My solution : Let $ I $ be the incenter of $ \triangle ABC $ (i.e. $ I $ is the center of $ \omega $) . Let $ E \equiv \odot (K) \cap BC $ and $ F $ be the antipode of $ E $ in $ \odot (K) $ . From homothety with center $ A $ that maps $ \omega \mapsto \odot (K) $ we get $ A, D, F $ are collinear , so combine $ MD=ME \Longrightarrow AD $ passes through the reflection $ N' $ of $ N $ in the perpendicular bisector of $ BC $ . From $ ID=IL, NK=NE $ and $ \angle EKN=\angle EFD=\angle IDL \Longrightarrow \triangle ILD \sim \triangle NKE $ , so combine $ \triangle ICD \sim \triangle CKE \Longrightarrow DL \cdot DN'=DL \cdot NE=ID \cdot KE=CD \cdot CE=CD \cdot BD $ , hence $ B, C, N', L $ are concyclic $ \Longrightarrow B, C, N, N', L $ are concyclic ($\because BCNN' $ is an isosceles trapezoid) . Q.E.D

It's MEMO 2014: http://www.artofproblemsolving.com/community/c6h607047p3607088

Let $H$ be the projection of $K$ onto $BC$ and let the tangent to $\omega$ at $L$ cut $BC$ at $Z.$ Denote $P \equiv KM \cap ZL$ and let $Q$ be the midpoint of $\overline{MP}.$ Let $\mathcal{H}_1$ be the homothety that sends the $\omega$ to the $A$-excircle and let $\mathcal{H}_2$ be the homothety with center $H$ and ratio $\tfrac{1}{2}.$ Since $\mathcal{H}_2 \circ \mathcal{H}_1 \circ D = K$ and $\mathcal{H}_2 \circ D = M$, we obtain $AD \parallel KM.$ Therefore, $\triangle ZDL \sim \triangle ZMP.$ Because $\triangle ZDL$ is isoceles, we have $ZQ \perp QK.$ Therefore, $Z, Q, H, K$ are inscribed in the circle of diameter $\overline{ZK}$, and from Power of a Point we find $MH \cdot MZ = MK \cdot MQ = MN \cdot MP.$ Moreover, if $E, F$ are the points of tangency of $\omega$ with $\overline{AC}, \overline{AB}$, respectively, then quadrilateral $LEDF$ is harmonic because $EE, FF, LD$ are concurrent. Therefore, $Z \in EF$, and because the cevians $AD, BE, CF$ are concurrent, it is well-known that $(Z, D; B, C)$ is harmonic. Since $M$ is the midpoint of $\overline{BC}$, it is well-known that $MB \cdot MC = MD \cdot MZ = MH \cdot MZ = MP \cdot MN.$ Therefore, $B, C, P, N$ are concyclic. Finally, because $(Z, D; B, C)$ is harmonic, it follows that $ZB \cdot ZC = ZD \cdot ZM = ZL \cdot ZP$, implying that $B, C, P, L$ are concyclic. $\square$

Let the incircle $(I)$ be tangent to $AC$ and $AB$ at $E,F$. Let $EF$ meet $BC$ at $G$. Note that $GL$ is tangent to $(I)$, since $DELF$ is harmonic. Let $P$ be the point such that $GDL\sim GMP$. Note that $GDL$ and $GMP$ are isosceles, and furthermore that $MP\parallel AD$ (with both orthogonal to $GI$). Additionally, $KBC\sim DEF$, and $KB\parallel DF, KC\parallel DE$, so $DL$, the $D$-symmedian of $DEF$, is parallel to $KM$, the $M$-median of $KBC$. Therefore $P,M,N,K$ are collinear. Let $Q$ be the midpoint of $MP$; then $\angle IQM=\angle IQK=90$, so $Q$ lies on $(BICK)$. Therefore \[ MP\cdot MN=2\cdot MQ\cdot \frac{1}{2}\cdot MK = MQ\cdot MK=MB\cdot MC \] implying $BPCN$ is cyclic. But $(G,D;B,C)$ is harmonic, so by a well-known property of harmonic divisions, \[ GL\cdot GP=GD\cdot GM = GB\cdot GC \] Thus $BLPC$ is cyclic as well, implying $BLCN$ is cyclic as desired.

2014MEMO，posted before http://www.artofproblemsolving.com/community/c6h607047p3607088

for the geometrically challenged folks.

Let $D'$ be the projection of $K$ onto $\overline{BC}$, and consider the negative inversion at $D$ fixing $\odot(ABC)$. It is not hard to see by projecting $K$ onto $ID$ that the image of the incircle is the perpendicular bisector of $\overline{KD'}$, so the image of $L$, $L'$, is the intersection of $AD$ with this perpendicular bisector. Since it is well-known that $AD\parallel MK$, dilation at $D'$ with ratio $2$ yields that $BL'NC$ is an isosceles trapezoid, which implies the conclusion.