Question

prove that the parallelogram circumscribing a circle is a rhombus​

Answer 1

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Answer 2

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) AP = AS (Tangents on the circle from point A) Adding all these equations, we obtain

BP = BQ (Tangents on the circle from point B)

DR + CR + BP + AP = DS + CQ + BQ + AS

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.