Not sure if I'm doing anything wrong, since the solution looks so simple... [asy][asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A,B,C); pair E = foot(B,A,C); pair F = foot(C,A,B); pair X = 0.6*B+0.4*C; pair H = orthocenter(A,B,C); path BXE = circumcircle(B,X,E); path CXF = circumcircle(C,X,F); pair P = intersectionpoints(BXE,CXF)[0]; pair Y = intersectionpoints(BXE,A--B)[0]; pair Z = intersectionpoints(CXF,10*A-9*C--10*C-9*A)[0]; draw(A--B--C--A--cycle); draw(BXE,blue); draw(CXF,blue); draw(C--F); draw(B--E); draw(P--X,dashed+red); draw(circumcircle(A,F,E),rgb(0.1,0.6,0.1)); draw(circumcircle(A,Y,Z)); draw(circumcircle(B,C,E),rgb(1.0,0.6,0.6)+dashed); draw(A--Z); draw(A--D); 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("$H$",H,dir(H)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$P$",P,dir(P)); [/asy][/asy] We use directed angles modulo $180^{\circ}$. Let $P$ be the intersection of $(BEX)$ and $(CFX)$. By the radical axis theorem on $(BFEC), (BEX), (CFX)$, we have $BE, CF, PX$ are concurrent, which means that $P, H, X$ are collinear. Next, we have $P$ lies on $(AEHF)$ because \[\measuredangle FPH=\measuredangle FPX=\measuredangle FCX=\measuredangle FCD=\measuredangle FAD\]where $D$ is the foot of the perpendicular from $A$ to $BC$. Lastly, we have $A,Y,P,Z$ are concyclic because \[\measuredangle ZPY=\measuredangle ZPX+\measuredangle XPY=\measuredangle ZCX+\measuredangle XBY = \measuredangle ACB+\measuredangle CBA=\measuredangle CAB=\measuredangle ZAY.\] To finish, \[\measuredangle PYA = \measuredangle PYB = \measuredangle PEB = \measuredangle PEH = \measuredangle PAH\]which implies that $(AYZ)$ is tangent to $AH$.

ACGNmath wrote: Not sure if I'm doing anything wrong, since the solution looks so simple... [asy][asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A,B,C); pair E = foot(B,A,C); pair F = foot(C,A,B); pair X = 0.6*B+0.4*C; pair H = orthocenter(A,B,C); path BXE = circumcircle(B,X,E); path CXF = circumcircle(C,X,F); pair P = intersectionpoints(BXE,CXF)[0]; pair Y = intersectionpoints(BXE,A--B)[0]; pair Z = intersectionpoints(CXF,10*A-9*C--10*C-9*A)[0]; draw(A--B--C--A--cycle); draw(BXE,blue); draw(CXF,blue); draw(C--F); draw(B--E); draw(P--X,dashed+red); draw(circumcircle(A,F,E),rgb(0.1,0.6,0.1)); draw(circumcircle(A,Y,Z)); draw(circumcircle(B,C,E),rgb(1.0,0.6,0.6)+dashed); draw(A--Z); draw(A--D); 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("$H$",H,dir(H)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$P$",P,dir(P)); [/asy][/asy] We use directed angles modulo $180^{\circ}$. Let $P$ be the intersection of $(BEX)$ and $(CFX)$. By the radical axis theorem on $(BFEC), (BEX), (CFX)$, we have $BE, CF, PX$ are concurrent, which means that $P, H, X$ are collinear. Next, we have $P$ lies on $(AEHF)$ because \[\measuredangle FPH=\measuredangle FPX=\measuredangle FCX=\measuredangle FCD=\measuredangle FAD\]where $D$ is the foot of the perpendicular from $A$ to $BC$. Lastly, we have $A,Y,P,Z$ are concyclic because \[\measuredangle ZPY=\measuredangle ZPX+\measuredangle XPY=\measuredangle ZCX+\measuredangle XBY = \measuredangle ACB+\measuredangle CBA=\measuredangle CAB=\measuredangle ZAY.\] To finish, \[\measuredangle PYA = \measuredangle PYB = \measuredangle PEB = \measuredangle PEH = \measuredangle PAH\]which implies that $(AYZ)$ is tangent to $AH$. Or this problem is really simple itself Yeah I think the point $P$ you have is crucial, but with that point there are so many ways to finish up the problem.

