quadrilateral inside a parallelogram 
the diagonals of the parallelogram creates four equal trianges, so △ ABD and △ BOC together becomes 3/4 s.  But stuck here, how do I find area of △DEM and △MFC, so as to solve to find out the area of MEOF??
 A: Let $DM=xDC$.
Thus, $$S_{\Delta ADM}=xS_{\Delta ADC}=\frac{1}{2}xS$$ and
$$\frac{S_{\Delta ADE}}{\frac{1}{2}xS}=\frac{AE}{AM}=\frac{AE}{AE+EM}=\frac{1}{1+\frac{EM}{AE}}=\frac{1}{1+\frac{DM}{AB}}=\frac{1}{1+x}.$$
Hence, $$S_{\Delta ADE}=\frac{\frac{1}{2}xS}{1+x}.$$
In another hand, 
$$S_{\Delta BCM}=(1-x)S_{\Delta BDC}=\frac{1}{2}(1-x)S$$ and
$$\frac{S_{\Delta BCF}}{\frac{1}{2}(1-x)S}=\frac{BF}{BM}=\frac{BF}{BF+FM}=\frac{1}{1+\frac{FM}{BF}}=\frac{1}{1+\frac{CM}{AB}}=\frac{1}{2-x}.$$
Hence, $$S_{\Delta ADE}=\frac{\frac{1}{2}(1-x)S}{2-x}.$$
Id est,
$$\frac{\frac{1}{2}xS}{1+x}+\frac{\frac{1}{2}(1-x)S}{2-x}=\frac{1}{3}S,$$
which gives $$x=\frac{1}{2},$$
$$S_{\Delta DEM}=S_{\Delta MFC}=\frac{1}{12}S$$ and
$$S_{EMFO}=\frac{1}{4}S-2\cdot\frac{1}{12}S=\frac{1}{12}S.$$
A: Notice that the area of △AMB is equal to the sum of the areas of △DMA and △BMC, or $\frac{1}{2}S$. And since we're given △AED and △BFC together have an area of $\frac{1}{3}S$, we can deduce that the combined area of △DME and △MFC is $\frac{1}{6}S$. But the area of △DOC is $\frac{1}{4}S$, hence we're left with $\frac{1}{12}S$ for the quadrilateral.
