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?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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]