Area of a Triangle Inside a Trapezoid

Solution 1:

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 $EF \parallel AB$.