Solved with @Blastoor

3. Here are two different proofs which need the same lemma: Switch $A$ for $H$ and the orthocenter with $A$. Obviously $AE=AF$. Lemma: Let $EC\cap BF=S$. Claim: $\angle SHC = \gamma$ and $\angle SHB = \beta$.

Now two ways: $1)$ Let $A'$ be the reflection of $A$ over $EF$. It's easy to get $\angle A'HC = \angle SHC$ so $(H,S,A')$ collinear. Notice the cyclic quadrilaterals: $EHMA'$ and $FHNA'$. By converse of radical axis on these two circles (since $HS,BF,CE$ concur) $EFNM$ cyclic $\blacksquare$ Another way to finish (without radical) is by angle chasing we get $NSMA'$ cyclic and we're done by angles. (here we need $\angle HMA' = \angle HNA'$ since $EHMA'$ and $FHNA'$ are congruent circles. $\blacksquare$ $2)$ Let $EC$ and $BF$ cut circle $BHC$ at $P$ and $Q$ respectively. By spiral similarity, some cyclic quads pop up: $PNSH$, $QMSH$. By lemma we can also get $EPNB$ and $FQMC$. Now simple angles gives us $EFMN$ cyclic $\blacksquare$ I wonder if the lemma could be proven without trig or any bashing.

Let $X=BF\cap CE$ and $H=BE\cap CD$ and let $O$ be the circumcenter of $\triangle ABC$ Now, observe that $AE,AF$ are isogonal lines and so are $AB,AC$ in $\angle BAC$. Thus, we have by the isogonality lemma that $AX, AH$ are also isogonal but as $AO, AH$ are isogonal, we have $A,O,X$ are collinear. Now, observe that $\triangle EAB\sim \triangle FAC$ and \begin{align*} \angle EBX=\angle EBC-\angle XBC=\angle BAX-(\angle ABC-\angle ABX)\\ =\angle BAX-(\angle ANX-\angle ABX) =\angle BAX-\angle BAN=\angle NAX \end{align*} Similarly, we also have $\angle FCX=\angle MAX$. Thus, \begin{align*} \frac{XE}{XF} & =\frac{\frac{XE}{BE}\cdot BE}{\frac{XF}{CF}\cdot CF}\\ & =\frac{BE}{CF}\cdot \frac{\sin \angle EBX}{\sin \angle FCX}\\ & =\frac{AC}{AB}\cdot \frac{\sin \angle NAX}{\sin \angle MAX}\\ & =\frac{\sin \angle AMX}{\sin \angle MAX}\cdot \frac{\sin \angle NAX}{\sin \angle ANX}\\ & =\frac{AX}{MX}\cdot \frac{NX}{AX}\\ & =\frac{NX}{MX} \end{align*}Thus, $MX\cdot XE=NX\cdot XF$ and we are done.

Let line parallel to $BC$ through $E$ meet $AC$ at $M'$, and let $AC$ meet $HB$ at $D$ where $H$ is the orthocenter of $\triangle ABC$. Also let $EF$ meet $BC$ at $T$. So \begin{align*} (HB,HC;HA,HM') &= (D,C;A,M') \\ &= (B,C;T,\infty) \\ &= (C,B;\infty,T) \\ &= (AC,AB;\ell,AT) \end{align*}where $\ell$ be line through $A$ parallel to $BC$. But we have $HB \perp AC$, $HC \perp AB$, $HA \perp \ell$ and so $HM' \perp AT$.

Now by easy angle chasing, we get $$\angle FHM' =\frac12 \angle BHC = \angle FAM'$$and so $A,H,F,M'$ are concyclic. But we also have $$\angle AM'E =\angle ACB = \angle AME$$and so $A,E,M,M'$ are concylic. Now radical axis theorem gives $E,F,M,H$ are concyclic. Similarly, $E,F,N,H$ are concylic as well, and so $E,F,M,N$ are concyclic. $\blacksquare$

Too easy anser wrote: Let $ABC$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $ABC$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $EC$ and $FB$ respectively such that $\angle EMA = \angle BCA$ and $\angle ANF = \angle ABC$. Prove that the points $E, F, N, M$ lie on a circle. Let $H$ be orthocenter and let $G$ be its reflection in $EF$ By isogonality lemma (or some cross ratio chase) we have $AG,BF,CE$ concurrent at say $K$. Moreover $\angle AGE = \angle AHE = \angle AHB = \angle ACB = \angle AME$ and hence $\square AGME$ is cyclic. Similarly $AGNF$ is cyclic and hence $KN\times KF = KA \times KG = KM \times KE \implies E,F,M,N$ are concyclic. Hence Proved !!

