Question

prove that the parallelogram circumscribing a circle is a rohmbus

Answer 1

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.