Prove that the paralleogram circumscribing a …

Given:- a circle with center O. a perellelogram ABCD touching the circle at points P, Q, R and S. To prove:- ABCD is rhombus. Proof :- a rhombus is a perellelogram with all sides equal. So, we have to prove all sides equal. In perellelogram ABCD, AB=CD & AD=BC, ( opposite sides of perellelogram are equal.)..................(01) From theorem 10.2, length of tangent drown from external point of circle are equal. Hence, AP = AS,........... (02) BP = BQ,............(03) CR = CQ, ............(04) DR = DS. ............(05) Adding equation (02) + (03) + (04) + (05) AP + BP + CR + DR = AS + BQ + CQ + DS (AP + BP)+(CR + DR) = (AS + DS)+(BQ + CQ) AB + CD = AD + BC But from equation (01) : AB=CD & AD=BC AB + AB = AD + AD 2AB = 2AD AB = AD So, AB = AD & AB=CD & AD=BC. So, AB = CD = AD = BC. So, ABCD is a perellelogram with all sides equal, There fore. ABCD is rhombus. Hence proved.