Call the altitudes $\overline{BZ}$ and $\overline{CY}$, and $H$ the orthocenter. Let $W$ be the midpoint of $\overline{BC}$. Then according to IMO Shortlist 2005 G5, the line $AW$ is concurrent with $(HYZ)$, $(HEF)$, $(HBC)$ at a point $Q$. [asy][asy] import graph; size(9.98895cm); pen ffxfqq = rgb(1.,0.49803,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.); pair H = (0.2,3.2), B = (-1.,0.2), C = (2.2,0.2), A = (0.2,1.), Y = (-0.55862,1.30344), Z = (1.21538,1.67692), F = (1.57146,1.14279), Q = (-0.68,2.76), M = (0.87710,0.51931); draw(H--B--C--cycle, linewidth(0.6) + zzttqq); draw(circle((0.34465,1.81068), 1.39682), linewidth(0.6) + ffxfqq); draw(circle((0.2,2.1), 1.1), linewidth(0.6) + yqqqqq); draw(circle((0.6,1.3), 1.94164), linewidth(0.6) + yqqqqq); draw(H--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--H, linewidth(0.6) + zzttqq); draw(B--Z, linewidth(0.6)); draw(C--Y, linewidth(0.6)); draw(B--F, linewidth(0.6) + blue); draw(C--(-0.71824,0.90439), linewidth(0.6) + blue); draw((0.6,0.2)--Q, linewidth(0.6) + green); draw(circle((-1.,1.5), 1.3), linewidth(0.6) + qqwuqq); draw((-0.71824,0.90439)--F, linewidth(0.6) + yqqqqq); dot("$H$", H, dir((-5.899, 6.902))); dot("$B$", B, dir((-8.675, -9.665))); dot("$C$", C, dir((5.858, -4.103))); dot("$A$", A, dir((1.801, 3.594))); dot("$Y$", Y, dir((-13.028, -3.649))); dot("$Z$", Z, dir((6.784, -1.212))); dot("$E$", (-0.71824,0.90439), dir((-10.327, -2.673))); dot("$F$", F, dir((7.966, -3.412))); dot("$Q$", Q, dir((-8.162, 8.978))); dot("$N$", (-0.15969,0.50808), dir((1.837, -9.672))); dot("$W$", (0.6,0.2), dir((-1.409, -11.804))); dot("$M$", M, dir((-1.312, -9.939))); [/asy][/asy] Since $WA \cdot WQ = WB^2$, it follows that $(AQB)$ is tangent to $\overline{BC}$, ergo $N \in (AQB)$. Then \[ \measuredangle QNF = \measuredangle QNB = \measuredangle QAB = \measuredangle QAZ = \measuredangle QHZ = \measuredangle QHF \]and hence $N$ lies on $(HQEF)$. Similarly, so does $M$.

My solution on the test: Relabel $A$ with $H$, and let the orthocenter of $HBC$ be $A$. Now $\angle BHC > 90$, $\angle EMH = \angle BCH = 90 - \angle B$, and $\angle HNF = \angle HBC = 90 - \angle C$. Let $A', B', C'$ be the feet of the $A, B, C$-altitudes of $ABC$, and let $X$ be the midpoint of $BC$. [asy][asy] import olympiad; unitsize(1.7inch); pair A, B, C, H, S, X, M, N, E, F, P, Q, R, I, J, Y, Z, O, L, n, m; A = dir(120); B = dir(210); C = dir(330); I = dir(270); P = foot(A, B, C); Q = foot(B, C, A); R = foot(C, A, B); H = extension(A, P, B, Q); J = foot(H, A, I); X = .5B + .5C; S = foot(A, H, X); E = extension(H, J, A, B); F = extension(H, J, A, C); Y = foot(C, E, F); Z = foot(B, E, F); O = circumcenter(Z, H, Q); n = 2 * foot(O, X, Q) - Q; L = circumcenter(Y, H, R); m = 2 * foot(L, X, R) - R; M = extension(m, H, C, E); N = extension(n, H, B, F); draw(A--B--C--cycle); draw(circumcircle(A, B, C), dashed+lightblue); draw(circumcircle(A, E, F), dotted+lightblue); draw(E--C, purple); draw(B--F, purple); draw(C--R, magenta); draw(B--Q, magenta); draw(A--P, magenta); draw(Y--Z); draw(B--Z); draw(C--Y); draw(circumcircle(m, n, X), dotted+lightblue); draw(X--n, dotted+lightred); draw(X--m, dotted+lightred); draw(X--S, dotted+lightred); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E'$", Y, dir(Y)); dot("$F'$", Z, dir(Z)); dot("$H$", H, dir(120)); dot("$A'$", P, dir(260)); dot("$B'$", Q, dir(50)); dot("$C'$", R, dir(110)); dot("$X$", X, dir(X)); dot("$E$", E, dir(135)); dot("$F$", F, dir(60)); dot("$S$", S, dir(S)); dot("$N'$", n, dir(n)); dot("$M'$", m, dir(m)); dot("$N$", N, dir(255)); dot("$M$", M, dir(M)); [/asy][/asy] First, $EF$ is the $H$ external angle bisector, so $\angle C'HE = \angle B'HF = \angle A/2$ and $\frac{C'E}{C'H} = \tan \angle A/2 = \frac{B'F}{B'H}$, or \begin{align*} \frac{C'E}{B'F} &= \frac{C'H}{B'H}\\ &= \frac{C'B}{B'C} \quad(\text{by } \triangle C'HB\sim\triangle B'HC). \end{align*}Let $S$ be the second intersection of $(ABC)$ and $(AB'HC')$, the center of spiral similarity from $BC'$ to $CB'$. From the above, $E$ is sent to $F$ under this spiral similarity as well, so $S$ lies on $(AEF)$ too. Also note that $S$ lies on $HX$ because $\angle ASH = 90$. Invert about $H$ with radius $\sqrt{HA\cdot HA'}$ and take a $-1$ homothety. $A, B, C$ are sent to $A', B', C'$ and $E$ is sent to the point $E'$ that is the second intersection of $HE$ and $(HA'CB')$, so $E'$ is the foot from $C$ to $EF$. Similarly $F$ is sent to the point $F'$ that is the foot from $B$ to $EF$. $S$ is sent to $X$ because $S$ lies on $(HB'C')$ while $X$ lies on $BC$. $N'$, the image of $N$, satisfies $\angle HF'N' = \angle HNF = 90 - \angle C$ and $F'HB'N'$ cyclic. Then $\angle N'B'H = 90 + \angle C$, and since $\angle BB'X = 90 - \angle C$, $N'B'X$ is collinear. Also $\angle F'N'X = \angle F'N'B' = 180 - \angle F'HB' = \angle A / 2$. Similarly $M'C'X$ is collinear and $\angle E'M'X = \angle A / 2$. Since $ASEF$ is cyclic, $A'XE'F'$ is cyclic and \begin{align*} \angle E'XF' &= \angle E'A'F\\ &= 180 - \angle F'A'B - \angle E'A'C\\ &= 180 - \angle F'HB' - \angle E'HC\\ &= \angle BHC = 180 - \angle A. \end{align*}Now $XE' = XF'$ so $\angle XE'F' = \angle XF'E' = \angle A / 2$. Since $\angle F'N'X = \angle F'E'X = \angle A/2$ and $\angle E'M'X = \angle E'F'X = \angle A/2$, $M'$ and $N'$ lie on $(XE'F')$. Inverting back, $EFNM$ is cyclic, as desired.

