From an external point, two tangents are drawn to a circle. Prove that the line joining the external point to the centre of the circle bisects the angle between the two tangents.

Given : PA and PB are tangents drawn from an external point P to the circle with centre O. To prove: \(\displaystyle \angle \mathrm{OPA}=\angle \mathrm{OPB}\) Construction: Join OA, OB Proof: In \(\displaystyle \Delta \mathrm{OPA}\) and \(\displaystyle \Delta \mathrm{OPB}\) \[\begin{aligned} & \mathrm{OP}=\mathrm{OP} \text { (common) } \\ & \mathrm{OA}=\mathrm{OA} \text { (radii) } \\ & \angle \mathrm{OAP}=\angle \mathrm{OBP} \text { (each } 90^{\circ}, \text { radius ⟂ tangents) } \\ & \therefore \Delta \mathrm{OPA} \cong \triangle \mathrm{OPB} \text { (RHS) } \\ & \Rightarrow \angle \mathrm{OPA}=\angle \mathrm{OPB} \text { (CPCT) } \end{aligned} \]