2. Given that the circumcenter of $\triangle A B C$ is $O$, the orthocenter is $H, O \neq H$, and the midpoints of sides $B C, C A$ are $D, E$ respectively. Let $D^{\prime}, E^{\prime}$ be the points symmetric to $D, E$ with respect to $H$. If the line $A D^{\prime}$ intersects the line $B E^{\prime}$ at point $K$, find the value of $\frac{|K O|}{|K H|}$.
As shown in the figure, construct the symmetric point $H^{\prime}$ of $H$ with respect to $D$, then the quadrilateral $H B H^{\prime} C$ is a parallelogram $\Rightarrow H^{\prime} B / / C H \Rightarrow H^{\prime} B \perp A B$, thus $A H^{\prime}$ is the diameter of circle $O$.
Also, the quadrilateral $D E D^{\prime} E^{\prime}$ is a parallelogram $\Rightarrow D^{\prime} E^{\prime} \parallel D E \cong \frac{1}{2} A B$
$\Rightarrow D^{\prime}$ is the midpoint of segment $A K$.
In $\triangle D^{\prime} A H^{\prime}$, applying Menelaus' theorem, we get
$\frac{D^{\prime} K}{K A} \cdot \frac{A O}{O H^{\prime}} \cdot \frac{H^{\prime} H}{H D^{\prime}}=\frac{1}{2} \cdot 1 \cdot 2=1 \Rightarrow K, H, O$ are collinear.
Thus, in $\triangle K A O$, applying Menelaus' theorem to line $D^{\prime} H H^{\prime}$, we get $\frac{K D^{\prime}}{D^{\prime} A} \cdot \frac{A H^{\prime}}{H^{\prime} O} \cdot \frac{O H}{H K}=1$
$$
\Rightarrow \frac{O H}{H K}=\frac{1}{2} \Rightarrow \frac{K O}{K H}=\frac{3}{2} .
$$