Let points $A=(0,0)$, $B=(2,3)$, $C=(6,4)$, and $D=(7,0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $\left(\frac pq,\frac rs\right)$, where these fractions are in lowest terms. What is $p+q+r+s$?

1. By the shoelace formula (15),\[[ABCD]=\frac12\left|(0-0)+(8-18)+(0-28)+(0-0)\right|=19.\]
2. Let the line through $A$ meet $\overline{CD}$ at $E=(x,y)$. Then $\triangle ADE$ must have area $\frac{19}2$: $\frac12\cdot7\cdot y=\frac{19}2$, so $y=\frac{19}7$. On line $CD$, $x=6+\frac{4-y}4=6+\frac9{28}$.
3. The intersection point is $\left(\frac{177}{28},\frac{19}7\right)$, so $p+q+r+s=177+28+19+7=\boxed{231}$.