61
Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2\angle OPQ$ .
._q_1.png)
Show SolutionHide Solution↓
$TP = TQ$
$\Rightarrow \angle TPQ = \angle TQP$
Let $\angle PTQ$ be $\theta$
$\Rightarrow \angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$
Now $\angle OPT = 90^\circ$
$\Rightarrow \angle OPQ = 90^\circ - (90^\circ-\frac{\theta}{2}) = \frac{\theta}{2}$
$\angle PTQ = 2 \angle OPQ$
$\Rightarrow \angle TPQ = \angle TQP$
Let $\angle PTQ$ be $\theta$
$\Rightarrow \angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$
Now $\angle OPT = 90^\circ$
$\Rightarrow \angle OPQ = 90^\circ - (90^\circ-\frac{\theta}{2}) = \frac{\theta}{2}$
$\angle PTQ = 2 \angle OPQ$