4. On the side $BC$ of the equilateral $\triangle ABC$, take a point $D$ such that $CD=2BD$. Draw $CH \perp AD$ at point $H$, and connect $BH$. Then $\angle DBH - \angle DAB$ is ( ).
(A) 0
(B) negative
(C) positive
(D) cannot be determined
4. A.

As shown in Figure 7, draw $A E \perp B C$ at point $E$. Suppose $B C=$
6a. Then
$$
\begin{array}{l}
E C=E B \\
=3 a, \\
B D=2 a, \\
D E=a, \\
D C=4 a .
\end{array}
$$

Therefore, $A, H, E, C$ are concyclic
$$
\Rightarrow D H \cdot D A=D C \cdot D E=B D^{2}
$$
$\Rightarrow B D$ is the tangent to the circumcircle of $\triangle A B H$
$$
\Rightarrow \angle D B H=\angle D A B
$$