area of quadrilateral! only areas given! Find the 
$$ar(CEF)+ar(FGB) =\;\;?$$
I am really stuck on this ... spend some hours... did not what to solve and how to proceed? Any suggestions ? Hints also work :)

 A: Hint: Try to show that $S(ABCD) =8 \,S(FEH)$
Full Solution:
You have
$$ \vec{FH} = \frac12 \vec{FG} = \frac12 (\vec{BG} - \vec{BF}) = \frac14 (\vec{BA} - \vec{BC})$$
$$ \vec{FE} = \vec{FC} + \vec{CE} = \frac12\vec{BC} + \frac12\vec{CD} = \frac12\vec{BD}$$
so
$$ S(FEH) = \frac12|\vec{FE}\times\vec{FH}| = \frac{1}{16}|\vec{BD}\times(\vec{BA} - \vec{BC})| $$
We also have
$$ S(ABCD) = S(BCD)+S(BDA) = \frac12|\vec{BC}\times\vec{BD}|+ \frac12|\vec{BD}\times\vec{BA}| = \frac12|\vec{BD}\times(\vec{BA}-\vec{BC})|$$
So $$S(ABCD) =8 \,S(FEH) = 64$$
Therefore 
$$ S(CEF) + S(FGB) = S(ABCD) - S(AGHED) - S(HFE) = 64 - 36 -8 = 20$$
A: Analytic geometry to the rescue?
We have $\triangle GHE=\triangle FHE$, so we can rewrite the data as
$$ \triangle EFG=16 \qquad \Box ADEG=28 $$
and forget about $H$ completely.
Next I would select a non-rectangular coordinate system with $G$ at $(0,0)$, $E$ at $(-1,0)$ and $D$ at $(0,-1)$. The coordinates of everything now become simple linear functions of, say, $x_A$ and $y_A$, and solving
$$ \frac{\triangle EFG}{\Box ADEG} = \frac{y_F}{1+x_A} = \frac{16}{28} $$
(where the middle fraction rewrites both areas in units of $\triangle EGC$ which is half the fundamental parallelogram of the coordinate system) gives us a line that $A$ must lie on.
Now we can compute the area $\triangle CEF + \triangle FGB$ for an arbitrary point on that line (in the same ad-hoc units). Hopefully it becomes a constant multiple of $y_F$ which we know to represent the area $16$.
A: Make the labels as indicated on the figure:
$\hspace{3cm}$
Note: 
$$\begin{align}&\text{1) $JE$ is the middle line of $\Delta ACD$ and $JE||AC$.}\\
&\text{2) $FG$ is the middle line of $\Delta ABC$ and $FG||AC$.}\\
&\text{3) 1) and 2) imply $JE||FG$}  \\
&\text{4) similarly, $FE||GJ$, hence $EFGJ$ is a parallelogram, whose area is $32$ (why?)}\\
&\text{5) $S_{ACD}=4S_{DEJ}$, $S_{ABD}=4S_{AGJ}$,$S_{ABC}=4S_{BFG}$,$S_{BCD}=4S_{CEF}$}\\
&\text{6) $S_{ABCD}=\frac12(S_{ACD}+S_{ABD}+S_{ABC}+S_{BCD})=2(\underbrace{S_{DEJ}+S_{AGJ}}_{12}+S_{BFG}+S_{CEF})$}\\
&\text{7) $S_{BFG}+S_{CEF}=S_{ABCD}-44=2(12+S_{BFG}+S_{CEF})-44\Rightarrow S_{BFG}+S_{CEF}=20.$}\\
 \end{align}$$
A: In any quadrangle $Q$ the midpoints of the four edges form a parallelogram $P$ whose area is ${1\over2}{\rm area}(Q)$, and the  four outer triangles make up the other half of ${\rm area}(Q)$. In the case at hand ${1\over4}{\rm area}(P)=8$, so that $P$ as well as the four outer triangles together have area $32$. The quadrangle $AGED$ in the figure has area $36-8=28$. This quadrangle consists of the left two outer triangles and one half of $P$. These triangles together then have area $28-{1\over2}{\rm area}(P)=12$, so that the right two outer triangles together have area $32-12=20$.

