Triangles and Similarity. ADEN is a square. BMDF is a square such that F lies on AD and M lies on the extension of ED. C is the point of intersection of AD and BE. If the area of triangle CDE is 6 square units, what is the area of triangle ABC?

I tried solving this as :
tria.CDE is similar to tria.BME is similar to tria.CFB.
So, the squares of the ratios of the sides must be equal as :
(CD/BM) = (FC/BM) = (DE/ME) = (BF/ME)
But, this solution is just making me go around in circles with countless equations.   Is there a smarter way of solving this problem?   Please advise.

Let DE = a, DM = b 


*

*Area of $\triangle ABF = \frac{1}{2} \cdot b \cdot (a-b)$

*Area of $\triangle CBF = \frac{1}{2} \cdot b \cdot CF$

*Area of $\triangle CDE = \frac{1}{2} \cdot a \cdot CD$

*$CF + CD = b$ and $CF = \frac{b}{a} CD$. Thus, $CD = \frac{ab}{a+b}$ and $CF = \frac{b^2}{a+b}$
Now, from 1, 2 & 4: area of $\triangle ABC = \frac{1}{2} \cdot (b \cdot (a-b) + b \cdot \frac{b^2}{a+b}) = \frac{1}{2} \cdot b \cdot \frac{a^2}{a+b}$
From 3 & 4: Area of $\triangle CDE = \frac{1}{2} \cdot \frac{a^2}{a+b}$
Hence, Area of $\triangle ABC =$ Area of $\triangle CDE$ = 6
===
Edit: 
A more intuitive way of doing this without much algebra is to notice that if you take away $\triangle CBF$ from $\triangle CDE$ you are left with a trapezoid. Area of trapezoid is the average of the two parallel sides multiplied by the height. Average of the two parallel sides (which are equal in length to CD and CF) is half the length of side DF. The height is the same as the difference in length of sides of the two squares. Thus, area of trapezoid is nothing but area of triangle ABF.
