In the overlapping [triangles](https://artofproblemsolving.com/wiki/index.php/Triangle) $\triangle{ABC}$ and $\triangle{ABE}$ sharing common [ side](https://artofproblemsolving.com/wiki/index.php/Edge) $AB$, $\angle{EAB}$ and $\angle{ABC}$ are [right angles](https://artofproblemsolving.com/wiki/index.php/Right_angle), $AB=4$, $BC=6$, $AE=8$, and $\overline{AC}$ and $\overline{BE}$ intersect at $D$. What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$? 

$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$
Solution 1
Since $AE \perp AB$ and $BC \perp AB$, $AE \parallel BC$. By alternate interior angles and $AA\sim$, we find that $\triangle ADE \sim \triangle CDB$, with side length ratio $\frac{4}{3}$. Their heights also have the same ratio, and since the two heights add up to $4$, we have that $h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}$ and $h_{CDB} = 3 \cdot \frac 47 = \frac {12}7$. Subtracting the areas, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{\mathrm{(B)}\ 4}$.

Solution 2
Let $[X]$ represent the area of figure $X$. Note that $[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$ and $[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$.
$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}$

Solution 3 (coordbash)
Put figure $ABCDE$ on a graph. $\overline{AC}$ goes from (0, 0) to (4, 6) and $\overline{BE}$ goes from (4, 0) to (0, 8). $\overline{AC}$ is on line $y = 1.5x$. $\overline{BE}$ is on line $y = -2x + 8$. Finding intersection between these points,
$1.5x = -2x + 8$. 
$3.5x = 8$
$x = 8 \times \frac{2}{7}$
$= \frac{16}{7}$
This gives us the x-coordinate of D. 
So, $\frac{16}{7}$ is the height of $\triangle ADE$, then area of $\triangle ADE$ is 
$\frac{16}{7} \times 8 \times \frac{1}{2}$
$= \frac{64}{7}$
Now, the height of $\triangle BDC$ is $4-\frac{16}{7} = \frac{12}{7}$
And the area of $\triangle BDC$ is $6 \times \frac{12}{7} \times \frac{1}{2} = \frac{36}{7}$
This gives us $\frac{64}{7} - \frac{36}{7} = 4$
Therefore, the difference is $4$

Solution 4
We want to figure out $Area(\triangle ADE) - Area(\triangle BDC)$.
Notice that $\triangle ABC$ and $\triangle BAE$ "intersect" and form $\triangle ADB$.
This means that $Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)$ because $Area(\triangle ADB)$ cancels out, which can be seen easily in the diagram.
$Area(\triangle BAE) = 0.5 * 4 * 8 = 16$
$Area(\triangle ABC) = 0.5 * 4 * 16 = 12$
$Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{\mathrm{(B)}\ 4}$