In triangle $$ABC$$, it is given that $$\angle CAB = 50^\circ$$ and $$\angle ABC = 70^\circ$$. Points $$D$$ and $$E$$ are chosen on sides $$BC$$ and $$AC$$, respectively, such that $$\angle ABE = \angle DAB = 30^\circ$$. If $$\angle DEB = x^\circ$$, what is the value of $$x$$?

Construct the equilateral triangle $$ABF$$ on the same side of $$AB$$ as $$C$$. Since $$\angle DAB = 30^\circ$$, the line $$AD$$ is the perpendicular bisector of $$BF$$, giving $$DB = DF$$, and $$BE$$ bisects $$\angle ABF = 60^\circ$$, giving $$EA = EF$$. The isosceles triangles $$AEF$$ and $$BDF$$ have base angles $$\angle EAF = 60^\circ - 50^\circ = 10^\circ$$ and $$\angle DBF = 70^\circ - 60^\circ = 10^\circ$$ with equal bases, so they are congruent and $$EF = DF$$, while $$\angle EFD = (60^\circ - 10^\circ) + 10^\circ = 60^\circ$$ makes triangle $$EFD$$ equilateral. Hence $$ED = DF = DB$$, so triangle $$BDE$$ is isosceles and $$\angle DEB = \angle DBE = 70^\circ - 30^\circ = 40^\circ$$, giving $$x = 40$$.