Prove that the lengths of the tangents from an external point to a circle are equal.
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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°)
∴ ΔOPTΔOQT (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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