In a convex quadrilateral $A B C D A B=B C$. The rays $B A$ and $C D$ intersect at point $E$, and the rays $A D$ and $B C$ intersect at point $F$. It is also known that $B E=B F$ and

$\angle D E F=25^{\circ}$. Find the angle $E F D$.
Triangles $B C E$ and $B A F$ are equal by two sides and the included angle, so $\angle B E C=\angle B F A$. Triangle $E B F$ is isosceles, so

$\angle B E F=\angle B F E$. Therefore, $\angle E F D=\angle E F A=\angle B F E-\angle B F A=\angle B E F-\angle B E C=\angle C E F=\angle D E F=25^{\circ}$.

## Answer

$25^{\circ}$.