Are you too tired to chase angles?

Construct a line parallel to $BC$ through $D$ and call its intersection with $AB$ , $F$ . Connect $FC$ and call its intersection with $BD$ , $G$ . Since $\triangle ABC$ is isosceles and $\angle DBC = 60^{\circ}$ , we know that $\triangle BGC$ and $\triangle FDG$ are equilateral triangles.Also notice that $\angle BEC = 50^{\circ}$ which implies $\triangle BEC$ is B-isosceles. Since $\triangle BGC$ is equilateral, this implies that $BE = BC = BG$ which implies that $\triangle BEG$ is isosceles which implies that $\angle BGE = 80^{\circ}$ . Due to supplementary angles, we find that $\angle FGE = 40^{\circ}$ . Also notice that $\angle BFC = 40^{\circ}$ which implies that $\triangle EFG$ is isosceles which implies that $FEGD$ is a kite which gives us an answer of $\boxed{30^{\circ}}$ since the diagonals are perpendicular.