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.**

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