Question

If all sides of a parallelogram touch a circle,show that the parallelogram is a rhombus

Answer 1

Hey mate this is my answer

Answer:

Step-by-step explanation:

Solution Given

Parallelogram ABCD touches a circle with centre O.

To prove

ABCD is a rhombus.

 

Proof

Since the length of the tangents from an external point to a given circle are equal

So,

AP=AS           (i)

BP=BQ           (ii)

CR=CQ and  (iii)

DR=DS          (iv)

 

Adding (i), (ii),(iii),(iv) we get

(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)

AB+CD=AD+BC

Since ABCD is a Parallelogram CD=AB and BC=AD

This implies, AB+AB=AD+AD

2AB=2AD

This implies AB=AD

But AB=CD and AD=BC as opposite sides of a Parallelogram are equal.

Therefore , AB=BC=CD=AD.

Hence ABCD is a rhombus.

Hope it helps.........

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