smc

$△BAC$ is a $6-8-10$ right triangle with hypotenuse $AB=10$. $△BDE$ is similar to $△BAC$ 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 $BD=\frac{BA\cdot DE}{AC}=10\cdot \frac{4}{6}=\frac{20}{3}\Rightarrow \boxed{\frac{20}{3}}$.