Question

Prove that the lengths of tangents drawn from an external point to a circle are equal? ​

Answer 1

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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Answer 2

Answer:

RTP: Angle OTP=AngleOTQ

PT & TQ Are equal,

AngleOTP=AngleOTQ=90 DEGREES

OT=OT[COMMON ANGLE]

OP=OQ

ANGLEOTP=ANGLEOQT

TriangleOPT& TriangleOTQ

PT=TQ

ANGLEOTP=ANGLEOTQ

PT=TQ

THEREFORE, ANGLEOTP=ANGLEOTQ

HENCE PROVED.

HOPE IT'S HELPS YOU.