Question

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

Answer 1

Suppose $\angle PTQ=\theta$

Since,"The lengths of tangents drawn from an external point to a circle are equal"

So, $△TPQ$ is an isosceles triangle.

$\therefore ,\angle TPQ=\angle TQP=\frac{1}{2}({180}^{\circ }-\theta )={90}^{\circ }-\frac{\theta }{2}$

Also,$The tangents at any point of a circle is perpendicular to the radius through the point of contact"

$\angle OPT={90}^{\circ }$

$\therefore ,\angle OPQ=\angle OPT-\angle TPQ$

$={90}^{\circ }-({90}^{\circ }-\frac{\theta }{2})$

$=\frac{\theta }{2}=\frac{1}{2}\angle PTQ$

Hence $\angle PTQ=2\angle OPQ$