8. In an acute-angled triangle $ABC$ with sides $AB=4, AC=3$, a point $N$ is marked on the median $AM$ such that $\angle BNM = \angle MAC$. Find the length of the segment $BN$.
Solution:

Let's make an additional construction, doubling the median $A M$ beyond point $M$, thereby obtaining point $K$ such that $K \in A M, K M=A M$. Triangles $K M B$

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and $A M C$ are equal by two sides and the angle between them, hence the corresponding sides and angles are equal: $B K=A C=3, \quad \angle M K B=\angle M A C$. Then we get that triangle $K B N$ is isosceles $\angle B N M=\angle B K N$ and $B N=B K=3$.

Answer: 3.