Example 1.2.6 (League 2002) 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 $(M H+N H) / O H$.
Analysis and Solution (1) Information Collection and Analysis 1: From $\angle A=60^{\circ}$, we deduce that $\angle B O C=120^{\circ}, \angle B H C=120^{\circ}$, thus points $B, O, H, C$ are concyclic.
The process of solving the problem is a process of collecting such fragmented information.
Recalling the Law of Sines, we have $B C=2 R \sin \angle A=\sqrt{3} R$.
Focusing on the conclusion: Points $M, N$ lie on segments $B H, H F$ respectively, $\therefore M H+N H=B H-B M+C N-$ $\mathrm{CH}=\mathrm{BH}-\mathrm{CH}$. Therefore, it boils down to finding the value of $(\mathrm{BH}-\mathrm{CH}) / \mathrm{OH}$.

How to connect “$B, O, H, C$ are concyclic” with “finding $(B H-C H) / O H$”? Noticing that the endpoints of the segments in the ratio all lie on the circle $B O H C$, we think of applying Ptolemy's theorem: $B H \cdot O C=O H \cdot B C+$ $C H \cdot O B$. Dividing both sides by $R$, and noting that $B C=\sqrt{3} R$, we get the solution: $(B H-C H) / O H=\sqrt{3}$.
(2) Information Collection and Analysis 2: Noticing that $B E$ and $C F$ are two altitudes, thus, $A, F, H, E$ and $B, F, E, C$ are each concyclic. $B, O, H, C$ are also concyclic, $\therefore \angle O B H=\angle O C H, O B=O C, B M=C N$, $\therefore \triangle B O M \cong \triangle C O N, \angle O M B=\angle O N C, O M=O N$, hence, $O, M, H, N$ are concyclic.
This time, the endpoints of the segments in the original desired expression all lie on this circle. Applying Ptolemy's theorem, we get:
$$
\begin{array}{l}
M H \cdot O N+N H \cdot O M=O H \cdot M N, \text { and } \angle M O N=\angle B H C=120^{\circ}, \\
\therefore(M H+N H) / O H=M N / O M=\sqrt{3} .
\end{array}
$$
(3) Analysis 3: Utilizing $(M H+N H) / O H=(B H-C H) / O H$, and using $a, b, c, A, B, C, R$ as intermediate variables for calculation, from $\triangle B H C$ we can get: $B H=2 R \cos B, C H=2 R \cos C, B H-C H=2 \sqrt{3} \sin$ $\left(\angle C-60^{\circ}\right) R$,
and $O H=2 R \sin \left(\angle C-60^{\circ}\right)$. Therefore, $(B H-C H) / O H=\sqrt{3}$.