Let $O$ denote the circumcenter of $\triangle{BHC}$ and let $E$ denote the circumcenter of $\triangle{AHP}$. We begin with a crucial claim. Claim: If $H'$ is the antipode of $H$ w.r.t $\odot(BHC)$ then $\overline{AM} \cap \overline{PG}=H'$ Proof: Let $H_A$ denote the $A$ -humpty point. Then notice that $$\angle HH_AH'=\angle HH_AM=90^\circ$$and hence $H' \in \overline{AM}$. Now just note that $$\angle HGH'+\angle HGP=180^\circ$$and so $H' \in \overline{PG}$ $\qquad \blacksquare$ Now notice that its well known that $H'$ is the reflection of $A$ in $M$ (just angle chase) Also clearly $\overline{AP} \parallel \overline{BC}$ and hence we have that $\overline{AM}=\tfrac{1}{2}\overline{AP}$. But notice that clearly if $X=\overline{MH} \cap \overline{AP}$ then $X$ Is the midpoint of $\overline{AP}$ and hence we get that $MXPF$ is a parallelogram which implies that $\overline{MH} \parallel \overline{PG}$ with which we are done $\qquad \blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25.15639756385678cm); 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 = -14.693141418188027, xmax = 10.463256145668753, ymin = -10.172937052451232, ymax = -0.3526248182641615; /* image dimensions */ pen qqttcc = rgb(0.,0.2,0.8); /* draw figures */ draw((-2.685950600670045,-2.6026295254967415)--(-3.624326490091042,-5.644785490957004), linewidth(0.8)); draw((-3.624326490091042,-5.644785490957004)--(0.33825647657857444,-5.690281506463871), linewidth(0.8)); draw((0.33825647657857444,-5.690281506463871)--(-2.685950600670045,-2.6026295254967415), linewidth(0.8)); draw(circle((-1.63070852463975,-4.593929414274443), 2.2536217708548194), linewidth(0.8)); draw(circle((-1.655361488872718,-6.741137583146432), 2.253621770854819), linewidth(0.8)); draw(circle((-5.254956026365177,-3.6468793463522564), 2.77312938133218), linewidth(0.8)); draw((-7.799308487827341,-2.543920998335784)--(-2.685950600670045,-2.6026295254967415), linewidth(0.8)); draw((-1.6430350067562338,-5.667533498710437)--(-5.242629544248693,-2.5732752619162627), linewidth(0.8) + qqttcc); draw((-0.6001194128424232,-8.732437471924131)--(-7.799308487827341,-2.543920998335784), linewidth(0.8) + qqttcc); draw((-0.6001194128424232,-8.732437471924131)--(-2.685950600670045,-2.6026295254967415), linewidth(0.8)); draw((-0.5650721123368723,-5.67991002608484)--(-7.799308487827341,-2.543920998335784), linewidth(0.8)); draw((-2.720997901175595,-5.655156971336034)--(-2.685950600670045,-2.6026295254967415), linewidth(0.8)); /* dots and labels */ dot((-2.685950600670045,-2.6026295254967415),linewidth(2.pt) + dotstyle); label("$A$", (-2.6739736932342324,-2.4396740976359865), NE * labelscalefactor); dot((-3.624326490091042,-5.644785490957004),linewidth(2.pt) + dotstyle); label("$B$", (-3.885207650012522,-5.551613648127904), NE * labelscalefactor); dot((0.33825647657857444,-5.690281506463871),linewidth(2.pt) + dotstyle); label("$C$", (0.38206275156022085,-5.6261511223911835), NE * labelscalefactor); dot((-2.710603564903013,-4.749837694368731),linewidth(2.pt) + dotstyle); label("$H$", (-2.6367049561025926,-4.675798325534371), NE * labelscalefactor); dot((-2.720997901175595,-5.655156971336034),linewidth(2.pt) + dotstyle); label("$D$", (-2.6553393246684127,-5.588882385259544), NE * labelscalefactor); dot((-1.6430350067562338,-5.667533498710437),linewidth(2.pt) + dotstyle); label("$M$", (-1.574545947850862,-5.588882385259544), NE * labelscalefactor); dot((-0.5650721123368723,-5.67991002608484),linewidth(2.pt) + dotstyle); label("$D'$", (-0.4937525710333115,-5.607516753825364), NE * labelscalefactor); dot((-7.799308487827341,-2.543920998335784),linewidth(2.pt) + dotstyle); label("$P$", (-7.910231260229606,-2.4396740976359865), NE * labelscalefactor); dot((-3.782515686424972,-5.996808283772401),linewidth(2.pt) + dotstyle); label("$G$", (-4.220626284197279,-6.0920103365366804), NE * labelscalefactor); dot((-5.242629544248693,-2.5732752619162627),linewidth(2.pt) + dotstyle); label("$T$", (-5.1709790810540905,-2.4955772033334465), NE * labelscalefactor); dot((-0.6001194128424232,-8.732437471924131),linewidth(2.pt) + dotstyle); label("$Z$", (-0.549655676730771,-8.551746987224902), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $ T = HM \cap AP $ Now $ \triangle AHP \sim \triangle DHD' $, consider the homothety $\Omega$ centered at $H$ that maps $ DD' \mapsto AP $ Then $ \Omega : M \mapsto T $. Now as $M$ is midpoint of $DD' \Rightarrow \boxed{ T \mbox{ is midpoint of } AP } $ Now it is well known that $ ( BHC ) := $ reflection of $(ABC)$ about $BC$ Let $ AM \cap ( BHC ) = Z $, then $ Z$ is reflection of $A$ about $M$. Thus, we have $ \boxed{ M \mbox{ is midpoint of } AZ }$ Hence by midpoint theorem $ \boxed{ MT \parallel PZ }$ Claim : $ G,P,Z $ are collinear. Let $ \measuredangle := $ directed angle mod $ 180^{\circ} $ Now $B,C,Z,G$ are concyclic and $ ABZC $ is a parallelogram thus $$ \measuredangle BGZ = \measuredangle BCZ = \measuredangle CBA $$Also $ \measuredangle PGB + \measuredangle BGH = \measuredangle PGH = \measuredangle PAH = 90^{\circ} $ \begin{align*} \Rightarrow \measuredangle PGB & = 90^{\circ} - \measuredangle BGH \\ & = 90^{\circ} - \measuredangle BCH \\ & = \measuredangle ABC \\ & = - \measuredangle BGZ \end{align*} $\therefore \measuredangle BGP = \measuredangle BGZ $ Thus we get $\boxed{ G,P,Z \mbox{ are collinear. } } $ Hence $ PG \parallel HM $ $ \Rightarrow \measuredangle MHG = \measuredangle PGH = \measuredangle PAH = 90^{\circ} $ Hence, proved.

