Prove that the parallelogram circumscribing a circle is a rhombus.

\[\left.\begin{array}{rl} \therefore \mathrm{AP} & =\mathrm{AS} \\ \mathrm{BP} & =\mathrm{BQ} \\ \mathrm{CR} & =\mathrm{CQ} \\ \mathrm{DR} & =\mathrm{DS} \end{array}\right\} \] Adding, \[\begin{aligned} & (\mathrm{AP}+\mathrm{BP})+(\mathrm{CR}+\mathrm{DR})=(\mathrm{AS}+\mathrm{DS})+(\mathrm{BQ}+\mathrm{CQ}) \\ & \quad \Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC} \end{aligned} \] Now \(\displaystyle \mathrm{AB}=\mathrm{CD}\) and \(\displaystyle \mathrm{AD}=\mathrm{BC}\) \[\begin{aligned} & \Rightarrow 2 \mathrm{AB}=2 \mathrm{BC} \\ & \Rightarrow \quad \mathrm{AB}=\mathrm{BC} \end{aligned} \] \(\displaystyle \Rightarrow \mathrm{ABCD}\) is a rhombus