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I'm stuck on this bonus problem on a take home test in my math class:

Let G be a point in the interior of the triangle ABC, and let D, E, and F be points on sides BC, AC, AB, respectively, such that AD, BE, and CF all intersect at G. Given that triangles AFG, AGE, and BDG all have the same area, show that D,E and F are the midpoints of the sides BC,AC, and AB, respectively.

Any ideas?

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Let $[AFG] = [AGE] = [BDG] = x$, $[GDC] = y$, $[GEC] = z$, and $[BFG] = w$.

Then, because $\frac{BD}{DC} = \frac{x}{y} = \frac{2x+w}{x+y+z}$, and $\frac{AE}{EC} = \frac{x}{z} =\frac{2x+w}{x+y+z}$,$\frac{x}{y} = \frac{x}{z}$, so $y = z$.

Now, $\frac{AF}{FB} = \frac{x}{w} = \frac{2x+y}{w+x+y}$. Therefore, $x^2 + xy = wx+wy \implies x(x+y) = w(x+y) \implies x = w$.

Also, $\frac{BD}{DC} = \frac{x}{y} = \frac{3x}{x+2y}$, we can find that $x^2 + 2xy = 3xy \implies x^2 = xy \implies x = y$.

To finish, $\frac{AE}{EC} = \frac{CD}{DB} = \frac{BF}{FA} = \frac{x}{x} = 1$, so we are done.

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