Just happened to stumble upon this problem recently; I guess I'll post a quick solution. Also, I think this problem needs to be added as the 2013 ISL G1 in the contest section.

As $\angle BNC = \angle BMC = 90$, $B, N, M, C$ is cyclic. Let this circle be $\omega_3$. Let $Z(\not= W) = \omega_1 \cap \omega_2$. Let $D = AH \cap BC$. Now $BN$ is the radical axis of $\omega_1$ and $\omega_3$ and $CM$ is the radical axis of $\omega_2$ and $\omega_3$. This implies that $A=BN \cap CM$ is the radical center of these three circles, so $A, Z, W$ are colinear. Now by Power of a Point, we have $AN \cdot AB = AH \cdot AD= AZ \cdot AW = AM \cdot AC$, so we have $$\angle AEW = \angle AZH = \angle XZW = \angle WZY = \angle AZX = \angle AZY = 90$$by simple angle chasing - so $X, Y, Z, H$ are colinear.

Let $Z $ the second point of intersection of $\omega _1$ and $\omega _2$. $(BCMN)$ are concyclic and the radical axes of its circumcircle , $\cal{C}$ $\ \text{and}\ \omega _1,$ $\cal{C}$ $ \text{and} \ \omega _2,\omega _1\ \text{and}\ \omega _2,$ are concurrent at $A$ . $\widehat{WZY}=\frac{\pi}{2},\widehat{WZX}=\frac{\pi}{2}$ thus $X,Y$ and $Z$ are colinear .if $H'$ is the intersection of $XY$ and $AH$ then $H'H_1ZW$ is cyclic where $H_1$ is the foot of $AH$.hence $AZ\cdot AW=AH_1\cdot AH'$ but $AZ\cdot AW=AN\cdot AB$ then $AH'\cdot AH_1=AB\cdot AN$ besides $BHH_1N$ is cyclic implies $AN\cdot AB=AH\cdot AH_1$ so $H=H'$ and the result follows. R.HAS

The easiest way to do this is with Reim's theorem. [asy][asy] /* IMO 2014 Question 4, free script by liberator, 12 August 2014 */ unitsize(2cm); defaultpen(fontsize(10pt)); /* Initialize objects */ pair A = (-2.5, 2.5); pair B = (-3.5, -0.5); pair C = (0.5, -0.5); pair H = orthocenter(A,B,C); pair M = foot(B,C,A); pair N = foot(C,A,B); pair W = (-1.5, -0.5); pair X = rotate(180, circumcenter(B,W,N))*W; pair Y = rotate(180, circumcenter(C,W,M))*W; pair P = intersectionpoint(H--Y, circumcircle(B,W,N)); /* Draw objects */ draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1)); draw(X--Y, rgb(0.4,0.6,0.8)+dashed+linewidth(1)); draw(A--H, rgb(0.4,0.6,0.8)); draw(B--X, rgb(0.4,0.6,0.8)); draw(C--Y, rgb(0.4,0.6,0.8)); draw(circumcircle(B,W,N), red); draw(circumcircle(C,W,M), red); draw(circumcircle(A,M,N), red); /* Place dots on and label each point, label circles */ dot(A); label("$A$", A, dir(90)); dot(B); label("$B$", B, dir(200)); dot(C); label("$C$", C, dir(340)); dot(H); label("$H$", H, dir(-90)); dot(M); label("$M$", M, dir(0)); dot(N); label("$N$", N, dir(180)); dot(P); label("$P$", P, dir(-70)); dot(W); label("$W$", W, 2.5*dir(-80)); dot(X); label("$X$", X, dir(180)); dot(Y); label("$Y$", Y, dir(60)); label("$\omega_1$", B--W, 15*dir(-90)); label("$\omega_2$", C--Y, 10*dir(0)); [/asy][/asy] Let $P$ be the second intersection of $\omega_1, \omega_2$. Then $P$ is the Miquel point of $\triangle MNW$ w.r.t $\triangle ABC$, so $AMPHN$ is cyclic. $BX \parallel AH$, hence $X, H, P$ are collinear by Reim's theorem; similarly, $H,P,Y$ are collinear, and the result follows.

