jmc

Let the medians intersect at point $G$ as shown below. We include the third median of the triangle in red; it passes through the intersection of the other two medians.



Point $G$ is the centroid of $△ABC$, so $AG:GD=BG:GE=2:1$. Therefore, $AG=\frac{2}{3}(AD)=10$ and $BG=\frac{2}{3}(BE)=\frac{40}{3}$.

Drawing all three medians of a triangle divides the triangle into six triangles with equal area. In $△ABC$ above, $△ABG$ consists of two of these six triangles, so the area of $△ABC$ is 3 times the area of $△ABG$: $$[ABC]=3[ABG]=3\cdot \frac{1}{2}\cdot AG\cdot BG=\frac{3}{2}\cdot 10\cdot \frac{40}{3}=\boxed{200}.$$