Let $P$ denote the reflection of $C$ over $D$. Note by homothety through $B$ with scale factor 2, $G = CP \cap (ABP)$. Note $AGBC$ is a parallelogram. Note $EF$ is a midline of $\triangle ABC$. $\angle BGC = 180 - \angle PGB = 180 - \angle PAB = 180 - \angle B$. $\angle BHC = 180 - \angle FHB = 180 - \angle FDB = 180 - \angle B$. Thus, $(BHGC)$.

Solution. Let $X$ be the intersection of $BG$ with $\Gamma$ . Claim 1. It is enough to prove that $FX\parallel DC$. Proof. $FX\parallel DC\implies \angle CGB=\angle GXF=180-\angle FXB=180-\angle FHB=\angle BHC\implies BHGC$ is cyclic. Claim 2. $FX\parallel DC$. Proof. Let $Y$ be the midpoint of $BC$. Since $YC\parallel FD$ and $YC=FD\implies DFYC$ is parallelogram $\implies YF\parallel CD$ we know from the midsegment theorem we know that that $XY\parallel DC\implies F-X-Y$ are collinear. Now we can say that $FX\parallel DC.\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -33.58780030258491, xmax = 15.588139964582565, ymin = -16.424492264051015, ymax = 14.522061496311256; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); pen ffqqtt = rgb(1,0,0.2); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-9.306454141100053,12.19887981304907)--(-16.82935260658268,-9.28667816658334)--(7.193603619735551,-9.05441201522391)--cycle, linewidth(0.8) + rvwvcq); /* draw figures */ draw((-9.306454141100053,12.19887981304907)--(-16.82935260658268,-9.28667816658334), linewidth(0.8) + rvwvcq); draw((7.193603619735551,-9.05441201522391)--(-9.306454141100053,12.19887981304907), linewidth(0.8) + rvwvcq); draw(circle((-19.036107897020255,-2.4841088834766554), 7.151553520972966), linewidth(0.8) + dbwrru); draw((-25.079381487000486,1.3399677475531506)--(7.193603619735551,-9.05441201522391), linewidth(0.8) + sqsqsq); draw((-13.067903373841368,1.4561008232328652)--(7.193603619735551,-9.05441201522391), linewidth(0.8) + ffqqtt); draw(circle((-4.775947184126969,-13.507026795294333), 12.770901488381606), linewidth(0.8) + linetype("4 4") + dtsfsf); draw((-13.067903373841368,1.4561008232328652)--(-1.0564252606822508,1.5722338989125797), linewidth(0.8) + rvwvcq); draw((-13.067903373841368,1.4561008232328652)--(-25.079381487000486,1.3399677475531506), linewidth(0.8) + rvwvcq); draw((-8.029519397660007,-1.1575250775215364)--(-12.429436002121363,-5.222101622052457), linewidth(0.8) + sexdts); draw((-12.429436002121363,-5.222101622052457)--(-16.82935260658268,-9.28667816658334), linewidth(0.8) + sexdts); draw((-11.896996093441633,-2.9057727535738005)--(-16.82935260658268,-9.28667816658334), linewidth(0.8) + wrwrwr); draw((-4.8178744934236555,-9.170545090903627)--(7.193603619735551,-9.05441201522391), linewidth(0.8) + rvwvcq); draw((-4.8178744934236555,-9.170545090903627)--(-16.82935260658268,-9.28667816658334), linewidth(0.8) + rvwvcq); draw((-25.079381487000486,1.3399677475531506)--(-16.82935260658268,-9.28667816658334), linewidth(0.8) + wrwrwr); draw((-11.896996093441633,-2.9057727535738005)--(-8.029519397660007,-1.1575250775215364), linewidth(0.8) + wrwrwr); draw((-9.306454141100053,12.19887981304907)--(-8.029519397660007,-1.1575250775215364), linewidth(0.8) + wrwrwr); draw((-25.079381487000486,1.3399677475531506)--(-12.429436002121363,-5.222101622052457), linewidth(0.8) + ffqqtt); draw((-12.429436002121363,-5.222101622052457)--(-4.8178744934236555,-9.170545090903627), linewidth(0.8) + linetype("2 2") + ffqqtt); /* dots and labels */ dot((-9.306454141100053,12.19887981304907),dotstyle); label("$A$", (-9.151741723224552,12.61231624321737), NE * labelscalefactor); dot((-16.82935260658268,-9.28667816658334),dotstyle); label("$B$", (-17.87580526576706,-10.348030095115927), NE * labelscalefactor); dot((7.193603619735551,-9.05441201522391),dotstyle); label("$C$", (7.384916036520206,-8.611898046848758), NE * labelscalefactor); dot((-13.067903373841368,1.4561008232328652),linewidth(4pt) + dotstyle); label("$D$", (-13.492071843892466,2.4125404596477584), NE * labelscalefactor); dot((-1.0564252606822508,1.5722338989125797),linewidth(4pt) + dotstyle); label("$E$", (-0.8617111927488331,1.9351041463742873), NE * labelscalefactor); dot((-25.079381487000486,1.3399677475531506),linewidth(4pt) + dotstyle); label("$F$", (-25.688399482969306,1.674684339134212), NE * labelscalefactor); dot((-11.896996093441633,-2.9057727535738005),linewidth(4pt) + dotstyle); label("$H$", (-11.712536494418622,-2.1448061670535576), NE * labelscalefactor); dot((-8.029519397660007,-1.1575250775215364),linewidth(4pt) + dotstyle); label("$G$", (-7.8496426870241764,-0.7993038296465024), NE * labelscalefactor); dot((-12.429436002121363,-5.222101622052457),linewidth(4pt) + dotstyle); label("$X$", (-12.493795916138845,-6.398329685308119), NE * labelscalefactor); dot((-4.8178744934236555,-9.170545090903627),linewidth(4pt) + dotstyle); label("$Y$", (-4.637798397729919,-8.828914552882154), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

