ABC is an isosceles triangle in which $$AB = AC$$. EDF is an isosceles triangle in which $$EF = DE$$. FD is parallel to AC. The degree measure of marked angle $$x$$ is

Since $$AB = AC$$ and the figure marks $$\angle ABC = 50^\circ$$, we get $$\angle ACB = 50^\circ$$ and $$\angle BAC = 180^\circ - 50^\circ - 50^\circ = 80^\circ$$. As $$FD$$ is parallel to $$AC$$ with $$AB$$ as the transversal, the corresponding angles give $$\angle BDF = \angle BAC = 80^\circ$$, and since $$E$$ lies on $$DB$$, this is the angle $$\angle EDF$$ of the triangle $$EDF$$. In that triangle $$EF = DE$$, so the angles opposite these equal sides are equal, which gives $$x = \angle DFE = \angle EDF = 80^\circ$$.