Let $Z$ be the second intersection of $\omega_1$ and $\omega_2$. Let $A'$ be the foot of the $A$-altitude. We have that $A$ lies on the radical axis of $\omega_1$ and $\omega_2$ because $BCMN$ is cyclic. Also by power of a point, $AZ\cdot AW=AN\cdot AB = AH\cdot AA'$, so $A'WZH$ is cyclic. Therefore, $\angle XZW=\angle YZW=\angle HZW=90^{\circ}$ and the conclusion follows. $\boxed{}$

Was too noob after drawing in the Miquel point to think of proving that it lies on $XH$, so I needed Evan's hint from his book to do so.

$AN\cdot AB=AM\cdot AC = AP\cdot AW\cdots (1)$ we have $\angle XPW=90=\angle WPX$ so $XPY$ collinear let $D$ the perpendicular from $A$ to BC$ we have $NHDB$ and HDPW$ cylich from $(1)$ $\angle XHD=\angle DWP=180-\angle DHP$ so $XHP$ collinear

Let $Z$ be the second intersection point $\omega_1$ and $\omega_2$ . Now $\angle YZW$$+$$ \angle XZW $$=$$ 90^o+90^o=180^o$. Therefore $Y$, $Z$ and $X$ are collinear. So we are left to prove that $Z$, $H$ and $X$ are collinear. Now since $Z$ is the miquel point, we have $A$, $N$, $H$, $Z$, $M$ are concyclic. So，$\angle AZM$$ +$$\angle MZW$ $= C+(A+B)=180^o$. Therefore $A$, $Z$ and $W$ are collinear. Now, $\angle AZH$ $=90^o$ and thus $\angle HZW$$ =90^o$. But $\angle WZX$$ =90^o$, therefore $Z$, $H$ and $X$ are collinear and we are done.

A non standard way to do it... Let $\ell_b,\ell_c$ be the lines perpendicular to line $BC$ at points $B,C$. Notice that as $W$ varies, $\angle WMY=\angle WNX=90^{\circ}$ and points $X,Y$ vary on lines $\ell_b,\ell_c$ respectively. Therefore, since this implies that $N$ and $NW$ are related projectively, i.e., the pencil and range have same cross ratio and essentially this is the same cross ratio for $X$ as it varies on the line $\ell_b$ we see that points $X$ and $W$ are projectively related to each other. Similarly, points $W$ and $Y$ are projectively related to each other. Thus, points $X$ and $Y$ are projectively related, i.e., there exist a projectivity mapping $X$ to $Y$. We prove that it is in fact a perspectivity. Thus, we only need to consider three positions of $W$ for which $X,Y,H$ are collinear. These are clearly true for $W=B,C,D$ where $D$ is the feet of altitude from $A$. Hence it holds for all $W$.

Let $\omega_1 \cap \omega_2$ = $G$. $A$ has the same power with respect $\omega_1$ and $\omega_2$ so $A,G$ and $W$ are collinear. Then $\angle NGA$ = $\angle ABW$ = $\angle NHA$, so $H,N,A$ and $G$ are concyclic ($\angle HGA$ = $90^\circ$). Finally $X,H,G$ and $Y$ are collinear.

Dear Mathlinkers, just for history, this result comes from Mannhein in 1893... actually a rediscovery... Sincerely Jean-Louis

$K$ is the foot of the altitudes from $A$. We need to check that $\frac{XB \cdot KC+YC \cdot KB}{BC} = {HK}$ ... (*) From $NXB \sim NWC$ and $MYC \sim MWB$, (*) becomes $KH \cdot KA = KB \cdot KC$, which is true when $H$ is reflected in $BC$ by power of $K$ wrt the circumcircle.

