prove that the parallelogram circumscribing a circle is a rhombus.???
Since ABCD is a parallelogram, AB = CD …(1) BC = AD …(2) It can be observed that DR = DS (Tangents on the circle from point D) CR = CQ (Tangents on the circle from point C) BP = BQ (Tangents on the circle from point B) AP = AS (Tangents on the circle from point A) Adding all these equations, we obtain DR + CR + BP + AP = DS + CQ + BQ + AS (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) CD + AB = AD + BC On putting the values of equations (1) and (2) in this equation, we obtain 2AB = 2BC AB = BC …(3) Comparing equations (1), (2), and (3), we obtain AB = BC = CD = DA Hence, ABCD is a rhombus.