Omg. Here is the story... Here is the proposal that I sent: Quote: Let $ABC$ be an acute triangle with $AB = BC$. Let $P$ be any point on $AC$. Line passing through $P$ perpendicular to $AB$, intersects ray $BC$ in point $T$. Line $AT$ intersects the circumscribed circle of $\triangle$ $ABC$ the second time at point $K$. Prove that $\angle AKP = \angle ABP$. When I opened this problem, I didn't realize any similarity first, and then in a few minutes of solving realized that the EGMO problem can be solved with my problem as the main lemma. I was a bit frustrated, because this meant I wouldn't be able to use my proposal elsewhere, and thought that it's a sad coincidence... But it seems that in official solutions this problem is attributed to Ukraine, and other problems Ukraine sent don't look like this, so the only conclusion is that they took my problem and upgraded it to this! I am glad that it happened this way, the problem became harder and better in my opinion. But has something similar even happened before?

Let $H$ be the orthocenter of $ABC$ and $B_1$ be the reflection of $H$ across $AC$. Also let $X$ be the intersection of the angle bisector of $\angle{HBC}$ with $AC$. By angle chasing we have that $AXFH$ and $AXMB_1E$ are both cyclic. This indicates that $C$ has equal power to both $(HEF)$ and $(EAB_1MX)$. Thus, the second intersection of these two circles (other than $E$) must also lie on $EC$, and so $M$ lies on $(HEF)$. Similarly, $N$ lies on $(HEF)$, so we are done.

Let $H$ be the orthocenter of $ABC$, $H_1$, $F_1$ be the reflexion of $H$ and $F$ across $AC$. Then $H_1$ on $\odot ABC$ and $\angle H_1F_1A = \angle HFA = \angle AEH$, hence $A,E,H_1,F_1$ are concyclic. Let $M'$ be the second intersection of $\odot AEH_1F_1$ with $EC$. Then $\angle AM'E = \angle AH_1E = \angle ACB$. Hence $M'=M$. Finally, $CM \cdot CE = CF_1 \cdot CH_1 = CF \cdot CH $, implying that $M$ lies on $\odot (HEF)$. Similarly $N$ lies on it as well.

Switch $A$ and $H$. Let $G, I$ be the points on $AB, AC$ such that $BHCGI$ is cyclic. Let $J$ be the intersection of $AD, BF, CE$, where $HD$ is the internal bisector of $BHC$. Let the second intersection of $(GEH), (FIH)$ be $T$. Then it suffices to show that $H, J, T$ are collinear. We have angle $\angle HTF = 90^\circ - C$, and so on, so $\angle FTE = A$. By the law of sines, $(GEH), (FIH)$ have the same radius, so $T$ is the reflection of $A$ in $EF$, and thus $AT$ bisects $\angle BAC,$ and thus $AT, HD$ are parallel. Note that the reflection of $AH$ in $EF$ is $HT$. Let $N$ be the intersection of $AD$ and $EF$, $D'$ be the intersection of $AH$ and $BC$, $W$ be the intersection of $HT$ and $BC$, and let $W'$ be the intersection of $HJ$ and $BC$. Then if $P = EF \cap BC$, $(WD'DP) = -1$ due to angle bisectors. We also have $(AJND) = -1$, so projecting this from $H$, we have $(D'W'PD) = -1$, so $W = W'$, and thus $H, J, J'$ are collinear, so $H, T, J$ are collinear, as needed.

