Let $W'$ be an arbitrary point on the angle bisector of $EDF$, and $X'Y'Z'$ be the anti-cevian triangle of $W'$ wrt $DEF$. We claim that the circumcircles of triangle $ABC,W'NX',Y'MZ'$ meet at a point. Let the second intersection of $(ABC)$ and $(W'NX')$ be $L_1$, $(ABC)$ and $(Y'MZ')$ be $L_2$. Note that $W'\to X'$ is an involution on line $DW$, so $(NX'W')$ passes through a fixed point. Let the midpoint of $W'X'$ be $P$, the line passing through $N$ parallel to $DW$ intersect $(NX'W')$ again at $R$, then $L_1\mapsto R\mapsto P$ is a projective map. Similarly the midpoint of $Y'Z'$ is $Q$, then $L_2\mapsto Q$ is also a projective map. Consider the newton line of $WYXZ$, $PQ$ passes through the midpoint of $EF$, so $L_1\mapsto L_2$ is projective. It suffices to check three points. It is not difficult to verify that when $W'=X'=Y'=Z'=D$, or when $(W'NX')$ or $(Y'MZ')$ becomes a line, $L_1=L_2$, so $L_1$ and $L_2$ are always the same. Take $W'=W$ we are done.

Let $B'',B'$ be the middle points of $BA,BCA$ and define $C',C''$ similarly. Claim: $\odot(B'ZW),\odot(C'WY),\odot(XNW),\odot(ABC)$ meets on a common point. Proof: It's enough to show that $T=(B'ZW)\cap (XNW)\in (ABC)$. Let $I,I_C,I_B$ be the incenter,$C,B$- ecxenters of $ABC$, due to homothety of $DEF,ABC$ (centered at their common barycenter ) we have that $YZ=I_BI_C/2$. Note that $ZY\perp DW\parallel AM\Rightarrow ZY\parallel B'C'$ and $B'C'=I_CI_B/2=YZ$ Similarly we get that $XZNB'$ is a parallelogram and thus $\angle NTB'=\angle NTW+\angle WTB'=180-\angle WXN+180-\angle B'ZW=\angle ZWX=90+\angle FED=90+\angle B=180-\angle B'BN$ and thus $T\in \odot(BAC)$. Claim: $\odot(YMZ),\odot(C''ZX),\odot(B''YZ),\odot(ABC)$ meets on a common point. Proof: It's enough to show that $T'=(MYZ)\cap (B''XY)\in (ABC)$ Note that $B''C''M,XYZ$ are homothetic and equal so $\angle B''T'M=B''T'Y+\angle YT'M=\angle B''XY+\angle YZM=\angle ZYX=\angle B'C'N=\angle B''AM$ and thus $T'\in \odot(ABC)$. Now it remains to show that $(YMZ)\cap (WZB')=U\in (ABC)$. $\angle MUB'=\angle MUZ+\angle ZUB'=\angle MYZ+\angle ZWB'$. Note that $MB'\perp B'N\parallel XZ\perp YW$ and $YW=IIB_B/2=IC''=MB'$ and thus $WB'\parallel YM$ and thus $\angle MYZ+\angle ZWB'=WZY=90-ZYX=90-B'C'N=\angle NMB'$ and hence $U\in \odot(ABC)$ and we are done.

