Rectangle $ABCD$ has $AB=8$ and $BC=6$. Points $E$ and $F$ lie on $\overline{AD}$ and $\overline{DC}$, respectively, and $\triangle BFE$ is an isosceles triangle with $\angle BFE=90^\circ$, as shown. What is the ratio of the area of $\triangle DEF$ to the area of $\triangle ABE$?

1. Set $A=(0,0)$, $B=(8,0)$, $C=(8,6)$, $D=(0,6)$, $E=(0,e)$, and $F=(f,6)$.
2. The conditions $BF=FE$ and $BF\perp FE$ give $(8-f)^2+36=f^2+(6-e)^2$ and $(8-f,-6)\cdot(-f,e-6)=0$. Solving with $0\le e\le6$ and $0\le f\le8$ gives $e=4$ and $f=6$.
3. The coordinates give $[DEF]=\frac12(2)(6)=6$ and $[ABE]=\frac12(8)(4)=16$. Therefore, $\dfrac{[DEF]}{[ABE]}=\boxed{\dfrac38}$.