10.3. Point $F$ is the midpoint of side $B C$ of square $A B C D$. A perpendicular $A E$ is drawn to segment $D F$. Find the angle $C E F$.

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Translation provided as requested, maintaining the original format and line breaks.
Answer: $45^{\circ}$.

Solution. Let the line $A E$ intersect the side $C D$ of the square at point $M$ (see Fig. 10.3). Then triangles $A D M$ and $D C F$ are equal (by the leg and acute angle). Therefore, point $M$ is the midpoint of side $C D$. Then triangle $C F M$ is a right and isosceles triangle, so $\angle C M F=45^{\circ}$.

Since $\angle M E F=\angle M C F=90^{\circ}$, a circle can be circumscribed around quadrilateral $M C F E$. The inscribed angles $C E F$ and $C M F$ subtend the same arc of this circle, hence $\angle C E F=\angle C M F=45^{\circ}$.

There are also "computational" methods of solving, for example, using coordinates or vectors.

Evaluation criteria:

![](https://cdn.mathpix.com/cropped/2024_05_06_c991e5a02c6db2b67b15g-1.jpg?height=381&width=397&top_left_y=1914&top_left_x=1637)

Fig. 10.3

+ a complete and justified solution is provided
- only the answer is provided