Prove that the lenghts of tangents drawn from an external point to a circle are equal.

**Given:** A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.

To prove: PA = PB

Construction: Join OA, OB, and OP.

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OA⊥PA and OB⊥PB … (1)

In ∆OPA and ∆OPB:

∠OAP =∠OBP (Using (1))

OA = OB (Radii of the same circle)

OP = OP (Common side)

Therefore,∆OPA OPB (RHS congruency criterion)

∴ PA = PB

(Corresponding parts of congruent triangles are equal)

Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.