Example 2 As shown in Figure 4-3, 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 line 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)
Extend $B E$ to intersect $\odot O$ at $L$. By the property of the orthocenter of a triangle, we know that $L$ is the reflection of $H$ over $A C$, so $L C = C H$.

Let the radius of $\odot O$ be $R$, $O H = d$, $C H = x$, and $B H = y$. Since $\angle C L B = \angle A = 60^{\circ}$, we have $L H = L C = C H = x$. Extend $O H$ to intersect $\odot O$ at $T$ and $S$, as shown in Figure 4-3. By the intersecting chords theorem, we have $T H \cdot H S = B H \cdot H L$, which is $(R + d)(R - d) = x \cdot y$, or $R^{2} = d^{2} + x y$.
In $\triangle B C L$ and point $H$ on side $B L$, applying Stewart's theorem and noting that $B C = 2 R \cdot \sin \angle A = \sqrt{3} R$, we get
$$
B C^{2} \cdot L H + L C^{2} \cdot B H = L H \cdot B H \cdot B L + C H^{2} \cdot B L,
$$

which is $(\sqrt{3} R)^{2} \cdot x + x^{2} \cdot y = x \cdot y \cdot (x + y) + x^{2} \cdot (x + y)$,
or $R^{2} = \frac{1}{3}(x^{2} + x y + y^{2})$.
Thus, we have $\frac{1}{3}(x^{2} + x y + y^{2}) = d^{2} + x y$.
This simplifies to $\frac{(x - y)^{2}}{d^{2}} = 3$, or $\frac{|x - y|}{d} = \sqrt{3}$.
When $A B > A C$, $M H + N H = B H - B M + C N - C H = B H - C H = y - x = |x - y|$,
so $\frac{M H + N H}{O H} = \frac{|x - y|}{d} = \sqrt{3}$ is the desired result.