Area of a triangle given an inside triangle and quadrilateral area. Draw triangle ABC.  Place D on AB and E on AC so that DE is parallel to BC, but do not draw segment DE.  Draw segments BE and CD, and let them intersect in F.  If triangle BCF has area 4, and quadrilateral ADFE has area 4, find the area of triangle ABC.  (Solve for XC)
I am so close.  I found the midpoint of BC and connected it A, which cuts directly through triangle BCF and quadrilateral ADFE, cutting those areas in half.  I know the the heights of triangles ECB and BCD are the same, making those areas the same.  I also know from playing with geogebra that the area of the unknown areas is 2 each.  I know the area of the whole triangle is 12 total, but I just don't know how to prove it.  Any help is appreciated and I uploaded a diagram from geogebra.  Thank you!!
geogebra picture
 A: Let us denote by $A[MNP]$ the area of triangle $MNP$ and by $L(MP)$ the length of segment $MP$.
Let $x:=[BDE]$ and $y:=[CEF].$
We are going to establish that 
$$x=y=2 \ \ \text{giving area} \ \ [ABC]=12 \tag{*}.$$
2 steps for the proof of (*):
1) 1st step : Let us establish that $D$ and $E$ are the midpoints of line segments $AB$ and  $AC$ resp.
Proof : let us recall the following result : if two triangles $MNP$ and $MQR$ share a same vertex $M$, and if points $N,P,Q,R$ all lie on the same line, then $A[MNP]/A[MQR]=L(NP)/L(QR)$.
Let us apply this result to triangles whose areas we know :
$A[BAE]/A[BCE]=\begin{cases}=L(AE)/L(CE)\\=(4+x)/(4+y)\end{cases}\tag{1a}$
$A[CDB]/A[CAD]=\begin{cases}=L(DB)/L(AD)\\=(4+x)/(4+y)\end{cases} \tag{1b}$
Equating (1a) and (1b) : $L(AE)/L(CE)=L(DB)/L(AD) \tag{2}$
But the fact that $DE$ and $CB$ are parallel gives (using similarity of triangles) the following proportionality : 
$$L(AE)/L(CE)=L(AD)/L(DB)\tag{3}$$
Using the product of (2) and (3), one concludes that $L(AD)=L(CD)$ and $L(AE)=L(EB)$, therefore establishing that $D$ and $E$ are midpoints of their resp. sides, and, moreover, using (1), that 
$$x=y \tag{4}$$ 
2) 2nd step : Therefore, $F$ being the intersection of medians, is the center of gravity of triangle $ABC$, situated at a 2 thirds of the median issued from $A$. As a consequence, the altitude of triangle $ABC$ is  $3$ times the altitude of triangle $FCB$ ; as both triangles share the same basis $BC$ : 
$$[ABC]=3[FBC] \ \iff \ (x+y+4+4)=3 \times 4 \ \text{with} \ x=y$$
from which one deduces result (*).
A: Let [.] denote areas. Due to DE || BC, we have $[DBC]=[EBC]$, which leads to $[DBF]=[ECF]=x$. Also, let $[DEF]=y$. Since $[ADC]=[BDC]=4+x$, D is the midpoint, which means $[ADE]=[BDE]$, or
$$4-y=x+y\tag{1}$$
Also, from the area ratios,
$$ \frac{[DBF]}{[CBF]}=\frac{[DBE]}{[CBE]} \implies 
\frac{x}{4}=\frac{x+y}{4+x}$$
or,
$$x^2=4y\tag{2}$$
Combine (1) and (2) 
$$x^2+2x-8=0$$
Solve to obtain
$$x=2$$
Thus, $[ABC] = 4+4+2x=12$.
