Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that PTQ = 2 OPQ.
Step-by-Step Solution
Key Concept: Use the fact that a radius drawn to a point of tangency is perpendicular to the tangent, and that the two tangents from an external point are equal. Hence triangles \(OTP\) and \(OTQ\) are congruent, giving equal base angles. Apply the exterior angle theorem in triangle \(OPQ\).
1. Draw the radii \(OP\) and \(OQ\) to the points of tangency \(P\) and \(Q\).
2. Since a radius is perpendicular to a tangent at the point of contact, we have
\[ OP \perp TP \quad \text{and} \quad OQ \perp TQ. \]
3. Hence \(\angle OPT = 90^{\circ}\) and \(\angle OQT = 90^{\circ}\).
4. From a point outside a circle, the two tangents drawn are equal; therefore
\[ TP = TQ. \]
5. Also, the radii are equal: \(OP = OQ\).
6. In triangles \(\triangle OTP\) and \(\triangle OTQ\) we have
\[ OP = OQ, \quad TP = TQ, \quad OT = OT. \]
Thus, by SSS (Side‑Side‑Side) congruence, \(\triangle OTP \cong \triangle OTQ\).
7. Consequently, the corresponding angles are equal: \(\angle POT = \angle QOT\).
8. Consider triangle \(OPQ\). The exterior angle at \(T\) for this triangle is \(\angle PTQ\). By the Exterior Angle Theorem,
\[ \angle PTQ = \angle POT + \angle QOT. \]
9. Since \(\angle POT = \angle QOT\) (from step 7), let each be \(x\). Then
\[ \angle PTQ = x + x = 2x. \]
10. But \(x = \angle OPQ\) (the angle subtended by the chord \(PQ\) at the centre). Hence
\[ \boxed{\angle PTQ = 2 \angle OPQ}. \]
Correct Answer: ∠PTQ = 2∠OPQ