In the adjoining figure, lines $$\ell_1$$ and $$\ell_2$$ are parallel lines. $$ABC$$ is an equilateral triangle. $$AD$$ bisects $$\angle EAB$$. Then $$x$$ is

The figure marks $$\angle EAD = 20^\circ$$, and since $$AD$$ bisects $$\angle EAB$$, we get $$\angle EAB = 2 \times 20^\circ = 40^\circ$$. The triangle $$ABC$$ is equilateral, so $$\angle BAC = 60^\circ$$ and hence $$\angle EAC = 40^\circ + 60^\circ = 100^\circ$$. As $$\ell_1$$ and $$\ell_2$$ are parallel with $$AC$$ as the transversal, $$x$$ and $$\angle EAC$$ are alternate angles, so $$x = 100^\circ$$.