Solution: Let $AH\cap BC$ = $P$. Result is trivial if $W$ = $P$. So, let $W$ ≠ $P$. Let $l_1$ and $l_2$ be the lines perpendicular to $BC$ at $B$ and $C$ respectively. Let $l_3$ be the line through $H$ parallel to $BC$. Let $l_3\cap l_2$ = $E$ and $l_3\cap l_1$ = $D$. Join $DN$, $PN$, $EM$ and $PM$. Now, $DHPB$, $HECP$ and $DECB$ are rectangles. Also, $DNHPB$ and $PHMEC$ are cyclic pentagons. Since, $N$ is the center of the spiral similarity that takes $XD$ to $WP$, so, $\frac{XD}{WP}$ = $\frac{ND}{NP}$ = tan$\angle NPD$ = tan$\angle DBN$ = tan$\angle BAP$ = $\frac{BP}{AP}$ = $\frac{DH}{AP}$ ......($1$) Similarly, $\frac{EY}{WP}$ = $\frac{EH}{AP}$ ....($2$) ($1$) and ($2$) gives $\frac{XD}{EY}$ = $\frac{DH}{EH}$ Thus, $\Delta DXH$ ~ $\Delta EYH$, or, $\angle DHX$ = $\angle EHY$. Hence, $X$, $H$ and $Y$ are collinear.

2013 IMO Shortlist wrote: Comment. The original proposal also included a second statement: Let $P$ be the point on $\omega_1$ such that $WP$ is parallel to $CN$, and let $Q$ be the point on $\omega_2$ such that $WQ$ is parallel to $BM$. Prove that $P, Q$, and $H$ are collinear if and only if $BW=CW$ or $AW\perp BC$. The Problem Selection Committee considered the first part more suitable for the competition. Does anyone have a nice solution to this extension from the shortlist?

liberator wrote: Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear. Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand Not sure but I think complex numbers will solve this and I think only some basic complex plane geometry theorems need to be used.

Let $P$ be the Miquel point of $ABC$. Hence, we have $\angle APH = \frac{\pi}{2}$ by properties of the orthocenter. Since $\angle XBW=\pi-\angle XHW \implies \angle XPW = \frac{\pi}{2}.$ Similarly, we have $\angle APY = \angle WPY = \frac{\pi}{2}.$ This implies that $X,H,$ and $Y$ are colinear. $\blacksquare$

Let $P$ and $Q$ be the centers of $\omega_1$ and $\omega_2$, respectively. Let $\omega_1$ and $\omega_2$ intersect at a second point $Z \neq W$. Claim: $Z$, $H$, $W$ are collinear. Proof. Since $AN \cdot AB = AM \cdot AC$, $A$ lies on the radical axis of $\omega_1$ and $\omega_2$, which implies that $W,Z,A$ are collinear, and $WA \perp PQ$. Since $PQ \parallel XY$, it follows that $\angle AZX = 90^{\circ}$. Furthermore, $(AH)$, which goes through $M$ and $N$, will go through $Z$, the Miquel point, which means that $\angle AZH = 90^{\circ}$, implying the claim. $\blacksquare$ Similarly, $Y,Z,H$ are collinear, completing the proof. $\blacksquare$

By orthic triangle lemma $(ANHM)$ is cyclic. Let $Z$ be the Miquel's point of circle $(ANM), (BWN)$ and $(CWM)$. $\overline{AH}$ is the diameter of $(ANHZM)$. $\rightarrow$ $\angle AZH=90^\circ$ . $\overline {XW}$ is the diameter of $(BWX)$. $\angle XZW=90^\circ$. So,$\angle XZA=(180-90)^\circ=90^\circ$. $\angle AZX+\angle AXY=90^\circ+90^\circ=180^\circ$. and we have $\angle AZH=90^\circ$. Hence, $X$, $H$ and $Y$ are collinear.

JoudTabsi wrote: •Let D be the intersection point of the Circumcircles of NWB and MWC other than W •MDN=360-MDW-NDW=B+C. Hence, ANHDM is cyclic. •MDA=MHA=C. Therefore, A,D,W are collinear. •ADH=AMH=90 and XDW=90 .So D lies on XH. •WDY=90 and WDX=90 So X,D,Y are collinear. But H lies on XD. Hence X,H,Y are collinear.

