4. Given $S_{\triangle M B C}=4, 3 A B=2 B C$, draw the perpendicular from point $C$ to the angle bisector $B E$ of $\angle A B C$, and let the foot of the perpendicular be $D$. Then $S_{\triangle B D C}=$ $\qquad$
4.3.

Construct the reflection of point $C$ about $BD$ as point $F$, thus, $BC=BF$. Connect $FA$ and $FD$, then point $F$ lies on the extension of $BA$, and points $C$, $D$, and $F$ are collinear.
$$
\text{Therefore, } \frac{S_{\triangle ABC}}{S_{\triangle FBC}}=\frac{AB}{FB}=\frac{AB}{BC}=\frac{2}{3} \text{.}
$$

Since $S_{\triangle FBC}=2 S_{\triangle DBC}$, we have $\frac{S_{\triangle ABC}}{S_{\triangle DBC}}=\frac{4}{3}$. Thus, $S_{\triangle BDC}=3$.