Here https://artofproblemsolving.com/community/u800085h2807853p24762198

We Complex Bash Obviously the Circumcircle is the unit circle and the Complex Affix of each point is in lower-cases.

Similarly $n = \frac{ef(d-a)}{ed-fa}$. We now find $\bar{m} = \frac{a-d}{ba-cd}$ and $\bar{n} = \frac{d-a}{ed-fa}$ Equation of $\overline{BE}$ is $w + \bar{w} be - (b+e) = 0$ and Equation of $\overline{CF}$ is $w + \bar{w} cf - (c+f) = 0$ Equation of line $\overline{MN}$ can be found by the determinant $$\begin{vmatrix} w & \bar{w} & 1 \\ m & \bar{m} & 1 \\ n & \bar{n} & 1 \end{vmatrix} = 0$$ which comes out as $w (ef - da + ba - cd) + \bar{w} ((cd-ba)ef + (da-ef)bc) + (bc(d-a) + ef(a-d)) = 0$ Now the only thing we need to check is that $$\begin{vmatrix} ef - da + ba - cd & (cd-ba)ef + (da-ef)bc & bc(d-a) + ef(a-d) \\ 1 & be & -(b+e) \\ 1 & cf & -(c+f) \end{vmatrix} = 0$$ Which can be done in less than a dozen lines of algebra (I hope I haven't made any mistake in calculations )

Let $T=BE \cap CF$ We use pascal in $ABEDCF$ and we obtain $G, T$ and $H$ are collinear. Now observe, $\angle MGT= \angle BGM-\angle BGT=90-\angle BCG-\angle ABT+ \angle ETH=90-\angle TED-\angle HFE+\angle ETH=\angle NHE-\angle THD=\angle NHT$ More over, $$\frac{HT}{TG}.\frac{BG}{HE}=\frac{\sin{\angle TED}}{\sin{\angle HTG}}.\frac{\sin{\angle BTG}}{\sin{\angle ABT}}=\frac{\sin{\angle BCG}}{\sin{\angle EFH}}$$ $$\Rightarrow \frac{HT}{TG}=\frac{\sin{\angle BCG}}{BG.}\frac{HE}{\sin{\angle EFH}}=\frac{HN}{GM}$$ Combining the angle relation and the ration of the lengths we get $\Delta NHT \sim \Delta MGT$. That leads us to $N, T$ and $M$ being collinear.

We can let $X=\overline{BE}\cap\overline{CF}\cap\overline{GH}$ by Pascal on $ABEDCE.$ Notice $$\begin{align*}\angle XHN&=\angle XHE-\angle NHE\\&=(\angle BED-\angle EXH)-(90-\angle EFH)\\&=(\angle EBA-\angle BXG)-(90-\angle BCG)\\&=\angle XGB-\angle MGB\\&=\angle XGM.\end{align*}$$ Then by LoS on $\triangle CGX,\triangle FHX$ and extended LoS on $\triangle BCG,\triangle EFH,$ $$\frac{GX}{GM}=\frac{2\sin\angle GBC\sin\angle GCD}{\sin\angle GCX}=\frac{2\sin\angle XFH\sin\angle FEH}{\sin\angle HXF}=\frac{HX}{HN}$$ as $\angle CBG=\angle AFC=180-\angle EFH$ and $\angle GCX=\angle FED=180-\angle FEH.$ Hence, $\triangle GMX\sim\triangle HNX.$ $\square$

