Verified answer

Answer:

Given: A circle with centre O and an external point T from which tangents TP and TQ are drawn to touch the circle at P and Q.

To prove: ∠PTQ = 2∠OPQ.

Proof: Let ∠PTQ = xº. Then,

∠TQP + ∠TPQ + ∠PTQ = 180º

[∵ sum of the ∠s of a triangle is 180º]

⇒ ∠TQP + ∠TPQ = (180º – x) ….(i)

We know that the lengths of tangent drawn from an external point to a circle are equal.

So, TP = TQ.

Now, TP = TQ

⇒ ∠TQP = ∠TPQ

= 1/2(180° - x )

= 90° - x/2

=180°-x

∴ ∠OPQ = (∠OPT–∠TPQ)

= 90° - ( 90° - x/2)

angle OPQ = 1/2 angle PTQ

= 2angleOPQ = angle PTQ