4. Let $O$ be the intersection of diagonals $A C$ and $B D$ in quadrilateral $A B C D$. If $\angle B A D + \angle A C B = 180^{\circ}$, and $B C = 3, A D = 4, A C = 5, A B = 6$, then $\frac{D O}{O B}=(\quad)$.
(A) $\frac{10}{9}$
(B) $\frac{8}{7}$
(C) $\frac{6}{5}$
(D) $\frac{4}{3}$
4. A.

As shown in Figure 4, draw $B E / / A D$, intersecting the extension of $A C$ at point $E$.
$$
\begin{array}{l}
\text { Then } \angle A B E=180^{\circ}-\angle B A D=\angle A C B \\
\Rightarrow \triangle A B C \backsim \triangle A E B \Rightarrow \frac{A C}{A B}=\frac{B C}{E B} \\
\Rightarrow E B=\frac{A B \cdot B C}{A C}=\frac{18}{5} .
\end{array}
$$

Furthermore, since $B E / / A D \Rightarrow \frac{D O}{O B}=\frac{A D}{B E}=\frac{4}{\frac{18}{5}}=\frac{10}{9}$.