2004 AMC 12A Problem 8

Let $x,$ $y,$ and $z$ be the areas of $\triangle ADE,$ $\triangle BDC,$ and $\triangle ABD,$ respectively. Then $\triangle ABE$ has area $\tfrac12 \cdot 4 \cdot 8 = 16 = x + z,$ and $\triangle ABC$ has area $\tfrac12 \cdot 4 \cdot 6 = 12 = y + z.$ The requested difference is $$\begin{aligned} x - y &= (x + z) - (y + z) \\ &= 16 - 12 = 4. \end{aligned}$$ Thus, the correct answer is B.