Find areas using ratio of areas 
In the following figure, $ABCD$ is a rectangle. $M$ and $N$ are the mid-points of $DC$ and $AB$ respectively and $AE:EN=BF:FN=1:2$. $EM$ and $FM$ intersect $DB$ at $G$ and $H$ respectively. If the areas of the rectangle $ABCD$ and the triangle $GHM$ are $96$ and $S$ respectively, find the value of $S$.


So the area of $\triangle EFM$ is $32$, but how do I proceed to find the area of $GHM$? Hints are welcomed. Thanks in advance.
 A: We can coordinatize the system as follows: Let $D$ be the origin, $A$ on $y$-axis, $C$ on $x$-axis. Let $AB=a$, $BC=b$. Then $ab=96$.

Using the two-point form for equation for a line (and intersecting lines to get G and H) we get
$$G\equiv \left(\frac{3a}{8}, \frac{3b}{8}\right)$$
$$H\equiv \left(\frac{3a}{4}, \frac{3b}{4}\right)$$
$$M\equiv \left(\frac{a}{2}, 0\right)$$
Using formula for area of a triangle,
$$ar(\Delta GMH)=\frac{9ab}{32}=\frac{9\times 96}{32}=27$$
$$\therefore S=27$$
A: 
Here is a geometric solution to find $S=9$.
Set  $AE=FB=x,EN=NF=2x,DM=MC=3x$.
Draw HK||AB.
$\triangle AHM \sim \triangle BHF$ ($\alpha 1 = \alpha 2,\theta 1= \theta 2$. )
\begin{align}\frac{FH}{HM}=\frac{FB}{MD}=\frac{1}{3}, \frac{MH}{MF}=\frac{MK}{KE}=\frac{3}{4}\end{align}
$\triangle DGM \sim \triangle HGK$ ($\alpha 2 = \alpha 3,\beta 1= \beta 2$. )
\begin{align}\frac{KG}{GM}=\frac{KH}{DM}=1\end{align}
$S_{\triangle HGM}$ and $S_{\triangle HGK}$ have same area.
\begin{align}\frac{S_{\triangle HGM}}{S_{\triangle HKM}}=\frac{1}{2},
\frac{S_{\triangle HKM}}{S_{\triangle MFE}}=(\frac{3}{4})^2, \frac{S_{\triangle MFE}}{S_{\square ABCD} }=\frac{1}{3}\end{align}
Multiply the above 3 equations on both sides,
\begin{align} \frac{S_{\triangle HGM}}{S_{\square ABCD} }=\frac{3}{32}\end{align}
\begin{align} S_{\triangle HGM} =\frac{3}{32}S_{\square ABCD}=\frac{3}{32}\cdot 96=9\end{align}
