Question

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

Hint:

Remember that \( \triangle TPQ \) is isosceles and the radius is perpendicular to the tangent to link the angles.

Answer 1

Step 1: Understanding the Concept:
We use the fact that tangents from an external point are equal and properties of isosceles triangles.
Step 2: Detailed Explanation:
Let \( \angle PTQ = \theta \).
Since \( TP = TQ \) (tangents from an external point), \( \triangle TPQ \) is an isosceles triangle.
Therefore, \( \angle TPQ = \angle TQP \).
In \( \triangle TPQ \):
\[ \angle PTQ + \angle TPQ + \angle TQP = 180^\circ \]
\[ \theta + 2\angle TPQ = 180^\circ \]
\[ \angle TPQ = \frac{1}{2}(180^\circ - \theta) = 90^\circ - \frac{\theta}{2} \]
We know that the radius OP is perpendicular to the tangent TP at point P.
\[ \angle OPT = 90^\circ \]
From the figure, \( \angle OPT = \angle OPQ + \angle TPQ \).
\[ 90^\circ = \angle OPQ + \left( 90^\circ - \frac{\theta}{2} \right) \]
\[ \angle OPQ = \frac{\theta}{2} \]
\[ \theta = 2\angle OPQ \]
\[ \angle PTQ = 2\angle OPQ \]
Step 3: Final Answer:
\( \angle PTQ = 2 \angle OPQ \). Hence Proved.