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 OB⊥PB In △OPA and △OPB ∠OPA=∠OPB (Using (1)) OA=OB (Radii of the same circle) OP=OP (Common side) Therefor △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. The length of tangents drawn from any external point are equal. So statement is correct
