South African Mathematics Olympiad Second Round

Let $BE=2x$; then with $BE=\frac{2}{5}BC$ we must have $EC=3x$. Now for $△DEC$, $\frac{1}{2}(3x)(DC)=9$, giving $DC=\frac{6}{x}$, and since $BC=5x$, the area of rectangle $ABCD$ is $5x\cdot \frac{6}{x}=30$.