We animate $W$ on $BC$. Observe that $X$ and $Y$ have degree of $1$. Then $X , H , Y$ collinear has degree $2$. It is enough to check $3$ values of $W$. We check for $B$ , $C$ and $AH \cap BC$

peace09 wrote:

Silly. Here's an actual solution: let $P$ be the Miquel point of $(AMN)$, $(BNW)$, and $(CMW)$. Clearly $A$ is the radical center of $(BCMN)$, $(BNPW)$, and $(CMPW)$; in particular $A,P,W$ are collinear. Then $HP\perp AW$ since $P\in(AMN)$ of diameter $AH$, $XP\perp AW$ since $P\in(BNW)$ of diameter $WX$, and $YP\perp AW$ since $P\in(CMW)$ of diameter $WY$; and so $H,X,Y$ are collinear. $\square$

Consider $D = (BNW) \cap (WMC)$, and let us demonstrate how powerful this point is. - $X, D, Y$ are collinear : it's because $\angle XDY = \angle XDW + \angle WDY = \frac{\pi}{2} + \frac{\pi}{2} = \pi$. - $A, D, W$ are collinear : we just need to show that $A$ lies on the radical axis of $(BNM)$ and $(WMC)$. That's true because $AN \cdot AB = AM \cdot AC$ by power of point (here, $B, N, M, C$ are concyclic). Hence $\angle ADH = \angle HDW = \frac{\pi}{2}$. Consider finally $(AHD)$ and $(HDW)$. If we show that $(XY)$ is their radical axis than we are done. Let us remember that a radical axis is the line perpendicular to the line formed by the centers, and, if a common point exist between the circles, passing through this common point. Here, we already showed that $(XY)$ contains $D$, a common point between the circles. In addition, it is perpendicular to $(AW)$, which is parallel to the line formed by the centers (why : the centers, since there are only right triangles, are midpoints, so the parallelism is not mysterious anymore, by Thales) $\huge \blacksquare$.

let $\omega_3$ be the circumcircle of $\triangle ANM$ donate $E$ be the miquel point now we have that $A-E-W$ as a fast result we get $X-E-H-Y$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -42.64193212363468, xmax = 37.62142919514159, ymin = -12.456287979858324, ymax = 25.85389619279325; /* image dimensions */ /* draw figures */ draw((-10.4950997206116,22.60018473975595)--(15.812330657344743,-5.12607334397776), linewidth(0.4)); draw((15.812330657344743,-5.12607334397776)--(-20.190422376757308,-4.594012954311232), linewidth(0.4)); draw((-20.190422376757308,-4.594012954311232)--(-10.4950997206116,22.60018473975595), linewidth(0.4)); draw((-10.4950997206116,22.60018473975595)--(-10.899013648982507,-4.731324413342385), linewidth(0.4)); draw((-16.298569222789236,6.322160526330897)--(15.812330657344743,-5.12607334397776), linewidth(0.4)); draw((-20.190422376757308,-4.594012954311232)--(-1.5099931362833348,13.130488137396515), linewidth(0.4)); draw((-5.47591954759328,-4.811468661146167)--(-1.5099931362833348,13.130488137396515), linewidth(0.4)); draw(circle((5.274466321965228,2.2215409038274205), 12.84655673265849), linewidth(0.4)); draw((-16.298569222789236,6.322160526330897)--(-5.47591954759328,-4.811468661146167), linewidth(0.4)); draw(circle((-12.779695673279125,-1.0842462590880841), 8.199837347688193), linewidth(0.4)); draw((16.024852191523735,9.254550468801007)--(-1.5099931362833348,13.130488137396515), linewidth(0.4)); draw((-5.47591954759328,-4.811468661146167)--(16.024852191523735,9.254550468801007), linewidth(0.4)); draw((-20.083471798964982,2.6429761429699887)--(-5.47591954759328,-4.811468661146167), linewidth(0.4)); draw((-20.083471798964982,2.6429761429699887)--(-16.298569222789236,6.322160526330897), linewidth(0.4)); draw((-20.083471798964982,2.6429761429699887)--(16.024852191523735,9.254550468801007), linewidth(0.4)); draw(circle((-10.629958786198769,13.474721301690662), 9.126459879221372), linewidth(0.4)); draw((-10.4950997206116,22.60018473975595)--(-5.47591954759328,-4.811468661146167), linewidth(0.4)); /* dots and labels */ dot((-10.4950997206116,22.60018473975595),dotstyle); label("$A$", (-11.030154386078584,23.32730429183617), NE * labelscalefactor); dot((15.812330657344743,-5.12607334397776),dotstyle); label("$B$", (16.057261110228783,-4.523964569876786), NE * labelscalefactor); dot((-20.190422376757308,-4.594012954311232),dotstyle); label("$C$", (-21.78285945294246,-5.052786130542223), NE * labelscalefactor); dot((-10.899013648982507,-4.731324413342385),linewidth(4pt) + dotstyle); label("$D$", (-10.677606678968292,-4.288932765136593), NE * labelscalefactor); dot((-10.76481785178594,4.34925786362537),linewidth(4pt) + dotstyle); label("$H$", (-10.501332825413147,4.818549668545914), NE * labelscalefactor); dot((-16.298569222789236,6.322160526330897),linewidth(4pt) + dotstyle); label("$M$", (-16.083338187992755,6.816320008837561), NE * labelscalefactor); dot((-1.5099931362833348,13.130488137396515),linewidth(4pt) + dotstyle); label("$N$", (-1.27633448936053,13.57348439511813), NE * labelscalefactor); dot((-5.47591954759328,-4.811468661146167),dotstyle); label("$W$", (-5.21311721875878,-4.230174813951544), NE * labelscalefactor); dot((16.024852191523735,9.254550468801007),linewidth(4pt) + dotstyle); label("$X$", (16.233534963783928,9.695459616904934), NE * labelscalefactor); dot((-20.083471798964982,2.6429761429699887),linewidth(4pt) + dotstyle); label("$Y$", (-21.665343550572363,2.1156839140336863), NE * labelscalefactor); dot((-7.270440253264129,4.989091970055546),linewidth(4pt) + dotstyle); label("$E$", (-7.034613705495285,5.464887131581447), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

