(A)OR(B)

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\displaystyle \angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}$.

\[\begin{aligned} & \mathrm{TP}=\mathrm{TQ} \\ & \Rightarrow \angle \mathrm{TPQ}=\angle \mathrm{TQP} \\ & \text { Let } \angle \mathrm{PTQ} \text { be } \theta \\ & \Rightarrow \angle \mathrm{TPQ}=\angle \mathrm{TQP}=\frac{180^{\circ}-\theta}{2}=90^{\circ}-\frac{\theta}{2} \\ & \text { Now } \angle \mathrm{OPT}=90^{\circ} \\ & \Rightarrow \angle \mathrm{OPQ}=90^{\circ}-\left(90^{\circ}-\frac{\theta}{2}\right)=\frac{\theta}{2} \\ & \angle \mathrm{PTQ}=2 \angle \mathrm{OPQ} \end{aligned} \]

\[\begin{aligned} & \mathrm{AQ}=\mathrm{AR} \\ & 2 \mathrm{AQ}=\mathrm{AQ}+\mathrm{AR} \\ & =\mathrm{AB}+\mathrm{BQ}+\mathrm{AC}+\mathrm{CR} \\ & =\mathrm{AB}+\mathrm{AC}+(\mathrm{BP}+\mathrm{CP}) \\ & =\mathrm{AB}+\mathrm{AC}+\mathrm{BC} \\ & \mathrm{AQ}=\frac{1}{2}(\mathrm{AB}+\mathrm{AC}+\mathrm{BC})=\frac{1}{2}(\text { Perimeter of } \triangle \mathrm{ABC}) \end{aligned} \]