Q. 114.4( 426 Votes )

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer :

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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