Prove that a parallelogram circumscribing a circle is a rhombus.

1.	Prove that a parallelogram circumscribing a circle is a rhombus.
Answer» Since ABCD is a parallelogram,AB = CD ---- i)BC = AD ---- ii)\xa0It can be observed thatDR = 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 obtainDR + CR + BP + AP = DS + CQ + BQ + AS(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)CD + AB = AD + BCOn putting the values of equations (1) and (2) in this equation, we obtain2AB = 2BCAB = BC …(3)Comparing equations (1), (2), and (3), we obtainAB = BC = CD = DAHence, ABCD is a rhombus.

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