# Problem 5. (3 points)

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

## Solution:

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

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

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

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

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

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

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