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

(N/A) We are given a circle with centre $O$,an external point $T$,and two tangents $TP$ and $TQ$ to the circle,where $P$ and $Q$ are the points of contact (see figure).
We need to prove that $\angle PTQ = 2 \angle OPQ$.
Let $\angle PTQ = \theta$.
Now,$TP = TQ$ (tangents from an external point are equal in length). So,$\triangle TPQ$ is an isosceles triangle.
Therefore,$\angle TPQ = \angle TQP = \frac{1}{2} (180^{\circ} - \theta) = 90^{\circ} - \frac{1}{2} \theta$.
Also,the radius is perpendicular to the tangent at the point of contact,so $\angle OPT = 90^{\circ}$.
Thus,$\angle OPQ = \angle OPT - \angle TPQ = 90^{\circ} - (90^{\circ} - \frac{1}{2} \theta)$.
$\angle OPQ = \frac{1}{2} \theta = \frac{1}{2} \angle PTQ$.
This gives $\angle PTQ = 2 \angle OPQ$.