For reference: Calculate $x~ if AE = BC, and AD = BE$
My progress: I tried for auxiliary lines...parallel to BC by A and parallel to BE by A... forming the parallelograms...I completed the Angles but I was unsuccessful...but I believe the solution is in the auxiliary lines
For illustration...DR SK's resolution
Let the circumcentre of $\triangle ABC$ be $O$. $\triangle OBC$ is equilateral.
$\Rightarrow AE=BC=AO$
$\triangle AOE$ is isosceles. Through angle chasing, $\angle BOE=\angle COE=30^{\circ}$ and thereafter $BE=EC$ and $\angle BEA=40^{\circ}$.
Let the circumcentre of $\triangle ABE$ be $P$. Again, $\triangle PBE$ is equilateral, and through angle chasing, $PB$ passes through $D$. Hence, $x=60^{\circ}$.