Prove that the parallelogram circumscribing a circle is a rhombus.

(N/A) Since $ABCD$ is a parallelogram,
$AB = CD \dots(1)$
$BC = AD \dots(2)$
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:
$2AB = 2BC$
$AB = BC \dots(3)$
Comparing equations $(1), (2),$ and $(3),$ we obtain:
$AB = BC = CD = DA$
Hence,$ABCD$ is a rhombus.