Introduce the following points : $O_1$ is the center of $\odot(ABC)$ $O_2$ is the center of $\odot(HBC)$ $O_3$ is the center of $\odot(AHP)$ $O_4$ is the center of $\odot(HDD')$ We know that $O_3$ and $O_4$ are the midpoints of $PH$ and $HD'$ respectively.

By Lemma $O_2$ is the reflection of $O_1$ over $BC$. Since $M$ is the midpoint of $DD'$ and $O_1O_2 \perp DD',O_1, O_4, O_2$ are collinear. $O_2O_3 \perp GH$ since $GH$ is the radical axis of $\odot(AHP)$ and $\odot(BHC)$ Hence it suffices to show that $MH \parallel O_2O_3$ $AP \parallel BC \Longrightarrow \frac{DH \div 2}{HP \div 2} = \frac{O_4H}{HO_3} = \frac{D'H}{HP} = \frac{AH}{HD}$ $O_4M = HD \div 2$ since $O_4M$ is the $D'$ midline wrt $\triangle D'HD$ $O_2M =O_1M = O_1B \cos A = AH \div 2$ The last equality above follows from LOS in $\triangle AHB$. Hence, $\frac{O_4H}{HO_3} = \frac{O_4M}{MO_2} \Longrightarrow MH \parallel O_2O_3$

A four page coordinate bash also does the trick.

[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(105), B = dir(210), C = dir(330), H = orthocenter(A,B,C), D = foot(A,B,C), M = (B+C)/2, R = dir(75), D1 = 2M-D, A1 = 2M-H, Q = 2*D1-R, P = extension(A,R,H,D1), G = foot(H,P,Q); draw(unitcircle^^circumcircle(H,G,Q)); draw(B--C^^A--P--Q--R--A--D^^A1--H--D1^^P--H--G); dot("$A$", A, dir(95)); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$H$", H, dir(50)); dot("$D$", D, dir(270)); dot("$M$", M, dir(240)); dot("$D'$", D1, dir(315)); dot("$A'$", A1, dir(320)); dot("$P$", P, dir(150)); dot("$G$", G, dir(180)); dot("$Q$", Q, dir(285)); dot("$R$", R, dir(75)); [/asy][/asy] Let $A'$ be the antipode of $A$ on $(ABC)$, $R$ be the point on $(ABC)$ such that $\overline{AR} \parallel \overline{BC}$, and $Q$ be the reflection of $A$ over $M$. It is well known that $M$ is the midpoint of $\overline{A'H}$, so by symmetry $Q$ is the antipode of $H$ on $(BHC)$, and $R$, $D'$, $A'$, $Q$ are collinear with $D$ the midpoint of $\overline{RQ}$. We can redefine $G \ne Q = \overline{PQ} \cap (BHC)$, and it suffices to prove $\overline{HA'} \parallel \overline{PQ}$. But this is immediate from $$\dfrac{D'A'}{D'H} = \dfrac{DH}{D'H} = \dfrac{D'R}{D'P} = \dfrac{D'Q}{D'P}.$$$\blacksquare$

In the following solution $(XYZ)$ denote the circumcircle of $XYZ$. Let $AH$ cut $(BHC)$ again at $X$.Let perpendicular on $BC$ at $D'$ cut $(BHC)$ at $R$ and $Y$ where $R,H$ lies same side on $BC$.As the perpendicular line on $BC$ at $D$ and $D'$ are symmetric about the diametre of $(BHC)$ passing through $M$, so $HRYX$;$HRD'D$;$D'DXY$ are rectangle. $\textcolor{blue}{CLAIM1:}$$P,G,Y$ are collinear. $\textcolor{red}{Proof:}$ Let $PG$ cut $(BHC)$ again at $Y'$.As $G,H$ are common point of $(BHC)$ and $(PAH)$ and $AH\cap (BHC)= X$,so by Reim's theorem $AP||XY'$.On the other hand $XY|| DD'||AP$ and $Y$ lies on $(BHC)$.So $Y'\equiv Y$.$\square$ Now as $PAHG$ is concyclic so $\angle PGH=180^{\circ}-\angle PAH=90^{\circ}$.So it is enough to show that $PG||HM$ which will prove that $90^{\circ}=\angle PGH=\angle MHG$. We will prove that $PY||HM$ Let the perpendicular line on $BC$ at $D'$ cut $PA$ at $S$.Also let $MH\cap PS=Q$ and $HM\cap YS=V$ $\textcolor{blue}{CLAIM2:}$ $D'S=D'Y$ $\textcolor{red}{Proof:}$ Let $X'$ be the reflection of $A$ over $BC$.Then $\angle BXC=\angle BAC=180^{\circ}-\angle BHC$ So $B,H,C,X'$ are concyclic.So $X'\equiv X$ Now as $ASD'D$ and $DD'YX$ are rectangle so $SD'=AD=DX=D'Y$.$\square$ $\textcolor{blue}{CLAIM3:}$ $RD'=D'V$ $\textcolor{red}{Proof:}$$DH||D'V$ and $M= HV\cap DD'$ is the midpoint of $DD'$.SO $DHD'V$ is a parallegram.So $HD=D'V$.But since $HDD'R$ is rectangle $HD=RD'$$\square$ $\textcolor{blue}{CLAIM4:}$ $QV||PY$. $\textcolor{red}{Proof:}$ On one hand we have, $\frac{SY}{SP}=2\frac {SD'}{SP}=2\frac{RD'}{RG}$.[The first equality by $D'$=midpoint of $SY$] On the other hand , $\frac{SV}{SQ}=\frac{RV}{RH}=2\frac{RD'}{RH}$[AS,$D'$ is the midpoint of $RV$] SO $\frac{SY}{SP}=\frac{SV}{SQ}$ which implies $PY||QV$.So $PG||HM$.SO $\angle MHG=90^{\circ}$$\blacksquare$.

