4. In a right isosceles triangle $\mathrm{ABC}$

$(\mathrm{AC}=\mathrm{BC})$ points $\mathrm{D}$ and $\mathrm{F}$ are the midpoints of segments $\mathrm{AB}$ and $\mathrm{BC}$

respectively. On the ray DC

point $\mathrm{E}$ is marked such that

$\mathrm{AF}=\mathrm{FE}$. Find the angles of triangle

AFE.
Answer: $45^{\circ}, 45^{\circ}, 90^{\circ}$.

Solution. Triangle CDB is a right isosceles triangle.

$(CD-$ is the median and altitude

![](https://cdn.mathpix.com/cropped/2024_05_06_5019dfdf7468c3a8e22fg-1.jpg?height=175&width=175&top_left_y=1628&top_left_x=2180)

of the right isosceles

triangle $\mathrm{ACB}$, and therefore, $\mathrm{CD}=\mathrm{DB}$). Drop
perpendiculars from $\mathrm{F}$ to $\mathrm{CD}$ and $\mathrm{DB}$. Their feet are the midpoints of segments CD and DB (denote them as M and $\mathrm{N}$, respectively. DMFN is a square. Triangles ANF and EMF are equal by hypotenuse and leg. Therefore, $\angle \mathrm{AFN}=\angle \mathrm{EFM}$. Further, $\angle \mathrm{AFE}=\angle \mathrm{EFM}+\angle \mathrm{MFA}=\angle \mathrm{AFN}+\angle \mathrm{MFA}=90^{\circ}$. Triangle $\mathrm{AFE}$ is a right isosceles triangle, its angles: $90^{\circ}, 45^{\circ}, 45^{\circ}$.

Criteria. Any correct solution: 7 points.

Answer only: 0 points.