anser wrote: Let $ABC$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $ABC$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $EC$ and $FB$ respectively such that $\angle EMA = \angle BCA$ and $\angle ANF = \angle ABC$. Prove that the points $E, F, N, M$ lie on a circle. Anton Trygub is back with another wonderful geo! Gives me vibe of IMO 2019 P2 somehow. Here's my solution. Claim 01. $BC$ is tangent to both $(ANB)$ and $(AMC)$. Proof. The angle condition gives us $\angle ABN + \angle NAB = \angle ANF = \angle ABC = \angle ABN + \angle NBC$, therefore $\angle NAB = \angle NBC$, forcing $BC$ tangent to $(ANB)$. By symmetry, $BC$ is tangent to $(AMC)$ as well. Let us define $B'$ and $C'$ as the feet of altitude from $B$ and $C$ in $\triangle ABC$, $D = BE \cap CF$, $X = (DB'C') \cap (DBC)$ and $L$ midpoint of $BC$. Claim 02. $X$ lies on both $(ANB)$ and $(AMC)$. Proof. To prove this, we'll prove that $ANBX$ is cyclic. Notice that $A$ is the orthocenter of $\triangle DBC$. It is well known that $L,A,X$ are collinear. Furthermore, let the reflection of $A$ wrt $L$ be $A'$, and $A' \in (DBC)$ by the reflection of orthocenter lemma. Therefore, $LA \cdot LX = LA' \cdot LX = LB \cdot LD = LB^2$. Therefore, $LB \equiv BC$ is tangent to $(ABX)$. This forces $X \in (ANB)$. Similarly, $X \in (AMC)$. Claim 03. $X$ lies on $(DEF)$. Proof. This is well-known. Note that $\frac{BE}{EB'} = \frac{CF}{FC'}$. Therefore since $(DXB'C')$ and $(DXBC)$ are cyclic, we obtain $(DXEF)$ being cyclic as well. Claim 04. $N,M$ lies on $(DXEF)$, proving the desired result. Proof. We'll now proceed with phantom point. Let us redefine $M$ and $N$ as $(AXC) \cap (DXEF)$ and $(AXB) \cap (DXEF)$ respectively. We'll prove that $B,N,F$ and $C,M,E$ collinear. By symmetry, it suffices to prove that $B,N,F$ are collinear. Notice that \begin{align*} \angle BNX = \angle BAX &= 180^{\circ} - \angle BAA' \\ &= 180^{\circ} - \angle AA'C \\ &= 180^{\circ} - \angle XA'C \\ &= \angle XDC \equiv \angle XDF = 180^{\circ} - \angle XNF \end{align*}and we are done.

MS_Kekas wrote: I am glad that it happened this way, the problem became harder and better in my opinion. But has something similar even happened before? While not from an olympiad, 2021 AIME I #15 is another example of this phenomenon (see post #11). See also 2021 AIME I #10, in which I turned a easy (and boring :/) AMC proposal into something much harder.

@MS_Kekas It's a very good sign that there is some thought process going behind the scene. Some people are saying that geometry is dead. I've never agreed to that, I think it yet has many more to give.

Let $H$ be the orthocenter of $\triangle ABC$. Let $X, Y$ be the feet of altitudes from $B, C$ respectively. Also, let $D=CE\cap BF$ and $h$ be the axial symmetry wrt line $EF$. Set $h(M,N,D)=(K,L,P)$. Note that $K$ lies on circle $HAE$ and $L$ lies on circle $HAF$. Now, it suffices to show that $(ELKF)$ cyclic, or $PL\cdot PF=PK\cdot PE$. Or, $P\in HA$ (since $HA$ is the radical axis of those two circles). Let $h(H)=S$. We will instead prove that $A,D,S$ are collinear or equivalently that cevians of $\triangle ABC$ $AS, BF, CE$ concur. By the law of sines: $$\frac{CB}{CX}\cdot\frac{EX}{EB}=\frac{\sin\angle XCE}{\sin\angle ECB}$$and $$\frac{BY}{BC}\cdot\frac{FC}{FY}=\frac{\sin\angle CBF}{\sin\angle FBY}$$and $$\frac{\sin\angle BAD}{\sin\angle DAC}=\frac{\sin\angle HAX}{\sin\angle YAH}=\frac{\sin\angle HYX}{\sin\angle YXH}=\frac{\sin\angle CBH}{\sin\angle HCB}=\frac{HC}{HB}$$So, $$\frac{\sin\angle XCE}{\sin\angle ECB}\cdot\frac{\sin\angle CBF}{\sin\angle FBY}\cdot\frac{\sin\angle BAD}{\sin\angle DAC}=\frac{CB}{CX}\cdot\frac{EX}{EB}\cdot \frac{BY}{BC}\cdot\frac{FC}{FY}\cdot \frac{HC}{HB}=\frac{BY}{CX}\cdot\frac{AX}{AB}\cdot \frac{AC}{AY}\cdot\frac{HC}{HB}=\frac{BY}{CX}\cdot\frac{HC}{HB}=1$$and the desired lines concur by trig Ceva.

Sketch: Let $U$ the point of intersection of $BF$ and $EC$, and$H$ the orthocenter. Let $W$ the reflection of $H$ on $EF$. By Desargues involution $\angle HAE= \angle EAU$ then $W, A, U$ are collinear. Later $\angle EMA = \angle BCA = \angle EHA = \angle EWA$ so $E,A,W,M$ are concyclic then $AU.UW=EU.UM$. Analogously: $AU.UW=NU.UF$, then $NU.UF=EU.UM$ so $E,F,N,M$ lie on a circle.

Cute Notation. Let $X,Y:=\{CE\cap BF, BE\cap CF\}$. Also, let $D:=\odot(AFN)\cap AX$. Proof. Note that $\angle ABC=\angle AYF=\angle ANF=\angle ADF$ and as $\{AE,AF\}$ and $\{AB,AC\}$ are isogonal, we get, $\{BE\cap CF, BF\cap CE\}$ are isogonal $\implies\angle ACB=\angle EYA=\angle EDA=\angle EMA\implies AD$ is radical axis of $\{\odot(AEM),\odot(AFN)\}$ and we are done.

Let $H$ be the ortocenter and $E_1= CA \cap BH, F_1= AB \cap CH$, $P$ the midpoint of $BC$ and $S= (HEF) \cap (HBC)$. It's well known that $A$ is the $S-$humpty point WRT $SBC$, so $A,P,S$ are collinear. Furthermore, $SAMC$ is cyclic, and it's well known that $S$ lies on $(HE_1F_1)$, so $\angle E_1HS= \angle SAC = \angle SMC$, so $M$ lies on $(EHS)=(HEF)$. Similarly, $N$ lies on $(HEF)$, so $ENFM$ is cyclic, as desired.

