1. 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 $B E, C F$ intersect at point $H$, points $M, N$ lie on segments $B H, H F$ respectively, and satisfy $B M=C N$. Find the value of $\frac{M H+N H}{O H}$.
As shown in the figure, $M H+N H=B H-B M+C N-C H$
$=B H-C H$. Take a point $D$ on $B H$ such that
$B D=C H \Rightarrow D H=M H+N H$. Connect $O D, O B$,
$O C$, then $\angle O B D=\angle O B C-\angle E B C$
$=30^{\circ}-\left(90^{\circ}-\angle A C B\right)=\angle A C B-60^{\circ}$,
$\angle O C H=\angle A C B-\angle O C B-\angle A C F=\angle A C B-60^{\circ} \Rightarrow \angle O B D=\angle O C H$,
thus $\triangle O B D \cong \triangle O C H \Rightarrow O D=O H$ and $\angle D O H=\angle B O C=120^{\circ}$, hence
$\triangle D O H$ is an isosceles triangle with a vertex angle of $120^{\circ}$, so $\frac{D H}{O H}=\frac{M H+N H}{O H}=\sqrt{3}$.