From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of chord AB.

\[\begin{gathered} \triangle \mathrm{ACP} \cong \triangle \mathrm{BCP} \\ \mathrm{AC}=\mathrm{BC} \\ \angle \mathrm{ACP}=\angle \mathrm{BCP} \\ \mathrm{CP}+\angle \mathrm{BCP}=180^{\circ} \\ \angle \mathrm{ACP}=90^{\circ} \end{gathered} \] \(\displaystyle \Rightarrow \mathrm{OP} \perp \mathrm{AB}\) Hence OP is perpendicular bisector of chord AB .