Quadrilateral $ABCD$ is a trapezoid with $AB$ parallel to $CD$. We know $AB = 20$ and $CD = 12$. What is the ratio of the area of triangle $ACB$ to the area of the trapezoid $ABCD$? Express your answer as a common fraction.
I notice that the area of the trapezoid is the average of the bases times the height, so I can write an expression for that: $A_{ABCD} = \frac{AB + CD}{2} \cdot h$. I also notice that the area of the triangle $ACB$ is half the base times the height, so I can write another expression for that: $A_{ACB} = \frac{AB}{2} \cdot h$. To find the ratio of the areas, I can divide the second expression by the first one: $\frac{A_{ACB}}{A_{ABCD}} = \frac{\frac{AB}{2} \cdot h}{\frac{AB + CD}{2} \cdot h}$. I can simplify this fraction by canceling out the common factors of $h$ and $2$: $\frac{A_{ACB}}{A_{ABCD}} = \frac{AB}{AB + CD}$. Now I can plug in the values of $AB$ and $CD$ that are given: $\frac{A_{ACB}}{A_{ABCD}} = \frac{20}{20 + 12}$. I can simplify this fraction by dividing the numerator and denominator by $4$: $\frac{A_{ACB}}{A_{ABCD}} = \frac{5}{8}$. 