jmc

Let the length of the height of trapezoid $ABCD$ be $h$; note that this is also the length of the height of triangle $ACB$ to base $AB$. Then the area of $ABCD$ is $\frac{20+12}{2}\cdot h=16h$. On the other hand, the area of triangle $ACB$ is $\frac{1}{2}\cdot 20\cdot h=10h$. Thus the desired ratio is $\frac{10}{16}=\boxed{\frac{5}{8}}$.