Find the area of a figure in a triangle Triangle $ABC$ in the figure has area $10$ . Points $D,E,$ and $F$, all distinct from $A,B,$ and $C$, are on sides $AB,BC,$ and $CA$ respectively, and $AD=2$ , $DB=3$ .
If triangle $ABE$ and quadrilateral $DBEF$ have equal areas,then what is that area?

Efforts made: I've tried to add some extra lines to see if i could get something usefull but ,guess, i didnt get anything . It seems like the problem is asking  some crazy creative thing to be done,i cant see what.

$[DBEF]=[ABE]$ is equivalent, by subtracting $[DBE]$ to both sides, to $[DEF]=[DEA]$.
These triangles share the $DE$-side, hence $[DEF]=[DEA]$ implies $DE\parallel AF$, so:
$$ \frac{BE}{BC}=\frac{BD}{BA}=\frac{3}{5} $$
and the area of $[BDE]$, consequently, equals $\frac{9}{25}[ABC]=\frac{18}{5}$. Since $[ABE]=\frac{5}{3}[DBE]$,
$$ [ABE]=[DBEF]=\color{red}{6} $$
follows.
