Indija TS 2008

Draw a line through $N$ which is parallel to $ED$. This passes through the midpoints $P$ of $DF$ and $Q$ of $FE$. Similarly the line through $M$ parallel to $FD$ passes through the mid-point $Q$ of $FE$ and $R$ of $ED$; the line through $L$ parallel to $FE$ passes through the mid-point $P$ of $FD$ and $R$ of $ED$. We thus obtain the medial triangle $PRQ$ of $ABC$. Since $AF=AE$, the line $AQ$ is also the bisector of $\angle A$. Similarly $BP$ bisects $\angle B$ and $CR$ bisects $\angle C$.

Thus $AQ$, $BP$, $CR$ concur at $I$, the in-centre of $ABC$. Now Desargues' theorem is applicable to the triangles $ABC$ and $QPR$. It follows that $PR\cap BC$, $RQ\cap CA$ and $QP\cap AB$ are collinear. Thus $L$, $M$ and $N$ are collinear.