Problem 2. (Option 1). Given an acute triangle $\mathrm{ABC}(\mathrm{AB}=\mathrm{BC})$ and $\mathrm{BC}=12$. $A N \perp B C$. On the side $\mathrm{BC}$, a point $M$ (M lies between B and $\mathrm{N}$) is marked such that $\mathrm{AN}=\mathrm{MN}$ and $\angle \mathrm{BAM}=\angle \mathrm{NAC}$. Find $\mathrm{BN}$.
Solution: 1). Let $\angle \mathrm{BAM}=\angle \mathrm{NAC}=\alpha, \angle \mathrm{MAN}=\angle \mathrm{AMN}=\beta=45^{\circ}$, (since triangle MAN is isosceles) $\prec=\succ \angle \mathrm{MAC}=\alpha+\beta$ and $\angle \mathrm{MCA}=2 \alpha+\beta=\succ$ $(\square \mathrm{AMC}) 2 \beta+\alpha+2 \alpha+\beta=180^{\circ}=\succ \alpha+\beta=60^{\circ}=\succ \angle \mathrm{MAC}=60^{\circ}$ and $\angle \mathrm{MAN}=45^{\circ}=\succ$ $\angle \mathrm{NAC}=15^{\circ}$.

2) $\angle \mathrm{BAM}=\angle \mathrm{NAC}=15^{\circ}=\succ \angle \mathrm{BAN}=60^{\circ}$.
3) $(\square \mathrm{ABN})$ : $\mathrm{AB}=\mathrm{BC}=12 . \angle \mathrm{A}=60^{\circ}, \angle \mathrm{N}=90^{\circ}, \angle \mathrm{B}=30^{\circ}=\succ \mathrm{BN}=6 \sqrt{3}$.

Answer: $BN=6 \sqrt{3}$.