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

We know that, the lengths of tangents drawn from an external point to a circle are equal.
$\therefore TP=TQ$
In $\Delta TPQ$
TP = TQ
$\Rightarrow \angle TQP=\angle TPQ\dots \dots \left(1\right)$ (In a triangle, equal sides have equal angles opposite to them)
$\angle TQP+\angle TPQ+\angle PTQ={180}^{\circ }$ (Angle sum property)
$\therefore 2\angle TPQ+\angle PTQ={180}^{\circ }$ (Using (1))
$\Rightarrow \angle PTQ={180}^{\circ }-2\angle TPQ\dots \dots \left(1\right)$
We know that, a tangent to a circle is perpendicular to the radius through the point of contact.
$OP\perp PT\therefore \angle OPT={90}^{\circ }\Rightarrow \angle OPQ+\angle TPQ={90}^{\circ }\Rightarrow \angle OPQ={90}^{\circ }-\angle TPQ\Rightarrow 2\angle OPQ=2({90}^{\circ }-\angle TPQ)={180}^{\circ }-2\angle TPQ\dots \dots \left(2\right)$
From (1) and (2), we get
$\angle PTQ=2\angle OPQ$