Solution: Let $M$ be the point of intersection of segments $B F$ and $C E$. Note that $\angle C F B = 90^{\circ}$, since $F A$ is a median equal to half the side to which it is drawn. From this, it follows that $\angle F B C = 180^{\circ} - 90^{\circ} - \angle F C B = 30^{\circ}$. From the isosceles right triangle $C A E$, it follows that $\angle E C A = 45^{\circ}$.

Then, in triangle $C M B$, the angle $\angle C M B = 180^{\circ} - \angle M B C - \angle M C B = 180^{\circ} - 30^{\circ} - 45^{\circ} = 105^{\circ}$, which is the answer.

Criteria: The angle $\angle C B F$ is found - 2 points.

The angle $\angle E C A$ is found - 1 point.

It doesn't matter whether the angle is found to be $105^{\circ}$ or $75^{\circ}$.
8.4. Point $A$ is located midway between points $B$ and $C$. The square $A B D E$ and the equilateral triangle $C F A$ are in the same half-plane relative to the line $B C$. Find the angle between the lines $C E$ and $B F$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f17ae90027cfb7d17226g-2.jpg?height=739&width=1216&top_left_y=727&top_left_x=434)