In triangle $ABC$, $AB=10$, $BC=12$, $AC=14$. Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$. Find the area of triangle $ADE$.

1. By the angle bisector theorem, $BE:EC=AB:AC=10:14=5:7$, so $BE=5$; since $BD=6$, $DE=1$.
2. $s=\frac{10+12+14}2=18$, so $[ABC]=\sqrt{18\cdot8\cdot6\cdot4}=24\sqrt6$ by Heron's formula.
3. $[ADE]=\frac{DE}{BC}[ABC]=\frac1{12}\cdot24\sqrt6=\boxed{2\sqrt6}$ by theorem 10.