Prove that a parallelogram circumscribing a circle is a rhombus.

Tangents from an external point are equal in length \[\begin{aligned} \therefore A P & =A S, B P=B Q \\ C Q & =C R, D R=D S \end{aligned} \]Hence $\displaystyle A B+C D=B C+A D$ $\displaystyle \therefore A B C D$ is a $\displaystyle \parallel^{\mathrm{gm}}, \Rightarrow A B=C D$ and $\displaystyle B C=A D$ \[\Rightarrow 2 A B=2 B C \text { or } A B=B C \] $\displaystyle \therefore A B C D$ is a rhombus