We use directed angles $\measuredangle$ modulo $\pi$. Let $Z$ be the second point of intersection between $\omega_1$ and $\omega_2$. Then observe that $X, Z, Y$ are collinear as $\measuredangle XZW = \measuredangle YZW = 90^{\circ}$. Let $D$ be the point where the altitude from $A$ meets $\overline{BC}$. Then make the observation that $BX \parallel HD$, as $\measuredangle XBW = \measuredangle 90^{\circ} = \measuredangle HDW$. We now make a claim: Claim: $X, H, Z$ are collinear. Proof: We have the following angle chase: \[\measuredangle BXH = \measuredangle DHZ = \measuredangle DWZ = \measuredangle BWZ = \measuredangle BXZ. \square\] Hence as $XZ$ is the same line passing simultaneously through both $H$ and $Y$, we find that $X, H, Z, Y$ are collinear, as desired. $\blacksquare$

Draw in $(BWN) \cap (CWM) = K$, then Miquel tells us $(AMNHK)$ is cyclic. Direct angles. We then see $\angle NKX = \angle NBX = 90 - \angle ABC = \angle NAH = \angle NKH$, so $K,H,X$ are collinear, and so are $K,H,Y$, finishing.

Let $Z$ be the second intersection point of $(BWN)$ and $(CWN)$. Claim 1. Points $X$, $Y$ and $Z$ are collinear. Proof. We see by constructions that $BXZW$ and $CYZW$ are cyclic. This implies \[ \measuredangle WZX = \measuredangle WBX = 90^{\circ} = \measuredangle WCY = \measuredangle WZY, \]which means $X$, $Y$ and $Z$ are indeed collinear. Claim 2. Quadrilateral $AHZM$ is cyclic. Proof. By Miquel's theorem, $ANZM$ is cyclic. However, by the properties of the orthocenter, $ANHM$ is cyclic. From this we deduce that in fact, $ANHZM$ is cyclic, thereby $AHZM$ is cyclic. Let $K$ be the intersection point of lines $AH$ and $BC$. Claim 3. Quadrilateral $KHZW$ is cyclic. Proof. From cyclic quadrilaterals, \[ \measuredangle KHZ = \measuredangle AHZ = \measuredangle AMZ = \measuredangle CMZ = \measuredangle CWZ = \measuredangle KWZ, \]proving our claim. Finally, $\measuredangle WZH = \measuredangle WKH = \measuredangle CKA = 90^{\circ}$. Hence $X$, $Y$, $Z$ and $H$ are collinear.

