In acute triangle $ABC$ ($\triangle ABC$ is not an isosceles triangle), $I$ is its incentre, and circle $ \omega$ is its inscribed circle. $\odot\omega$ touches $BC, CA, AB$ at $D, E, F$ respectively. $AD$ intersects with $\odot\omega$ at $J$ ($J\neq D$), and the circumcircle of $\triangle BCJ$ intersects $\odot\omega$ at $K$ ($K\neq J$). The circumcircle of $\triangle BFK$ and $\triangle CEK$ meet at $L$ ($L\neq K$). Let $M$ be the midpoint of the major arc $BAC$.Prove that $M, I, L$ are collinear.

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Very beautiful problem! $\textbf{Solution:}$ Let the line $IM$ intersects $(BIC)$ at $S$. We will show that $S$ lie on both $(CEK),(BFK)$. It is well known that $IS$ is isogonal to $IN$ with $N$ is the midpoint of $BC$. To be simpler, we will only show $(CEK)$ passes through $S$. Consider an inversion $\mathcal{I}$ centered at $I$ with radius $ID$ the problem turned into: $\textbf{Problem:}$ Given triangle $\triangle{ABC}.$ $D,E,F$ are midpoints of $BC,CA,AB$ . $Z$ is the midpoint of $EF$. The $A-$symmedian cut $(ABC)$ again at $X$. $(XEF)$ cut $(ABC)$ again at $K$. Prove that: $K$ lie on $(CEZ)$. (Notice that $S$ becomes the midpoint of $B'C'$ where $B',C'$ are midpoints of $DF,DE$ since $BB'C'C$ is cyclic and $IN,IS$ are isogonal) $\textbf{Solution:}$ We need $\angle{CZE} = \angle{CKE} \Leftrightarrow \angle{CZE} = \angle{EKX} - \angle{CKX} \Leftrightarrow \angle{CZE} = \angle{EFX} - \angle{CAX}.$ We will prove $\angle{CZE} = \angle{EFX} - \angle{CAX}$ Since $\triangle{ABX} \overset{+}{\sim} \triangle{ADC}$ and $F,Z$ are midpoints of $AB,AD$ so it is not hard to see $XF,CZ$ intersects at a point $L$ lie on $(ABC)$. So that $\angle{CZE} = \angle{LZF} = \angle{EFX} - \angle{ZLF} = \angle{EFX} - \angle{CLX} = \angle{EFX} - \angle{CAX}$ and we are done. $Q.E.D.$

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We claim that $BICL$ is harmonic, which is well-known to imply $M,I,L$ collinear. Invert w.r.t. the incircle. The inverted problem reads: Inverted problem wrote: Let $ABKC$ be a harmonic quadrilateral, $E$ be the midpoint of $AC$, $F$ be the midpoint of $AB$, $G$ be the midpoint of $EF$, and $X=(ABC) \cap (CEG) \cap (BFG)$. Then $E,F,X,K$ concyclic. As $(AEF)$ and $(ABC)$ are tangent at $A$, by radical center it suffices to show that $XK$ and the tangent at $A$ intersect on line $EF$. Now let $A', K'$ be on $(ABC)$ such that $AA' \parallel KK' \parallel BC$, and let $D$ be the midpoint of $BC$. By Reim's we have $X,G,A'$ collinear. Finally by Pascal on $AAK'KA'X$ we get that $AA \cap KX$ lies on the line through $G$ parallel to $AA'$ and $KK'$, which is just $EF$, hence proved. Remark. The proof also works for any $E \in AC$, $F \in AB$ such that $EF \parallel BC$.