In the figure, $ \angle EAB$ and $ \angle ABC$ are right angles. $ AB \equal{} 4, BC \equal{} 6, AE \equal{} 8$, and $ \overline{AC}$ and $ \overline{BE}$ intersect at $ D$. What is the difference between the areas of $ \triangle ADE$ and $ \triangle BDC$?
[asy]unitsize(4mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));

pair A=(0,0), B=(4,0), C=(4,6), Ep=(0,8);
pair D=extension(A,C,Ep,B);

draw(A--C--B--A--Ep--B);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$E$",Ep,N);
label("$D$",D,2.5*N);
label("$4$",midpoint(A--B),S);
label("$6$",midpoint(B--C),E);
label("$8$",(0,3),W);[/asy]$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 8\qquad
\textbf{(E)}\ 9$
1. **Calculate the area of $\triangle ABE$:**
   \[
   [ABE] = \frac{1}{2} \times AB \times AE = \frac{1}{2} \times 4 \times 8 = 16
   \]
   Let the area of $\triangle ADE$ be \( x \) and the area of $\triangle ABD$ be \( z \). Therefore,
   \[
   x + z = 16
   \]

2. **Calculate the area of $\triangle ABC$:**
   \[
   [ABC] = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 4 \times 6 = 12
   \]
   Let the area of $\triangle BDC$ be \( y \) and the area of $\triangle ABD$ be \( z \). Therefore,
   \[
   y + z = 12
   \]

3. **Express the areas in terms of \( z \):**
   From the equations above, we have:
   \[
   x + z = 16 \quad \text{and} \quad y + z = 12
   \]

4. **Subtract the second equation from the first:**
   \[
   (x + z) - (y + z) = 16 - 12
   \]
   Simplifying, we get:
   \[
   x - y = 4
   \]

Thus, the difference between the areas of $\triangle ADE$ and $\triangle BDC$ is \( 4 \).

The final answer is \(\boxed{4}\)