Question

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

Answer 1

Refer image

We know that length of taughts drawn from an external point to a circle are equal

$\therefore TP=TQ---\left(1\right)$

$4\therefore \angle TQP=\angle TPQ$ (angles of equal sides are equal)$---\left(2\right)$

Now, $PT$ is tangent and $OP$ is radius.

$\therefore OP\perp TP$ (Tangent at any point pf circle is perpendicular to the radius through point of cant act)

$\therefore \angle OPT={90}^o$

or, $\angle OPQ+\angle TPQ={90}^o$

or, $\angle TPQ={90}^o-\angle OPQ---\left(3\right)$

In $△PTQ$

$\angle TPQ+\angle PQT+\angle QTP={180}^o$ ($\therefore$ Sum of angles triangle is ${180}^o$)

or, ${90}^o-\angle OPQ+\angle TPQ+\angle QTP={180}^o$

or, $2({90}^o-\angle OPQ)+\angle QTP={180}^o$ [from $\left(2\right)$ and $\left(3\right)$]

or, ${180}^o-2\angle OPQ+\angle PTQ={180}^o$

$\therefore 2\angle OPQ=\angle PTQ----$ proved