10.2. In triangle $A B C$, point $M$ is the midpoint of side $B C$. On the bisector of angle $B A C$, point $N$ is taken such that $B N \perp A N$. Determine the length of side $A C$, if $A B=10 \mathrm{~cm}$ and $M N=2 \mathrm{~cm}$.
Solution. Let $B N \cap A C=\{L\}$.

We distinguish the following potential cases:

1. If $N \in(B C)$, then $A N$ is the bisector and altitude in $\triangle A B C$, which means it is also the median. Therefore, $N \equiv M$, which contradicts the hypothesis $M N=2 \mathrm{~cm}$. This case is impossible.
2. Suppose point $N$ is inside $\triangle A B C$, i.e., $L \in(A C)$. The right triangles $A B N$ and $A L N$ are congruent by the (HL) criterion, so $B N=L N$ and

![](https://cdn.mathpix.com/cropped/2024_06_06_f05c41a97656664dbd2fg-1.jpg?height=460&width=642&top_left_y=1432&top_left_x=1324)
$A L=A B=10 \mathrm{~cm}$.

Since $N$ is the midpoint of segment $B L$ and $M$ is the midpoint of segment $B C$, then $[M N]$ is the midline of $\triangle B C L$, so $C L=2 \cdot M N=4 \mathrm{~cm}$, and $A C=A L+L C=14 \mathrm{~cm}$.

3. Suppose point $N$ is outside $\triangle A B C$, i.e., $C \in(A L)$, then similarly to case 2, we have $\triangle A B N \equiv \triangle A L N$, $C L=2 \cdot M N=4 \mathrm{~cm}$, and $A C=A L-C L=6 \mathrm{~cm}$.

Case 3 is possible only if $M N<\frac{1}{2} A L=\frac{1}{2} A B$, which is

![](https://cdn.mathpix.com/cropped/2024_06_06_f05c41a97656664dbd2fg-1.jpg?height=442&width=486&top_left_y=1955&top_left_x=1433)
satisfied in this problem.

Therefore, both cases, 2 and 3, are possible and the problem has two solutions.

Answer: $14 \mathrm{~cm}$ or $6 \mathrm{~cm}$.