一, (50 points) As shown in Figure 5, in $\triangle A B C$, $\angle A=$ $60^{\circ}, A B>A C$, point $O$ is the circumcenter, 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 $\frac{M H + N H}{O H}$.
As shown in Figure 9, take $B K = C H$ on $B E$, and connect $O B, O C, O K$. By the properties of the circumcenter of a triangle, we know $\angle B O C = 2 \angle A = 120^{\circ}$. By the properties of the orthocenter of a triangle, we know $\angle B H C = 180^{\circ} - \angle A = 120^{\circ}$.
Thus, $\angle B O C = \angle B H C$. Therefore, $B, C, H, O$ are concyclic. Hence, $\angle O B H = \angle O C H$. Also, $O B = O C, B K = C H$, so $\triangle B O K \cong \triangle C O H$.
$$
\begin{array}{c}
\because \angle B O K = \angle C O H, O K = O H, \\
\therefore \angle K O H = \angle B O C = 120^{\circ}, \\
\angle O K H = \angle O H K = 30^{\circ}.
\end{array}
$$

Observing $\triangle O K H$, $\frac{K H}{\sin 120^{\circ}} = \frac{O H}{\sin 30^{\circ}}$, thus $K H = \sqrt{3} O H$.
Also, $B M = C N, B K = C H$, so $K M = N H$.
$$
\begin{array}{l}
\therefore M H + N H = M H + K M = K H = \sqrt{3} O H. \\
\text{Therefore, } \frac{M H + N H}{O H} = \sqrt{3}.
\end{array}
$$