7.5. In triangle $A B C$, we construct $A D \perp B C$, where $D \in (B C)$. Points $E$ and $F$ are the midpoints of segments $A D$ and $C D$, respectively. Determine the measure in degrees of angle $B A C$, knowing that $B E \perp A F$.
## Solution:

Let $B E \cap A F=\{M\}$. According to the problem, $m(\angle B M F)=90^{\circ}$. Since in $\triangle B A F$ we have $A D \perp B F$, $B M \perp A F$, and $A D \cap B M=\{E\}$, it follows that $F E \perp A B$. Let $F E \cap A B=\{N\}$. Thus, $m(\angle B N F)=90^{\circ}$.

Since points $E$ and $F$ are the midpoints of segments $A D$ and $C D$ respectively, $E F$ is the midline in $\triangle A D C$. Since $E F$ is the midline, it follows that $E F \| A C$, which implies $F N \| A C$.

Since $F N \| A C$ and $B N \perp N F$, it follows that $B A \perp A C$. Thus, $m(\angle B A C)=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_43fc600879e0e9a806ceg-1.jpg?height=544&width=835&top_left_y=842&top_left_x=645)

Answer: $m(\angle B A C)=90^{\circ}$.