smc

Draw $\overline{AE}$, as seen in the diagram. From the problem, we know that $\overline{EB}$ and $\overline{AD}$ are medians of $△ACE$. Let $G$ be the midpoint of $\overline{AE}$. Then, $\overline{CG}$ is also a median of $△ACE$, and it goes through $△ACE$'s centroid, $F$. Because medians divide their triangle into $6$ smaller triangles of equal area, we know that $[△DFE]=\frac{1}{6}[△ACE]$. Because $[△ACE]=\frac{1}{2}\ast (15+15{)}^2=\frac{900}{2}=450$, $[△DFE]=\frac{450}{6}=75$. Thus, our answer is $\boxed{75}$.