smc

Let $G$ be the intersection point of $AE$ and $DF$. Since $[DBEF]=[ABE]$, we have $[DBEG]+[EFG]=[DBEG]+[ADG]$, i.e. $[EFG]=[ADG]$. It therefore follows that $[ADG]+[AGF]=[EFG]+[AGF]$, so $[ADF]=[AFE]$. Now, taking $AF$ as the base of both $△ADF$ and $△AFE$, and using the fact that triangles with the same base and same perpendicular height have the same area, we deduce that the perpendicular distance from $D$ to $AF$ is the same as the perpendicular distance from $E$ to $AF$. This in turn implies that $AF\parallel DE$, and so as $A$, $F$, and $C$ are collinear, $AC\parallel DE$. Thus $△DBE\sim △ABC$, so $\frac{BE}{BC}=\frac{BD}{BA}=\frac{3}{5}$. Since $△ABE$ and $△ABC$ have the same perpendicular height (taking $AB$ as the base), $[ABE]=\frac{3}{5}\cdot [ABC]=\frac{3}{5}\cdot 10=6$, and hence the answer is $\boxed{6}$.