#1**+10 **

Note that triangle FBE ≈ triangle ABC

And similar triangles are similar in all respects

And since BE/EC = 1/2, then BE/BC = [ BE] / [ BE + EC] = 1 / [1 +2] = 1/3

Thus, the base of triangle FBE = 1/3 the base of triangle ABC

So......the height of triangle FBE will also be 1/3 of that of triangle ABC

This means that the area of traingle FBE = (1/3)^2 area of triangle ABC = 1/9 the area of triangle ABC

And triangle DEC ≈ triangle ABC

And since DC = 2/3 of AC, by similar reasoning as above......each of its dimensions will be 2/3 of those of triangle ABC

So.......the area of triangle DEC = (2/3)^2 the area of triangle ABC = 4/9 the area of triangle ABC

Thus.....area of triangle ABC - area of triangle FBE - area of triangle DEC = the area of AFED

So

ABC - (1/9) ABC - (4/9) ABC = ABC - (5/9)ABC = (4/9)ABC

Thus, AFED is (4/9) the area of triangle ABC

CPhill
Jan 12, 2016

#1**+10 **

Best Answer

Note that triangle FBE ≈ triangle ABC

And similar triangles are similar in all respects

And since BE/EC = 1/2, then BE/BC = [ BE] / [ BE + EC] = 1 / [1 +2] = 1/3

Thus, the base of triangle FBE = 1/3 the base of triangle ABC

So......the height of triangle FBE will also be 1/3 of that of triangle ABC

This means that the area of traingle FBE = (1/3)^2 area of triangle ABC = 1/9 the area of triangle ABC

And triangle DEC ≈ triangle ABC

And since DC = 2/3 of AC, by similar reasoning as above......each of its dimensions will be 2/3 of those of triangle ABC

So.......the area of triangle DEC = (2/3)^2 the area of triangle ABC = 4/9 the area of triangle ABC

Thus.....area of triangle ABC - area of triangle FBE - area of triangle DEC = the area of AFED

So

ABC - (1/9) ABC - (4/9) ABC = ABC - (5/9)ABC = (4/9)ABC

Thus, AFED is (4/9) the area of triangle ABC

CPhill
Jan 12, 2016