Prove that the lengths of the tangents drawn from an external point to a circle are equal.
Prove that the lengths of two tangents drawn from an external point to a circle are equal.
Consider the following diagram.
Let P be an external point and PA and PB are tangents to the circle. We need to prove that PA = PB Now consider the triangles ΔOAP and ΔOBP
m∠A = m∠B = 90°
OP = OP [common]
OA = OB = radii of the circle
Thus, by Right Angle-Hypotenuse-Side criterion of congruence we have,
ΔOAP ≅ ΔOBP
The corresponding parts of the congruent triangles are congruent.
Thus, PA = PB
Given: TP and TQ are two tangent drawn from an external point T to the circle C (O, r).
To prove: TP = TQ
Construction: Join OT.
Proof: We know that a tangent to the 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)
⇒ TP = TQ (CPCT)
Hence, the lengths of the tangents drawn from an external point to a circle are equal.