Since $BA = B\Gamma = BE$, point $B$ is the circumcenter of the triangle $A\Gamma E$. The angle $A\hat{E}\Gamma$ is inscribed to the circle $C(B, BA)$ and so:
$$A\hat{E}\Gamma = \frac{1}{2} A\hat{B}\Gamma = 30^{\circ}.$$

From $BA = BE$ and $BA = B\Gamma$, we have $B\Gamma = BE$, and the triangle $B\Gamma E$ is isosceles. We draw the altitude from $B$, let $BZ$, $Z \in \Gamma E$.

Then $BZ$ is also median and bisector of the angle $E B \Gamma$ of the triangle $B\Gamma E$. Let $BZ$ meet $AE$ at $K$. Then the triangles $BK\Gamma$ and $BKE$ are equal because they have:

$$B\Gamma = BE,\ BK \text{ common side and } KB\Gamma = KBE.$$
Therefore we have: $B\Gamma K = B\hat{E}K$ and since
$$B\hat{E}K = B\hat{A}K \text{ it follows that } B\Gamma K = B\hat{A}K.$$
Thus quadrilateral $ABK\Gamma$ is inscribable, and hence:
$$ZKE = BKA = B\Gamma A = 60^{\circ}$$
Finally we get:
$$A\hat{E}\Gamma = K\hat{E}Z = 90^{\circ} - EKZ = 90^{\circ} - 60^{\circ} = 30^{\circ}.$$