Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ = 2 \angle OPQ$. [Figure required: circle with centre O, external point T, tangents TP and TQ touching the circle at points P and Q.]

Step 1: Establish tangent properties

According to Theorem 10.1 in the NCERT textbook, the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, the radius $OP$ is perpendicular to the tangent $TP$, and the radius $OQ$ is perpendicular to the tangent $TQ$.

Step 2: Equal tangents from external point

According to Theorem 10.2 in the NCERT textbook, the lengths of tangents drawn from an external point to a circle are equal. Thus, the tangents $TP$ and $TQ$ from point $T$ to the circle are equal in length.

Step 3: Identify isosceles triangle

Since $TP = TQ$, triangle $TPQ$ is an isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are equal. Therefore, $\angle TPQ = \angle TQP$.

Step 4: Apply angle sum property in $\triangle TPQ$

The sum of angles in any triangle is $180^{\circ}$. For triangle $TPQ$, the sum of its internal angles is $180^{\circ}$. Substituting $\angle TPQ = \angle TQP$, we get $2 \angle TPQ + \angle PTQ = 180^{\circ}$.

Step 5: Express $\angle PTQ$ in terms of $\angle TPQ$

From the angle sum property of $\triangle TPQ$, we can express $\angle PTQ$ as $180^{\circ}$ minus two times $\angle TPQ$. This relation will be used later.

Step 6: Relate $\angle TPQ$ to $\angle OPQ$

We know that $OP \perp TP$, so $\angle OPT = 90^{\circ}$. This angle can be split into $\angle OPQ$ and $\angle TPQ$. Hence, $\angle OPQ + \angle TPQ = 90^{\circ}$.

Step 7: Express $\angle TPQ$ in terms of $\angle OPQ$

Rearranging the equation from the previous step, we can express $\angle TPQ$ as $90^{\circ} - \angle OPQ$. This will allow us to substitute into our equation for $\angle PTQ$.

Step 8: Substitute and simplify

Substitute the expression for $\angle TPQ$ from equation (2) into equation (1). This combines the relationships we've established to solve for $\angle PTQ$ in terms of $\angle OPQ$.

Step 9: Final simplification

Distribute the $2$ and simplify the expression. The $180^{\circ}$ terms cancel out, leaving us with $\angle PTQ = 2 \angle OPQ$, which proves the required statement.

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