Let $X,Y$ be the feet of altitudes from $B,C$ to the respective opposite sides of $\triangle ABC$. Let $BX \cap CY=G$ and $(GEF)\cap FB=N'$. We will show that $N=N'$, i.e. $\angle AN'F=\angle ABC$. Similar will hold for $M$ and we would be done. Invert about $A$. Henceforth we will denote the image of a point $X$ by $X$ only (and forget what $X$ was initially, except for point $A$). Inverted diagram is attached at the bottom of the post. Let $K$ be the center of circumcircle of cyclic quadrilateral $XBCY$ which is the midpoint of $XY$. We have to show $\angle AFN'=\angle ACB$. We prove this below: Claim 1: $K$ lies on $(EGF)$ Proof: Note that $$\angle GKE=\angle XKE=(1/2)\cdot \angle XKB=\angle XYA=\angle GYA=\angle GFE$$(here we have used the fact that both $E,K$ lie on the perpendicular bisector of $XB$)Which proves our claim. Claim 2: $N',B,K$ are collinear Proof: Let $N''$ denote $KB\cap (EGF)$, we will show that $N''BAF$ is cyclic which will let us conclude our claim. Note that $$\angle EAB=\angle FAY=\angle FGY=\angle FGK=\angle FEK=\angle FN''K=\angle FN''B$$as desired. Now to complete, note that $$\angle AFN'=\angle N'FE=\angle N'KE=\angle BKE=\angle XKE=\angle GKE$$$$=\angle GFE=\angle GFA=\angle GYA=\angle XYB=\angle ACB$$which finishes off the proof. $\blacksquare$

Very beautiful problem! Replace $A$ with $H$ and let the orthocenter of $HBC$ be $A$, so that the diagram looks nicer. Let $(AEF)$ and $(ABC)$ meet for the second time at $K$. Here is a long and boring proof that $(KHB)$ is tangent to $BC$. Let $KD$ meet $(KHB)$ for the second time at $P$ and let $Q = EF \cap BC$. Since $Q$ is the miquel point of $EFCB$, $QBEK$ is cyclic. Also, let $KP \cap (ABC) = R$ Claim 1: $CR || EF$ Proof: This is just angle chasing. $\angle KRC = \angle KBC = 180 - \angle KBQ = 180 - \angle KEQ = \angle KEF$. $\square$ Claim 2: $BP || EF$ Proof: Also just angle chasing. $\angle PEH = \angle KEQ = \angle KBQ = \angle KAC = 180 - \angle KRC$, so $BP || CR || EF$. $\square$ So, now we have $\angle PKB = \angle EKB = \angle EQB = \angle PBC$ since $PB || EF$. So, $(KHB)$ is tangent to $BC$. Similarly, $(KHC)$ is also tangent to $BC$. Now, finally, let $BH \cap AC = V$. Since $\angle KNF = 180 - \angle KNB = 180 - \angle KHB = \angle KHV = 180 - \angle KAV$, $N \in (KAFE)$. Similarly, $M \in (KAFE)$ and so $EFNM$ is cyclic, as desired. $\blacksquare$

Remake the problem changing $A$ and $H$, the orthocenter. Notice that to construct $P,Q$ you must reflect $A$ through $EF$ to get $A'$, and by definition their intersections with $FB,EC$ are $P,Q$. By pop we just need to have $HA',FB,EC$ concurring in one point. By desargue's theorem in $HBC,EFA'$ we just need to have $A'F\cap BH=I, A'E\cap CH=G$, $GI,BC,EF$ concurring. For this, we'll prove that $F$ is in the perpendicular bisector of $CI$, and analougsly for $E$ we'll be done. As $BH\perp AC$, we just need to have $F$ the orthocenter of $HCI$, but this is analogous as saying $A'F || AB$, which is true, because easily we get $A'EAF$ is a rhombus. $\blacksquare$

Let $X,Y,Z$ be foots of $B,C,A$ altitudes in triangle $ABC$, which has orthocenter $H$. From angle conditions we get $(ANB)$ and $(AMC)$ tangent to $BC$ at $B$ and $C$, respectively. Let $R=(ANB)\cap (AMC)$ and $RA\cap BC=K$. $KB^2=KA\cdot KR=KC^2 \implies K$ is midpoint of $BC$. $\angle BRC=\angle BRA+\angle CRA=\angle ABC+\angle ACB=\angle AHC+\angle AHB=\angle BHC \implies BCHR$ is cyclic. Since $RA$ bisects $BC$ and $R\in (HBC)$ we get $RHXY$ is cyclic. So $R$ is spiral similarity $\mathcal{S}$ that maps $BX$ to $CY$. $$\frac{EX}{EB}=\frac{XA}{AB}=\frac{AY}{AC}=\frac{YF}{FC} \implies \mathcal{S}:E\to F$$$\implies RHEF$ is cyclic. Since $RANB$ is cyclic, $\angle RNF=\angle RAY=180-\angle RHY \implies N\in (RHF)\implies N\in (RHEF)$. Similarly we get $M\in (RHEF)$. So $RHEFMN$ is cyclic, which gives $EFMN$ is cyclic. So we are done.

