jmc

Let $x,$ $y,$ and $z$ be the areas of $△ADE,$ $△BDC,$ and $△ABD,$ respectively. The area of $△ABE$ is $$\frac{1}{2}\cdot 4\cdot 8=16=x+z,$$and the area of $△BAC$ is $$\frac{1}{2}\cdot 4\cdot 6=12=y+z.$$Subtracting these equations gives $$(x+z)-(y+z)=16-12⟹x-y=\boxed{4}.$$