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Question
Prove that a parallelogram circumscribing a circle is a rhombus.
Solution
Let ABCD be the rhombus circumscribing the circle with centre O, such that AB, BC, CD and DA touch the circle at points P, Q, R and S respectively. We know that the tangents drawn to a circle from an exterior point are equal in length.
∴ AP = AS ...(i) [tangents from A]
BP = BQ ...(ii) [tangents from B]
CR = CQ ...(iii) [tangents from C]
DR = DS ...(iv) [tangents from D]
∴ AB + CD = AP + BP + CR + DR
= AS + BQ + CQ + DS ...[From (i), (ii), (iii), (iv)]
= (AS + DS) + (BQ + CQ)
= AD + BC
Hence, (AB + CD) = (AD + BC)
⇒ 2AB = 2AD
[∵ opposite sides of a parallelogram are equal]
⇒ AB = AD
∴ CD = AB = AD = BC
Hence, ABCD is a rhombus.
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