How am I supposed to find the area of the shaded quadrilateral? In the diagram (which is not drawn to scale) the small triangles each have the area shown. Find the area of the shaded quadrilateral.

 A: A POSSIBLE PATH
Consider the Figure below and let $x$ be the desired area.



*

*By Menelaus's Theorem on $\triangle BCE$ cut by $AD$ you get
\begin{equation}\frac{\overline{EP}}{\overline{PB}}\cdot\frac{\overline{BD}}{\overline{DC}}\cdot\frac{\overline{AC}}{\overline{AE}}=1.\tag{*}\label{eq1}\end{equation}

*Observe that the ratio $\frac{\overline{EP}}{\overline{PB}} = \frac12$. Why?

*Similarly, you can find $\frac{\overline{BD}}{\overline{DC}}=\frac{18}{7+x}$, and $\frac{\overline{AC}}{\overline{AE}}=\frac{25+x}{21}$.

*This info in \eqref{eq1} will give the equation in $x$ $$7(7+x) = 3(25+x),$$and the final result $x=\frac{13}2$.

A: 
It can be deduced that the area $[FED] = 7\cdot 4/14=2$. 
Furthermore, examine the ratios below,
$$\frac{[FDC]}{[BDC]}=\frac{9}{18}=\frac{\frac{a}{a+b} \frac{d}{c+d}[ABC]  }{ \frac{b}{a+b}[ABC]}  =\frac ab  \frac{1}{\frac cd +1} \tag{1}$$
$$\frac{[FDB]}{[FCB]}=\frac{6}{21}=\frac{\frac{c}{c+d} \frac{b}{a+b} [ABC]  }{ \frac{d}{c+d} [ABC]}  =\frac cd  \frac{1}{\frac ab +1} \tag{2}$$
From (1) and (2), 
$$ \frac cd = \frac 12$$
Then, the area of the quadrilateral is
$$[AFED]= 2+ [AFD] = 2+\frac 12\cdot 9 = \frac {13}{2}$$
A: There is a simple solution obtained by splitting the area to be found into two areas $x$ and $y$ by diagonal $CM$ (see figure).

This diagonal allows to use the fact that $\dfrac{[ACM]}{[MCA']}=\dfrac{[BAM]}{[MAB']}=\dfrac{AM}{MA'}$ (ratio of areas equal to the ratio of bases because of a common height) :
$$\dfrac{7+x}{y}=\dfrac{14}{4}\left(=\dfrac{AM}{MA'}\right)\tag{1}$$
For a similar reason, $\dfrac{[B'CM]}{[MCB]}=\dfrac{[B'AM]}{[MAB]}=\dfrac{BM}{MB'}$ giving :
$$\dfrac{x}{y+4}=\dfrac{7}{14}\left(=\dfrac{BM}{MB'}\right)\tag{2}$$
(1) and (2) give rise to a linear system of 2 equations in the 2 unknowns $x$ and $y$ ; its solutions are :

$$x=\dfrac72 \ \ \ \text{and} \ \ \ y=3, $$
  giving finally total area 
$$x+y=\dfrac{13}{2}.$$

Remark : one can find in book 6 of Euclides : "Triangles sharing the same altitude are between themselves [understand : their resp. areas] as their bases."
A: I tried to build a general formula for the shaded area.
To simplify, assume base area = 1.  We can scale it back later.
Let left and right triangle area be $x, y$. Solving for shaded area, $z$
Using @dfnu setup, with symbols instead of actual numbers:
$${EP \over PB}{BD \over DC}{AC \over AE} = {x \over 1}{1+y \over x+z}{1+x+y+z \over 1+x} = 1$$
$$\large z = {x + y + 2 \over {1 \over x\,y} - 1}$$
Note: the formula implied $x\,y < 1$
For this example, $\large x={7\over14}, y={4\over14}$, we have $\large z={13\over28}$ 
Scaled back with actual base area of 14, we have shaded area = $\large{13\over2}$ 
Lets get the area of ΔABC (still, assume base area = 1):
$$ΔABC = 1+x+y+z = {(1+x)(1+y) \over 1-x\,y}$$
