Prove that the parallelogram circumscribing a circle is a rhombus.

ABCD is a parallelogram, circumscribing a circle is a rhombus.
ABCD is a parallelogram, therefore
AB = CD …(1)
BC = AD …(2)
We know that the tangent drawn from some external point to circle are equal. so,
DR = DS [Tangents drawn from point D] …(3)
CR = CQ [Tangents drawn from point C] …(4)
BP = BQ [Tangents drawn from point B] …(5)
AP = AS [Tangents drawn from point A] …(6)
Adding (3), (4), (5) and (6), we have
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CB) + BP + AP ) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC …(7)
From (1), (2) and (7), we have
AB = BC = CD = DA
Hence, ABCD is a rhombus.