Find area of a quadrilateral inside a rectangle I am asked to find the area of the shaded portion of the rectangle ABCD, whre $AD=5$, $AB=4$ and $E$ is the midpoint of $BC$.
I know that if I subtract $\triangle ABC+\triangle AFD - \triangle AFG$ to the total area of the rectangle I get the answer but I am having troubles figuring out the area of  $\triangle AFG$. Can someone point me in the right direction or a simpler way to solve it?

 A: I hope you mean that $F$ is a midpoint of $AB$ and we need to calculate $S_{DGEC}$.
Let $AE\cap DC=\{K\}$.
Thus, since $DC=CK=4$, we obtain
$$FG:GD=AG:GK=AF:DK=2:8=1:4.$$
Hence, $$S_{\Delta ADG}=\frac{4}{5}S_{\Delta ADF}=\frac{4}{5}\cdot\frac{4\cdot5}{4}=4.$$
Also, $$S_{\Delta ABE}=\frac{1}{4}S_{ABCD}=5.$$
Id est, $$S_{DGEC}=S_{ABCD}-S_{\Delta ABE}-S_{\Delta AEG}=20-5-4=11.$$
A: 
\begin{equation}
\text{You need to subtract } (\triangle ABE +\triangle AFD -\triangle AFG)\text{ to the total area of the rectangle.}
\end{equation}

let T be the total area of the rectangle:
\begin{equation} T=AB\times BC=20\end{equation}
\begin{equation}
\triangle ABE=(AB\times BE) /2=AB\times BC /4=T/4=5\\
\triangle AFD=(AD\times AF) /2=AD\times AB /4=T/4=5
\end{equation}
\begin{equation}
\triangle AFG=(base\triangle AFG\times height\triangle AFG) /2=AF\times GH/2 \\
\text{where H is the projection of G on AB}\\
GH=BC/5 \text{ (to be proved)}\\
\text{so }\triangle AFG=AB\times BC/20=T/20=1\\
\end{equation}
\begin{equation}
\text{then: } \triangle ABE +\triangle AFD -\triangle AFG=5+5-1=9 \\
\text{and the area of the shaded part is: } T-9=20-9=11
\end{equation}

\begin{equation}
\text{Why } GH=BC/5 \text{ ??}
\end{equation}

let K be the intersection of DF With CB
\begin{equation}
\end{equation}
to under stud the following steps you must help your self by drawing the figure!
\begin{equation}
\text{we have: } CK=2AB \text{ and } CE=AB/2\\
\text{that gives } EK=3AB/2\\
\text{then } GE=3AG/2\\
\text{then } AE=5AG/2\\
\text{then } EB=5GH/2\\
\text{then } CB/2=5GH/2\\
\text{then } CB/5=GH\\
\text{so } GH=CB/5
\end{equation}
A: 
Let the area $I=[ABCD]= 20$ and evaluate
$$\frac{GE}{AE} = \frac{[ADF]}{[ADEF]}=\frac{[ADF]}{[ABCD]-[CDE]-[BEF]}
=\frac{\frac14I}{I-\frac14I-\frac18I} = \frac35
$$
Then $[DEG]=\frac{GE}{AE}[ADE] = \frac35\cdot \frac12I = \frac3{10}I$
and
$$[CDGE] = [DER]+[CDE] = (\frac3{10}+\frac14)I = 11$$
