Or consider AED the base line and then the two triangles have the same height and bases in the ratio 2 to 1.
One way to make the problem easier is to translate point A rightwards and parallel to line BC until it lies directly above the point D. If you simultaneously translate F so that line BF continues to bisect AD, then you will not change the areas of any of the regions. (In other words, treat D as the origin in R[sup]2[/sup] and let BC lie on the x-axis. Then, for some real number r, multiplication by the matrix
[ 1 r ]
[ 0 1]
preserves all areas (because its determinant is 1) and maps the point A to a point directly above the origin.)
So, we may as well assume that AD is perpendicular to BC. It is then easy to see that triangles ABE and BDE have the same area A[sub]1[/sub]. Set x[sub]0[/sub] to be the length of DC and y[sub]0[/sub] to be the length of AD. Then
Area of ABC = 5 cm[sup]2[/sup] = 3/2 x[sub]0[/sub] y[sub]0[/sub], and
Area of ABD = 2 A[sub]1[/sub] = x[sub]0[/sub] y[sub]0[/sub],
so you can solve for A[sub]1[/sub] to find that A[sub]1[/sub] = 5/3 cm[sup]2[/sup], which is part (a).
To solve for part (b), you can write down the equations for the lines containing AC and BE in terms of x[sub]0[/sub] and y[sub]0[/sub] to get the coordinates of their intersection at F. You find that F = (2/5 x[sub]0[/sub], 3/5 y[sub]0[/sub]). Hence, triangle AEF has “base” 1/2 y[sub]0[/sub] and “height” 2/5 x[sub]0[/sub]. Thus, its area A[sub]2[/sub] is 1/10 x[sub]0[/sub] y[sub]0[/sub]. But we saw earlier that x[sub]0[/sub] y[sub]0[/sub] = 2 A[sub]1[/sub] = 10/3 cm[sup]2[/sup], so this gives us A[sub]2[/sub] = 1/3 cm[sup]2[/sup], solving part (b).
I stand corrected. But that is what I meant by some clever argument. I haven’t read the rest of it, but I guess it is correct.
The question is underdetermined in the sense that there are many dissimilar triangles that satisfy the conditions, but these answers are always the same. I made one of them, not entirely at random and sure enough the areas were 5/3 and 1/3. It was done so that the first was 5/3 and the second had quadratic irrationalities and I used a high precision program to calculate the area of the small triangle. So I was getting convinced the answer was right.
I think we got it!
Thanks for your help! I can’t believe I was stumped by triangles.
Absolutely. We’re in complete ignorance about the lengths of the sides, and the size of the angles, and none of the formulae above do anything to address this (nor can). But it turns out that we need surprisingly little information to work out the areas of the small triangles. (Given the area, we can specify some constraints on the sides by working backwards through the semiperimeter formula (Heron’s formula), but that doesn’t get us very close to the actual size of the sides.)
My hat’s off to CJJ* - I reckon myself a pretty fair trigonometrician, but I was a little stuck until I read his excellent solution.
Thx…I too was stumped for a little while. Abandoning attempts to determine congruent lengths/angles was a key step for me (I spent quite a bit of time trying to pointlessly prove the angle at B was bisected by BF, convinced somehow that the two shaded triangles were similar). Once I started manipulating everything in terms of area and the (1/2)bh equation, the problem unravelled (drawing the line FD was also a breakthru).
garysoh
I think we got it? LOL
Omphaloskeptic, CJJ*, Tyrrell McAllister and many others were the major contributors don’t you think?
By the way Malacandra, I was going along a similar path. My great “discovery” was constructing a rectangle around triangle ABC such that its width would be AC and its height would be ‘h’ - the height of triangle ABC. Therefore, the rectangle’s area would be 2 x ABC = 10. Then I went on to Heron’s formula, the 1/2 b x h formula and realized I was not going to find the solution that way and gave up. 
CJJ*, I think a nice way to close in on the finish is this:
BED = 5/3
BEA = 5/3
AEF = x
DEF = x
FDC = y
BFC = 3y
Then we get a couple of easy-to-solve simultaneous equations for x and y:
5/3 + 5/3 + 2x + y = 5
5/3 + x + y = 3y
reducing down to
2x + y = 5/3
2y - x = 5/3
giving y = 1 and x = 1/3.
Yup!
I guess this problem’s been solved to death now. But since I find it aesthetically preferable to solve geometric problems “geometrically”, rather than with algebra and solutions to simultaneous linear equations, here’s my small contribution:
The following is a solution for part two that doesn’t require getting your hands dirty with even the simple algebra of CJJ*'s/Malacandra’s. You don’t even have to assume that the area of a large region is decomposable into the sum of the areas of its subregions. Everything is established directly in terms of distance ratios.
First, we establish the ratio AF/AC, by way of establishing AF/FC. [In the usual overloading of notation, I’ll use YZ to indicate the length of the line segment YZ; in my own idiosyncratic notation, I’ll also use X_YZ to mean the distance of the point X from the line containing YZ.]. Now, note that AF/AC = F_AB/C_AB and FC/AC = F_BC/A_BC. Therefore, AF/FC = (F_AB/F_BC)/(C_AB/A_BC).
Furthermore, note that E_AB = F_AB * BE/BF while E_BC = F_BC * BE/BF. Therefore, E_AB/E_BC = F_AB/F_BC. Furthermore, E_AB = 1/2 D_AB and D_AB = 2/3 C_AB. Finally, E_BC = 1/2 A_BC. Putting this all together, we find that F_AB/F_BC = (1/3 C_AB)/(1/2 A_BC) = 2/3 C_AB/A_BC. That is to say, (F_AB/F_BC)/(C_AB/A_BC) = 2/3, telling us that AF/FC = 2/3, from which we can naturally conclude that AF/AC = 2/5.
Having done the hard work, the rest is easy. Consider the triangle ADC; it has 1/3 the area of ABC [because DC is 1/3 of BC; you can view this as shrinking the base by a factor of 1/3 while holding the height the same or as shrinking the height by a factor of 1/3 while holding the base the same, depending on whether you originally took the perspective of BC or AC as “the base”]. And, in turn, AEC has 1/2 the area of ADC [because AE is 1/2 of AD; same reasoning]. Finally, AEF has 2/5 the area of AEC. Thus, since ABC has area 5 cm^2, we see that AEF has area 2/5 * 1/2 * 1/3 * 5 = 1/3 cm^2. Ta-da!
(I got this solution by starting with Omphaloskeptic’s barycentric approach and reworking it into a more directly geometric intuition, though the resemblances to his solution may now be tortured and lost).