# Problem 5. (3 points)

In trapezoid $A B C D$, the lateral side $C D$ is equal to the diagonal $A C$. On the smaller arc $B C$ of the circumscribed circle of triangle $B C D$, a point $E$ is chosen such that $C D=C E$. Find the angle $\angle A E B$.
Answer: $90^{\circ}$.

Solution:

From the isosceles triangle $A C D$ and the parallelism of $B C$ and $A D$, we get $\angle A D C = \angle C A D = \angle A C B = \alpha$.

Let the line $A D$ intersect the circumcircle of triangle $B C D$ at point $F$. Then $C D F B$ is an inscribed, i.e., isosceles, trapezoid, from which the arcs $C D, C E$, and $B F$ are equal. Hence, $\angle C D E = \angle B D F = \beta$.

$\angle E C B = \angle E D B = \gamma$, since these angles subtend the same arc.

$\angle C D B = \angle C D E + \angle E D B = \beta + \gamma$.

$\angle E C A = \angle E C B + \angle A C B = \alpha + \gamma$. Therefore, in the isosceles triangle $E C A$, the equality $\angle C A E = \angle A E C = \frac{180^{\circ} - \alpha - \gamma}{2}$ holds.

Moreover, $\alpha = \angle A D C = \angle C D F = \angle C D B + \angle B D F = 2 \beta + \gamma$.

$\angle A E B = \angle C E B - \angle A E C = (180^{\circ} - \angle C D B) - \frac{180^{\circ} - \alpha - \gamma}{2} = 180^{\circ} - \beta - \gamma - 90^{\circ} + \frac{2 \beta + 2 \gamma}{2} = 90^{\circ}$