AHSME 1995 Problem 8

In $\triangle ABC$, $\angle C = 90^\circ, AC = 6$ and $BC = 8$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{BC}$, respectively, and $\angle BED = 90^\circ$. If $DE = 4$, then $BD =$ $\mathrm{(A) \ 5 } \mathrm{(B) \ \frac {16}{3} } \mathrm{(C) \ \frac {20}{3} } \mathrm{(D) \ \frac {15}{2} } \mathrm{(E) \ 8 }$