Question

Two tangents TP and TQ are drawn to a circle with centre ‘O’ from an external point T, then prove that ∠PTQ = 2. ∠OPQ.

Answer 1

Given a circle with centre O. 

Two tangents TP, TQ are drawn to the circle from an external point T. 

We need to prove ∠PTQ = 2∠OPQ 

Let ∠PTQ = θ

TP = TQ (The lengths of tangents drawn from an external point to a circle are equal) 

So ΔTPQ is an isosceles triangle 

∴ ∠TPQ + ∠TQP + ∠PTQ = 180° 

(Sum of three angles in a triangle) 

∠TPQ = ∠TQP = \(\frac{1}{2}\)(180° – θ) 

= 90°- \(\frac{θ}{2}\)

∠OPQ = ∠OPT – ∠TPQ 

= 90° – 0(90° – \(\frac{θ}{2}\) ) = \(\frac{θ}{2}\)

= \(\frac{1}{2}\) ∠PTQ 

∴∠PTQ = 2∠OPQ