Let D be the A-foot on BC and O be the center of $\omega$. Clearly $AFH \sim BHC$. Let $M$ be the midpoint of AH. Then from the similarity we can say $M$ and $N$ correspond each other in this similar figures(the parallelograms). Thus $\angle FEA=\angle CHZ$. Thus a rotation of $ABC$ by angle of $\angle CHZ$ maps the lines AB and CH to XY and ZH. Thus $ZH \perp XY$. Also from Midpoint theorem, $MO || ZH \implies MO \perp EF \implies \text{M is the midpoint of XY}$. As $H$ is the reflection of $A$ about the midpoint of $XY$ and whose antipode is $Z$, we conclude that $H$ has to be the orthocenter of $XYZ$. @below u r sniped!

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -8.088271803475516, xmax = 17.71361703979325, ymin = -7.856625809574939, ymax = 7.328723304938418; /* image dimensions */ /* draw figures */ draw((-4.000000000000000,3.750000000000004)--(-6.160000000000007,-3.990000000000005), linewidth(1.200000000000002)); draw((-6.160000000000007,-3.990000000000005)--(5.120000000000005,-3.930000000000005), linewidth(1.200000000000002)); draw((5.120000000000005,-3.930000000000005)--(-4.000000000000000,3.750000000000004), linewidth(1.200000000000002)); draw((-5.156830546881834,-0.3953094596598884)--(-3.972646184340936,-1.392517343904857), linewidth(1.200000000000002)); draw((-3.972646184340936,-1.392517343904857)--(-2.815815637459105,2.752792115755036), linewidth(1.200000000000002)); draw(circle((-0.5336769078295348,-1.388741328047574), 6.198552913051435)); draw((-6.648889958473672,-2.401770232437794)--(-1.323756225867269,4.759252888532941), linewidth(1.600000000000002)); draw((-1.323756225867269,4.759252888532941)--(2.932646184340935,-6.527482656095152), linewidth(1.600000000000002)); draw((2.932646184340935,-6.527482656095152)--(-6.648889958473672,-2.401770232437794), linewidth(1.600000000000002)); draw((-0.5336769078295348,-1.388741328047574)--(-5.156830546881834,-0.3953094596598884), linewidth(1.200000000000002)); draw((-0.5336769078295348,-1.388741328047574)--(-2.815815637459105,2.752792115755036), linewidth(1.200000000000002)); draw((-6.160000000000007,-3.990000000000005)--(2.932646184340935,-6.527482656095152), linewidth(1.200000000000002)); draw((2.932646184340935,-6.527482656095152)--(5.120000000000005,-3.930000000000005), linewidth(1.200000000000002)); draw((-0.5336769078295348,-1.388741328047574)--(-6.648889958473672,-2.401770232437794), linewidth(1.200000000000002)); draw((-0.5336769078295348,-1.388741328047574)--(-1.323756225867269,4.759252888532941), linewidth(1.200000000000002)); draw((-4.000000000000000,3.750000000000004)--(-3.972646184340936,-1.392517343904857), linewidth(1.200000000000002)); /* dots and labels */ dot((-4.000000000000000,3.750000000000004),dotstyle); label("$A$", (-3.889192657027477,3.895513939918007), NE * labelscalefactor); dot((-6.160000000000007,-3.990000000000005),dotstyle); label("$B$", (-6.530122937812407,-4.502644352978076), NE * labelscalefactor); dot((5.120000000000005,-3.930000000000005),dotstyle); label("$C$", (5.222016811680533,-3.763183874358294), NE * labelscalefactor); dot((-3.972646184340936,-1.392517343904857),dotstyle); label("$H$", (-4.100467079490270,-1.835304769385294), NE * labelscalefactor); dot((-5.156830546881834,-0.3953094596598884),dotstyle); label("$E$", (-5.473750825498435,-0.1187000868750890), NE * labelscalefactor); dot((-2.815815637459105,2.752792115755036),dotstyle); label("$F$", (-2.964867058752751,3.103234855682528), NE * labelscalefactor); dot((-0.5336769078295348,-1.388741328047574),linewidth(5.000000000000000pt) + dotstyle); label("$O$", (-0.4295739891992180,-1.122253593573363), NE * labelscalefactor); dot((-6.648889958473672,-2.401770232437794),dotstyle); label("$X$", (-7.243174113624338,-2.521946642389377), NE * labelscalefactor); dot((-1.323756225867269,4.759252888532941),dotstyle); label("$Y$", (-1.221853073434697,4.925476749424130), NE * labelscalefactor); dot((2.932646184340935,-6.527482656095152),dotstyle); label("$Z$", (3.162091192668287,-7.011528119723761), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Using, $\triangle HFC\sim \triangle HEP$ and PoP relations, $\overline{OE}=\overline{OF}$. Let $M$ be the midpoint of $\overline{EF}$. So, $M$ is also the midpoint of $\overline{XY}$ because $OM\perp EF\implies OM\perp XY$. SInce, $AEHF$ is a parallelogram, using orthocenter-reflection, $H$ is the orthocenter of $\triangle XYZ$

