In $$\triangle ABC$$, $$AB = BC$$. Point $$D$$ is on side $$AC$$ and point $$E$$ is on side $$AB$$ such that $$BD = BE = AD$$. It is known that $$\angle ADE = 30^\circ$$. Then find the measure of $$\angle ABC$$, $$\angle ACB$$, $$\angle BDC$$, $$\angle DEA$$.

Since $$AB = BC$$, let $$\angle BAC = \angle ACB = x$$. As $$AD = BD$$, $$\angle ABD = \angle BAD = x$$. In $$\triangle ADE$$, the exterior angle $$\angle DEB = \angle EAD + \angle ADE = x + 30^\circ$$, and since $$BE = BD$$, also $$\angle BDE = x + 30^\circ$$. The angle sum of $$\triangle BDE$$ gives $$(x + 30^\circ) + (x + 30^\circ) + x = 180^\circ$$, so $$x = 40^\circ$$. Therefore $$\angle ABC = 180^\circ - 2x = 100^\circ$$, $$\angle ACB = 40^\circ$$, $$\angle BDC = \angle BAD + \angle ABD = 80^\circ$$ and $$\angle DEA = 180^\circ - (40^\circ + 30^\circ) = 110^\circ$$.