Question

Hola Mates!!! ❤️

Question:

Prove that Parallelogram circumscribing a circle is a rhombus .

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Answer 1

hey mate here is your answer

Given: ABCD be a parallelogram circumscribing a circle with centre O.

To prove: Abcd is a rohmbus

We know that the tangents drawn to a circle from an exterior point are equal in length.

Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.

Adding the above equations,

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

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

AB + CD = AD + BC

2AB = 2BC

(Since, ABCD is a parallelogram so AB = DC and AD = BC)

AB = BC

Therefore, AB = BC = DC = AD.

Hence, ABCD is rohmbus hope it help you

Answer 2

$\huge\mathfrak{Answer:}$

Given:

• A parallelogram ABCD which circumscribes a circle with centre O.

To Prove:

• We need to prove that a Parallelogram circumscribing a circle is a rhombus .

Proof:

We have been given a parallelogram ABCD which circumscribes a circle with centre O.

We know that the tangents drawn to a circle from an external point are equal in length.

AP = AS

BP = BQ

CR = CQ

DR = DS

On adding, we get

(AP + BP) + (CR + DR)

= (AS + DS) + (BQ + CQ)

=> AB + CD = AD + BC

But, AB = CD [Opposite sides of parallelogram]

and BC = AD

∴ AB + CD = AD + BC

=> 2(AB) = 2(BC)

=> AB = BC

Similarly, AB = DA and DA = CD.

Thus AB = BC = CD = DC

Hence, ABCD is a rhombus.