In the triangle ABC, D and E are the midpoints of AC and BC. AE and BD intersect at F. How to show that the area of CDFE equals the area of AFB? In the triangle ABC, D and E are the midpoints of AC and BC. AE and BD intersect at F. Show that the area of CDFE equals the area of AFB.
 A: Since your configuration is affinely equivalent to the special case where the triangle is equilateral it is sufficient to look at this special case. But here the claim is obvious.
A: Consider the following figure:

$AB,DE,$ and the white lines are parallel. The centroid ($F$) cuts the  median ($GF$) in the ratio 2:1. $F$ cuts $GH$ in 2:1 again because of the similarity of $ABF$ and $DFE$. The height of $ABC$ is cut to $6$ equal parts. $AG=2DE$. Parts of the said height are the heights of the triangles involved. It is easy see the required equality.
A: Useful lemma: in triangle XYZ let W be a point on side YZ located portion $p$ of the way from Y to Z (I.e. YW/YZ = $p$). Then area(XYW)= $(p)$ area(XYZ).
Proof: take YZ as the base, then triangles have same height and bases have ratio $p$.
Now wolog area(ABC)=1. By the lemma, area(ABE) = area(ADB) = 1/2.
Since medians intersect 2/3 of the way from vertex to opposite side, then by lemma area(AFB) = (2/3) area(ABE) = 1/3. 
Finally by Venn diagram or inclusion-exclusion,
Area(CDFE) = 1 - ( 1/2 + 1/2 - 1/3) = 1/3 as desired.
A: Key point is:- If two triangles have the same base and the same altitude, then their areas are equal.

Therefore, we have 
a) [$\triangle BAD$] = ([5] + [6]) = ([3] + [4]) = [$\triangle BDC$]; and 
b) [5] = … = [3].
Then,  [6] = ([5] + [6]) – ([5]) = ([3] + [4]) – ([5]) = ([3] + [4]) – [3] = [4]
