In the adjoining figure, XY and X'Y' are parallel tangents to a circle with centre O. Another tangent AB touches the circle at C intersecting XY at A and $\displaystyle \mathrm{X}^{\prime} \mathrm{Y}^{\prime}$ at B . Prove that AB subtends right angle at the centre of the circle; or $\displaystyle \angle \mathrm{AOB}=90^{\circ}$.

Join OC. \[\begin{aligned} & \triangle \mathrm{POA} \cong \triangle \mathrm{COA} \\ & \angle \mathrm{POA}=\angle \mathrm{COA} \end{aligned} \] Similarly, \(\displaystyle \angle \mathrm{QOB}=\angle \mathrm{COB}\) \[\angle \mathrm{POA}+\angle \mathrm{QOB}+\angle \mathrm{COA}+\angle \mathrm{COB}=180^{\circ} \] \[\begin{aligned} & \Rightarrow 2(\angle \mathrm{COA}+\angle \mathrm{COB})=180^{\circ} \\ & \Rightarrow \angle \mathrm{COA}+\angle \mathrm{COB}=90^{\circ} \\ & \therefore \angle \mathrm{AOB}=90^{\circ} \end{aligned} \]