# Problem 5. (3 points)

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

## Solution:

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

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

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

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

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

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

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