Let $H$ be the orthocenter of $\triangle ABC$. Let $P,Q$ be the feet of the altitudes from $C,B$ in $\triangle ABC$. Let $S$ be the midpoint of $\overline{BC}$. Let $K$ be the $H-$Queue point in $\triangle HBC$. [asy][asy] size(10cm); import olympiad; void my_dot(pair A) { fill(circle(A, .01),black); } pair H = dir(110); pair B = dir(200); pair C = dir(340); pair Q=foot(B,H,C); pair P=foot(C,H,B); pair A=extension(B,Q,C,P); pair I=incenter(A,B,P); pair E=extension(A,I,B,H); pair F=extension(A,I,C,H); pair S=(1B+1C)/(1+1); pair L=extension(P,Q,B,C); pair K=extension(L,H,S,A); pair N = intersectionpoints(circumcircle(A,B,K), circumcircle(K,E,F))[1]; pair M= intersectionpoints(circumcircle(A,C,K), circumcircle(K,E,F))[1]; draw(H--B--C--H); draw(B--A--P); draw(C--A--Q); draw(E--A--F); draw(B--F); draw(C--E); draw(arc(circumcenter(A,C,K), circumradius(A,C,K), 200, 270),red); draw(arc(circumcenter(A,B,K),circumradius(A,B,K),270,420),orange); draw(S--A--K); draw(A--N); draw(A--M); draw(circumcircle(H,B,C),blue); draw(circumcircle(H,E,F),dotted+darkgreen); label("$B$", B, dir(150)); label("$A$", A, dir(90)); label("$C$", C, dir(340)); label("$P$",P,dir(170)); label("$Q$",Q,dir(30)); label("$E$",E,dir(160)); label("$F$",F,dir(0)); label("$H$",H,dir(100)); label("$S$",S,dir(270)); label("$K$",K,dir(120)); label("$M$",M,dir(250)); label("$N$",N,dir(300)); my_dot(H); my_dot(B); my_dot(C); my_dot(P); my_dot(Q); my_dot(A); my_dot(E); my_dot(F); my_dot(S); my_dot(K); my_dot(N); my_dot(M); [/asy][/asy] Since $APEB\sim AQFC$, then $(HPQ),(HEF),(HBC)$ concur on $K$, which is the center of the spiral similarity mapping $BEP$ to $CFQ$. Also, $K$ is the projection of $H$ on line $SA$, so it is the $A-$ Humpty point in $\triangle ABC$. Notice that $$ \measuredangle CBN=\measuredangle CBA-\measuredangle NBA=\measuredangle FNA-\measuredangle NBA=\measuredangle BAN $$Hence $(ABN)$ is tangent to $BC$. Similarly, $(ACN)$ is tangent to $BC$. Therefore $(ABN)$ and $(ACN)$ intersect on the $A-$ Humpty point of $\triangle ABC$, which is precisely $K$, and so $ABNK$ and $ACMK$ are cyclic. Also, $$ SA\cdot SK=SB^2 \Longrightarrow \triangle SAB\sim \triangle SBK \Longrightarrow \measuredangle BAS=\measuredangle SBK $$$$ \triangle KBC\sim \triangle KEF \Longrightarrow \measuredangle CBK=\measuredangle FEK $$Finally, $$ \measuredangle FNK=180^o-\measuredangle KNB=180^o-\measuredangle KAB=\measuredangle BAS=\measuredangle SBK=\measuredangle CBK=\measuredangle FEK $$Hence $EFKN$ cyclic, and so $HEFKMN$ is cyclic, as desired. By angle chasing, $$ \measuredangle FNK=180^o-\measuredangle KNB=180^o-\measuredangle KAB=\measuredangle BAS=\measuredangle SBK=\measuredangle CBK=\measuredangle FEK $$

Let $H,G$ denotes orthocenter of $ABC$ and it's reflection over $EF.$ Note that by Isogonal theorem $BF,CE,AG$ concur. Furthermore $\angle AGE=\angle BHA=\angle ACB=\angle AME$ implies that $AMGE$ is cyclic, and analogously for $ANGF.$ Then by radical axis $MFEN$ is cyclic too.

Let $P_A$ denote the $A$-humpty point of $\triangle ABC$. Let $N_A$ be the midpoint of minor arc $BC$, and $M_A$ major arc $BC$, i.e. $E, F$ lie on $AN_A$. Let $G$ be the midpoint of $BC$, $R = AN_A \cap BC$, $S = AM_A \cap BC$. Let $X, Y, Z$ be the feet of the altitudes from $A, B, C$ onto $BC, CA, AB$ respectively. Consider the negative inversion $\phi$ centred at $A$ radius $\sqrt{AH \cdot AX}$. It's well-known that $\phi$ swaps $(H, X), (Z, C), (Y, B), (P_A, G)$. Let $U, V$ be the projections of $B, C$ onto $AN_A$ respectively. Note that $\measuredangle FUB = 90 = \measuredangle FYB$, hence $UBFY$ is cyclic. Now we have $AF \cdot AU = AY \cdot AB$, so $\phi$ swaps $(U, F)$ and analogously $(V, E)$. Claim 1: $HP_AEF$ cyclic. Proof: It suffices to prove in the inverted figure that $UXGV$ is cyclic. $F, Y, H$ collinear means $AUBX$ cyclic. Similarly, $E, Z, H$ collinear means $ADCV$ cyclic. Thus, by PoP $RU \cdot RA = RB \cdot RX$, and $RA \cdot RV = RX \cdot RC$. Mutliplying, we get $RU \cdot RV \cdot RA^2 = RB \cdot RC \cdot RX^2$. It's well-known that $(R, S; B, C) = -1$. Thus, under inversion at $R$ radius $\sqrt{RB \cdot RC}$, we get $(P_\infty, S'; C, B) = -1$. Thus $S'$ is the harmonic conjugate of $P_\infty$, the midpoint of $BC$: $G$. Now we have $RB \cdot RC = RS \cdot RG$. It's also well-known that $AR \perp AS$. Thus, $\measuredangle SAR = \measuredangle AXS$ and by converse of alternate segment, $RA$ is tangent to $(AXS)$. By tangent-secant, we have $RA^2 = RX \cdot RS$. Finally, we get $RU \cdot RV \cdot RX \cdot RS = RS \cdot RG \cdot RX^2 \Rightarrow RU \cdot RV = RG \cdot RX$. By converse of PoP, we have $UXGV$ cyclic. $\square$ Since $\measuredangle BMA = \measuredangle CBA$, $CB$ is tangent to $(AMB)$ at $B$. It's well-known that $P_A$ also lies on $(AMB)$. Furthermore, $P_A, A, G$ collinear, and $HP_A \perp AP_A$. Thus, we have $\measuredangle BMP_A = \measuredangle BAP_A = \measuredangle ABC + \measuredangle BGA$. Yet we also have $\measuredangle FHP_A = \measuredangle CHP_A = \measuredangle HCG + \measuredangle CGP_A + \measuredangle GP_AH = \measuredangle YCB + \measuredangle BGP_A + 90 = 90 + \measuredangle YBC + \measuredangle BGA + 90 = \measuredangle ABC + \measuredangle BGA$. $\measuredangle FMP_A = \measuredangle BMP_A = \measuredangle FHP_A$ means $P_AMFH$ is cyclic by converse of angles in a cyclic quad. Thus, $M$ lies on $HP_AEF$. Analogously, $N$ does too, so $HP_AEMNF$ is cyclic. $\blacksquare$