Okay, I actually over-write every solution billion more times than others ;-; . Let $O$ denote the circumcenter of $\odot(ABC)$, and $M$ be the midpoint of $AH$. Firstly, as $AEHF$ is a parallelogram, $$\measuredangle HEB = \measuredangle HEA = \measuredangle AFH = \measuredangle CFH = - \measuredangle HFC$$. Also, as $EH \parallel AC$, $AC \perp BH \implies EH \perp BH \implies \measuredangle EHB = 90^{\circ}$. Similarly, $\measuredangle CHF = 90^{\circ} \implies -\measuredangle FHC = -90^{\circ} \equiv 90^{\circ}$. Now, $\measuredangle HEB = - \measuredangle HFC$ and $\measuredangle EHB = - \measuredangle FHC$ together imply, $\triangle EHB \stackrel{-}{\sim} \triangle FHC$. So, $\triangle EHB \sim \triangle FHC$ from which we get, \begin{align*} &\dfrac{EH}{FH} = \dfrac{EB}{FC}\\ \implies &EH \cdot FC = EB \cdot FH\\ \implies &FA \cdot FC = EB \cdot EA\\ \implies &\operatorname{Pow}_{\odot(ABC)}(E) = \operatorname{Pow}_{\odot(ABC)}(F)\\ \implies &OE^2 - r^2 = OF^2 - r^2\\ \implies &OE = OF\\ \implies &\triangle OEF \text{ is isoceles.} \end{align*} But, as $AEHF$ is a parallelogram, $M$ is also the midpoint of $EF \implies OM \perp EF \implies OM \perp XY \implies M$ is also midpoint of $XY$. Now, finally, as $A$ is the $Z-$antipode in $\odot(XYZ)$, the reflection of $A$ over the midpoint of $XY$ is the orthocenter of $\triangle XYZ$. But the midpoint of $XY$ is $M$ itself, and the reflection of $A$ over $M$ is the point $H$ itself and so $H$ is the orthocenter of $\triangle XYZ$ and we are done.

Same solutions, mostly.

Let $M$ be the midpoint of $EF$. Since $\angle EHF+\angle CHB =\pi$ then $H$ has an isogonal conjugate wrt $EFCB$ but $H$ has the circumcenter $O$ as isogonal wrt $\angle EBC $ and $\angle BCF $ so $ O$ and $ H$ are isogonal conjugates with respect $EBCF$ hence $\angle OEF =\angle BEH=\angle A ,\angle OFE =\angle CFH=\angle A$ thus $OM\perp EF$ and $OM\perp XY\implies M$is the midpoint of $XY$. More $A$ is the antipode of $Z$ and $H$ is the reflection of $A$ in the midpoint of $XY$ therefore $H$ is the orthocenter of $XYZ$. Best regards RH HAS

Let $N$ be the midpoint of $AH$ and $O$ be the center of $\omega$. Notice that since $H$ is the reflection of $A$ through $N$ and $A$ is the antipode of $Z$ it suffices to prove that $N$ is the midpoint of $XY$. We rewrite this with phantom points: let $E'\in AB$ and $F'\in AC$ be such that $E',F',N$ are collinear and $E'F'\perp ON$. Then, we now want to show that $N$ is the midpoint of $E'F'$. Let $K$ be the second intersection of $\omega$ and the circle of diameter $AH$ and $H'$ be the reflection of $H$ through $BC$. Notice that $ON$ contains two points in the perpendicular bisector of $AK$, hence $AK\parallel E'F'$. It is well-known that $KCH'C$ is harmonic, so projecting through $A$ we get \[-1 = (K,H';B,C) = (\infty_{E'F'},N,E',F')\]So $N$ is the midpoint of $E'F'$, therefore $AE'HF'$ is a parallelogram, hence $E'=E$, $F'=F$. $\square$

Let $ZH$ hit the circumcircle again at $T$. If we show that $H$ is the reflection of $T$ over $XY$, we are done. This is equivalent to showing $AT \parallel XY$. We will do this by showing $ATEF$ is an isosceles trapezoid, which follows from proving $OM$ is the common perpendicular bisector of $AT$ and $EF$. Indeed, $ME = MF$ by definition and $MT = MA = MH$ by right triangle $ATH$. Next, $OA = OT$ by definition, and\[EA \cdot EB = FC \cdot FC \implies R^2 - OE^2 = R^2 - OF^2 \implies OE = OF\]follows from ratios of $\triangle EHB \sim \triangle FHC$. Done.

