The above is the famous 1/7-th triangle (constructed with C.a.R.). To get it, divide the sides 1:2, and connect to the opposite corners. The area of the green triangle is 1/7 of the area of the full triangle. This result and some discussions and proofs can be found in the net easily. How to prove it?
The geometry proof is sketched above. Red, blue and brown lines are parallel. However, you will have a hard time inserting the details. First of all, it is not obvious that the intersections of the brown lines are on the blue lines. If you have that, it is easy to see that the the green triangle is 1/16 of the triangle formed by the blue lines. However, you will still have to prove that the gray triangle is a fixed proportion of the blue one.
In this case, analytic geometry is very handy. For a start, you observe that the problem is independed of affine distortions, since the proportions of distances, and the proportions of areas are perserved by affine mappings. So you can prove the result for any specific triangle of your choice, which is not hard to do.
You can use trilinear coordinates. You find that the corners of the green triangle have coordinates (4,2,1)/7, (2,1,4)/7, and (1,4,2)/7. How? Note, the the divisions on the side obey D=1/3A+2/3B, and that one of the corners of the green triangle has to be P=xD+(1-x)C. The same corner can be written as a convex combination P=yE+(1-y)A, where E=1/3B+2/3C. Comparing the coefficients, you find x and y, and finally the trilinear coordinates of P.
Using the determinant formula for areas, you get the result. The details involve some facts about determinants, but are not really difficult to do.
So the analytic proof is certainly easier here. Of course, the result is wrong in non-Euclidean geometry.