Let $T = BE \cap CF$. Consider the inversion $\Phi$ at $T$ fixing the circle. Angle chase shows $\angle BH^*C = 180^\circ - \angle BGC$ (where $H^* = \Phi(H)$), i.e. $H \in \odot(BCG)$. So $\Phi$ just swaps $\odot(EFH)$ and $\odot(BCG)$, implying points $N,T,M$ are collinear. $\blacksquare$ [asy][asy] size(200); pair A=dir(110),B=dir(50),C=dir(-20),D=dir(-60),E=dir(-140),F=dir(170),T=extension(F,C,E,B),G=extension(A,B,C,D),H=extension(A,F,E,D),N=circumcenter(F,H,E),M=circumcenter(G,B,C); draw(unitcircle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$G$",G,dir(G)); dot("$H$",H,dir(H)); dot("$T$",T,dir(-90)); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); draw(A--G--D--H--A,red); draw(E--B^^F--C,blue); draw(N--F--E--N^^M--B--C--M,brown); draw(N--M,dotted); [/asy][/asy] Here's another different proof: Claim 1: $\angle NFE + \angle MBC = \angle FTE$. Proof: Here by $\widehat{XY}$ we will mean angle subtended by minor arc $XY$ on circumference. $$\begin{align*} \angle FHE + \angle BGC &= \angle AHD + \angle AGD \\ &= 180^\circ - (\angle DAH + \angle ADH ) + 180^\circ - (\angle DAG + \angle ADG) \\ &= 180^\circ - (\widehat{DF} + \widehat{EA}) + 180^\circ - (\widehat{DB} + \widehat{AC}) \\ &= 360^\circ - (180^\circ + \widehat{FE} + \widehat{BC}) \\ &= 180^\circ - (\angle TBF + \angle TFB) \\ &= 180^\circ - \angle FTE \end{align*}$$ Since $\angle NFE = 90^\circ - \angle FHE$ and $\angle MBC = 90^\circ - \angle BGC$, so our Claim follows. $\square$ Now we can basically ignore points $A,D,H,G$ and only keep in mind that $N,M$ lie on perpendicular bisectors of segments $FE,BC$ (respectively) and our Claim 1 is satisfied. Let $K=FN \cap BM$, $X=OM \cap FC$, $Y = ON \cap BE$ and $O$ be circumcenter of $FBCE$. Observe $X,Y \in \odot(BOF)$ as $$\angle FYB = \angle FOB= \angle FXB = \widehat{FB}$$ [asy][asy] size(200); pair F=dir(110),B=dir(70),C=dir(0),E=dir(-150),O=(0,0),T=extension(E,B,C,F),N=1.2*(E+F),M=extension(O,1/2*(B+C),N,T),K=extension(N,F,M,B),X=extension(O,M,F,C),Y=extension(O,N,E,B); draw(unitcircle); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$O$",O,dir(-90)); dot("$N$",N,dir(N)); dot("$M$",M,dir(M)); dot("$T$",T,dir(T)); dot("$K$",K,dir(K)); dot("$X$",X,dir(-90)); dot("$Y$",Y,dir(-90)); draw(E--F--C--B--E^^F--Y^^B--X,red); draw(B--O--N--K--M--O--F,brown); draw(circumcircle(F,O,B),blue); draw(N--M,dotted); [/asy][/asy] Claim 2: $K \in \odot(BOF)$. Proof: This is just angle chase combined with Claim 1, $$\begin{align*} \angle FKB &= \angle NFT + \angle MBT - \angle FTB \\ &= (\angle NFE + \angle EFT) + (\angle MBC + \angle CBT) - \angle FTB \\ &= (\angle NFE + \angle MBC) + (\angle EFT + \angle CBT) - \angle FTB \\ &= \angle ETF + 2 \angle EFT - \angle FTB \\ &= \angle ETF + \angle EFT - \angle FET \\ &= 180^\circ - 2 \angle FET \\ &= 180^\circ - \angle FOB \end{align*}$$ proving our Claim. $\square$ By Pascal on $FKBYOX$ w.r.t. $\odot(FOB)$, we obtain points $N,T,M$ are collinear, as desired. $\blacksquare$ Remark: Angle chase shows that this problem is equivalent to Polish Mathematical Olympiad finals 2021 P5

Suppose that $K = BE \cap CF$, and let $BE$ and $CF$ intersect $(BCG)$ at $E'$ and $F'$. By Reim's theorem, we have $GE' \parallel HE$, $GF' \parallel HF$ and $E'F' \parallel EF$, so $K$ is the homothetic center of $\triangle GE'F'$ and $\triangle HEF$. Since $M$ and $N$ are their respective circumcenters, it follows that $M$, $N$, $K$ are collinear.

