As shown in the figure, fold △BCD along BD to △BFD, and connect AF and CF.

Because AC=BC, △ABC is an isosceles triangle.

From ∠ ACB = 40 ∠ CAB = ∠ CBA = 70,

And because BA=BD, that is, △ABD is an isosceles triangle,

So ∠ cab = ∠ ADB = 70, ∠ Abd = 40, ∠ CBD = 30,

Because △BFD is converted from △BCD, there are BC=BF and CD=FD.

∠ CBD =∠ FBD = 30, that is ∠ CBF = 60, so △CDF is an isosceles triangle.

And △CBF is an equilateral triangle with ACF = CFD = 60-40 = 20.

CA=CB=CF, ∠ BCF = 60, that is, △CAF is an isosceles triangle,

Then ∠ CFA = ∠ CAF = 80, ∠ FDA = ∠ ACF+∠ CFD = 40, ∠ DFA = 80-20 = 60,

And because there is BE=CD in AC=BC, we know that CE=DA,

Therefore, from CD=FD, ∠ ACB = ∠ FDA = 40, CE=DA, it is proved that △ CDE △ DFA (SAS).

There is ∠ CDE = ∠ DFA = 60, so ∠ EDB = 180-∠ ADB -∠ CDE = 50.