1. As shown in the figure, 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 $\frac{M H+N H}{O H}$.
1.【Analysis and Solution】As shown in the figure, 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 $\angle B O C=2 \angle A=120^{\circ}$. By the property of the orthocenter of a triangle, we know $\angle B H C=180^{\circ}-$
$$
\begin{aligned}
\angle A= & 120^{\circ} . \\
& \therefore \angle B O C=\angle B H C . \\
& \therefore B, C, H, O \text{ are concyclic.} \\
& \therefore \angle O B H=\angle O C H . \\
& \text{Also, } O B=O C, B K=C H . \\
& \therefore \triangle B O K \cong \triangle C O H . \\
& \because \angle B O K=\angle C O H, O K=O H, \\
& \therefore \angle K O H=\angle B O C=120^{\circ}, \angle O K H=\angle O H K=30^{\circ} .
\end{aligned}
$$

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