MarkBcc168 wrote: Let acute scalene triangle $ABC$ have orthocenter $H$ and altitude $AD$ with $D$ on side $BC$. Let $M$ be the midpoint of side $BC$, and let $D'$ be the reflection of $D$ over $M$. Let $P$ be a point on line $D'H$ such that lines $AP$ and $BC$ are parallel, and let the circumcircles of $\triangle AHP$ and $\triangle BHC$ meet again at $G \neq H$. Prove that $\angle MHG = 90^\circ$. Proposed by Daniel Hu. Another $MH$ line configuration!

Claim 1: $\overline{MH}$ passes through the midpoint of $AP.$

Now, let us denote the centers of $\odot(APH)$ and $\odot(ABC)$ by $X$ and $O$ respectively. Also, we note $X$ is the midpoint of $PH$, since $\triangle APH$ is right-angled at $A$. Claim 2: $O'$, the reflection of $O$ over $BC$ is the circumcenter of $\triangle BHC.$

Now, let us denote the midpoint of $GH$ by $L$. Since $\overline{GH}$ is the radical axis of $\odot(APH)$ and $\odot(BHC)$, so $\overline{O'X}$ must pass through $L$ and $\overline{O'X} \perp \overline{GH}.$ Next, in $\triangle APH$, we have $XF \parallel AH$ and $XF = \frac{AH}{2}$ by Midpoint theorem. Also, we have $O'M \parallel AH$ and $O'M = \frac{AH}{2}.$ Hence, $XFMO'$ is a parallelogram. So, $XO' \parallel FM\implies \angle GLO' = \angle GHM.$ By Radical Axis theorem, $\angle GLO' = 90^{\circ} \implies \angle MHG = 90^{\circ}. \quad\blacksquare$

Let $O$ denote the circumcenter of $\odot(ABC)$ and $O'$ the circumcenter of $\odot(BHC)$. It is well - known that $\odot(BHC)$ is reflection of $\odot(ABC)$ over $BC$, therefore $O'$ is reflection of $O$ over the midpoint of $BC$. Let $J$ denote the midpoint of $HP$, which is obviously the center of $\odot(AHP)$. Claim: $HM \parallel JO'$ Proof: Assume that $OO'$ intersects $HD'$ at point $K$. Clearly $KM$ is the midline in $\triangle HDD'$. It is well - known that $AH=OO'$ (this can be simply proved using complex numbers). Also note that $\triangle APH \sim \triangle DD'H$. Therefore: $$ \frac{PH}{HD'}=\frac{2JH}{2HK} =\frac{AH}{HD}=\frac{OO'}{HD}=\frac{2MO'}{2KM}$$As a result: $$\frac{JH}{HK}=\frac{O'M}{MK} $$This implies that $HM \parallel JO'$ as desired. Now to finish note that $GH$ is radical axis of $\odot(BHC)$ and $\odot(AHP)$. Therefore $GH \perp JO'$. Since $HM \parallel JO'$, we conclude that $GH \perp HM $, which implies that $\angle MHG=90$ as desired.

The $A-Queue$ point is the orthocenter of triangle $APG$. The problem statement follows easily from this fact.