$\text{simple solution with angle chasing, the critical step is to construct the point}\; Q$

GLIDING PRINCIPLE means that the external angle bisector condition is mostly superficial or something Restate the problem as follows: Restated Problem wrote: Let $\triangle ABC$ be acute and let $H$ be its orthocenter. Let $X$ and $Y$ be the feet of the altitudes from $C$ and $B$ respectively and let $E$ and $F$ be on $\overline{AB}$ and $\overline{AC}$ respectively such that $\overline{EF}$ bisects $\angle BHC$ externally. Let $M$ be the point on $\overline{EC}$ such that $\angle EMH=\angle BCH$ and let $N$ be the point on $\overline{FB}$ such that $\angle FNH=\angle CBH$. Prove that $E,F,M,N$ are concyclic. Let $(AXY) \cap (ABC):=Q\neq A$ be the so-called $A$-queue point. It is well-known that $(QHB)$ and $(QHC)$ are tangent to $\overline{BC}$. I claim that $M$ is infact $\overline{EC} \cap (QMC)$. Indeed, if we let this intersection be $M'\neq C$, then we should have $$\measuredangle EM'H=\measuredangle HM'C=\measuredangle HCB,$$which is exactly what we want. Similarly, $N=\overline{FB} \cap (QNB)$. Since $Q$ is the center of the spiral similarity sending $\overline{XY}$ to $\overline{BC}$, and $\frac{XE}{EB}=\frac{YF}{FC}$ by angle bisector theorem/similar triangles, by the gliding principle $Q$ is also the center of spiral similarity sending $\overline{EF}$ to $\overline{BC}$, so it is the Miquel point of $BCFE$ and thus $AQEF$ is cyclic. Furthermore, a simple angle chase yields $$\measuredangle QAF=\measuredangle QAC=\measuredangle QBC=\measuredangle QNB=\measuredangle QNF,$$hence $QANF$ is cyclic as well. Likewise, $QAME$ is cyclic, so $A,Q,E,F,M,N$ all lie on a circle, as desired. $\blacksquare$

Cross-ratio FTW!

Claim: $N$, $E$ and $Z$ are collinear. Proof. We have \[(B,H;N,X)\stackrel{B}{=}(C,P;F,A)\stackrel{\triangle HXB\sim\triangle HAC}{=}(B,Y;E,X)\stackrel{H}{=}(B,Z;Q,X)=(X,Q;Z,B)\stackrel{E}{=}(B,H;N',X).\]Thus, $\angle ENF=180^\circ-\angle BHY=180^\circ-\angle BAC$ which is symmetric in $B$ and $C$.

Here's a different solution motivated by JMO 2023/6. (In particular, spoiler force radical axis and use symmetry .) Let $H$ be the orthocenter of $ABC$. We claim that $HENMF$ is cyclic. We will prove that $HEMF$ is cyclic; the proof for $HENF$ is analogous. Since the bisector of $\angle BAC$ forms equal angles with $\overline{AC}$ and $\overline{AB}$, it forms equal angles with $\overline{BH}$ and $\overline{CH}$. Thus, we have $\angle HEF=\angle HFE$. Let the angle bisector of $\overline{EF}$ intersect $\overline{AB}$ and $\overline{AC}$ at $P$ and $Q$, respectively. Since the corresponding sides of $HEF$ and $AQP$ are perpendicular, the triangles are similar isosceles triangles. Hence, $AEHP$ and $AHFQ$ are cyclic. Since $P$ and $Q$ are symmetric about $\overline{EF}$, we have \[\angle AQE=\angle APE=\angle AHE=\angle ACB=\angle AME,\]so $AEMQ$ is cyclic. Consequently, \[CH \cdot CF=CA \cdot CQ=CE \cdot CM,\]so $HEMF$ is cyclic, as desired.

