101
Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2 \angle OPQ$.
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$TP = TQ$
$\Rightarrow \angle TPQ = \angle TQP$
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 $\angle OPT = 90^{\circ}$
$\Rightarrow \angle OPQ = 90^{\circ} - (90^{\circ} - \frac{\theta}{2}) = \frac{\theta}{2}$
$\angle PTQ = 2 \angle OPQ$