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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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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Accepted Answer

Step 1: $TP = TQ$ (tangents from external point are equal). $\Rightarrow \angle TPQ = \angle TQP$.
Step 2: Let $\angle PTQ = 	heta$. Then $\angle TPQ = \angle TQP = \frac{180° - 	heta}{2} = 90° - \frac{	heta}{2}$.
Step 3: Now $\angle OPT = 90°$ (radius $\perp$ tangent). $\Rightarrow \angle OPQ = 90° - (90° - \frac{	heta}{2}) = \frac{	heta}{2}$.
Step 4: $	herefore \angle PTQ = 	heta = 2 	imes \frac{	heta}{2} = 2 \angle OPQ$. Hence proved.

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