Question
Prove that the parallelogram circumscibing a circle, is a rhombus.
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Answer
Given: A parallelogram ABCD circumscribes a circle with centre O, To prove: AB = BC = CD = AD Proof: We know that tangents drawn from an external point to the circle are equal.$\therefore$ AP = AS ....(tangent from A)BP = BQ ....(tangent from B) CR = CQ ....(tangent from C) DR = DS ....(tangent from D) ⇒ AB + CD = AP + BP + CR + DR = AS + BQ + CQ + DS = (AS + DS) + (BQ + CQ) = AD + BC Since opposite sides of a parallelogram are equal, 2AB = 2AD ⇒ AB = AD So, AB = BC = CD = AD Hence proved.
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