We will basicaly use Pascal's Theorem. Let $A^{*}$ and $D^{*}$ be the antipodals of $A$ and $D,$ respectively, and let $P = {BA^{*}}\cap {D^{*}C}$ and $Q = {D^{*}E} \cap {FA^{*}}.$ Then $M$ and $N$ are the midpoints of $PG$ and $QH,$ respectively. [asy][asy] import geometry; import graph; size(250); pair A,B,C,D,E,F,G,H, Ar,Dr,P,Q,M,N,R; A = dir(130); B=dir(73); C = dir(11.1); D = dir(-70); E=dir(-115.69); F = dir(179.97); G = extension(A,B,C,D); H = extension(A,F,D,E); Ar = -A; Dr = -D; Q = extension(F,Ar,E,Dr); P = extension(B,Ar,C,Dr); M= P*0.5+G*0.5; N=Q*0.5+H*0.5; R=extension(B,E,C,F); draw(unitcircle); draw(circumcircle(H,F,E),blue); draw(circumcircle(B,C,G),green); draw(A--H); draw(D--H); draw(A--G); draw(C--G); draw(H--Q, dotted+cyan); draw(G--P, dotted + cyan); draw(Q--P, dotted + red); draw(G--H, dotted + red); draw(E--Dr, dotted +magenta); draw(F--Ar, dotted +magenta); draw(B--Ar, dotted +magenta); draw(C--Dr,dotted +magenta); draw(B--E); draw(C--F); dot("$R$", R, dir(R)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(-125)); dot("$D$", D, dir(D)); dot("$E$", E, dir(-70)); dot("$F$", F, dir(135)); dot("$M$", M, dir(135)); dot("$N$",N, dir(-80)); dot("$A^{*}$", Ar, dir(Ar)); dot("$D^{*}$", Dr, dir(Dr)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); [/asy][/asy] Now, let $R = BE \cap CF.$ Applying Pascal's Theorem at $ABCDEF$ yelds that $R,G$ and $H$ are colinear. On the other hand, by applying Pascal's Theorem again at $BA^{*}FCD^{*}E$ we get that $P,R$ and $Q$ are colinear. Hence, in order to show that $M,R$ and $N$ are colinear, it is sufficient(and necessary) to show that $GM \parallel NH.$ That is just angle chasing; we have $$\measuredangle (NH, ED) = \frac{\pi}{2} + \measuredangle (HF, FE) = \frac{\pi}{2} + \measuredangle AFE$$ similarly, $$\measuredangle (GM, AB) = \frac{\pi}{2} - \measuredangle DCB$$ Hence $$\measuredangle (NH, ED) + \measuredangle (GM, AB) \measuredangle AFE - \measuredangle DCB = \measuredangle (AB,DE)$$ and indeed $GM \parallel HN. \square$

Solution : Let $X=BE \cap CF$ By using Pascal's theorem in $ABEDCF$ and we shall get $G, X$ and $H$ as collinear. Now observe that, $\angle MGX= \angle BGM-\angle BGX=90-\angle BCG-\angle ABX+ \angle EXH=90-\angle XED-\angle HFE+\angle EXH=\angle NHE-\angle XHD=\angle NHX$ Again, $$\frac{HX}{XG}.\frac{BG}{HE}=\frac{\sin{\angle XED}}{\sin{\angle HXG}}.\frac{\sin{\angle BXG}}{\sin{\angle ABX}}=\frac{\sin{\angle BCG}}{\sin{\angle EFH}}$$ $$\Rightarrow \frac{HX}{XG}=\frac{\sin{\angle BCG}}{BG.}\frac{HE}{\sin{\angle EFH}}=\frac{HN}{GM}$$ Combining the angle relation and the ration of the lengths we get $\Delta NHX \sim \Delta MGX$. That leads us to $N, X$ and $M$ being collinear. Therefore $BE,CF$ and $MN$ are concurrent.

