10. 2. 21 * In $\triangle A B C$, $\angle A=60^{\circ}, A B>A C$, point $O$ is the circumcenter, the two altitudes $B E$ and $C F$ intersect at point $H$, points $M$ and $N$ lie on segments $B H$ and $H F$ respectively, and satisfy $B M=C N$. Find the value of $\frac{M H+N H}{O H}$.
Given $\angle A=60^{\circ}, B M=C N$, thus $M H+N H=$ $B H-H C=2(F H-H E)$. Therefore, $\overrightarrow{O H}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=$ $\overrightarrow{O A}+\overrightarrow{A H}, \angle B O C=120^{\circ}$. So $|\overrightarrow{A H}|=|\overrightarrow{O B}+\overrightarrow{O C}|=$ $R$. (where $R=O A=O B=O C$)

From the cyclic quadrilateral, it is easy to know, $\angle A H F=\angle A B C, \angle A H E=$ $\angle A C B$. Hence $F H-H E=R(\cos B-\cos C)=\sqrt{3} R \sin \frac{C-B}{2}$. Since $\angle O A B=90^{\circ}-\angle C=\angle H A E, \angle O A H=2 \angle C-120^{\circ}=\angle C-\angle B . O H=$ $2 R \sin \frac{\angle O A H}{2}=2 R \sin \frac{C-B}{2}$. Therefore, $\frac{M H+N H}{O H}=\frac{2(F H-H E)}{O H}-\sqrt{3}$.