4. In an acute triangle $\triangle A B C$, it is known that $A B=c, A C=b$ $(b>c), \angle B A C=60^{\circ}$, its orthocenter is $H$, circumcenter is $O, O H$ intersects $A B$ and $A C$ at points $D$ and $E$ respectively. Then the length of $D E$ is $($.
(A) $\frac{b+c}{2}$
(B) $\frac{b+c}{3}$
(C) $2(b-c)$
(D) $3(b-c)$
4. B.

As shown in Figure 4, it is easy to see that
$$
\begin{array}{l}
\angle B O C \\
=2 \angle B A C \\
=120^{\circ}, \\
\angle B H C=\angle G H F \\
=180^{\circ}-\angle A \\
=120^{\circ} .
\end{array}
$$

Therefore, points $B, H, O, C$ are concyclic.
Hence, $\angle D H C=\angle O B C=30^{\circ}$.
Also, $\angle D C H=90^{\circ}-\angle B A C=30^{\circ}$.
Thus, $D H=D C$.
Similarly, $E H=E B$.
Therefore, $D C+E B=D E$.
Given that $\angle A D E=\angle D C H+\angle D H C=60^{\circ}$, we know $A D =A E=D E$.
Thus, $D E=\frac{1}{3}(A C+A B)=\frac{b+c}{3}$.