Quadrilateral geometry problem, couldn't solve it. So, I got this question a little while ago and couldn't see how to solve it. The problem follows as such: "In the following figure, G is the midpoint of CD and I is the midpoint of GE. BE:EA = 4:1 and CF:FB = 2:5. Find DH:HA."

Any ideas? Am I forgetting some trick of geometry that's required to solve this? I got a ratio of 8:5 using a geometry program, but cannot see how to prove it analytically.
 A: I'm going to use a property of linear parameterization (without proving anything here) which would allow us to deform the figure so that the lines GE and DA are parallel, CD and BA are parallel, and line CB is inclined so that the distance from B to E is 4 times that from E to A.  I will add a line parallel to GE and DA which meets line BA at point K.  I will call L the distance from C to K , G to E , and D to A  .  The proportions of the lengths BK : KE : EA  are  3 : 1 : 1  .

We will only be using proportions here, which are preserved under the kind of deformation we are making.  Since point F is $ \ \frac{2}{7} \ $ of the way from C to B , the perpendicular projections down to line GE (point F'') and that to line DA (point F') are at $ \ \frac{2}{7} L \ $ .  The distance of point F above line CK is  $ \ \frac{2}{7} \ $ of the distance from B to K , or  $ \ \frac{2}{7} \ \cdot \ 3 \ = \ \frac{6}{7} \ . $
We are given that point I lies at the midpoint of GE, or at $ \ \frac{1}{2}L \ ; $ the distance from F'' to I is then  $ \ \frac{1}{2}L \ - \ \frac{2}{7}L \ = \ \frac{3}{14}L \ . $  Triangle FIF'' is similar to triangle FHF'  .  The altitude FF'' is  $ \ \frac{6}{7} \ + \ 1 \ = \ \frac{13}{7} \ , $ and the altitude FF' is  $ \ \frac{6}{7} \ + \ 2 \ = \ \frac{20}{7} \ . $  We can thus establish the proportionality
$$ \ \frac{x}{\frac{20}{7}} \ = \   \frac{\frac{3}{14}L}{\frac{13}{7}} \ , $$
with  $ \ x \ $ being the distance from  F' to H  .  
We conclude that
$$ x \ = \ \frac{20}{7} \ \cdot \ \frac{7}{13} \ \cdot \ \frac{3}{14}L \ = \   \frac{30}{91}L \ .$$
Since point F' lies at  $ \ \frac{2}{7}L \ $ , point H lies at $ \ \frac{2}{7}L \ + \ \frac{30}{91}L \ = \ \frac{26 + 30}{91}L \ = \ \frac{56}{91}L \ . $  Hence, the distance from H to A is  $ \ L \ - \ \frac{56}{91}L \ = \ \frac{91 - 56}{91}L \ = \ \frac{35}{91}L \ , $ which is to say that the lengths DH and HA lie in the proportion of 56:35 , or 8:5 .
[I had a variant discussion involving similar triangle HIH', then found I didn't need it (but it will corroborate the result found here).]
A: This is a coordinate geometry approach:
Identify the line segment $DA$ with line segment $(0, 0)$ to $(1, 0)$ in the Cartesian plane. Write the vector $D$ to $C = (a, b)$, so that the coordinate of $C$ is at $(a, b)$; write the vector $C$ to $B = (c, d)$ so that the coordinate of $B$ is $(a+c, b+d)$. Using these information allows you to find coordinates of $F$, $I$, $E$, and finally $H$. For instance, $F$ should have coordinate $\left(\frac{a+2c}{7}, \frac{b+2d}{7}\right)$.
For your reference, E has coordinate $\left(1+\frac{a+c-1}{5}, \frac{b+d}{5}\right)$, and I has coordinate $\left(\frac{a}{4}+\frac{1}{2}+\frac{a+c-1}{10}, \frac{b}{4}+\frac{b+d}{10}\right)$.
Set up a line equation through $FI$, and find the $x$-intercept of that line will reveal that $H$ has a coordinate of $\left(\frac{8}{13}, 0\right)$. Meaning that the ratio of $DH:HA = 8:5$.
[Note: the algebra to solve for the x-intercept  is rather nasty, I admit I used WolframAlpha for that: Click for WolframAlpha's work.]
