ACGNmath wrote:

Not sure if I'm doing anything wrong, since the solution looks so simple...
$[asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A,B,C); pair E = foot(B,A,C); pair F = foot(C,A,B); pair X = 0.6*B+0.4*C; pair H = orthocenter(A,B,C); path BXE = circumcircle(B,X,E); path CXF = circumcircle(C,X,F); pair P = intersectionpoints(BXE,CXF)[0]; pair Y = intersectionpoints(BXE,A--B)[0]; pair Z = intersectionpoints(CXF,10*A-9*C--10*C-9*A)[0]; draw(A--B--C--A--cycle); draw(BXE,blue); draw(CXF,blue); draw(C--F); draw(B--E); draw(P--X,dashed+red); draw(circumcircle(A,F,E),rgb(0.1,0.6,0.1)); draw(circumcircle(A,Y,Z)); draw(circumcircle(B,C,E),rgb(1.0,0.6,0.6)+dashed); draw(A--Z); draw(A--D); dot("$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$",P,dir(P)); [/asy]$

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.