In the figure, $\angle EAB$ and $\angle ABC$ are right angles. $AB=4, BC=6, AE=8$, and $AC$ and $BE$ intersect at $D$. What is the difference between the areas of $\triangle ADE$ and $\triangle BDC$? $\mathrm{(A) \ } 2 \qquad$ $ \mathrm{(B) \ } 4 \qquad$ $ \mathrm{(C) \ } 5 \qquad$ $ \mathrm{(D) \ } 8 \qquad$ $ \mathrm{(E) \ } 9$
Solution
Adding $[ABD]$ to both areas: $[ADE]+[ABD]=[ABE]$ and $[BCD]+[ABD]=[ABC]$. So $[ADE]-[BCD]=[ABE]-[ABC]=\frac12\cdot4\cdot8-\frac12\cdot4\cdot6=16-12=4$.
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 $\triangle BDC$?
$\mathrm{(A) \ } 2 \qquad$ $ \mathrm{(B) \ } 4 \qquad$ $ \mathrm{(C) \ } 5 \qquad$ $ \mathrm{(D) \ } 8 \qquad$ $ \mathrm{(E) \ } 9$ 
