One, (50 points)
As shown in the figure, in $\triangle A B C$, $\angle A=60^{\circ}, A B>A C$, point $O$ is the circumcenter, and 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 $\frac{M H+N H}{O H}$.
-, (50 points)
Take $B K = C H$ on $B E$, and connect $O B, O C, O K$. By the property of the circumcenter of a triangle, we know $\angle B O C - 2 \cdot \angle A = 120^{\circ}$, and by the property of the orthocenter of a triangle, we know
$\angle B H C = 180^{\circ} - \angle A = 120^{\circ}$,
so $\angle B O C = \angle B H C$.
Therefore, $B, C, H, O$ are concyclic.
Thus, $\angle O B H = \angle O C H$, and since $O B = O C, B K = C H$, we have $\triangle B O K \cong \triangle O O H$.
Since $\angle B O K = \angle C O H, O K = O H$,
we have $\angle K O H = \angle B O C = 120^{\circ}, \angle O K H = \angle O H K = 30^{\circ}$.
Observing $\triangle O K H$, $\frac{K H}{\sin 120^{\circ}} = \frac{O H}{\sin 30^{\circ}}$, so $K H = \sqrt{3} \cdot O H$;

Since $B M = C N, B K = C H$, we have $K M = N H$,
so $M H + N H = M H + K M = K H = \sqrt{3} \cdot O H$, hence $\frac{M H + N H}{O H} = \sqrt{3}$.