I didn't find a synthetic solution until after the time, so I will write the (fairly horrible) barybash I did (I am sorry). We use Barycentric coordintates wrt triangle $ABC$. Use Conway's Notation: $S_A=\frac {b^2+c^2-a^2}{2}, S_B= \frac {a^2+c^2-b^2}{2}, S_C= \frac {a^2+b^2-c^2}{2}$, and let $S_{AB}=S_AS_B$ and $S$ be twice the area of triange ABC. We have $B=(0, 1, 0), C=(0, 0, 1), H=(S_{BC} : S_{CA} : S_{AB})$. Let circle $BHC$ have equation $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0$ From points $B$ and $C$, we get $v=0, w=0$ respectively, and then using $H$, we obtain $$-S_AS_BS_C(a^2S_A+b^2S_B+c^2S_C)+uS_BS_C(S_{BC}+S_{CA}+S_{AB})=0$$Now $S_B+S_C=a^2$, with cyclic varations, and so it follows that $\frac{1}{2} \cdot (a^2S_A+b^2S_B+c^2S_C)=S_{BC}+S_{CA}+S_{AB}$, (and also by the Conway identites, both are equal to $S^2$) and we get $u=2S_A$. Since $H=(S_{BC} : S_{CA} : S_{AB}), D(0: S_{CA} : S_{AB})=(0: S_C: S_B)$, meaning $D'=(0: S_B: S_C)$. Let $D'H$ have equation $ux+vy+wz$=0. By using $D'$ and $H$, we can find the ratio $u: v: w$ and find that the line has equation $S_A(S_C^2-S_B^2)x-S_BS_C^2y+S_B^2S_Cz=0$. The equation of $BC$ is $x=0$, so the point at infinity on $BC$ is $(0: -1: 1)$. Since $AP$ is parallel to $BC$, it meets it at $(0: -1: 1)$. Additionally using $A=(1, 0, 0)$, we find that $AP$ has equation $y+z=0$ P is the intersection of $S_A(S_C^2-S_B^2)x-S_BS_C^2y+S_B^2S_Cz=0$ and $y+z=0$ and we can solve this to find that $P=(-S_{BC} : S_A(S_B-S_C): S_A(S_C-S_B))$ Let circle $AHP$ have equation $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0$ Using $A=(1, 0, 0)$ gives $u=0$. Using $H=(S_{BC} : S_{CA} : S_{AB})$ and the identity $\frac{1}{2} \cdot (a^2S_A+b^2S_B+c^2S_C)=S_{BC}+S_{CA}+S_{AB}$ gives $S_Cv+S_Bw=0$. Using $P=(-S_{BC} : S_A(S_B-S_C): S_A(S_C-S_B))$, we get (after some simplifications) that $v-w=\frac{a^2S_A(S_B-S_C)}{S_{BC}}+c^2-b^2$. We can solve these equations to find that $v=\frac{S_AS_B}{S_C}-S_A+S_B$ and $w=S_C-S_A+\frac{S_AS_C}{S_B}$. We have that circle $BHC$ is $$-a^2yz-b^2zx-c^2xy+2S_A(x+y+z)=0$$and that cirlce $AHP$ is $$-a^2yz-b^2zx-c^2xy+((\frac{S_AS_B}{S_C}-S_A+S_B)y+(S_C-S_A+\frac{S_AS_C}{S_B})z))(x+y+z)=0$$. These circles meet at $H$ and $G$, so we have that $GH$ is their radical axis. By the Barycentric Radical Axis theorem the two circles $-a^2yz-b^2zx-c^2xy+(u_1x+v_1y+w_1z)(x+y+z)=0, -a^2yz-b^2zx-c^2xy+(u_2x+v_2y+w_2z)(x+y+z)=0$ have radical axis $(u_1-u_2)x+(v_1-v_2)y+(w_1-w_2)z=0$ So GH is $$2S_Ax+(S_A-S_B-\frac{S_AS_B}{S_C})y+(S_A-S_C-\frac{S_AS_C}{S_B})z=0$$which we write as $$2S_AS_BS_Cx+(S_AS_BS_C-S_B^2S_C-S_AS_B^2)y+(S_AS_BS_C-S_BS_C^2-S_AS_C^2)z=0$$ Choose the point $P'$ on this with $x=0$. We find $P'=(0 : S_AS_C^2+S_BS_C^2-S_AS_BS_C : S_AS_BS_C-S_B^2S_C-S_AS_B^2)$ $$=(0 , \frac{S_AS_C^2+S_BS_C^2-S_AS_BS_C}{(S_A+S_B)S_C^2-(S_A+S_C)S_B^2} , \frac{S_AS_BS_C-S_B^2S_C-S_AS_B^2)}{(S_A+S_B)S_C^2-(S_A+S_C)S_B^2}=(0 : \frac{c^2S_C^2-S_AS_BS_C}{c^2S_C^2-b^2S_B^2} , \frac{S_AS_BS_C-b^2S_B^2}{c^2S_C^2-b^2S_B^2}) $$. Shift the circumcenter of the triangle to the zero vector so that $\vec H= \vec A+\vec B+ \vec C$. We have $\overrightarrow{P'H}=\vec A+ \frac{S_AS_BS_C-b^2S_B^2}{c^2S_C^2-b^2S_B^2}\vec B + \frac{c^2S_C^2-S_AS_BS_C}{c^2S_C^2-b^2S_B^2}\vec C$ We also have $\vec H= \frac{S_{BC}}{S_{BC}+S_{CA}+S_{AB}}\vec A + \frac{S_{CA}}{S_{BC}+S_{CA}+S_{AB}}\vec B+ \frac{S_{AB}}{S_{BC}+S_{CA}+S_{AB}}\vec C, \vec M =\frac{1}{2}\vec B+\frac{1}{2}\vec C$ giving $$\overrightarrow{MH}=\frac{S_{BC}}{S_{BC}+S_{CA}+S_{AB}}\vec A+\frac{S_{CA}-S_{BC}-S_{AB}}{2S_{BC}+2S_{CA}+2S_{AB}}\vec B+\frac{S_{AB}-S_{BC}-S_{CA}}{2S_{BC}+2S_{CA}+2S_{AB}}\vec C$$ The sum of the coefficients of this vector is $0$, so by the Generalised perpendicularity criterion, $MH\perp HP' \Leftrightarrow a^2(z_1y_2+y_1z_2)+b^2(x_1z_2+z_1x_2)+c^2(y_1x_2+x_1y_2)=0$ where $$(x_1,y_1,z_1)=(1, \frac{S_AS_BS_C-b^2S_B^2}{c^2S_C^2-b^2S_B^2}, \frac{c^2S_C^2-S_AS_BS_C}{c^2S_C^2-b^2S_B^2}), (x_2,y_2.z_2)=(\frac{S_{BC}}{S_{BC}+S_{CA}+S_{AB}}, \frac{S_{CA}-S_{BC}-S_{AB}}{2S_{BC}+2S_{CA}+2S_{AB}},\frac{S_{AB}-S_{BC}-S_{CA}}{2S_{BC}+2S_{CA}+2S_{AB}}) $$. It is sufficient to show that $$a^2((c^2S_C^2-S_AS_BS_C)(S_{CA}-S_{BC}-S_{AB})+(S_AS_BS_C-b^2S_B^2)(S_{AB}-S_{BC}-S_{CA}))$$$$+b^2((c^2S_C^2-b^2S_B^2)(S_{AB}-S_{BC}-S_{CA})+(c^2S_C^2-S_AS_BS_C)(2S_{BC}))$$$$+c^2((S_AS_BS_C-b^2S_B^2)(2S_{BC})+(c^2S_C^2-b^2S_B^2)(S_{CA}-S_{BC}-S_{AB}))=0$$ Using $S_B+S_C=a^2$ and cyclic variations, we can express everything in terms of $S_A, S_B, S_C$, and we show that this expression is equal to 0: Replace $S_A, S_B, S_C$ with $A, B, C$ respectively. The above is $$(B+C)(((A+B)C^2-ABC)(CA-BC-AB)+(ABC-(A+C)B^2)(AB-BC-CA))$$$$+(A+C)(((A+B)C^2-(A+C)B^2)(AB-BC-CA)+((A+B)C^2-ABC)(2BC)) $$$$+(A+B)((ABC-(A+C)B^2)(2BC)+((A+B)C^2-(A+C)B^2)(CA-BC-AB)) $$$$=(B+C)(C(AC+BC-AB)(CA-BC-AB)+B(AC-AB-BC)(AB-BC-CA))$$$$+(A+C)((AC^2+BC^2-AB^2-B^2C)(AB-BC-CA)+2BC^2(AC+BC-AB))$$$$+(A+B)((2B^2C(AC-AB-BC)+((AC^2+BC^2-AB^2-B^2C)(CA-BC-AB))$$ Using $AC^2+BC^2-AB^2-B^2C=C(AC+BC-AB)+B(AC-AB-BC)$, we can factorise the first line in the above expression as $$(B+C)(AC+BC-AB)(AC-BC-AB)(C-B)$$, the second line as $$(A+C)(AC+BC-AB)(AC-BC-AB)(-B-C)$$and the third as $$(A+B)(AC+BC-AB)(AC-BC-AB)(B+C)$$. So the expression is the sum of these, which is $(B+C)(AC+BC-AB)(AC-BC-AB)(C-B-A-C+A+B)=0$ This implies $MH\perp HP' $ which shows $MH\perp HG$ as required.

