Pentagon Geometry 

$ABCDG$ is a pentagon, such that $\overline{AB} \parallel \overline{GD}$ and $\overline{AB}=2\overline{GD}$. Also, $\overline{AG} \parallel \overline{BC}$ and $\overline{AG}=3\overline{BC}$.
We need to find ratio of areas of $\triangle BEC$ and $\triangle DFG$.

Using vector algebra, I was somehow able to calculate the ratio. My calculation suggests that the ratio is $\dfrac{6}{21}$. I am not sure of this though.
How do we find this ratio without resorting to the use of vector algebra? I am not able to figure out the suitable construction required to solve this problem.
Please help me out.
Thank you.
 A: If you accept without proof that the given data determine this ratio uniquely we can set up the figure in such a way that the areas of the two triangles can be calculated easily. 
Choose $$A=(0,0), \quad B=(2,0),\quad C=(2,1),\quad D=(1,3),\quad G=(0,3)\ .$$
Intersecting the  diagonals using analytic geometry then gives
$$E=\left({12\over7},{6\over7}\right), \quad F=\left({2\over3},2\right)\ .$$
The areas of the two triangles can now immediately be read off:
$$|\triangle(BEC)|={1\over2}\cdot 1\cdot{2\over7}={1\over7},\quad |\triangle(DFG)|={1\over2}\cdot 1\cdot1={1\over2}\ .$$
The required ratio therefore comes to ${\displaystyle{2\over7}}$, as you have found out yourself using vector algebra.
A: 
Let $H$ complete parallelogram $\square ABHG$. Also, let $J$ and $K$ complete "double-size" parallelogram $\square ABKJ$ as shown; note that diagonal $BJ$ contains segment $BE$.
Since $GD\parallel AB$, we have $\triangle GDF \sim BAF$. (Why?) Since $|GD|= \frac12|AB|$, we have $|DF| = \frac12|FA| = \frac13|DA|$; thus, the area of $\triangle GDF$ (with "base" $DF$) is one-third the area of $\triangle GDA$ (with "base" $DA$). Moreover, four copies of $\triangle GDA$ fill $\square ABHG$. (Why?) Thus:
$$|\triangle GDF| = \frac13|\triangle GDA|=\frac13\cdot\frac14|\square ABHG|$$
Likewise, since $BC\parallel AJ$ and $|BC| = \frac16|AJ|$, and since six copies of $\triangle BCA$ fill $\square ABHG$, we have
$$|\triangle BCE| = \frac1{6+1}|\triangle BCA| = \frac17\cdot\frac16|\square ABHG|$$
Therefore,
$$\frac{|\triangle BCE|}{|\triangle GDF|} = \frac{1/42}{1/12} = \frac{12}{42} = \frac{6}{21} = \frac{2}{7}$$
A: If I remember right, any affine transformation preserves the ratio of distances and areas. If so, you can re-set your problem in an easy configuration, where $GD=BC=1$ and $AB\perp AG$, as in the picture below (sorry for the messy sketch).
You need to find the areas of the two triangles in this setting, which I guess is easier. Then all you need is an affinity to get the "simple" pentagon to the general one of your choice.

A: Several observations:


*

*$AG$ is parallel to $BC$.

*$AB$ is parallel to $DG$. 


This means that extending $BC$ and $DG$ gives us a rectangle $AGXB$. (We let $X$ be the point where $BC$ and $DG$ meet, when sufficiently extended.) This is the key observation.
This tells us that the line $BG$ bisects the rectangle into equal-area triangles. The bottom triangle is $BXG$. 
It follows from the square-extension that the line $DG$ has length half of the line $XG$. 
Consequently, the line $BD$ bisects $BXG$ into two equal-area triangles: $BDG$ and $BDX$. 
There is an unmarked point on the line $AC$ that intersects $BG$. We call that point $Y$. Note that because of the bisection, the area of $BEC$ is just half of the area of $BYC$. 
Now we ask what the ratio between triangles $BYC$ and $DFG$ is. Then we will just divide that by two to find the desired result.
I'm currently not in a position to complete this answer -- the way to do it is to work out the areas of triangles $ABC$ and $ADG$, and from those find $BYC$ and $DFG$ by subtracting $AYB$ and $AFG$. Some use of the Pythagorean Theorem or Trigonometric rules may be necessary.
I'll complete this within 48 hours from now. (Or if someone wants to jump in and edit, please do so.)
