What is the value of $\alpha$ and $\beta$ in a triangle?

Question

On triangle $ABC$, with angles $\alpha$ over $A$, $\beta$ over $B$, and $\gamma$ over $C$. Where $\gamma$ is $140^\circ$.

On $\overline{AB}$ lies point $D$ (different from $A$ and $B$).

On $\overline{AC}$ lies point $E$ (different from $A$ and $C$).

$\overline{AE}$, $\overline{ED}$, $\overline{DC}$, $\overline{CB}$ are of same length.

What is the value of α and β?

EDIT: Here's my "solution", however the sum of the angles is greater than $180^\circ$.

Answer 1

I drew a simple diagram from which one can deduce that $$\alpha=10^\circ$$ $$\beta=40^\circ-\alpha=30^\circ$$ by noticing that angle $DEC=2\alpha$ and hence angle $DCB=140^\circ-2\alpha$ giving the value of the angle $DBC=20^\circ+\alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that $$40^\circ-\alpha=20^\circ+\alpha\implies \alpha=10^\circ$$

Answer 2

The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...

And $$\angle ACB\not= \angle DCA$$