Serious one? Notice D is the intersection point of the angle bisector of $\angle ABC$ and $\Gamma_1$, then $OD$ is perpendicular to $GK$, thus $GK$ is parallel to $AC$ Finally, $\angle DNM=\angle DCA= \angle DBC$ ,so $BCNM$ is concyclic

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Sourorange wrote: Serious one? Notice D is the intersection point of the angle bisector of $\angle ABC$ and $\Gamma_1$, then $OD$ is perpendicular to $GK$, thus $GK$ is parallel to $AC$ Finally, $\angle DNM=\angle DCA= \angle DBC$ ,so $BCNM$ is concyclic This is incomplete. Review the definition of $M$ in the problem statement. You need to show something about it.

We complex bash with $\Gamma_2$ as the unit circle, and let $O$ be the center of $\Gamma_1$, so that $$d = 0$$$$|e|=|f|=|g|=1$$$$b = \frac{2ef}{e+f}$$$$o = \frac{bg(\overline{b}-\overline{g})}{\overline{b}g-b\overline{g}} = \frac{efg(e+f-2g)}{(e+f)(ef-g^2)}$$$$k = \frac{o}{g\overline{o}} = \frac{ef(e+f-2g)}{2ef-eg-fg}$$$$a = 2f + o - f^2\overline{o} - b = \frac{fg(e+f-2g)}{ef-g^2}$$$$c = 2e + o - e^2\overline{o} - b = \frac{eg(e+f-2g)}{ef-g^2}$$$$m = \frac{kg(e+f) - ef(k+g)}{kg-ef} = \frac{e+f}2$$$$n = \frac{c(k+g)}{c+kg\overline{c}} = \frac{e(ef-g^2)}{2ef-eg-fg}$$Then we have the vectors $$b-c = \frac{e(2ef^2-e^2g-2efg-f^2g+2eg^2)}{(e+f)(ef-g^2)}$$$$m-n = \frac{2ef^2-e^2g-2efg-f^2g+2eg^2}{2(2ef-eg-fg)}$$$$m-b = \frac{(e-f)^2}{2(e+f)}$$$$n-c = \frac{e(e-g)^2(f-g)^2}{(ef-g^2)(2ef-eg-fg)}$$So $$\frac{(b-c)(m-n)}{(m-b)(n-c)} = \frac{(2ef^2-e^2g-2efg-f^2g+2eg^2)^2}{(e-f)^2(e-g)^2(f-g)^2}$$which is real. $\blacksquare$

Sourorange wrote: Serious one? Notice D is the intersection point of the angle bisector of $\angle ABC$ and $\Gamma_1$, then $OD$ is perpendicular to $GK$, thus $GK$ is parallel to $AC$ Finally, $\angle DNM=\angle DCA= \angle DBC$ ,so $BCNM$ is concyclic OK, I didn't mention why $M$ lies on $BD$, but it's quite simple. Assume $EF$ intersects $BD$ at $L$, we need to prove $L=M$ Since $p(L, (ABC))=DL \cdot BL= FL^2=DF^2-DL^2=p(L, (D))$, Thus $L$ lies on the radical axis of $(ABC)$ and $(D)$, which indicates that $L=M$

This was a very interesting yet slightly painful problem. The fact that $N$ can be dealt similarly to $M$ is not immediately obvious. We go about this the hard way. Let $\Gamma$ and $\omega$ denote the circumcircle of $\triangle ABC$ and circle $(D,E)$ respectively. We let $G$ be the second tangency point from $C$ to $\omega$. Then, the following key claim appears. Claim : $M$ lies on $BD$ and $N$ lies on $EG$. Proof : Let $M' = \overline{BD} \cap \overline{EF}$. Note that since $DEBF$ is cyclic, \[M'B \cdot M'D = M'E\cdot M'F\]which implies that $\text{Pow}_{\Gamma} (M') = \text{Pow}_{\omega}(M')$. Thus, $M'$ lies on the radical axis of these two circles, which implies that it lies on $KL$ which implies that $M'=M$ and indeed $M$ must lie on $BD$. Similarly, $N$ lies on $EG$ (by noting that $EDGC$ is cyclic and repeating the same argument). Now, we also make the following observation. Claim : Triangles $DBF$ and $DEM$ and triangles $NED$ and $GCD$ are similar. Proof : We note that \[\measuredangle DEM = \measuredangle DEF = \measuredangle DBF \]which since $BD \perp EF$ (since $BFDE$ is clearly a cyclic kite) implies the desired similarity. Similarly, we can also show that $\triangle NED \sim \triangle GCD$. Now, we are almost there. To finish, simply note that by the above similarities, \[DB \cdot DM = DE \cdot DF = DE \cdot DG = DC \cdot DN\]which implies that indeed $BMNC$ is cyclic and we are done.

Let $\Omega$ be the circumcircle of cyclic quadrilateral $EDFB$. Applying the Radical Axis Theorem on circles $\Gamma_1$, $\Gamma_2$, and $\Omega$ yields that lines $EF$, $GK$, and $DB$ concur at $M$. Since $D$ is the midpoint of $\overset\frown{AC}$ $$AD^2=CD^2\Rightarrow pow(A,\Gamma_2)=pow(C,\Gamma_2)\Rightarrow pow(A,\Gamma_2)-pow(A,\Gamma_1)=pow(C,\Gamma_2)-pow(C,\Gamma_1)$$This implies that $AC\parallel GK$. Let $DB$ and $AC$ intersect at $X$. Applying Reim's Theorem it is sufficient to show that the circumcircle of triangle $\Delta CXB$ is tangent to $CD$. This follows from $$\angle XBC=\angle CAD=\angle XCD$$

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Clearly $BEDF$ is cyclic, so by the Radical Axis Theorem we get that $B,D,M$ are collinear. Now the inversion wrt $\Gamma _2$ maps $\Gamma _1$ to $KG$, so it maps $B$ to $M$ and $C$ to $N$. Hence $BCNM$ is cyclic.