Let $X=BE\cap CF$ and let $O$ be the center of the circle, so our goal is to show $M,X,N$ collinear. We use moving points. Begin by keeping all points but $A$ fixed, and extend all definitions to degenerate cases through continuity. As we slide $A$ projectively on the circle, we define $M$ to be the intersection of the perpendicular bisector of $BC$ and of $CG$. Note that $G$ varies projectively, so the perpendicular bisector of $CG$ varies projectively in the pencil through $\infty_{CD^\perp}$, so $M$ varies projectively on the perpendicular bisector of $BC$. A similar argument shows that $N$ varies projectively on the perpendicular bisector of $EF$. Thus, $XM$ and $XN$ both vary projectively on the pencil through $X$, so it suffices to check the problem for three values of $A$. The problem is trivial when $A=D$, as $M=N=O$. Therefore, it suffices to solve it when $A=C$, as the case $A=E$ will then follow by symmetry. Thus, we have entirely reduced to the case of $A=C$. The problem when $A=C$ reads as follows. Let $BCEF$ be a cyclic quadrilateral with center $O$, and let $X=BE\cap CF$. Let $D$ vary on the circle. Let $M$ be the intersection of the perpendicular bisector of $BC$ and the line through $C$ perpendicular to $CD$. Let $H=CF\cap DE$ and let $H$ be the intersection of the perpendicular bisectors of $EF$ and $FH$. Show that $X,M,N$ collinear. We solve this problem by varying $D$ projectively. By the same logic as before, $N$ varies projectively, and $M$ varies projectively as the line through $C$ perpendicular to $CD$ varies projectively on the pencil through $C$. Thus, it suffices to check three values of $D$. As before, the case $D=C$ is trivial, since $M=N=O$. When $D=B$, we see that $M=\infty_{BC^\perp}$ and $N$ is the circumcenter of $XFE$ (as $H=X$), so the problem is true by the fact that $XEF\sim XCB$ and that the circumcenter and orthocenter are isogonal conjugates. When $D=F$, we see that $N$ is the intersection of the line through $F$ perpendicular to $CF$ and the perpendicular bisector of $EF$, and $M$ is the intersection of the line through $C$ perpendicular to $CF$ and the perpendicular bisector of $BC$. Thus, by the same similarity idea as above, the problem reduces to the following: Let $XBC$ be a triangle, and let $M$ be the intersection of the line through $C$ perpendicular to $XC$ and the perpendicular bisector of $BC$, and let $N'$ be defined similarly with $B$ and $C$ swapped. Show $XM$ and $XN'$ are isogonal in $\angle BXC$. This follows by $\sqrt{bc}$ inversion at $X$, since the circle through $B$ and $C$ tangent to $XB$ gets mapped to the circle through $B$ and $C$ tangent to $XC$, so their corresponding circumcenters, i.e. $M$ and $N'$, are isogonal, as desired.