[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 = -9.713265938717898, xmax = 13.883674226637623, ymin = -10.427071492316575, ymax = 8.290048185585707; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((0.9488075920872171,4.513200822008997)--(-4.48,-0.66), linewidth(1.2) + wrwrwr); draw((-4.48,-0.66)--(2.42,-0.56), linewidth(1.2) + wrwrwr); draw((2.42,-0.56)--(0.9488075920872171,4.513200822008997), linewidth(1.2) + wrwrwr); draw((0.9488075920872171,4.513200822008997)--(1.0226257460407102,-0.5802518007820187), linewidth(1.2) + wrwrwr); draw(circle((-1.0040302689482739,-2.4019114425691224), 3.888009908030206), linewidth(1.2) + wrwrwr); draw(circle((-1.0559697310517269,1.1819114425691228), 3.888009908030206), linewidth(1.2) + wrwrwr); draw((1.00074705419067,0.9293779368707521)--(2.7449862391248985,-1.371622033162286), linewidth(1.2) + wrwrwr); draw((-4.753046777021447,-3.432200851975958)--(2.6930467770214452,2.212200851975957), linewidth(1.2) + wrwrwr); draw(circle((5.847659023735562,2.7919108533615655), 5.192454574238403), linewidth(1.2) + wrwrwr); draw((-4.753046777021447,-3.432200851975958)--(8.950331808346226,6.955443739885415), linewidth(1.2) + wrwrwr); draw((0.9488075920872171,4.513200822008997)--(10.694570993280454,4.654443769852377), linewidth(1.2) + wrwrwr); draw((-3.0826257460407103,-0.6397481992179815)--(10.694570993280454,4.654443769852377), linewidth(1.2) + wrwrwr); /* dots and labels */ dot((0.9488075920872171,4.513200822008997),dotstyle); label("$A$", (0.8151138801022,4.7471648179827755), NE * labelscalefactor); dot((-4.48,-0.66),dotstyle); label("$B$", (-4.966868879261028,-0.6005836614178768), NE * labelscalefactor); dot((2.42,-0.56),dotstyle); label("$C$", (2.603267277901804,-0.650718803412258), NE * labelscalefactor); dot((-1.03,-0.61),linewidth(4pt) + dotstyle); label("$M$", (-1.223715227669311,-0.48360166343098754), NE * labelscalefactor); dot((1.0226257460407102,-0.5802518007820187),linewidth(4pt) + dotstyle); label("$D$", (1.0825013040722342,-0.45017823543473345), NE * labelscalefactor); dot((-3.0826257460407103,-0.6397481992179815),linewidth(4pt) + dotstyle); label("$D'$", (-3.011868625468915,-0.5003133774291146), NE * labelscalefactor); dot((1.00074705419067,0.9293779368707521),linewidth(4pt) + dotstyle); label("$H$", (0.7148435961134372,1.1207228803892082), NE * labelscalefactor); dot((-1.0559697310517269,1.1819114425691228),linewidth(4pt) + dotstyle); label("$O$", (-1.357408939654328,0.467588844154296), NE * labelscalefactor); dot((-3.06074705419067,-2.1493779368707515),linewidth(4pt) + dotstyle); label("$A'$", (-3.1789857654501863,-2.5725659131968674), NE * labelscalefactor); dot((2.7449862391248985,-1.371622033162286),linewidth(4pt) + dotstyle); label("G", (2.8038078458793296,-1.2356287933467043), NE * labelscalefactor); dot((-4.753046777021447,-3.432200851975958),linewidth(4pt) + dotstyle); label("$T'$", (-5.300832875234808,-3.508421897091982), NE * labelscalefactor); dot((2.6930467770214452,2.212200851975957),linewidth(4pt) + dotstyle); label("$T$", (2.703537561890567,2.40752485824499), NE * labelscalefactor); dot((10.694570993280454,4.654443769852377),linewidth(4pt) + dotstyle); label("$P$", (10.758583708987848,4.78058824597903), NE * labelscalefactor); dot((5.821689292683837,4.583822295930688),linewidth(4pt) + dotstyle); label("$N$", (5.728357795551579,4.78058824597903), NE * labelscalefactor); dot((8.950331808346226,6.955443739885415),linewidth(4pt) + dotstyle); label("$L$", (9.020565453182625,7.086804777720561), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $A'$ be the antipode of $A$ and let $T'$ be the reflection of $T$ wrt $M$. Claim. If $HG\perp HT$, $HG=AT$. pf) note that $(BHC)$ has the same radius with $(ABC)$. Since $HT'=TA'$, by Pythagorean thm, we get $AT=HG$. Now, let $N=MH\cap AP, L=MH\cap (AHP)$ note that $N$ is the midpoint of $AP$ since $\triangle AHP\sim \triangle DHD'$ Since $\angle ATN=\angle NLP=90$ and $AN=NP$, $ATPL$ is a parallelogram. Redefine $G$ as the intersection of $(AHP)$ and a line perpendicular to $AP$ passing through $T$. Note that $AN^2=NP^2=NT\cdot NH$. Therefore $\angle ATP=180-\angle AHP=180-\angle AGP$. Since $GT\perp AP$, $T$ is the orthocenter of $\triangle AGP$. Let $O'$ be the midpoint of $HP$. (the circumcenter of $(AHP)$) It is well-known that $GT=2O'N$. Since $AH=2O'N$ ($N, O'$ are midpoints of $AP, PH$), $AH=GT$. Hence, $AHGT$ is a parallelogram. Then, we obtain that $AT=HG$. Note that since $L$ is the reflection of $T$ wrt $N$, $L$ is the antipode of $G$. $\therefore$ $GL$ is the diameter of $(AHP)$. $\angle GHL=90$ Now by Claim, we obtain that $B, H, C, G$ are cyclic. We are done.

Complex numbers work here very well

Wrong

Am I the only one that used that the radical center of the two circles and the circumcenter?

