Area of a Triangle Inside a Trapezoid I am stuck on the following math contest problem:
Let $ABCD$ be a trapezoid with $AB \parallel CD$. The bisectors of $\angle CDA$ and $\angle DAB$ meet at $E$, the bisectors of $\angle ABC$ and $\angle BCD$ meet at $F$, the bisectors of $\angle BCD$ and $\angle CDA$ meet at $G$, and the bisectors of $\angle DAB$ and $\angle ABC$ meet at $H$. Quadrilaterals $EABF$ and $EDCF$ have areas $24$ and $36$, respectively, and triangle $ABH$ has area $25$. Find the area of triangle $CDG$
Diagram (by the user Moti):

I don't know what the source is, so if anyone knows, please let me know. I searched for a solution on Art of Problem Solving (AoPS) but I could not find one.
So far, I have only figured out that $\angle AED = 90^{\circ} = \angle BFC$ and that $[CDG] = [EDCF]+[EFG] = 36+[EFG]$. Does anyone know how to solve the problem?
Note: [X] denotes the area of the polygon X.
 A: 
If $M$ and $N$ are midpoints of sides $AD$ and $BC$ respectively, we know $MN$ is parallel to sides $AB$ and $CD$. You can try and show that points $M, E, F$ and $N$ are collinear (otherwise see the analysis at the end of the answer).
As $S_{\triangle EFH}:S_{\triangle ABH} = 1:25$,
$AB = 5 EF$
Also comparing the area of quadrilaterals $EABF$ and $EDCF$,
$(CD + EF):(AB+EF) = 36:24 \implies CD = 8 EF$
$S_{\triangle EFG} = \frac{1}{64} \cdot S_{\triangle CDG}$
Or, $ \displaystyle 64 \cdot (S_{\triangle CDG} - 36) = S_{\triangle CDG} \implies S_{\triangle CDG} = \frac{256}{7}$


 If $M$ and $N$ are midpoints of $AD$ and $BC$, $MN \parallel AB$. But given $DE$ is angle bisector of $\angle ADK$ and $DE \perp AK$, $AE = EK$ and $\angle AKD = \angle A / 2$. Also given $\triangle AED$ is right at $E$, $M$ is the circumcenter of $\triangle AED$ and $AM = ME$. That leads to $\angle AEM = \angle A/2$. As $\angle AEM = \angle AKD$, $ME \parallel CD$ or $ME \parallel AB$. Similarly we conclude $FN \parallel AB$. So points $M, E, F$ and $N$ must be collinear and hence $EF \parallel AB$.

