10.4. In triangle $A B C$, points $D$ and $F$ are marked on sides $A B$ and $A C$ respectively such that lines $D C$ and $B F$ are perpendicular to each other and intersect at point $E$ inside triangle $A B C$. It turns out that $A D=D C$ and $D E \cdot E C=F E \cdot E B$. What degree measure can angle $B A C$ have? (7 points)
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_89eb40249b21b160d5e6g-3.jpg?height=243&width=437&top_left_y=2083&top_left_x=250)

$D E \cdot E C=F E \cdot E B \Rightarrow \frac{D E}{E F}=\frac{E B}{E C}, \angle D E B=\angle F E C=90^{\circ}$, so triangles $\triangle D E B$ and $\triangle F E C$ are similar. From the similarity, we get: $\angle D B E=\angle F C E=\alpha$. (This equality of angles can also be justified by the fact that quadrilateral $D B C F$ is inscribed in a circle, from the condition $D E \cdot E C=F E \cdot E B$)

$\triangle A D C$ is isosceles $(A D=D C) \Rightarrow \angle D A C=\angle D C A(\angle F C E)=\angle D B E=\alpha$.

$\angle B D E$ is the exterior angle for $\triangle A D C \Rightarrow \angle B D E=\angle D C A+\angle D A C=2 \alpha$.

$\angle D B E+\angle B D E=2 \alpha+\alpha=3 \alpha=90^{\circ}$ (sum of the acute angles in a right triangle), hence $\alpha=30^{\circ}$.

Answer: $\angle B A C=30^{\circ}$.

| Criteria | Points |
| :--- | :---: |
| Complete solution of the problem. | 7 |
| Proved that $\angle D B E=\angle F C E$, without further progress. | 4 |
| Determined that quadrilateral $D B C F$ is inscribed in a circle. | 2 |
| Incorrect solution. | 0 |