, 2014

Solution:

Answer: $\boxed{\dfrac{256}{7}}$

Let $M, N$ be the midpoints of $AD, BC$ respectively. Since $AE$ and $DE$ are bisectors of supplementary angles, the triangle $AED$ is right with right angle $E$. Then $EM$ is the median of a right triangle from the right angle, so triangles $EMA$ and $EMD$ are isosceles with vertex $M$. But then $\angle MEA = \angle EAM = \angle EAB$, so $EM \parallel AB$. Similarly, $FN \parallel BA$. Thus, both $E$ and $F$ are on the midline of this trapezoid. Let the length of $EF$ be $x$. Triangle $EFH$ has area $1$ and is similar to triangle $ABH$, which has area $25$, so $AB = 5x$. Then, letting the heights of trapezoids $EABF$ and $EDCF$ be $h$ (they are equal since $EF$ is on the midline), the area of trapezoid $EABF$ is $\dfrac{6xh}{2} = 24$. So the area of trapezoid $EDCF$ is $36 = \dfrac{9xh}{2}$. Thus $DC = 8x$. Then, triangle $GEF$ is similar to and has $\dfrac{1}{64}$ times the area of triangle $CDG$. So the area of triangle $CDG$ is $\dfrac{64}{63}$ times the area of quadrilateral $EDCF$, or $\dfrac{256}{7}$.