Maybe this solution has already been posted, but anyway... [asy][asy] import olympiad; import geometry; pair A = dir(110), B = dir(210), C = dir(-30), H=A+B+C, D=foot(A, B, C), M=(B+C)/2, D1=2*M-D; real r = abs(A-H)/abs(A-D); pair N = (H-r*M)/(1-r), P = 2*N-A; pair O1 = circumcenter(A, P, H), O2 = circumcenter(B, H, C); pair P = intersectionpoint(line(A, N), line(D1, H)); string[] l = {"$A$", "$B$", "$C$", "$D$", "$D'$", "$M$", "$N$", "$H$", "$O_1$", "$O_2$", "$P$"}; pair[] p = {A, B, C, D, D1, M, N, H, O1, O2, P}; pair[] ll = {dir(80), dir(215), dir(-10), D, dir(-40), dir(-55), dir(90), dir(45), dir(250), O2, P}; draw(A--B--C--cycle); draw(A--D, grey); draw(O1--N--M--O2--cycle, blue); draw(circumcircle(A, P, H)^^circumcircle(B, H, C), magenta); draw(D1--P--A); for (int i=0; i<l.length; ++i) {dot(l[i], p[i], dir(ll[i]));} [/asy][/asy] Since $O_1N = \frac12 AH = O_2M$ and $O_1N \parallel O_2M$, $O_1NMO_2$ is a parallelogram. In particular, $NM \parallel O_1O_2$. Since $\overline{GH}$ is the radical axis of $(APH)$ and $(BHC)$, it is perpendicular to $\overline{O_1O_2}$ which implies the result.

I know this is similar to many of the above solutions. But let's try to write this up as concisely as possible. Construct the parallelogram $ABA'C$. Since $AP\parallel BC$, line $HM$ passes through the midpoint of $AP$. Hence, the homothety $\mathcal{H}(A,2)$ maps $HM$ to $PA'$; these two lines are then parallel. However, it's evident that $P$ and $A'$ are the antipodes of $H$ w.r.t. $\odot(AHP)$ and $\odot(BHC)$ respectively. Thus, $G$ is the projection of $H$ onto $PA'$, done.

Let $A'$ and $H'$ be the reflections of $A$ and $H$ over $M$, respectively, so that $ABA'C$ and $HBH'C$ are both parallelograms. Then $A'$ lies on $(BHC)$ by orthocenter reflections, and $\overline{A'D'}$ and $H'$ are the $A'$-altitude and orthocenter of $\triangle A'BC$, respectively, by taking a homothety at $M$. Now, let $O_1$ and $O_2$ denote the centers of $(AHP)$ and $(BHC)$, respectively. Since $GH$ is the radical axis of $(AHP)$ and $(BHC)$, we have $\overline{O_1O_2} \perp \overline{GH}$; it now suffices to show that $\overline{O_1O_2} \parallel \overline{HH'}$. But $O_1$ is the midpoint of $\overline{HP}$ since $\angle HAP = 90^{\circ}$, and $O_2$ is the midpoint of $\overline{A'H}$ by orthocenter reflections; thus $\overline{O_1O_2} \parallel \overline{AP}$. To finish, note that $$\frac{D'H}{D'P} = \frac{DH}{DA} = \frac{D'H'}{D'A'},$$so $\overline{AP} \parallel \overline{HH'}$, and $\overline{O_1O_2} \parallel \overline{HH'}$ as well.

Denote the reflection of $A$ over $M$ as $A'$. Because $A$ is the orthocenter of $\triangle BHC$, we must have that $A'$ is the antipode of $H$ with respect to $(BHC)$. Then, notice that \[\angle PGA' = \angle PGH + \angle A'GH = \angle PAH +\angle A'GH = 180^\circ .\] Hence, the points $A'$, $B$, and $P$ are collinear. Notice now that extending $MH$ to intersect $AP$ at point $N$ shows that $MN$ is the midline in triangle $AA'P$, so $\overline{MH} \parallel \overline{A'P}$. This immediately gives us our desired result. $\square$ [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 = -19.076698703296444, xmax = 33.06965747135947, ymin = -15.954717961850513, ymax = 16.311332984343537; /* image dimensions */ /* draw figures */ draw((5,12)--(0,0), linewidth(1)); draw((0,0)--(14,0), linewidth(1)); draw((14,0)--(5,12), linewidth(1)); draw((-3.8,12)--(5,12), linewidth(1)); draw(circle((0.6,7.875), 6.031220854851861), linewidth(1)); draw(circle((7,-4.125), 8.125), linewidth(1)); draw((-3.8,12)--(9,-12), linewidth(1)); draw((0.6,12)--(7,0), linewidth(1)); draw((1.5743944636678182,1.9230103806228354)--(5,3.75), linewidth(1)); draw((5,12)--(9,-12), linewidth(1)); draw((5,3.75)--(5,12), linewidth(1)); /* dots and labels */ dot((5,12),dotstyle); label("$A$", (5.142562031377331,12.306273028752814), NE * labelscalefactor); dot((0,0),dotstyle); label("$B$", (0.0024857242618652,-0.67553276152887205), NE * labelscalefactor); dot((14,0),dotstyle); label("$C$", (14.175478494309406,0.27553276152887205), NE * labelscalefactor); dot((5,3.75),linewidth(4pt) + dotstyle); label("$H$", (4.8,2.633367466837235), NE * labelscalefactor); dot((7,0),linewidth(4pt) + dotstyle); label("$M$", (7.1542595842573995,0.22664266141592815), NE * labelscalefactor); dot((-3.8,12),linewidth(4pt) + dotstyle); label("$P$", (-3.953684131215914,12.32738292863987), NE * labelscalefactor); dot((1.5743944636678182,1.9230103806228354),linewidth(4pt) + dotstyle); label("$G$", (1.450287726520743,2.229450114183053), NE * labelscalefactor); dot((9,-12),linewidth(4pt) + dotstyle); label("$A'$", (9.06595713713747,-11.644652555751542), NE * labelscalefactor); dot((0.6,12),linewidth(4pt) + dotstyle); label("$N$", (0.7641614751089444,12.32738292863987), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

