prove that the length of the tangents drawn from an external point to a circle are equal, hence show that the centre lies on the bisector of the angle between the two tangents?

Given: PT and TQ are two tangent drawn from an external point T to the circle C (O, r).

To prove: 1. PT = TQ

 2. ∠OTP = ∠OTQ 

Construction: Join OT.

Proof: We know that, a tangent to circle is perpendicular to the radius through the point of contact.

∴ ∠OPT = ∠OQT = 90°

In ΔOPT and ΔOQT,

OT = OT  (Common)

OP = OQ  ( Radius of the circle)

∠OPT = ∠OQT  (90°)

∴ ΔOPT ΔOQT  (RHS congruence criterion)

⇒ PT = TQ  and ∠OTP = ∠OTQ (CPCT)

PT = TQ,

∴ The lengths of the tangents drawn from an external point to a circle are equal.

∠OTP = ∠OTQ,

∴ Centre lies on the bisector of the angle between the two tangents.