Official Solution: Denote $G, I, I_a, I_b, I_c$ to be the centroid, incenter, $A$-excenter, $B$-excenter, and $C$-excenter of triangle $ABC$ respectively. Let $P$ be the midpoint of major arc $ABC$, and let $T$ be the second intersection of circles $(ABC)$ and $(WNX)$. Then we wish to show that $TYMZ$ is cyclic. We will prove this in $4$ steps. Step 1: $PYXN$ is a parallelogram. Proof: We will prove that $XY\parallel PN$ and $XY=PN$. Then $N, P$ are the midpoints of $I_cI_b$ and $I_cI_a$ respectively. Thus a homothety centered at $G$ with scale factor $-2$, followed by a homothety centered at $I_c$ with scale factor $\frac{1}{2}$, sends $XY$ to $I_aI_b$ to $PN$. Thus $XY\parallel I_aI_b \parallel PN$ with $XY=\frac{1}{2}I_aI_b=PN$. Step 2: $PWZM$ is a parallelogram. Proof: We will prove that $ZW\parallel PM$ and $ZW=PM$. Then $M, P$ are the midpoints of $I_aI$ and $I_aI_c$ respectively. Thus a homothety centered at $G$ with scale factor $-2$, followed by a homothety centered at $I_a$ with scale factor $\frac{1}{2}$, sends $ZW$ to $I_cI$ to $PM$. Thus $ZW\parallel I_cI \parallel PM$ with $ZW=\frac{1}{2}I_cI=PM$. Step 3: $TWPY$ is cyclic Proof: We will use directed angles modulo $2\pi$. The key step is to prove $\angle(NT, TP)=\angle(YW, WX)$. Using results from Step $1$ and Step $2$, we get $$\angle(NT, TP)=\angle(NM, MP)=\angle(NM, WZ)=\pi+\angle(EF, FW)=\angle(EW, WD)=\angle(YW, WX)$$To prove $TWPY$ cyclic, we angle chase that $$\angle(PY, YW)=\angle(NX, YW)=\angle(NX, XW)-\angle(YW, WX)$$$$=\angle (NT, TW)-\angle(NT, TP)=\angle(PT, TW)$$ Step 4: $TYMZ$ is cyclic Proof: We will use directed angles modulo $2\pi$. Using the previous step and the fact that $WX\perp YZ$, we get $$\angle(PT, TM)=\angle(PN,NM)=\pi- \angle(NM, MP)=\pi-\angle(YW, WX)=\angle(ZY, YW)$$To prove $TYMZ$ cyclic, we angle chase that $$\angle(YT, TM)=\angle(YT, TP)+\angle(PT, TM)=\angle(YW, WP)+\angle(ZY, YW)$$$$=\angle(ZY, WP)=\angle(YZ, ZM)$$which the last step holds because $WP\parallel ZM$ by Step $2$. This completes the proof. $\blacksquare$ (PS: I think this is one of the best geom I ever created xD - though the number of points made it daunting to attempt..)

Yet another solution from the official packet, bashy but shows that the problem can be solved in a (very) different way:

