Slovenija 2008

Let $T$ be the intersection of the chords $DE$ and $AB$. We know that $DTB$ is a right triangle. Let $\angle CAB=\angle ACB=\alpha$. The line $AC$ is tangent to the circumcircle, so the angle $\angle CAB$ is equal to the corresponding angle $\angle AEB$ over the chord $AB$. Thus, $\angle AEB=\alpha$.

We have $\angle ABD=\frac{\pi }{2}-\alpha$, so $\angle TDB=\alpha$ and $\angle EDB=\alpha$. This angle is equal to $\angle EAB$ because they are both the inscribed angles over $BE$. So, $\angle BAE=\alpha =\angle BEA$ and the triangle $ABE$ is isosceles with the apex at $B$. Triangles $ABE$ and $ABC$ have congruent angles and a common side $AB$ next to the corresponding two angles, so they are congruent.