Find the angle $\angle BAC$ in the following picture .
My attempt : I tried to apply the relationships in the both circles between different angles and arcs in many ways but it didn't work . Also joining the point $A$ , $E$ and the line $d$ didn't help (Note that if the intersection point of the $AE$ and the line $d$ is $D$ then $BD = DC$)
Note that $\angle BAE = \angle EBC$ because they both correspond to the circular arc $\overset{\frown}{BE}$
Similarly, $\angle CAE = \overset{\frown}{CE} = \angle ECB$
Now, the total internal angle of $\triangle ABC$ consists of\begin{align} 180^{\circ} &= \color{magenta}{\angle BAC} + 30^{\circ} + \angle EBC + \angle ECB + 40^{\circ} \\ &= \color{magenta}{\angle BAC} + 30^{\circ} + \color{magenta}{\big(}\angle BAE + \angle CAE\color{magenta}{\big)} + 40^{\circ} \\ &= \color{magenta}{2\angle BAC} + 30^{\circ} + 40^{\circ} \end{align} Thus the desired $\displaystyle\angle BAC = \frac{180^{\circ} - 30^{\circ} - 40^{\circ}}2 = 55^{\circ}$