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Solution


Q.

Prove that a parallelogram circumscribing a circle is a rhombus

Ans.

We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle) with centre O.

Since tangents to a circle from an external point are equal in length,

∴ AP = AS, BP = BQ, CR = CQ and DR = DS

Adding, we get

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

⇒ AB + CD = AD + BC

But AB = CD [opposite sides of ABCD]

and BC = AD

∴ AB + CD = AD + BC ⇒ 2 AB = 2 BC

⇒ AB = BC

Similarly AB = DA and DA = CD

Thus, AB = BC = CD = AD

Hence ABCD is a rhombus.