Example 8 As shown in Figure 7, 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 $N$ are
on segments $B H$ and $H F$ respectively,
and satisfy $B M=C N$. Find
the value of $\frac{M H+N H}{O H}$.
(2002, National High School Mathematics Competition)
Given $\angle A=60^{\circ}, B M=C N$, we know
$$
M H+N H=B H-H C=2(F H-H E) .
$$

Since $O H=O A+O B+O C=O A+A H$,
thus, $|\boldsymbol{A H}|=|\boldsymbol{O B}+\boldsymbol{O C}|=R$, where $R$ is the circumradius of $\triangle A B C$.
Also, $\angle A H F=\angle A B C, \angle A H E=\angle A C B$, so $F H-H E=R(\cos B-\cos C)=\sqrt{3} R \sin \frac{C-B}{2}$.
Since $\angle O A B=90^{\circ}-\angle C=\angle H A E$, therefore, $\angle O A H=2 \angle C-120^{\circ}=\angle C-\angle B$.
Since $A O=A H=R$, we have
$$
O H=2 R \sin \frac{\angle O A H}{2}=2 R \sin \frac{C-B}{2} \text {. }
$$

Thus, $\frac{M H+N H}{O H}=\frac{2(F H-H E)}{O H}=\sqrt{3}$.