Question

From an external point $T,$ two tangents $TP$ and $TQ$ are drawn to a circle having it's center at O. Prove that $\angle PTQ=2\angle OPQ$

Answer 1

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