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

Step 1: Identify properties of tangents and radii

When a tangent is drawn to a circle, the radius drawn to the point of tangency is perpendicular to the tangent. Therefore, $\angle OPT = 90^\circ$ and $\angle OQT = 90^\circ$. Also, the lengths of tangents from an external point to a circle are equal, so $TP = TQ$.

Step 2: Analyze triangle TPQ

Since $TP = TQ$ (tangents from an external point), triangle $TPQ$ is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal. Let $\angle PTQ = \theta$. Then, the sum of angles in $\triangle TPQ$ is $180^\circ$, so $\angle TPQ + \angle TQP + \angle PTQ = 180^\circ$. This implies $2\angle TPQ + \theta = 180^\circ$, so $\angle TPQ = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{1}{2}\theta$.

Step 3: Relate angles using perpendicularity

We know that $\angle OPT = 90^\circ$. From the diagram, we can see that $\angle OPQ = \angle OPT - \angle TPQ$. Substituting the values we found: $\angle OPQ = 90^\circ - (90^\circ - \frac{1}{2}\theta) = 90^\circ - 90^\circ + \frac{1}{2}\theta = \frac{1}{2}\theta$.

Step 4: Conclude the proof

Since we defined $\angle PTQ = \theta$ and found that $\angle OPQ = \frac{1}{2}\theta$, we can conclude that $\angle PTQ = 2\angle OPQ$. This completes the proof.

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