Question

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

Answer 1

We know that length of tangents drawn from an external point to a circle are equal

∴ TP = TQ  −−−(1)

∴ ∠TQP =∠TPQ (angles of equal sides are equal)  −−−(2)

Now, PT is tangent and OP is radius.

∴ OP⊥TP (Tangent at any point pf circle is perpendicular to the radius through point of cant act)

∴ ∠OPT = 90°

or, ∠OPQ + ∠TPQ = 90°

or, ∠TPQ = 90° −∠OPQ   −−−(3)

In △PTQ

∠TPQ + ∠PQT + ∠QTP = 180° (∴  Sum of angles triangle is 180°)

or, 90° − ∠OPQ + ∠TPQ + ∠QTP = 180°

or, 2(90° − ∠OPQ) + ∠QTP = 180° [from (2) and (3)]

or, 180° − 2∠OPQ + ∠PTQ = 180°

∴ ∠PTQ = 2∠OPQ

Hence Proved.

Answer 2

Join OQ. 

∠OPQ = ∠OQP {OP = OQ} 

∠OPQ + ∠OQP + ∠POQ = 180° {Angle sum property} 

2∠OPQ = 180° – ∠POQ ...(i) 

Also, ∠PTQ + ∠POQ = 180° 

∠PTQ = 180° – ∠POQ ...(ii) 

From (i) and (ii), 

∠PTQ = 2∠OPQ 

Hence Proved.