Find area of CED and BDF.
ABDC is a quadrangle. Sorry for my drawing skills, but CBEF is also quadrangle.
CB = EF
AC = 2
AB = 4
We get that CB is . Now I am a bit confused, as I can't move from this on. What should I do next?
Are all angles in the quadrilateral BCEF right angles? (And how about the angles in ABDC for that matter?)
One idea is to find the area of the quadrilateral BCEF (which I hope is a rectangle), and then subtract the area of the triangle BCD. This gives you the sum of the areas of CED and BDF.
Or did you want the individual areas of the two grey triangles?
You probably know already that the area of a rectangle is base times height. You can take BC as being the base (of BCEF). Then if you draw the line segment from the point D to the diagonal BC (and call the point of intersection for G), you have that DG is the height of BCEF.
See if you can find this height. Notice that you have two right angled triangles BDG and CDG, where the sought-for height is a leg of both.
Actually quite easy...
let u = DF and v = BF ; leaves DE = 2sqrt(5) - u
right triangle BDF: u^2 + v^ 2 = 4 
right triangle CDE: (2sqrt(5) - u)^2 + v^2 = 16 ; simplify:
20 - 4usqrt(5) + u^2 + v^2 = 16 
Substitute  in :
20 - 4usqrt(5) + 4 = 16
4usqrt(5) = 8 ; u = 2 / sqrt(5)
Substitute that in  to get v
OK? I'll let you finish...
Nice "catch", BJ!
So we really don't need point A here; problem is same as:
rectangle CBFE with D on FE such that angle BDC = 90 degrees.
In other words, the rectangle is divided into 3 right triangles.
Smallest case with integer solution: BD = 15, CD = 20; so CB = EF = 25.
Makes CE = BF = 12, DE = 16 and DF = 9.