Tedious, but I think the most straightforward way is to set up a coordinate system so that B is at the origin and C is on the x-axis, at (c, 0). You can take A to be at (a, b) and then find the coordinates of the other points.
Put A at (0,0), B at (0,3), C at (4,0).
According to the problem statement, D is at (3,0), E is at (0,2) and G (midpoint of CE) is at (2,1).
x - 2y = 0 defines the line AG.
3x + 4y = 12 defines the line BC.
Solve for the intersection and you get F at (12/5,6/5) which is 2/5 of the way from C to B.
So AF and CE intersect at G for the 3-4-5 triangle with A at the origin.
Stretching the axes and varying the angle between them preserves the proportional placement of the points and demonstrates the relation for any triangle.