jmc

We first compute the area of $ABCD.$ A quick way to do so (besides the shoelace formula) is to draw the rectangle with vertices $A=(0,0),$ $(0,3),$ $(4,3),$ and $(4,0),$ and divide the part of the rectangle outside $ABCD$ into squares and right triangles, as shown:Then $$[ABCD]=12-2\cdot 1-1-\frac{3}{2}=\frac{15}{2}.$$Therefore, the two pieces of $ABCD$ must each have area $\frac{1}{2}\cdot \frac{15}{2}=\frac{15}{4}.$

Let $E$ be the point where the line through $A$ intersects $\overline{CD},$ as shown: Triangle $△AED$ must have area $\frac{15}{4}.$ We have $AD=4,$ so letting $h$ denote the length of the altitude from $E$ to $\overline{AD},$ we must have $$\frac{1}{2}\cdot 4\cdot h=[△AED]=\frac{15}{4}.$$Thus, $h=\frac{15}{8}.$ Therefore, $E=(t,\frac{15}{8})$ for some value of $t.$

Since $C=(3,3)$ and $D=(4,0),$ the slope of $\overline{CD}$ is $\frac{0-3}{4-3}=-3,$ so the point-slope form of the equation of line $CD$ is $y-0=-3(x-4),$ or simply $y=-3x+12.$ When $y=\frac{15}{8},$ we get $\frac{15}{8}=-3x+12,$ and so $x=\frac{27}{8}.$ Therefore, $E=\boxed{(\frac{27}{8},\frac{15}{8})}.$