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.