Four. (Full marks 14 points) As shown in the figure, $O$ is the intersection point of the diagonals of quadrilateral $A B C D$. It is known that $\angle B A D+\angle B C A$
$$
\begin{array}{l}
=180^{\circ}, A B=5, A C=4, A D \\
=3, \frac{B O}{D O}=\frac{7}{6} . \text { Find } B C .
\end{array}
$$
Draw $B B^{\prime} \perp A C, C C^{\prime} \perp$ $A B, D D^{\prime} \perp A C$, with the feet of the perpendiculars being $B^{\prime}, C^{\prime}, D^{\prime}$ (as shown in the figure).

It is easy to see that $\triangle B O B^{\prime} \backsim$ $\triangle D O D^{\prime}$, so $\frac{B O}{O D}=\frac{B B^{\prime}}{D D^{\prime}}$.
Since $S_{\triangle A B C}=\frac{1}{2} A B \cdot C C^{\prime}$
$$
=\frac{1}{2} A C \cdot B B^{\prime},
$$

we have $\frac{B B^{\prime}}{C C^{\prime}}=\frac{A B}{A C}$.
Also, in $\triangle A B C$, we have $\angle B A C+\angle A B C+\angle B C A=180^{\circ}$.
Thus, $\angle C A D=\angle A B C$.
Therefore, in the right triangles $\triangle B C C^{\prime} \subset \mathrm{Rt} \triangle A D D^{\prime}$, we have $\frac{B C}{A D}=\frac{C C^{\prime}}{D D^{\prime}}$.
So, $\frac{B O}{O D}=\frac{B B^{\prime}}{D D^{\prime}}=\frac{B B^{\prime}}{C C^{\prime}} \cdot \frac{C C^{\prime}}{D D^{\prime}}=\frac{A B}{A C} \cdot \frac{B C}{A D}$, which means $\frac{7}{6}=\frac{5}{4} \cdot \frac{B C}{3}$.
Thus, $B C=\frac{7}{6} \times \frac{12}{5}=\frac{14}{5}$.