Let $S,V,T$ be the midpoints of $AB,AC,AH$ respectively, Clearly $STOV$ is a parallelogram,and $STOV\sim AEHF$ So we can know that $\triangle EHF\sim \triangle TOV$ Also,as $FH\bot TV,EH\bot OV$ Hence $EF\bot TO\Rightarrow ZH\bot XY$ At last,since $EF\bot TO\Rightarrow FY=EX \Rightarrow XHAY$ is a parallelogram$\Rightarrow \angle XHY + \angle YAZ = 180^{\circ}$ Which means that $H$ is the orthocenter of triangle $XYZ$

Nice problem! We can easily see that H is the orthocenter by using the property that the reflection of the antipode across the midpoint is the orthocenter. A similar construction can be found in EGMO.

Notice that $AFHE\sim CZBH$, so $\angle AEF=\angle CHZ$, which means $ZH\perp XY$. It is well known that $ZH\cap (ABC)$ at the $HM$-point $S$, and $SM=MH$ where $M$ is the midpoint of $EF$, so $S$ is the reflection of $H$ about $XY$, meaning $H$ is the orthocenter.

What the heck is written about Sharky Devil in #13? I don't agree man... Anyways, here's a solution with some trigonometry and little harmonics. Solution: Let $M$ be the midpoint of $\overline{EF}$ and define $D \coloneqq ZH \cap \odot(ABC) \ne Z$. Denote $\odot(ABC)$ by $\omega$. Let $\ell$ be the line parallel to $EF$ through $A$. Finally call the reflection of $H$ across $BC$ as $H'$. [asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.1, xmax = 11.5, ymin = -5.997767906976736, ymax = 7.943683255813955; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); draw((1.561561860465115,5.83884604651163)--(-0.6152,-2.4178)--(10.4,-2.34)--cycle, linewidth(0.7) + qqwuqq); draw((1.6029481162912134,-0.02076687348364359)--(2.8366804491236888,4.658886594056553)--(1.561561860465115,5.83884604651163)--(0.3278295276326403,1.1591925789714344)--cycle, linewidth(0.7) + red); draw((8.181851883708786,-4.737033126516358)--(4.348258852634379,6.767446272426885)--(-1.1837488758780492,-0.9493670993988986)--cycle, linewidth(0.7) + yqqqyq); draw((-0.772139787279335,1.6818795435489826)--(-0.4660842983358079,2.108808283620569)--(-0.8930130384073942,2.4148637725640962)--(-1.1990685273509214,1.9879350324925098)--cycle, linewidth(0.7) + qqwuqq); /* draw figures */ draw((1.561561860465115,5.83884604651163)--(-0.6152,-2.4178), linewidth(0.7) + qqwuqq); draw((-0.6152,-2.4178)--(10.4,-2.34), linewidth(0.7) + qqwuqq); draw((10.4,-2.34)--(1.561561860465115,5.83884604651163), linewidth(0.7) + qqwuqq); draw(circle((4.871706872086951,0.5509064599976364), 6.238538696568829), linewidth(0.7)); draw((1.6029481162912134,-0.02076687348364359)--(2.8366804491236888,4.658886594056553), linewidth(0.7) + red); draw((2.8366804491236888,4.658886594056553)--(1.561561860465115,5.83884604651163), linewidth(0.7) + red); draw((1.561561860465115,5.83884604651163)--(0.3278295276326403,1.1591925789714344), linewidth(0.7) + red); draw((0.3278295276326403,1.1591925789714344)--(1.6029481162912134,-0.02076687348364359), linewidth(0.7) + red); draw((8.181851883708786,-4.737033126516358)--(4.348258852634379,6.767446272426885), linewidth(0.7) + yqqqyq); draw((4.348258852634379,6.767446272426885)--(-1.1837488758780492,-0.9493670993988986), linewidth(0.7) + yqqqyq); draw((-1.1837488758780492,-0.9493670993988986)--(8.181851883708786,-4.737033126516358), linewidth(0.7) + yqqqyq); draw((1.561561860465115,5.83884604651163)--(-1.1990685273509214,1.9879350324925098), linewidth(0.7)); draw((8.181851883708786,-4.737033126516358)--(-1.1990685273509214,1.9879350324925098), linewidth(0.7)); /* dots and labels */ dot((1.561561860465115,5.83884604651163),dotstyle); label("$A$", (1.660613023255814,6.086473953488374), NE * labelscalefactor); dot((-0.6152,-2.4178),dotstyle); label("$B$", (-1.2118706976744171,-2.7786051162790635), NE * labelscalefactor); dot((10.4,-2.34),dotstyle); label("$C$", (10.500929302325577,-2.08524697674418), NE * labelscalefactor); dot((8.181851883708786,-4.737033126516358),dotstyle); label("$Z$", (8.346566511627904,-5.353935348837202), NE * labelscalefactor); dot((1.6029481162912134,-0.02076687348364359),linewidth(4.pt) + dotstyle); label("$H$", (1.0663060465116283,-0.5251911627906923), NE * labelscalefactor); dot((0.3278295276326403,1.1591925789714344),linewidth(4.pt) + dotstyle); label("$E$", (0.02626883720930326,1.67869720930233), NE * labelscalefactor); dot((2.8366804491236888,4.658886594056553),linewidth(4.pt) + dotstyle); label("$F$", (2.626361860465116,5.021673953488375), NE * labelscalefactor); dot((-1.1837488758780492,-0.9493670993988986),linewidth(4.pt) + dotstyle); label("$X$", (-1.880466046511626,-1.1937865116279014), NE * labelscalefactor); dot((4.348258852634379,6.767446272426885),linewidth(4.pt) + dotstyle); label("$Y$", (4.4588083720930225,6.953171627906979), NE * labelscalefactor); dot((1.5822549883781645,2.9090395865139937),linewidth(4.pt) + dotstyle); label("$M$", (1.3139339534883725,3.263515813953492), NE * labelscalefactor); dot((-1.1990685273509214,1.9879350324925098),linewidth(4.pt) + dotstyle); label("$D$", (-1.7814148837209285,1.802511162790702), NE * labelscalefactor); dot((1.6365854656824668,-4.783262130739834),linewidth(4.pt) + dotstyle); label("$H'$", (1.0910688372093027,-5.428223720930225), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Claim: $M$ is also the midpoint of $\overline{XY}$ and $DA \parallel XY$. Proof: For the midpoint part, we will use trigonometry and Power of a Point. By Power of a Point on, it suffices to show that $E$ and $F$ have same power with respect to $\odot(ABC)$. Observe that $\angle BHE = \angle CHF = 90^\circ$. From $\triangle BHE$, we clearly have \begin{align*} \overline{BE} = \frac{2R\cos(B)}{\sin(A)} \qquad \text{and} \qquad \overline{FA} = \overline{HE} = \frac{2R\cos(B)}{\tan(A)} \end{align*}Now from $\triangle CHF$ we would have. \begin{align*} \overline{EF} = \overline{HF} = \frac{2R\cos(C)}{\tan(A)} \qquad \text{and} \qquad \overline{CF} = \frac{2R\cos(C)}{\sin(A)} \end{align*}From here's its just a single line computation to check that $\mathrm{Pow}_{\omega}(E) = \mathrm{Pow}_{\omega}(F)$. This finishes the first part. For the second part of the solution, observe that $(E,M;F,\infty) = -1$. Project this on $\omega$ to get \[-1 = (E, M ; F, \infty) \overset{A}{=} (B, H'; C, \ell \cap \odot(ABC))\]By say Brazil 2011/5, we know $\ell \cap \odot(ABC)$ is precisely $D$. Therefore we even get that $DA \parallel XY$. $\square$ Now, with reflection of orthocenter across midpoint property and $ZH \perp EF$, we can conclude that $H$ is indeed the orthocenter of $\triangle XYZ$ and the solution is complete. $\blacksquare$

Nice problem but spent too much time solving it coz I thought it was simple angle chasing (and it was!).

Let $AH\cap EF=M$, the midpoint of of both segments, we have $O$ and $H$ are isogonal conjugates w.r.t $BEFC$, this is because $\measuredangle ACO=\measuredangle BCH$ and we need to show that $OM \perp XY$. We have $OE=OF$, this is because we have $\measuredangle EFO =\measuredangle HFC=\measuredangle BAC$ and $\measuredangle FEO =\measuredangle HEB=\measuredangle BAC $. This gives us $OM\perp EF\implies OM \perp XY$. Note that $A-O-Z$ is the diameter and $AM=HM$, so $H$ is the orthocenter of $\triangle XYZ$.