The 3 sides of triangle ABC are each divided in a ratio of 2 to 1. In other words, on side AB, AC'/AB = 1/3, or AC'/C'B = 1/2. This enables one to compute the barycentric coordinates x, y, z of any of the points E, F, G. In turn, it is possible to compute the ratios of the areas of subtriangles to the area of ABC.
F is the intersection of BB' and CC'. Thus we have 1/2 = AC'/C'B = y/x and 1/2 = CB'/B'A = x/z. Thus x = 2y and also z = 2x. Thus if y = 1 (an arbitrary choice), then x = 2 and z = 4. Since the sum x+y+z i supposed to be 1, we divide by 7 = 1+2+4 to get the true value x = 2/7, y = 1/7, z = 4/7.
This means that the areas of the 3 triangles divided by area ABC are equal to the same numbers 2/7, 1/7, 4/7 as shown in the figure, no matter where A, B and C are dragged.
Note: By Ceva, if the line AF intersects BC at Q, then BQ/QC = z/y = 4/1 = 4.
The same computations give barycentric coordinates for E and G (same numbers, different order). For example, the barycentric coordinates of D are (x,y,z) = (4/7,2/7,1/7).
Taking the area of ABC to be 1, the sum of the areas of triangles DAB + EBC + FCA = 3(2/7) = 6/7. Thus the area of DEF = 1/7.
James King, 1/15/2003