Question

prove that a parallelogram circumscribing a circle is a rhombus​

Answer 1

Answer:

$\huge\mathcal\red{\boxed{\boxed{Solution..}}}$

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

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 a rhombus.

#Aami..❤❤

Answer 2

$\huge\sf\blue{Given}$

✭ A parallelogram ABCD which circumscribes a circle with centre O

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$\huge\sf\gray{To\;Prove}$

➠ Parallelogram circumscribing a circle is a rhombus

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$\huge\sf\purple{Steps}$

❍ 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 Proved!!

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