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Question

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

Suppose $$\angle{PTQ}=\theta$$
Since,"The lengths of tangents drawn from an external point to a circle are equal"
So, $$\triangle{TPQ}$$ is an isosceles triangle.
$$\therefore,\angle{TPQ}=\angle{TQP}=\dfrac{1}{2}\left({180}^{\circ}-\theta\right)={90}^{\circ}-\dfrac{\theta}{2}$$
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}-\left({90}^{\circ}-\dfrac{\theta}{2}\right)$$
$$=\dfrac{\theta}{2}=\dfrac{1}{2}\angle{PTQ}$$
Hence $$\angle{PTQ}=2\angle{OPQ}$$

Mathematics

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