(A)OR(B)

Prove that a parallelogram circumscribing a circle is a rhombus.

ABCD is a parallelogram touching the circle at \(\displaystyle \mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}\) by sides AB, BC, CD, DA respectively.

We know that tangents drawn from the external point to a circle are equal.

\[\therefore \mathrm{AP}=\mathrm{AS} \quad-------(\mathrm{i}) \]

\[\begin{aligned} \mathrm{PB}=\mathrm{BQ} & \\ \mathrm{CR}=\mathrm{CQ} & \\ \mathrm{DR}=\mathrm{DS} & \\ & \text { Adding (i), (ii), (iii), (iv) } \\ & (\mathrm{AP}+\mathrm{PB})+(\mathrm{CR}+\mathrm{DR})=(\mathrm{AS}+\mathrm{DS})+(\mathrm{BQ}+\mathrm{CQ}) \\ & \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC} \\ & \mathrm{ABCD} \text { is a parallelogram } \\ & \Rightarrow \mathrm{AB}=\mathrm{CD}, \mathrm{AD}=\mathrm{BC} \\ & \Rightarrow 2 \mathrm{AB}=2 \mathrm{AD} \Rightarrow \mathrm{AB}=\mathrm{AD} \\ & \Rightarrow \mathrm{ABCD} \text { is a rhombus. } \end{aligned} \]

\[\begin{aligned} & (\mathrm{AP}+\mathrm{PB})+(\mathrm{CR}+\mathrm{DR})=(\mathrm{AS}+\mathrm{DS})+(\mathrm{BQ}+\mathrm{CQ}) \\ & \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC} \\ & \mathrm{ABCD} \text { is a parallelogram } \\ & \Rightarrow \mathrm{AB}=\mathrm{CD}, \mathrm{AD}=\mathrm{BC} \\ & \Rightarrow 2 \mathrm{AB}=2 \mathrm{AD} \Rightarrow \mathrm{AB}=\mathrm{AD} \\ & \Rightarrow \mathrm{ABCD} \text { is a rhombus. } \end{aligned} \]

PQ = PR (tangents drawn from an external point to the circle )

\[\therefore \angle \mathrm{PQR}=\angle \mathrm{PRQ} \]

In \(\displaystyle \triangle \mathrm{PQR}, \angle \mathrm{PQR}=\angle \mathrm{PRQ}=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}\)

Draw a perpendicular QL from Q to QP

Now, \(\displaystyle \angle \mathrm{PQL}=90^{\circ}\)

\(\displaystyle \therefore \angle \mathrm{RQL}=90^{\circ}-75^{\circ}=15^{\circ}\)

\(\displaystyle \Delta \mathrm{RQL} \cong \Delta \mathrm{SQL}(\mathrm{SAS})\)

\(\displaystyle \therefore \angle \mathrm{RQL}=\angle \mathrm{SQL}=15^{\circ}\)

\(\displaystyle \therefore \angle \mathrm{RQS}=15^{\circ}+15^{\circ}=30^{\circ}\)