In the given figure, AB and CD are tangents to a circle centred at O. Is $\displaystyle \angle \mathrm{BAC}=\angle \mathrm{DCA}$ ? Justify your answer.

Join OA and OC \[\begin{aligned} & \mathrm{OA}=\mathrm{OC} \\ & \angle \mathrm{OAC}=\angle \mathrm{OCA} \end{aligned} \] Also, \(\displaystyle \angle \mathrm{OAB}=\angle \mathrm{OCD}\) \[\begin{aligned} & \Rightarrow \angle \mathrm{OAC}+\angle \mathrm{OAB}=\angle \mathrm{OCA}+\angle \mathrm{OCD} \\ & \Rightarrow \angle \mathrm{BAC}=\angle \mathrm{DCA} \end{aligned} \]