Question

12. Prove that a cyclic parallelogram is a rectangle.
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Answer 1

Solution :-

Let the cyclic parallelograms be ABCD

Given :-

ABCD is a cyclic parallelogram .

[ So, here we prove by using properties of parallelogram ]

To prove :-

ABCD is a rectangle

Proof :-

As we know that all angles of rectangle are 90°

So here we have to prove that angles of parallelogram to 90°

Here, ABCD is a parallelogram

Therefore,

ΔA = ΔC ( Opposite angles of parallelogram) ...( 1 )

In a cyclic quadrilateral , Sum of opposite angles is 180°

Therefore,

ΔA + ΔC = 180°

ΔA + ΔA = 180° ( From ( 1 ) )

2ΔA = 180°

ΔA = 180°/2

ΔA = 90°

Now,

ΔB = ΔD ( Opposite sides of parallelogram) ...( 2)

Therefore,

ΔB + ΔD = 180°

ΔB + ΔB = 180°

2ΔB = 180°

ΔB = 180/2

ΔB = 90°

From ( 1 ) and ( 2 )

ΔA = ΔB = ΔC = ΔD = 90° $.$

[ Note :- Refer the above attachment ]

Answer 2

$\huge\fcolorbox{black}{lime}{Answer}$

Solution :-

Let the cyclic parallelograms be ABCD

Given :-

ABCD is a cyclic parallelogram .

[ So, here we prove by using properties of parallelogram ]

To prove :-

ABCD is a rectangle

Proof :-

As we know that all angles of rectangle are 90°

So here we have to prove that angles of parallelogram to 90°

Here, ABCD is a parallelogram

Therefore,

ΔA = ΔC ( Opposite angles of parallelogram) ...( 1 )

In a cyclic quadrilateral , Sum of opposite angles is 180°

Therefore,

ΔA + ΔC = 180°

ΔA + ΔA = 180° ( From ( 1 ) )

2ΔA = 180°

ΔA = 180°/2

ΔA = 90°

Now,

ΔB = ΔD ( Opposite sides of parallelogram) ...( 2)

Therefore,

ΔB + ΔD = 180°

ΔB + ΔB = 180°

2ΔB = 180°

ΔB = 180/2

ΔB = 90°

From ( 1 ) and ( 2 )

ΔA = ΔB = ΔC = ΔD = 90° ..

[ Note :- Refer the above attachment ]