jmc

$\angle CDF=\angle AEB$ and $\angle BAE=\angle CFD$, so we know that $△AEB\sim △FDC$. Thus, denoting the side length of $BEFC$ as $x$, we can create the ratios: $\frac{28}{x}=\frac{x}{58}\Rightarrow x^2=1624$, and since $x^2$ is the area of square $BCFE$, $\boxed{1624}$ is the answer.