In $\triangle ABC$, $AB=6$, $AC=8$, and $BC=10$. The bisector of $\angle A$ meets $\overline{BC}$ at $D$, and the line through $D$ parallel to $\overline{AB}$ meets $\overline{AC}$ at $E$. What is $DE$?

1. By the angle bisector theorem, $BD:DC=AB:AC=6:8=3:4$, so $DC=\frac47\cdot10=\frac{40}7$.
2. Since $DE\parallel AB$, $\triangle CDE\sim\triangle CBA$, so $\dfrac{DE}{AB}=\dfrac{DC}{BC}=\dfrac47$.
3. $DE=\frac47\cdot6=\boxed{\frac{24}7}$.