Example 3 (2002 National High School League) 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 $N$ are on segments $B H$ and $H F$, respectively, and satisfy $B M=C N$. Find the value of $\frac{M H+N H}{O H}$.
Analyzing by connecting $O B, O C$, from the properties of the circumcenter of a triangle, we know $\angle B O C=2 \angle A=120^{\circ}$, and from the properties of the orthocenter of a triangle, we know $\angle B H C=180^{\circ}-\angle A=120^{\circ}$. Therefore, we can conclude that $B, C, H, O$ are concyclic. Applying Ptolemy's theorem in this circle to study the properties of related segments.

Solution: Connect $O B, O C$. Let the circumradius of $\triangle A B C$ be $R$. From the properties of the circumcenter of a triangle, we know $\angle B O C=2 \angle A=120^{\circ}$, and from the properties of the orthocenter of a triangle, we know $\angle B H C=$ $180^{\circ}-\angle A=120^{\circ}$.
So $\angle B H C=\angle B O C$
Therefore, $B, C, H, O$ are concyclic.
By Ptolemy's theorem,
$$
\begin{aligned}
B H \cdot O C=O H \cdot B C+C H \cdot O B \\
\text { Also, } B C=2 R \sin 60^{\circ}=\sqrt{3} R
\end{aligned}
$$

So $(B H-C H) R=\sqrt{3} R \cdot O H$
So $(B M+M H)-(C N-N H)=\sqrt{3} \cdot O H$
Since $B M=C N$
Thus $\frac{M H+N H}{O H}=\sqrt{3}$.