# Problem 5. (3 points)

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

Solution:

From the isosceles property of triangle $ABC$ and the parallelism of $BC$ and $AD$, we get $\angle ABC = \angle ACB = \angle CAD = \alpha$.

Let the line $BC$ intersect the circumcircle of triangle $ABD$ at point $F$. Then $ABFD$ is an inscribed, i.e., isosceles, trapezoid, from which the arcs $AB, AE$, and $DF$ are equal. Hence, $\angle ABE = \angle DBF = \beta$.

$\angle EAD = \angle EBD = \gamma$, since these angles subtend the same arc.

$\angle ABD = \angle ABE + \angle EBD = \beta + \gamma$.

$\angle EAC = \angle EAD + \angle CAD = \gamma + \alpha$. Therefore, in the isosceles triangle $EAC$, the equality $\angle ACE = \angle AEC = \frac{180^{\circ} - \alpha - \gamma}{2}$ holds.

Moreover, $\alpha = \angle ABC = \angle ABF = \angle ABD + \angle DBF = 2\beta + \gamma$.

$\angle CED = \angle AED - \angle AEC = (180^{\circ} - \angle ABD) - \frac{180^{\circ} - \alpha - \gamma}{2} = 180^{\circ} - \beta - \gamma - 90^{\circ} + \frac{2\beta + 2\gamma}{2} = 90^{\circ}$