Here's a more braindead solution. Also, the first part of this solution might not be necessary. We begin by showing that the desired conclusion holds when $X$ is the midpoint of $\overline{BC}$. Since $X$ is the center of $(BFEC)$ in this case, we know that $\overline{XY}$ bisects $\angle BYE$ by the trillium theorem, and that, as a consequence, $\overline{XY}$ is the perpendicular bisector of $\overline{EF}$. By symmetry, $Z$ also lies on the perpendicular bisector of $\overline{EF}$, and a two-second angle chase finishes (consider the midpoint of $\overline{EF}$). Now, it suffices to show that the signed ratio $\tfrac{AY}{AZ}$ is independent of the position of $X$. To do this, let $T\equiv\overline{AC}\cap (BXE)\neq E$ and $S\equiv\overline{AB}\cap(CXF)\neq F$. By power of a point, we can obtain $CT = \tfrac{CX\cdot a}{CE}$, $BS = \tfrac{BX\cdot a}{BF}$, $$AY = \frac{AE \cdot (b - CT)}{c} = \frac{AE \cdot (b - \frac{CX\cdot a}{CE})}{c},$$and $$AZ = \frac{AF \cdot (BS - c)}{b} = \frac{AF \cdot (\frac{BX\cdot a}{BF} - c)}{b}.$$Dividing the two yields $$\frac{AY}{AZ} = \frac{\frac{AE \cdot (b - \frac{CX\cdot a}{CE})}{c}}{\frac{AF \cdot (\frac{BX\cdot a}{BF} - c)}{b}} = \frac{AE\cdot BF\cdot b}{AF\cdot CE \cdot c},$$where the last equality comes from the substituting $BX = a - CX$ and doing some standard manipulations. Since this expression contains no $X$, we're done.

Great problem! Here's a length bash solution (i.e. PoP, Pythagorean theorems and a bit of trig) Let $(BXE)$ intersect $AC$ at $Y'$ and let $(CXF)$ intersect $AB$ at $Z'$. It's easy to see that $\frac {AZ'}{AZ}=\frac {AC}{AF}=\frac {AB}{AE} =\frac {AY'}{AY}$, so $YZZ'Y'$ is cyclic (probably this could have been proven by radical axis, but I saw it that way). Note that $AH$ is tangent to $(AYZ)$ $\iff$ $Y'Z' \perp BC$. We can prove the last one using $BZ'^2-CZ'^2=BY'^2-CY'^2$. We calculate only the first one and we will see that it's symmetrical expression. I will put only a sketch: using $BZ'.BF=BX.BC$, Pythagorean theorem for $CFZ'$ and $BF=a.cos \beta$, $CF=a.sin \beta$ we obtain the first expression is $a(BX-CX)$, so the second is $-a(CX-BX)$ and we're done. Edit: this is similar to the above one - it again includes points $Y'$ and $Z'$

Almost the same as #1 but anyways, Let $G=\overline{HX} \cap (BEX)$ then by POP $GH \cdot HX = BH \cdot HE = CH \cdot HF$ so $G \in (CFX)$. $$\measuredangle YGZ = \measuredangle XGZ - \measuredangle XGY = \measuredangle XCA - \measuredangle XBA = \measuredangle YAZ \implies G \in (AYZ).$$$\measuredangle HGE = \measuredangle XGE = \measuredangle XBE = \measuredangle HAE$ so $G \in \odot(AH)$ and finally to finish off, note that $$\measuredangle GZA = \measuredangle GZC = \measuredangle GXB = \measuredangle GEB = \measuredangle GEH = \measuredangle GAH.$$

Nothing new I guess, Solved with Aayuu By Miquel's theorem, $(BEX), (CFX), (AYZ)$ concur at a point, say $P$. Then $\angle AZY = \angle APY = \angle EPA - \angle EPY = \angle APE - \angle ABE$, so to show that $\angle BAH = \angle AZY$, it suffices to show that $\angle APE = \angle AFE$ (or $P \in (AEF)$)since then $\angle AZY = \angle AFE - \angle ABE = C - (90 - A) = 90 - B = \angle BAH$. Note that by radax, $PX$ goes through $H$, so $\angle EPH = \angle EPX = \angle EBC = \angle EAH$ so $P \in (AEF)$, so we're done. $\blacksquare$

https://hackmd.io/@sine/TWTST_22_1_M5

A common method, We can know that $P$,$X$,and $H$ are collinear because $$FH\cdot HC=HE\cdot BH$$That is, H is on the root axis of the $(PXB)$and$(PXC)$,we have $$PH\cdot HX=HE\cdot HB=AH\cdot HD$$which means that $P$,$A$,$F$,$H$,$E$ lies on a same circle. Consider $$\angle ZPY=\angle ZPX-\angle YPX=\angle EHC=\angle FAC$$, so $A,P,Y,Z$ lies on a same circle. At last,$$\angle YZA=\angle APY=\angle APX-\angle YPX=\angle BAH$$which implies that $(AYZ)$ is tangent to $AH$.

