(a) Prove that the parallelogram circumscribing a circle is a rhombus.

We know that lengths of tangents drawn from an external point to a circle are equal

\[\begin{aligned} & \therefore \mathrm{AP}=\mathrm{AS} \quad ---(1) \\ & \mathrm{BP}=\mathrm{BQ} \quad ---(2) \\ & \mathrm{CR}=\mathrm{CQ} \quad ---(3) \\ & \mathrm{DR}=\mathrm{DS} \quad ---(4) \end{aligned} \]

Adding ($\displaystyle 1$), ($\displaystyle 2$), ($\displaystyle 3$) and ($\displaystyle 4$), we have

\[\begin{aligned} & (\mathrm{AP}+\mathrm{BP})+(\mathrm{CR}+\mathrm{DR})=\mathrm{AS}+(\mathrm{BQ}+\mathrm{CQ})+\mathrm{DS} \\ & \Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{BC}+\mathrm{AD} \\ & \because \mathrm{AB}=\mathrm{CD} \text { and } \mathrm{BC}=\mathrm{AD} \\ & \therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD} \end{aligned} \]

Therefore, ABCD is a rhombus.

PA and PB are tangents from the external point P to the circle with centre O.

\(\displaystyle \angle \mathrm{OAP}=\angle \mathrm{OBP}=90^{\circ}\)

In quadrilateral OAPB,

\[\begin{aligned} & \angle \mathrm{APB}+\angle \mathrm{OAP}+\angle \mathrm{OBP}+\angle \mathrm{AOB}=360^{\circ} \\ & \Rightarrow \angle \mathrm{APB}+90^{\circ}+90^{\circ}+\angle \mathrm{AOB}=360^{\circ} \\ & \Rightarrow \angle \mathrm{APB}+\angle \mathrm{AOB}=180^{\circ} \\ & \therefore \angle \mathrm{APB} \text { and } \angle \mathrm{AOB} \text { are supplementary. } \end{aligned} \]