easy to see that NF is tangent line to Г_1 Г_2 and so angle AFK + ANK = KDB + KDC = 180

Actually, I think that $MN$ is always tangent to $\Gamma_1,\Gamma_2$, since $BM=MD,DN=NC$. Anyway, let $DK$ intersect $MN$ at $P$. Then $PM^2=PK\cdot PD=PN^2$, so $P$ is the midpoint of $MN$. Note that $\angle ANK=\angle KDC=\angle KMB$ so $ANKM$ is cyclic. Also, since $\angle APM=\angle DPM$, $ANKM$ is also harmonic, thus $\frac{MK}{KN}=\frac{MA}{AN}$. Note that $\frac{AE}{AC}=\frac{BP}{BC}=\frac{BF}{BA}$, so $\frac{AE}{BF}=\frac{AC}{AB}=\frac{AN}{BM}$, so $\frac{EN}{FM}=\frac{AN}{BM}=\frac{AN}{AM}=\frac{KN}{KM}$. Thus we have $\triangle KEN\sim\triangle KFM$ so $\angle KEN=\angle KFM$ so $KEAF$ is cyclic and we are done.

oneplusone wrote: Actually, I think that $MN$ is always tangent to $\Gamma_1,\Gamma_2$, since $BM=MD,DN=NC$. Anyway, let $DK$ intersect $MN$ at $P$. Then $PM^2=PK\cdot PD=PN^2$, so $P$ is the midpoint of $MN$. Note that $\angle ANK=\angle KDC=\angle KMB$ so $ANKM$ is cyclic. Also, since $\angle APM=\angle DPM$, $ANKM$ is also harmonic, thus $\frac{MK}{KN}=\frac{MA}{AN}$. Note that $\frac{AE}{AC}=\frac{BP}{BC}=\frac{BF}{BA}$, so $\frac{AE}{BF}=\frac{AC}{AB}=\frac{AN}{BM}$, so $\frac{EN}{FM}=\frac{AN}{BM}=\frac{AN}{AM}=\frac{KN}{KM}$. Thus we have $\triangle KEN\sim\triangle KFM$ so $\angle KEN=\angle KFM$ so $KEAF$ is cyclic and we are done. Yes, that condition is alway true, so it could (and should) be removed from the problem. In Chinese version, it is given in the problem as a known fact.

Easy to see that the circumcircle of $AKC$ is tangent to $AB$ and the circumcircle of $AKB$ is tangent to $AC$. Let the centers $O_1$,$O_2$ resp. Then $AEKF$ is on a circle with center $O_3$, separating the segment $O_1O_2$ with the ratio $AF:FB$. We are done

Let $L$ be the midpoint of $BC, AL$ cuts $MN$ at the midpoint $H$ of $MN$. Easy to see that the common chord $DK$ of the two circles goes through $H$. $AP$ cuts $EF$ at $O$, the midpoint of $AP$, clearly $O$ in on $MN$. $\angle DHN = \angle KNC = 180^{\circ} - \angle KDC = 180^{\circ} - \angle BMK = \angle AMK$ (*), so $ANKM$ is cyclic, $\angle MAK = \angle MNK$ (**), hence $\Delta AMK\sim \Delta NHK$. $\frac{HO}{HN} = \frac{LP}{LC} = 1 - \frac{CP}{LC} = 1 - \frac{EP}{NL} = 1 - \frac{AF}{AM} = \frac{MF}{AM}$. Thus point $O$ in $\Delta NHK$ corresponds to point $F$ in $\Delta AMK$. Hence $\angle HOK = \angle MFK$. $MFOK$ is cyclic, thus $\angle AMK = \angle KOE$, but according to (*), $\angle AMK = \angle KNC$, hence $ONEK$ is cyclic and $\angle MNK = \angle FEK$, this together with (**) give that $AFKE$ is cyclic.

$AMKN$ is cyclic, this is a well known fact: for any three points on the sides of a triangle $\triangle ABC$ (points $D, N, M$ for us) the circles $(AMN), (BDM), (CPE)$ are concurrent, with very easy proof: let $K$ be the 2nd common point of $(BDM), (CDN)$, then $\angle DKM=180^\circ-\angle B, \angle DKN=180^\circ-\angle C\implies \angle MKN=180-\angle A$, so our claim has been proved. $MN$ is the common tangent of the 2 circles, hence $\angle KCN=\angle KNM=\angle KAM\ (\ 1 \ )$; also, from $AMDN$ cyclic, $\angle AMK=\angle CNK$, so $\triangle AMK\sim\triangle CNK \ (\ 2\ )$. As $\frac{AF}{AB}=\frac{PC}{BC}=\frac{CE}{CA}\implies\frac{AF}{AM}=\frac{CE}{CN}$, so $F$ and $E$ are homologous points in the 2 similar triangles, consequently $\frac{MF}{NE}=\frac{MK}{KN}\implies\triangle FMK\sim\triangle ENK\iff\angle MKN=\angle FKE$, so $AFKE$ is cyclic. Remark: from the 2 similar triangles at $(2)$ we get $\frac{AM}{NC}=\frac{MK}{NK}$; with $CN=AN\implies AM\cdot NK=AN\cdot MK$, i.e. $AMKN$ is harmonic. Best regards, sunken rock