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 $BD =$
$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ \frac {16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 }$
$\triangle BAC$ is a $6-8-10$ right triangle with hypotenuse $AB = 10$.
$\triangle BDE$ is similar to $\triangle BAC$ by angle-angle similarity since $E=C = 90^\circ$ and $B=B$, and thus $\frac{BD}{BA} = \frac{DE}{AC}$.
Solving the above for $BD$, we get $BD=\frac{BA\cdot DE}{AC} = 10\cdot \dfrac{4}{6}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}$.