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.**

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