Given:
PT and TQ are two tangents drawn from an external point T to the circle C(O,r).
To prove: PT=TQ
Construction: Join OT.
Proof:
We know that, a tangent to circle is perpendicular to the radius through the point of contact.
Therefore, ∠OPT=∠OQT=90o
In ΔOPT and ΔOQT,
OT=OT
Radius of the circle =OP=OQ
∠OPT=∠OQT=900
Therefore, ΔOPT≅ΔOQT (RHS congruence criterion)
Therefore, PT=TQ
So, the length of the tangents drawn from an external point to a circle are equal.