Prove that the parallelogram circumscribing a circle is a rhombus. 

Since ABCD is a parallelogram circumscribed in a circle

$AB=CD........\left(1\right)$

$BC=AD........\left(2\right)$

$DR=DS$ (Tangents on the circle from same point $D$)

$CR=CQ$(Tangent on the circle from same point $C$)

$BP=BQ$ (Tangent on the circle from same point $B$ )

$AP=AS$ (Tangents on the circle from same point $A$)

Adding all these equations we get

$DR+CR+BP+AP=DS+CQ+BQ+AS$

$(DR+CR)+(BP+AP)=(CQ+BQ)+(DS+AS)$

$CD+AB=AD+BC$

Putting the value of equation $1$ and $2$ in the above equation we get

$2AB=2BC$

$AB=BC...........\left(3\right)$

From equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$ we get

$AB=BC=CD=DA$

$\therefore ABCD$ is a Rhombus