One-liner (excluding constructions)[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 = -11.028184276513091, xmax = 13.651231623973653, ymin = -9.68127833226805, ymax = 5.291831456807111; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((1.5,2.51)--(-0.52,-2.77)--(9.84,-2.77)--cycle, linewidth(2) + zzttff); /* draw figures */ draw((1.5,2.51)--(-0.52,-2.77), linewidth(0.8) + zzttff); draw((-0.52,-2.77)--(9.84,-2.77), linewidth(0.8) + zzttff); draw((9.84,-2.77)--(1.5,2.51), linewidth(0.8) + zzttff); draw((-2.6384543058622527,2.51)--(1.5,2.51), linewidth(0.8)); draw((-2.6384543058622527,2.51)--(7.82,-2.77), linewidth(0.8)); draw((-0.52,-2.77)--(1.5,0.4206818181818154), linewidth(0.8)); draw((1.5,0.4206818181818154)--(9.84,-2.77), linewidth(0.8)); draw((1.5,2.51)--(1.5,-2.77), linewidth(0.8)); draw(circle((-0.5692271529311265,1.4653409090909082), 2.3179761919930204), linewidth(0.8) + ccqqqq); draw(circle((4.66,-3.8146590909090947), 5.284289225261901), linewidth(0.8) + ccqqqq); draw((4.66,-2.77)--(1.5,0.4206818181818154), linewidth(0.8) + ffvvqq); draw((1.5,0.4206818181818154)--(0.45538967495481386,-0.6138834613557778), linewidth(0.8)); draw((-2.6384543058622527,2.51)--(4.66,-2.77), linewidth(0.8)); draw((4.66,-2.77)--(0.45538967495481386,-0.6138834613557778), linewidth(0.8)); draw((0.45538967495481386,-0.6138834613557778)--(-2.6384543058622527,2.51), linewidth(0.8) + ffvvqq); draw((xmin, -1.670886075949367*xmin + 5.0163291139240505)--(xmax, -1.670886075949367*xmax + 5.0163291139240505), linewidth(0.8)); /* line */ draw((-0.52,-2.77)--(-2.6384543058622527,2.51), linewidth(0.8)); draw((9.84,-2.77)--(7.82,-8.05), linewidth(0.8)); draw((7.82,-8.05)--(-0.52,-2.77), linewidth(0.8)); draw((1.5,0.4206818181818154)--(2.420659623917179,0.9716826537080058), linewidth(0.8)); draw((0.45538967495481386,-0.6138834613557778)--(7.82,-8.05), linewidth(0.8) + ffvvqq); draw((1.5,-8.05)--(7.82,-8.05), linewidth(0.8)); draw(circle((4.66,-1.7253409090909093), 5.284289225261899), linewidth(0.8) + zzttff); draw((4.66,-2.77)--(7.82,-5.960681818181815), linewidth(0.8) + ffvvqq); draw((7.82,-5.960681818181815)--(7.82,-2.77), linewidth(0.8)); draw((7.82,-5.960681818181815)--(7.82,-8.05), linewidth(0.8)); draw((1.5,-2.77)--(1.5,-8.05), linewidth(0.8)); /* dots and labels */ dot((1.5,2.51),linewidth(1pt) + dotstyle); label("$A$", (1.5793801724103085,2.544820998024576), NE * labelscalefactor); dot((-0.52,-2.77),linewidth(1pt) + dotstyle); label("$B$", (-0.4431706239291322,-2.7330734609945218), NE * labelscalefactor); dot((9.84,-2.77),linewidth(1pt) + dotstyle); label("$C$", (9.919994408743431,-2.7330734609945218), NE * labelscalefactor); dot((1.5,0.4206818181818154),linewidth(1pt) + dotstyle); label("$H$", (1.5793801724103085,0.4644830360754426), NE * labelscalefactor); dot((1.5,-2.77),linewidth(1pt) + dotstyle); label("$D$", (1.5793801724103085,-2.7330734609945218), NE * labelscalefactor); dot((4.66,-2.77),linewidth(1pt) + dotstyle); label("$M$", (4.738411892407149,-2.7330734609945218), NE * labelscalefactor); dot((7.82,-2.77),linewidth(1pt) + dotstyle); label("$D'$", (7.89744361240399,-2.7330734609945218), NE * labelscalefactor); dot((-2.6384543058622527,2.51),linewidth(1pt) + dotstyle); label("$P$", (-2.5620333629514036,2.544820998024576), NE * labelscalefactor); dot((0.45538967495481386,-0.6138834613557778),linewidth(1pt) + dotstyle); label("$G$", (0.539211191435739,-0.5756859448991241), NE * labelscalefactor); dot((7.82,-8.05),linewidth(1pt) + dotstyle); label("$A'$", (7.89744361240399,-8.010967920013618), NE * labelscalefactor); dot((1.5,-8.05),linewidth(1pt) + dotstyle); label("$A''$", (1.5793801724103085,-8.010967920013618), NE * labelscalefactor); dot((2.420659623917179,0.9716826537080058),linewidth(1pt) + dotstyle); label("$H_A$", (2.5039748221654814,1.003829915099292), NE * labelscalefactor); dot((7.82,-5.960681818181815),linewidth(1pt) + dotstyle); label("$A'''$", (7.89744361240399,-5.930629958064485), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $A'''$ be the antipode of $A$ on $(ABC)$ Let $A'$ be the "parallelogram point" Let $A''$ be the reflection of $A$ in $BC$ By Reim's we have $P-G-A'$ now $\frac{DH}{HP}=\frac{DH}{HA}=\frac{DA'''}{DA'}$ implying $PG || HM$ implying the result.

12AM solves sus Let $O$, $O_{1}$, and $O_{2}$ denote the centers of $(ABC)$, $(AHP)$, and $(BHC)$ respectively. Let $M_{1}$ denote the midpoint of $HD'$. Claim: $O_{1}O_{2}$ and $HM$ are homothetic at $M_{1}$. Proof: It is well known that $2OM=AH$. But note that by considering a triangle $A'BC$ where $A$ is reflected about $M$ to $A'$, by orthocenter reflections $H$ lies on $(A'BC)$. So $O_2$ is the center of $(A'BC)$ and thus the reflection of $O$ over $BC$. So $2O_{2}M=2OM=AH$. Now $$\frac{HO_1}{HM_1}=\frac{PH}{HD'}=\frac{AH}{HD}=\frac{2O_{2}M}{2M_{1}M}=\frac{O_{2}M}{M_{1}M}$$Thus our claim is proven $\square$ Now Since $HG$ is the radical axis of $(APH)$ and $(HBC)$. $O_{1}O_{2} \perp HG$, but since $HM$ and $O_{1}O_{2}$ are homothetic, $HM \perp HG$ And we are finished

Still can't believe I tried to linpop this. Let $O_1$ and $O_2$ denote the centers of $(APH)$ and $(BHC)$, respectively. It is equivalent to show that $\overline{O_1O_2} \parallel \overline{MH}$. We prove the following stronger claim: Claim: Let $M'$ denote the midpoint of $AP$. Then $M'O_1O_2M$ is a parallelogram. Proof. First, note that $\overline{MO_2} \parallel \overline{M'O_1}$ as both lines are parallel to $\overline{AH}$. Furthermore, since $AHO_2O$ is a parallelogram (well-known), it follows that \[ M'O_1 = \frac{AH}{2} = \frac{OO_2}{2} = O_2M, \]so $M'O_1O_2M$ is a parallelogram, as desired. Done.

