IMO Hk TST

Suppose $AD$, $BE$, $CF$ meet at $P$. Since $P$ is the midpoint of $AD$, $BE$, $CF$, all of $ABDE$, $BCEF$ and $CDFA$ are parallelograms. Let $A'$ be the point such that $ACA'E$ is a parallelogram. Since $EA' = AC = FD$ and $EA' \parallel AC \parallel FD$, $FDA'E$ is also a parallelogram. Thus, we have $DA' = FE$. Together with $CD = AF$ and $CA' = AE$, we find that $\triangle DCA' \cong \triangle FAE$. By symmetry, we also have $\triangle DEA' \cong \triangle BAC$. It follows that
$$\begin{align*} [ABCDEF] &= [ACE] + [ABC] + [CDE] + [EFA] \\ &= [ACE] + [EDA'] + [CDE] + [A'DC] \\ &= [ACA'E] = 2[ACE]. \end{align*}$$