Japanese Junior Mathematical Olympiad

Let $|PQR|$ denote the area of triangle $PQR$. $AD$, $BE$, $CF$ cross at a point $G$ and $AG : GD = BG : GE = CG : GF = 2 : 1$. Take a point $C'$ on line $GC$ so that $G$ is the midpoint of $CC'$. From $C'G = GC$ and $CG : GF = 2 : 1$, it follows that $C'F = GF$. Since $AF = BF$, $C'F = GF$ and $\angle AFC' = \angle BFG$, triangles $AFC'$ and $BFG$ are congruent and so $AC' = BG$. Now we have $AC' = BG = \frac{8}{3}$, $AG = 2$, $GC' = GC = \frac{10}{3}$, so $AGC'$ is a right triangle with hypotenuse $GC'$. Therefore $|AGC'| = \frac{8}{3}$. By $C'G = GC$ we get $|AGC| = |AGC'| = \frac{8}{3}$, and by $BG : GE = 2 : 1$ we get $|ABC| = 3|AGC| = 8$.