Let $A'$ be the reflection of $A$ across $M$. Then, since the foot from $H$ to $\overline{PA'}$ lies on both $(AHP)$ and $(BHC)$, it follows that $G$ is the foot from $H$ to $\overline{PA'}$. A homothety of scale factor $\tfrac{1}{2}$ centered at $A$ sends line $PA'$ to line $HM$, so from $\overline{HG} \perp \overline{A'P}$ it follows that $\overline{HG} \perp \overline{HM}$.

Extend rays $AD$ and $PG$ to intersect $(BHC)$ at $S$ and $T$ respectively. Suppose $AS$ and $PT$ intersect at $N$. First, we claim that $\triangle HDM \sim \triangle NST$. Clearly we have $\angle NST = 90^\circ = \angle HDM$. Since $BC \parallel ST$, we get $TD' \perp BC$, so $DD' = ST$ and $\frac{DM}{ST} = \frac{DM}{DD'} = \frac{1}{2}$. Now by similar triangles from parallel lengths, we have $$\frac{ST}{AP} = \frac{DD'}{AP} \implies \frac{SN}{SA} = \frac{DH}{DA}.$$But $SA = 2(DA)$ since it is well known that $(BHC)$ is a reflection of $(ABC)$ about $D$, then $\frac{DH}{SN} = \frac{1}{2}$. Combining this with the earlier facts, by SAS Similarity yields $\triangle HDM \sim \triangle NST.$ Then $PT \parallel HM$, and $\angle MHG = \angle HGT = 90^\circ$ and we are finished.

By homothety $MH$ bisects $AP$ and since $PG \perp GH$ line $PG$ passes through $B+C-A$ and we are done by homothety centered at $A$ with scale factor $2$.

GrantStar wrote: By homothety $MH$ bisects $AP$ and since $PG \perp GH$ line $PG$ passes through $B+C-A$ and we are done by homothety centered at $A$ with scale factor $2$. Very slick solution

Let $O_1,O_2$ be the centers of circles $(AHP)$ and $(BHC)$. Let $N$ be the mid-point of $\overline{DH'}$. Then, we have $\triangle HAP \sim \triangle HDD'$. Since both triangles are right-angled, $O_1\in HP$ and $N$ is the circumcenter of $HDD'$. Hence, the homothety at $H$ with ratio $-\frac{AH}{HD}$ sends $HDD'$ to $HAP$. Hence, \[\frac{AH}{HD}=\frac{HO_1}{NH}=\frac{AH/2}{HD/2}=\frac{OM}{NM}=\frac{MO_2}{NM}\]and the conclusion follows by Thales' Theorem. $\blacksquare$. Note, we have $AH=2OM$ and $OM=MO_2$ as $O_2$ is the reflection of $O$ in $BC$.

7 minute speedrun b4 mom make me sleep Let $F=\frac{P+A}{2}$, $FHM$ collinear. $FM \perp HG \perp FD$, let $PG$ intersect $(AGM)$ at $I$, and $(BHC)$ at $J$. Let $FM$ intersect $(BHC)$ at $G'$ (sus why i named it this ur boutta see why). By reim $GHG'J$ rectangle so $\angle GHM = \angle GHG' = 90^{\circ}$ QED

Interesting.... Let $X$ be reflection of $A$ above midpoint $M$ with $H' = HM \cap XD'$. Observe $XD' \perp BC$ and as everything is just reflection above midpoint $M$, we get $H'$ is orthocenter of $\triangle XBC$. Hence $$\frac{D'H}{PH} = \frac{DH}{HA} = \frac{H'D'}{H'X} \Rightarrow HM \parallel PX$$Now if $G = PX \cap (BHC)$. Key observation is $XH$ is diameter in $(BHC)$. Because If $H_A$ be $A$-humpty point, then we know $X \in AH_A$ and $\measuredangle XH_AH = 90$. $$HG\perp GX \parallel HM \Rightarrow GH \perp HM \Rightarrow \measuredangle GHM = 90$$

Let $H'$ be the antipode of $H$ in $(BHC)$. Claim 1: $H'$ is the reflection of $A$ over $M$. Proof. Angle chasing shows that $\square{ABH'C}$ is a parallelogram. By a homothety centered at $M$, $MH$ bisects $AP$. Hence by midpoint theorem, $MH$ is parallel to $H'P$. Since $HP$ and $HH'$ are both diameters, $H'P$ is in turn parallel to the line joining the centers of $(BHC)$ and $(AHP)$. Therefore $MH$ is parallel to the line joining the centers of $(BHC)$ and $(AHP)$. However, it is well known that the line joining the centers of two circles is perpendicular to their radical axis. Hence $MH \perp HG$ and we are done!

Let $A', D', H'$ be the reflections of $A, D, H$ respectively over $M.$ It is well-known that $(ABC)$ passes through $H',$ therefore by homothety (reflections are basically homothety) $(\triangle BHC)$ passes through $A'.$ But it is also well-known that $A'H$ is its diameter because $AH'$ is the diameter of $(\triangle ABC),$ therefore $\angle PGA'=\angle PGH+\angle HGA'=180^\circ,$ so $G=(\triangle BHC) \cap PA'.$ It thus suffices to show that $PA' \parallel HM.$ Clearly by our definitions $HDH'D'$ is a parallelogram so $HM$ passes through $H'.$ Therefore $$\frac{D'H'}{D'H} = \frac{DH}{HD'} = \frac{AH}{PH} = \frac{H'A}{PH},$$so $\triangle D'H'H \sim \triangle D'A'P$ and the desired result follows. QED