I wasn't desperate enough to use inversion so I came up with this instead. Let $\measuredangle$ denote directed angles modulo $180^\circ$. We have $$\measuredangle BGC+\measuredangle EHF=\measuredangle BAF+\measuredangle EDC=\measuredangle BCX+\measuredangle XBC=\measuredangle BXC.$$ Since $XBC \sim XFE$, we can compose a reflection over a line through $X$ and a dilation at $X$ such that the resulting transformation maps $F$ to $B$ and $E$ to $C$. Suppose that this transformation maps $N$ to $N'$. It suffices to show that $\overline{XM}$ and $\overline{XN}$ are isogonal with respect to $\angle BXC$. Observe the following lemma. Lemma: Let $P$ and $Q$ be points on the perpendicular bisector of $\triangle ABC$. Then, $P$ and $Q$ are inverses with respect to the circumcircle of $ABC$ if and only if $\overline{AP}$ and $\overline{AQ}$ are isogonal with respect to $\angle BAC$. Proof: Let $O$ be the circumcenter of $ABC$. For the if direction, we will prove that $\measuredangle OPA=\measuredangle QAO$, which suffices. Since the $A$-altitude and $\overline{AO}$ are isogonal with respect to $ABC$, the directed angle between the $A$-altitude and $\overline{AP}$ is equal to the directed angle between $\overline{AQ}$ and $\overline{AO}$. Since $\overline{PQ}$ is parallel to the $A$-altitude, we have $\measuredangle OPA=\measuredangle QAO$, as desired. The converse is clear by uniqueness. $\square$ If $O$ is the circumcenter of $BCX$, we have $$\measuredangle BMO=\measuredangle BGC=\measuredangle BXC+\measuredangle FHE=\measuredangle BOM+\measuredangle ON'B=\measuredangle OBN',$$ so $M$ and $N'$ are inverses with respect to the circumcircle of $BCX$, as desired. $\square$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.846109032258052, xmax = 15.708729054374183, ymin = -7.27546163156413, ymax = 6.765737157794893; /* image dimensions */ pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((0.45269760075919907,-1.8833432923747244), 3.7801918556637735), linewidth(2)); draw(circle((0.45269760075919996,-1.8833432923747244), 3.7801918556637726), linewidth(0.4)); draw((xmin, -0.4614874235435215*xmin-5.131959835926818)--(xmax, -0.4614874235435215*xmax-5.131959835926818), linewidth(0.4)); /* line */ draw((xmin, 0.8516129032258101*xmin + 2.3321262456524723)--(xmax, 0.8516129032258101*xmax + 2.3321262456524723), linewidth(0.4)); /* line */ draw((xmin, -0.45745723368333807*xmin + 1.3680219871147685)--(xmax, -0.45745723368333807*xmax + 1.3680219871147685), linewidth(0.4)); /* line */ draw((xmin, 0.5740907879410247*xmin-6.218526721815619)--(xmax, 0.5740907879410247*xmax-6.218526721815619), linewidth(0.4)); /* line */ draw((-2.900396276482841,-0.13788864786840901)--(-2.774749537546841,-3.851447820865748), linewidth(0.4)); draw((3.9451313646220614,-0.43670689346261216)--(3.375447508457395,-4.280713402031743), linewidth(0.4)); draw(circle((-3.6958844426792545,-2.0237088502354235), 2.0467338024799657), linewidth(0.4)); draw(circle((5.03418858344353,-2.5623227488821447), 2.3883652130019946), linewidth(0.4) + fuqqzz); draw((-3.6958844426792545,-2.0237088502354235)--(5.03418858344353,-2.5623227488821447), linewidth(0.4)); draw((-2.900396276482841,-0.13788864786840901)--(3.375447508457395,-4.280713402031743), linewidth(0.4)); draw((-2.774749537546841,-3.851447820865748)--(3.9451313646220614,-0.43670689346261216), linewidth(0.4)); 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dot((-2.774749537546841,-3.851447820865748),dotstyle); label("$C$", (-2.6949809195399834,-3.6434174360507408), NE * labelscalefactor); dot((1.0492369130972348,-5.616169475638818),dotstyle); label("$D$", (0.9370632759733968,-5.286968161394788), NE * labelscalefactor); dot((3.375447508457395,-4.280713402031743),dotstyle); label("$E$", (3.4531162382284757,-4.069523179658457), NE * labelscalefactor); dot((3.9451313646220614,-0.43670689346261216),dotstyle); label("$F$", (4.021257229705429,-0.23457148718901272), NE * labelscalefactor); dot((-5.684322766062706,-2.5087163680267555),linewidth(4pt) + dotstyle); label("$G$", (-5.596558126011566,-2.3448094555319874), NE * labelscalefactor); dot((7.354527903590921,-1.9963600027088542),linewidth(4pt) + dotstyle); label("$H$", (7.430103178567149,-1.837540713141849), NE * labelscalefactor); dot((-3.6958844426792545,-2.0237088502354235),linewidth(4pt) + dotstyle); label("$M$", (-3.60806465584223,-1.8578314628374546), NE * labelscalefactor); dot((5.03418858344353,-2.5623227488821447),linewidth(4pt) + dotstyle); label("$N$", (5.116957713268125,-2.4056817046188037), NE * labelscalefactor); dot((0.45269760075919996,-1.8833432923747244),linewidth(4pt) + dotstyle); label("$O$", (0.5312482820612873,-1.715796214968216), NE * labelscalefactor); dot((0.33291750702024897,-2.272271344330264),linewidth(4pt) + dotstyle); label("$I$", (0.12543328814917776,-2.4259724543144094), NE * labelscalefactor); dot((-1.6755196242868575,-2.3511920693724626),linewidth(4pt) + dotstyle); label("$J$", (-1.5992804359772876,-2.1824834579671433), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] By Pascal, $G-I-H$ Let $J = (BGC) \cap GH$ $\measuredangle BJH = \measuredangle BCD = \measuredangle BEH$, so $BJEH$ is cyclic Invert at $I$ with radius $\sqrt{IA \cdot ID}$ the small circles swap and hence we are done.

Having seen 2021 IMO 3 makes this really easy... (the idea is literally identical making it a 1 minute solve) Pascal's on $AFCDEB$ means that $X=BE\cap CF$ is on $GH$. Define $Y=(EFH)\cap GH$, now $$\measuredangle FYG=\measuredangle FYH=\measuredangle FEH=\measuredangle FCD=\measuredangle FCG$$ hence $F,Y,C,G$ concyclic. The point is that there is a negative inversion at $X$ with radius $$\sqrt{XF\cdot XC}=\sqrt{XB\cdot XE}=\sqrt{XY\cdot XG}$$ hence circles $(FYE)=(EFH)$ and $(BCG)$ are swapped, so the circumcenters are collinear with $X$. Done!

Mimics a proof shared by math_comb01 for Pascals. Construct circumcircles of $HEF$ and $BCG$. Then, let the former intersect $CF$ and $BE$ again at $F'$ and $E'$ respectively. By Reims, $E'F'$ and $BC$ are parallel. Also, $\measuredangle F'E'H = \measuredangle F'FH = \measuredangle AFC = \measuredangle ABC = \measuredangle GBC$, and similarly, $\measuredangle E'F'H = \measuredangle GCB$, and so the two circles are homothetic, with homotethy centre $BE \cap CF$, and we are done.