Let $H$ denote the orthocenter of $\triangle ABC$. We will show that $HENF$ is cyclic, whence $HEMNF$ is cyclic by symmetrical reasons. First of all, it is well known that the $A$-Queue point $Q$ of $\triangle ABC$ lies on $(HEF)$. Let $K$ be the midpoint of side $BC$. By reflecting the orthocenter, if we let $A'=2K-A$, then $KA \cdot KQ = KA' \cdot KQ = KB \cdot KC = KB^2$ by power of a point in $(HBC)$, so $BC$ is tangent to $(ABQ)$. Thus by the given angle condition, $N$ lies on $(ABQ)$. Hence by Reim's theorem twice we conclude that $N$ also lies on $(HEF)$, as desired.

This can we convert as $H$ is main vertex and $A$ is orthocenter anyway here is solution From $\angle EAC = 90 + \frac{A}{2}$ we get $\angle HEF = \frac{A}{2}$ and similarly $\angle HFE \implies HE=HF$ Consider $B'$ and $C'$ be feet of altitude from $B$ and $C$.Let $D$ be intersection of angle bisector of $\angle BHC$ with $EF$, $Y = HD \cap BB'$ and $K$ be reflection of $H$ above $EF$ By angle chasing we get $H,E,A,Y$ are cyclic. From $\angle HYB' = \angle HFE \implies Y,F,K,A$ are cyclic $$\angle ANF = \angle ACB = \angle AHF = \angle AKF$$ $A,N,K,F$ are cyclic by power of point $$BN.BF=BA.BY = BE.BH$$$H,E,N,F$ lie on same circle and similarly from other side we get $H,E,F,M$ are cyclic $\blacksquare$

Let $H$ be the orthocentre of triangle $ABC$ so that $B$, $E$, $H$ and $C$, $F$, $H$ are collinear. Claim: The angle bisectors of $\angle BAC$ and $\angle BHC$ are parallel. Proof. Let $\Omega$ be the circumcircle of triangle $HBC$. Let $H'$ be the $H$-antipode in $\Omega$ and let $M_H$ and $N_H$ denote the midpoints of arcs $\widehat{BC}$ and $\widehat{BHC}$ of $\Omega$. Now note that $BACH'$ is a parallelogram, which implies that the angle bisectors of $\angle BAC$ and $\angle BH'C$ are parallel. But the latter is the line $H'N_H$, which is parallel to $HM_H$, which is the angle bisector of $\angle BAC$, since $HM_HH'N_H$ is a rectangle, so we're done. $\square$ It follows that $HE = HF$. Let $Y$ and $Z$ be points on lines $HB$ and $HC$, respectively, such that $AH = AY = AZ$. Now $AEYM$ is cyclic since $\angle EMA = \angle EYA = \angle BHA = 90^\circ - \angle HBC = \angle BCA$ as triangle $HAY$ is $A$-isosceles. Similarly, $AFZN$ is also cyclic; let the circumcircles of the two quadrilaterals be $\Omega_1$ and $\Omega_2$, respectively. Additionally, if we let $L$ be the point for which $HELF$ is a rhombus, $A$ lies on the perpendicular bisector of $\overline{HL}$, so $AH = AL$. Thus, $AY = AL$. Now the angle bisector of $\angle YEL$ is parallel to the angle bisector of $\angle BHC$ (since $EL \parallel HC$) which is perpendicular to $EF$, implying that line $EAF$ externally bisects angle $\angle YEL$. Since $AY = AL$, we must have that $AEYL$ is cyclic, and hence that $L \in \Omega_1$. Similarly, $L \in \Omega_2$ too. Claim: $BE$, $CF$ and $AL$ concur. Proof. Let $K = BE \cap CF$, and use Cartesian coordinates with $H = (0, 0)$, $E = (s, -1)$, $F = (s, 1)$ and $A = (s, a)$, where $a \notin \{-1, 1\}$ obviously. Now $L = E + F - H = (2s, 0)$. Since $AB \perp HF$, the slopes of lines $AB$ and $HF$ multiply to $-1$ and therefore line $AB$ has equation $y = -s(x - s) + a$. As $HE$ has equation $y = -\frac 1s x$, we may derive that $B = \left(\frac{s(s^2 + a)}{s^2 - 1}, -\frac{s^2 + a}{s^2 - 1}\right)$. Similarly, we may find that $C = \left(\frac{s(s^2 - a)}{s^2 - 1}, \frac{s^2 - a}{s^2 - 1}\right)$. Hence, line $BF$ has slope \[\frac{1 + \frac{s^2 + a}{s^2 - 1}}{s - \frac{s(s^2 + a)}{s^2 - 1}} = -\frac{2s^2 + a - 1}{(a + 1)s}\]and so has equation \[y = -\frac{2s^2 + a - 1}{(a + 1)s}(x - s) + 1.\]Similarly, line $CE$ has equation \[y = -\frac{2s^2 - a - 1}{(a - 1)s}(x - s) - 1.\]Hence, if we let $x$ be the $x$-coordinate of $BF \cap CE$, then \[\frac{1}{x - s} = \frac 12 \left(\frac{2s^2 + a - 1}{(a + 1)s} - \frac{2s^2 - a - 1}{(a - 1)s}\right) = \frac{a^2 - 2s^2 + 1}{(a^2 - 1)s}.\]Now line $AL$ has equation \[y = -\frac as (x - s) + a.\]Therefore, if we let $x$ be the $x$-coordinate of $BF \cap AL$, then \[\frac{1}{x - s} = \frac{1}{a - 1} \left(\frac as - \frac{2s^2 + a - 1}{(a + 1)s}\right) = \frac{a^2 - 2s^2 + 1}{(a^2 - 1)s}.\]Hence the two values of $x$ are equal, as desired. $\square$ Now noting that $KE \cdot KM = KA \cdot KL = KF \cdot KN$ finishes by Power of a Point.