Prove that a parallelogram circumscribing a circle is a rhombus.
Given: ABCD be a parallelogram circumscribing a circle with centre O.
To prove: ABCD is a rhombus.
We know that the tangents drawn to a circle from an exterior point are equal in length.
∴ AP = AS, BP = BQ, CR = CQ and DR = DS.
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
∴ AB + CD = AD + BC or 2AB = 2BC (since AB = DC and AD = BC)
∴ AB = BC = DC = AD.
Therefore, ABCD is a rhombus.