The point $F$, being the intersection of the two segments inside the triangle, can be exploited more.  Since your equations are based on areas, you can do some more area ratios by connecting $FC$.  You can get two more equations using area methods in triangles $FBC$ and $FAC$.  Then you can solve for $n$.

Alternatively, using the fact that $E$ is a midpoint, we find another midpoint to connect with.  Here we pick the midpoint $M$ of $CD$, and connect $EM$.  Now $MD=DB$ as well, and $EM\parallel AD$, and it means that $F$ is the midpoint of $BE$. Now there is another equation for $n$ based on the triangle $ABE$.