Question

In the following figure, two tangents TP and TQ are drawn to circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

Answer 1

Let ∠OPQ be θ, then ∠TPQ = 90° – θ

Since, TP = TQ

∴ ∠TQP = 90° – θ   ...(Opposite angles of equal sides)

Now, ∠TPQ + ∠TQP + ∠PTQ = 180°   ...(Angle sum property of a Triangle)

⇒ 90° – θ + 90° – θ + ∠PTQ = 180°

⇒ ∠PTQ = 180° – 180° + 2θ

⇒ ∠PTQ = 2θ

Hence, ∠PTQ = 2∠OPQ

Hence Proved.