We let $T$ denote the second intersection of circles $(BNW)$ and $(CMW)$. We start ff by noting that since, \[\measuredangle WTX = \measuredangle WBX = \frac{\pi}{2}\]and \[\measuredangle WTY = \measuredangle WCY = \frac{\pi}{2}\]which implies that points $X$ , $T$ and $Y$ are collinear. Next, by Radical Center Theorem on $(BNTW)$ , $(CMTW)$ and $(BNMC)$, lines $\overline{BN}$ , $\overline{TW}$ and $\overline{MC}$ concur. It is clear that $A= BN \cap CM$, so \[AT \cdot AW = AN \cdot AB = AH \cdot AD\]which implies that $DHTW$ is also cyclic. Now this finishes since, \[\measuredangle WTH = \measuredangle WDH = \frac{\pi}{2} = \measuredangle WTX\]which implies that $X$ , $H$ and $T$ are collinear. Combining this with our previous observation concludes that points $X$ , $Y$ and $H$ are collinear, as desired.

Let $Z$ be the other intersection of $\omega_1$ and $\omega_2$. By radax on $\omega_1,\omega_2,(BNMC)$ we get that $A,Z,W$ are collinear. Since $\angle XZW=\angle YZW=90^{\circ}$, we get that $X,Y,Z$ are collinear. Thus it suffices to show that $\angle HZW=90^{\circ}$, which is equivalent to $Z$ being on $(AH)$. Let $f$ denote an inversion centered at $A$ with radius $\sqrt{AN\cdot AB}$. Then $f$ swaps $Z$ and $W$. Unsurprisingly, $f$ also swaps $(AH)$ and $BC$, so we win. $\blacksquare$

Nothing new but posting it for storage. Let $D$ be the feet of the altitude from $A$ to $BC$, let $\omega_1 \cap \omega_2=\{Z\}$. Claim: Points $\overline{X-Z-Y}$ are collinear. Proof: $\angle XZW \stackrel{\omega_1}{=}=90$ and $\angle YZW \stackrel{\omega_1}{=}=90$ Hence $\angle XZY=\angle XZW+\angle YZW=180 \implies \angle XZY=180 \implies$ Points $\overline{X-Z-Y}$ are collinear. $\square$ Claim: Points $\overline{A-Z-W}$ are collinear. Proof: First note that $\omega_1 \cap \omega_2 =\{W,Z\}$ so $ZW$ is the radical axis of $\omega_1$ and $\omega_2$ Also its well known that points $B,N,M$ and $C$ are concyclic so by Power of the Point Theorem we get: $Pow(A,\omega_1)=AN \cdot AB=Pow(A, \odot(BNMC))=AM \cdot AC=Pow(A,\omega_2) \implies Pow(A,\omega_1)=Pow(A,\omega_2)$ So $A$ lies on the radical axis of $\omega_1$ and $\omega_2$ hence $A$ lies on the line $WZ \therefore$ Points $\overline{A-Z-W}$ are collinear. $\square$ Claim: Points $W,D,H$ and $Z$ are concyclic. Proof: It's also well known that points $D,N,B$ and $H$ are concyclic So by POP we get: $AH \cdot AD=Pow(A,\odot(DNBH))=AN \cdot AB=Pow(A,\omega_1)=AZ \cdot AW \implies AH \cdot AD= AZ \cdot AW$ which by the converse of POP we have that: Points $W,D,H$ and $Z$ are concyclic. $\square$ Let $\odot(WDHZ)=\Gamma$. Claim: Points $\overline{X-H-Y}$ are collinear. Proof: From $\Gamma \implies \angle HZW \stackrel{\Gamma}{=} 180-\angle HDZ=180-90=90=\angle XZW \implies \angle HZW=\angle XZW \implies$ Points $\overline{X-H-Z}$ are collinear, combining with the first claim that Points $\overline{X-Z-Y}$ are collinear we get that $\overline{X-H-Z-Y}$ are all collinear Hence points $\overline{X-H-Y}$ are collinear $\blacksquare$