Given: ABCD be a parallelogram circumscribing a circle with centre O.To prove: ABCD is a rhombus.We know that the tangents drawn to a circle from an exterior point are equal in length.Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.Adding the above equations,AP + BP + CR + DR = AS + BQ + CQ + DS(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)AB + CD = AD + BC2AB = 2BC(Since, ABCD is a parallelogram so AB = DC and AD = BC)AB = BCTherefore, AB = BC = DC = AD.Hence, ABCD is a rhombus.