In rectangle $ABCD$, points $E$ and $F$ are on sides $AB$ and $CD$, respectively, such that $AE = CF > AD$ and $\angle CED = 90^{\circ}$. Lines $AF$, $BF$, $CE$ and $DE$ enclose a rectangle whose area is 24% of the area of $ABCD$. Compute $\frac{BF}{CE}$ .