I found out this configuration while fooling around with Geogebra. I tried to solve it on my way back home from work but apparently it's not that easy. Is it a well-known theorem in elementary geometry?
Let $M$ is an arbitrary point inside the triangle $ABC$. $AM, BM, CM$ intersects $BC, CA, AB$ at $D, E, F$ respectively. The line passing through $A$ and parallel to $BM$ intersects $CM$ at $H$. Let $K$ on $AC$ such that $DK \parallel BM$. $DH$ intersect $AB$ at $I$. The parallel line to $CM$ passed through $A$ intersects $BE$ at $L$. $DL$ intersects $AC$ at $N$, $O\in AB$ such that $DO \parallel CM$. Let $P, Q$ be the intersections of $HL$ and $AB, AC$ respectively. $EQ$ intersects $PF$ at $R$; $FN$ intersects $IE$ at $S$. Let $T$ be the midpoint of $BC$. Prove that:
i- $K, I, M$; $O, M, N$ are colinear.
ii- $AH, EF, KI$; $AL, EF, ON$ are concurrent.
iii- $A, R, S, T$ are colinear.
