Find the area of quadrilateral $ABCD$ using the hexagon properties. Let $ABCDEF$ be a convex hexagon with the following properties.
(a) $AC$ and $AE$ trisect $\angle BAF$.
(b) $BE || CD$ and $CF ||DE$.
(c) $AB = 2AC = 4AE = 8AF$.
Suppose that quadrilaterals $ACDE$ and $ADEF$ have area $2014$ and $1400$, respectively. Find the area of quadrilateral $ABCD$.
The diagram was hard to make in geogebra. And the diagram I have attached might not be true for general case.

Not much progress, I got that triangles $AFE, AEC$ and $ACB$ are similar to one another, each being twice as large as the preceding one.
 A: To calculate areas, first we prove that $AD$ , $CF$ and $EB$ are concurrent. And to do that, first we state and prove a lemma about an exciting property of trapezoids.
Lemma: The line passing through the intersection of diagonals and the intersection of extensions of legs of a trapezoid, passes through midpoints of the bases of the trapezoid.
In the figure below, legs $EX$ and $CY$ of the trapezoid $XECY$ meet at $A$ and diagonals of the trapezoid intersect at $Z$ . $AZ$ intersects base $XY$ at $M$ and base $EC$ at $N$ . We show that $XM=YM$ and $EN=CN$ .

Triangles $XEC$ and $YEC$ share the same base ($EC$) and the same hight (the distance between the parallel bases), so they have equal areas: $ S_{XEC} = S_{YEC} $ . We can write this equality as:
$$ S_{ZXE} + S_{ZEC} = S_{ZYC} + S_{ZEC} $$
Therefore:
$$ S_{ZXE} = S_{ZYC} $$
Drawing $ZS$ and $ZT$ perpendicular to $EX$ and $CY$ , respectively, we have:
$$ ZS.EX = ZT.CY$$
$$ \frac{ZS}{ZT} = \frac{CY}{EX} $$
Now, noting that $XY$ and $EC$ are parallel and extensions of $EX$ and $CY$ meet at $A$ , we have:
$$ \frac{CY}{EX} = \frac{YA}{XA} $$
Therefore:
$$ \frac{ZS}{ZT} = \frac{YA}{XA} $$
$$ ZS.XA = ZT.YA $$
$$ S_{AZX} = S_{AZY} $$
Or, drawing $XH$ and $YK$ perpendicular to $AZ$ , we can write:
$$ XH.AZ = YK.AZ $$
$$ XH = YK $$
As a result, the right triangles $MHX$ and $MKY$ are equal. Therefore
$$ XM = YM $$
And it follows easily that $EN = CN$ .
That proves our lemma.
Next, we get back to our problem and show that $AD$ , $CF$ and $EB$ are concurrent.

First, note that the triangles $AFE$ , $AEC$ and $ACB$ are similar because they have equal angles at $A$ and equal ratios of sides:
$$ \widehat{FAE} = \widehat{EAC} = \widehat{CAB} $$
$$ \frac{AF}{AE} = \frac{AE}{AC} = \frac{AC}{AB} = \frac{1}{2} $$
These similarities have an important result, which we will used later on:
$$ S_{ACB} = 4S_{AEC} = 16S_{AFE} \qquad(1) $$
Triangles $AFE$ and $AEC$ are similar and they are placed such that they share the edge $AE$ , constructing the quadrilateral $AFEC$.
Similarly, triangles $AEC$ and $ACB$ are similar and they are placed such that they share the edge $AC$ , constructing the quadrilateral $AECB$.
The quadrilaterals $AFEC$ and $AECB$ are similar. Now, consider their respective diagonals, $FC$ and $EB$ , which intersect $AE$ and $AC$ at $X$ and $Y$ , respectively. Because of the similarity of the shapes it is easy to show that the triangles $AFX$ and $AEY$ are similar. As a result:
$$ \frac{AX}{AY} = \frac{AF}{AE} = \frac{1}{2} $$
Recalling that $\frac{AE}{AC}$ is also $\frac{1}{2}$ , we conclude that $XY$ is parallel to $EC$ .
We can now use the lemma we proved at the beginning: $XECY$ is a trapezoid. Its parallel bases are $XY$ and $EC$ . Its legs $EX$ and $CY$ meet at $A$ , and its diagonals intersect at $Z$ . $AZ$ intersects $EC$ at $N$ . We now know that $EN=CN$ .
Next, we consider that point D is at the intersection of two lines drawn from $E$ and $C$ , and parallel to $EB$ and $FC$ , respectively. This means that the quadrilateral $CZED$ , which is formed by these two pairs of parallel lines, is a parallelogram.
Now, we know that diagonals of a parallelogram bisect each other. We have also shown that the extension of $AZ$ bisects $EC$. Therefore, diagonal $DZ$ of the parallelogram should necessarily overlap $AN$ , as they both pass through points $N$ and $Z$ . This concludes our proof of the claim that $AD$ passes through $Z$ , the intersection of $CF$ and $EB$ .
We are now able to proceed with the calculation of the area of the quadrilateral $ABCD$ :
$$ S_{ABCD} = S_{ACB} + S_{ANC} + S_{NDC} $$
The area of $ANC$ is half of the area of $AEC$ , because $NC$ is half of $EC$. The areas of $AFE$ , $AEC$ and $ACB$ are related as we showed in equation (1). The area of $NDC$ is a quarter of the area of the parallelogram $CZED$. Therefore, if we have the areas of $AFE$ and the parallelogram $CZED$, we will be able to calculate the area of the shape as
$$ S_{ABCD} = (16+2)S_{AFE} + \frac{1}{4}S_{CZED} \qquad(2) $$
Note that we have two unknowns ($S_{AFE}$ and $S_{CZED}$) and we are given two pieces of information in the problem: areas of the quadrilaterals $ACDE$ and $ADEF$ . This allows us to build a system of two equations with two unknowns:
$$ S_{ACDE} = S_{AEC} + S_{EDC} = 4S_{AFE} + \frac{1}{2}S_{CZED}$$
$$ S_{ADEF} = S_{AFE} + S_{AEN} + S_{EDN} = 3S_{AFE} + \frac{1}{4}S_{CZED} $$
Note that because $N$ is the midpoint of $EC$ , we have $S_{AEN} = \frac{1}{2}S_{AEC}$ . The interested reader can solve the above system of equations for $S_{AFE}$ and $S_{CZED}$ and plug the results into equation (2) to calculate the requested area.
