Prove that the parallekogram circumscribing a …

Prove by equalling adjacent sides of a parallelogram

Given ABCD is a ||gm such that its sides touch a circle with centre O. ∴ AB = CD and AB || CD, AD = BC and AD || BC Now, P, Q, R and S are the touching point of both the circle and the ||gm We know that, tangents to a circle from an exterior point are equal in length. ∴ AP = AS [Tangents from point A] ... (1) BP = BQ [Tangents from point B] ... (2) CR = CQ [Tangents from point C] ... (3) DR = DS [Tangents from point D] ... (4) On adding (1), (2), (3) and (4), we get AP + BP + CR + DR = AS + BQ + CQ + DS ⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) ⇒ AB + CD = AD + BC ⇒ AB + AB = BC + BC [∵ ABCD is a ||gm . ∴ AB = CD and AD = BC] ⇒ 2AB = 2BC ⇒ AB = BC Therefore, AB = BC implies AB = BC = CD = AD Hence, ABCD is a rhombus