt\blue➦ ∠A + ∠C = 180° \\ \\ \large \tt\blue➦ ∠ A + ∠A = 180° \\ \\ \large \tt\blue➦ 2 ∠A = 180° \\ \\ \large \tt\blue➦ ∠ A = \frac{ \cancel{180° }} {\cancel{2} } \\ \\ \large \tt\blue➦ ∠ A = 90°$

$\huge\sf {\underline{ \underline \color{darkgreen}{proved}}} \huge ✔$

$\large \therefore \bf\: ABCD \: is \: a \: rectangle$