Let $K,X=(BXE)\cap (CXF)$. Claim: $AYKZ$ cyclic. Proof: $180^\circ - \angle AZK = \angle KZC=\angle KXC= \angle BYK = \angle AYK$. Claim: $K \in (AFHE)$. Proof: $\angle KFH = \angle KFC = \angle KXC = 180^\circ - \angle KXB = 180^\circ - \angle KEB = 180^\circ - \angle KEH$. Now $\angle AZY = \angle AKY = \angle YKE - \angle AKE = 180^\circ - \angle YBE - (180^\circ - \angle AFE) = 90^\circ + A - (180^\circ - C) = 90^\circ - B$. Let $T$ be a point on line $AH$ on the opposite side of $A$ from $H$. Then $\angle TAY = \angle HAF = 90^\circ - B = \angle AZY \implies AT$ tangent to $(AYZ) \implies AH$ tangent to $(AYZ)$.

Let $(BEX)$ intersect $\overline{AC}$ at $P$ and $(CFX)$ intersect $\overline{AB}$ at $Q$. I claim that $PQYZ$ is cyclic. Indeed, $\measuredangle PYQ=\measuredangle PYB=\measuredangle PEB=90^\circ$ and $\measuredangle PZQ=\measuredangle CFQ=90^\circ$, as desired. Furthermore, we actually have $90^\circ=\measuredangle PEB=\measuredangle PXB$ and likewise $90^\circ=\measuredangle QFC=\measuredangle QXC$, so $\overline{PQ}$ is the perpendicular to $\overline{BC}$ at $X$. This now implies that $$\measuredangle BAD=\measuredangle BQX=\measuredangle YQP=\measuredangle YZP=\measuredangle YZA,$$done. $\blacksquare$ Remark: Actually, the motivation to add $P$ and $Q$ came from trying to lengthbash and prove $\frac{BF}{CE}=\frac{AY}{AZ}$. Trying to use power of a point to relate $\frac{AY}{AZ}$ to other things naturally motivates the addition of $P$ and $Q$. Unfortunately a comedy of errors occurred, and instead of noticing that the PoP implies $PQYZ$ cyclic, I thought that it was sufficient to prove that $PQYZ$ was cyclic to finish the problem (mixed up which equations were true and which ones we wanted to prove), and then found the angle chase that actually finishes the problem anyways

Let $D = (CDX) \cap (BEX)$ Claim 1: $D-H-X$

Proof

Claim 2: $D$ lies on $(AEFH)$

Proof

Claim 3: $D$ lies on $(AYZ)$

Proof

Claim 4: $90^\circ-\measuredangle B = \measuredangle ADY$

Proof

By Claim 4, we're done

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(47.630266781564124cm); 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 = -15.468156288163454, xmax = 32.16211049340067, ymin = -10.461960997973392, ymax = 11.838373744345812; /* image dimensions */ pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); pen qqttzz = rgb(0.,0.2,0.6); pen ttffcc = rgb(0.2,1.,0.8); pen cczzff = rgb(0.8,0.6,1.); pen ffzzcc = rgb(1.,0.6,0.8); pen ffwwzz = rgb(1.,0.4,0.6); pen qqzzff = rgb(0.,0.6,1.); pen ccwwff = rgb(0.8,0.4,1.); pen zzwwff = rgb(0.6,0.4,1.); pen ffttcc = rgb(1.,0.2,0.8); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen ccccff = rgb(0.8,0.8,1.); /* draw figures */ draw(circle((9.5,-0.12), 5.368769111990194), linewidth(2.) + ffcqcb); 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[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.630266781564124cm); 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 = -15.468156288163454, xmax = 32.16211049340067, ymin = -10.461960997973392, ymax = 11.838373744345812; /* image dimensions */ pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); pen qqttzz = rgb(0.,0.2,0.6); pen ttffcc = rgb(0.2,1.,0.8); pen cczzff = rgb(0.8,0.6,1.); pen ffzzcc = rgb(1.,0.6,0.8); pen ffwwzz = rgb(1.,0.4,0.6); pen qqzzff = rgb(0.,0.6,1.); pen ccwwff = rgb(0.8,0.4,1.); pen zzwwff = rgb(0.6,0.4,1.); pen ffttcc = rgb(1.,0.2,0.8); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen ccccff = rgb(0.8,0.8,1.); /* draw figures */ draw(circle((9.5,-0.12), 5.368769111990194), linewidth(1.) + ffcqcb); 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