Here's a solution using linearity of PoP. Let the exterior, interior angle bisector of $\angle EDF$ intersects $EF$ at $G, H$ respectively. Let $\odot ABC, \odot WNX$ and $\odot YMZ$ be $c, c_1, c_2$ respectively. Let the radical axis of ${c}$ and $c_1$ cut $YZ$ at ${K}$ and that of ${c}$ and $c_2$ cut $WX$ at ${J}$. We denote the nine point circle of $\triangle ABC$ as $c_9$. Let $MN\cap EF=I$, and the in/ex-bisector of $\angle BAC$ cut $BC$ at $P,Q$resp. For any point ${V}$ and $i \in \left\{ 1, 2, 9 \right\} $ define $f_i(V)=Pow(V , c_i)-Pow(V, c)$. To prove $\odot ABC, \odot WNX$ and $\odot YMZ$ has a common point, we only need $NJ \perp MK$. It's easily shown that $f_1(D)=\frac{a^2-bc}{4}$ and $f_2(D)=\frac{a^2+bc}{4}$ by the properties of incenter and excenter. Moreover, $YZEF, XWEF$ are concyclic respectively, so $Pow(G , c_1)=Pow(G , c_9)$ and $Pow(H , c_2)=Pow(H , c_9)$, so $f_1(G)=f_9(G)$ and $f_2(H)=f_9(H)$. Since $\frac{\overline{FG}}{\overline{EG}}=-\frac{b}{c}$ and $\frac{\overline{FH}}{\overline{EH}}=\frac{b}{c}$, $$f_1(G)=f_9(G)=\frac{b}{b-c}f_9(E)+\frac{-c}{b-c}f_9(F)=\frac{b}{b-c}\frac{b^2}{4}+\frac{-c}{b-c}\frac{c^2}{4}=\frac{b^2+bc+c^2}{4}$$$$f_2(H)=f_9(H)=\frac{b}{b+c}f_9(E)+\frac{c}{b+c}f_9(F)=\frac{b}{b+c}\frac{b^2}{4}+\frac{c}{b+c}\frac{c^2}{4}=\frac{b^2-bc+c^2}{4}$$ Now we can compute $\frac{\overline{KD}}{\overline{DG}}=\frac{f_1(D)-0}{f_1(G)-f_1(D)}=\frac{a^2-bc}{(b+c-a)(b+c+a)}$, and the respective $\frac{\overline{JD}}{\overline{DH}}=\frac{a^2+bc}{(-b+c-a)(-b+c+a)}$. The next step is rotate and scale $\triangle JDN$ such that $JDN$ becomes $J'DM$; we would like to prove $KDJ'M$ concyclic, which gives us $KM\perp JN$ as desired. By ptolemy it is $$KD\sin\angle MDJ'+DJ'\sin\angle KDM=DM\sin\angle KDJ'=DM$$Notice $90-\angle KDM=\angle MDJ'=\angle NDJ=\angle HGD$. We have $\sin\angle HGD=\frac{DI}{DG}$ so we have a $\frac{KD}{DG}$ term; similarly $\frac{JD}{DH}$. By dividing $DM$ at both sides this equality turns into one involving only the basic lengths of $\triangle ABC$, which is then obvious.

Thank too much

Denote by $\mathcal{K}$ the Kiepert hyperbola of $ABC$. Suppose $P$ is the antipode of $N$ in $(NWX)$. Let $H$ be the orthocenter of $\triangle ABC$. Notice that $W$ and $X$ also lies on $\mathcal{K}$. Claim: $H$ is also the orthocenter of $\triangle XWP$. Proof: Since $\angle NXP= \angle NWP=90^{\circ}$, we only need $NXHW$ to be a parallelogram. Consider taking homothety at $G$ with factor $-2$ sending $W$ to the incenter $I$ and $X$ to the $A-$excenter $J$. Let $H'$ and $N'$ be the image of $H$ and $N$ resp. Then it suffices to show $H' M N'$ are collinear with $H'M=MN'$. Now take homothety at $G$ again with factor $-1/2$, we need to show the further intersection of $GM$ with 9pt circle is the midpoint of $HN$. This is well known, in fact any $(N,M)$ being antipodal will work. $\square$ Now, notice that $H \in \mathcal{K}$. This implies $P$ the orthocenter of $\triangle XWH$ also lies on $\mathcal{K}$. Claim: the intersection of $(WXN)$ and $(ABC)$ lies on $\mathcal{K}$. Proof: Let $T_1$ and $T_2$ be the fourth intersection of $\mathcal{K}$ with $(WXNP)$ and $(ABC)$ resp. Notice that in a rectangular circumhyperbola through $ABC$ with orthocenter $H$, the fourth intersection of $(ABC)$ with this hyperbola is the reflection of $H$ in the center of this hyperbola. Since $\triangle ABC$ and $\triangle XWP$ have the same orthocenter and these points lies on $\mathcal{K}$, we conclude $T_1=T_2$ as they both are the reflection of $H$ in the center of $\mathcal{K}$. $\square$ Till this end, we can also prove that the intersection of $(YMZ)$ and $(ABC)$ also lies on $\mathcal{K}$ by a similar method. Hence the three circles are concurrent at the fourth intersection of $\mathcal{K}$ with $(ABC)$. $\blacksquare$