+501 Points $A(0, 0)$, $B(6, 0)$, $C(6, 10)$ and $D(0, 10)$ are vertices of rectangle $ABCD$, and $E$ is on segment $CD$ at $(2, 10)$. What is the ratio of the area of triangle $ADE$ to the area of quadrilateral $ABCE$? Express your answer as a common fraction.
+605 $\triangle ADE$ has base $2$ and height $10$, so it has area $10$. The area of the rectangle is $6*10=60$, so the answer is $\frac{10}{60-10}=\frac{10}{50}=\boxed{\frac15}$.
From MATHCOUNTS? (only source that asks for common fraction, unless it's illegal AoPS homework)
