Example 5 As shown in Figure 3-9, in $\triangle A B C$, $\angle A=60^{\circ}, A B>A C$, point $O$ is the circumcenter, the two altitudes $B E, C F$ intersect at point $H$, points $M, N$ are on segments $B H, H F$ respectively, and satisfy $B M=C N$. Find the value of $\frac{M H+N H}{O H}$.
(2002 National High School League Question)
Solution 1: Connect $O B, O C$. By the properties of the circumcenter and orthocenter of a triangle, we know $\angle B O C = 2 \angle A = 120^{\circ}, \angle B H C = 180^{\circ} - \angle H B C - \angle H C B = 180^{\circ} - (90^{\circ} - \angle C) - (90^{\circ} - \angle B) = 180^{\circ} - \angle A = 120^{\circ}$. Thus, points $B, C, H, O$ are concyclic. Applying Ptolemy's theorem to quadrilateral $B C H O$, we have
$$
B O \cdot C H + O H \cdot B C = B H \cdot O C.
$$

Let the circumradius of $\triangle A B C$ be $R$, then $B O = O C = R$, and since $\angle A = 60^{\circ}$, we have $B C = \sqrt{3} R$, which gives $R \cdot C H + O H \cdot \sqrt{3} R = B H \cdot R$, or equivalently, $B H - C H = \sqrt{3} O H$.
Since $M H + N H = (B H - B M) + (C N - C H) = B H - C H$,
it follows that $\frac{M H + N H}{O H} = \sqrt{3}$ is the desired result.
Solution 2: As in Solution 1, we know that points $B, C, H, O$ are concyclic, so $\angle O B H = \angle O C H$. Given $B O = O C$ and $B M = C N$, we have $\triangle O B M \cong \triangle O C N$, thus $O M = O N$ and $\angle B M O = \angle C N O$. This implies that points $O, M, H, N$ are concyclic, and the vertex angle of isosceles $\triangle O M N$ is $\angle M O N = \angle N H E = 120^{\circ}$, hence $\frac{M N}{O M} = \frac{\sin 120^{\circ}}{\sin 30^{\circ}} = \sqrt{3}$.

Applying Ptolemy's theorem to quadrilateral $O M H N$, we have $M H \cdot O N + N H \cdot O M = O H \cdot M N$, thus $\frac{M H + N H}{O H} = \frac{M N}{O M} = \sqrt{3}$ is the desired result.