In the figure, $\angle EAB$ and $\angle ABC$ are right angles. $AB=5, BC=8, AE=14$, and $AC$ and $BE$ intersect at $D$. What is the difference between the areas of $\triangle ADE$ and $\triangle BDC$?

1. Adding $[ABD]$ to both areas: $[ADE]+[ABD]=[ABE]$ and $[BDC]+[ABD]=[ABC]$, so $[ADE]-[BDC]=[ABE]-[ABC]$.
2. $[ABE]=\frac12\cdot5\cdot14=35$, $[ABC]=\frac12\cdot5\cdot8=20$.
3. Difference $=35-20=\boxed{15}$.