My solution: Let $X, Y, I$ be the midpoints of $BD, BC, BG,$ respectively. Since $D, G, C$ are collinear, it follows that $X, Y, I$ are also collinear. Since $FD=DE=\frac{1}{2}BC=YC$ and $FD \parallel YC$, it follows that $FDCY$ is a parallelogram, so $FY \parallel DC.$ Furthermore, $XY \parallel DC$ since $XY$ is the midsegment of $\triangle BDC$, hence, $F, X, Y$ are collinear. Therefore four points $F, X, I, Y$ are collinear. Now, note that $\angle HBG = \angle HBI = \angle HFI = \angle HFY = \angle FCD = \angle HCG$, it follows that $BHGC$ is cyclic. [asy][asy] import graph; size(13cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -5.213047496984735, xmax = 4.747992444341872, ymin = -3.0156886925367217, ymax = 3.427756286011175; draw((-2.85,-1.92)--(-0.57,2.72), linewidth(2)); draw((-0.57,2.72)--(3.09,-1.96), linewidth(2)); draw((3.09,-1.96)--(-2.85,-1.92), linewidth(2)); draw((-4.68,0.42)--(1.26,0.38), linewidth(2)); draw(circle((-3.1998350980732506,-0.30801206387779917), 1.6495119587464153), linewidth(2)); draw((-4.68,0.42)--(3.09,-1.96), linewidth(2)); draw((-2.28,-0.76)--(0.12,-1.94), linewidth(2)); draw((3.09,-1.96)--(-1.71,0.4), linewidth(2)); draw((-2.85,-1.92)--(-0.5889217605279318,-0.15119680107376676), linewidth(2)); draw((-2.28,-0.76)--(-4.68,0.42), linewidth(2)); draw((-2.85,-1.92)--(-1.5658604606790387,-0.5338805796118258), linewidth(2)); /* dots and labels */ dot((-0.57,2.72),dotstyle); label("$A$", (-0.5366186796457587,2.803136619723369), NE * labelscalefactor); dot((-2.85,-1.92),dotstyle); label("$B$", (-2.8131929896684236,-1.8404174783372966), NE * labelscalefactor); dot((3.09,-1.96),dotstyle); label("$C$", (3.120693840065743,-1.8815108774351785), NE * labelscalefactor); dot((-1.71,0.4),linewidth(4pt) + dotstyle); label("$D$", (-1.6790151745668793,0.4690315509636716), NE * labelscalefactor); dot((1.26,0.38),linewidth(4pt) + dotstyle); label("$E$", (1.2961469201197804,0.4443755115049424), NE * labelscalefactor); dot((-4.68,0.42),linewidth(4pt) + dotstyle); label("$F$", (-4.645958589433962,0.48546891060282443), NE * labelscalefactor); dot((-2.28,-0.76),linewidth(4pt) + dotstyle); label("$X$", (-2.2461040821176517,-0.6980209834161771), NE * labelscalefactor); dot((0.12,-1.94),linewidth(4pt) + dotstyle); label("$Y$", (0.1537504251986596,-1.8732921976156023), NE * labelscalefactor); dot((-1.719460880263966,-1.0355984005368832),linewidth(4pt) + dotstyle); label("$I$", (-1.6872338543864558,-0.9692374174621983), NE * labelscalefactor); dot((-0.5889217605279318,-0.15119680107376676),linewidth(4pt) + dotstyle); label("$G$", (-0.5530560392849114,-0.08161999694794714), NE * labelscalefactor); dot((-1.5658604606790387,-0.5338805796118258),linewidth(4pt) + dotstyle); label("$H$", (-1.531078937814504,-0.4678979484680379), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

In short lines Let $I$ and $J$ be the midpoints of $BG$ and $BC$ respectively. We see that, $JI||CD$ and $JF||CD \Rightarrow JIF$ are collinear. So $\angle BGC=\angle BIJ=\angle BHC$ and we are done.

Let S be midpoint of BG and M be midpoint of BC. ∠GAH = ∠SFH so we need to prove FS || CD. FECB is parallelogram so FM || CD and SM || CD so F,S,M are collinear and FS || CD.

Let,$(BHC) \cap CD$ at $G$ and $BG \cap (FBD)$ at $M$ we will prove $FBMHD$ cyclic which equivalent to prove $BMHD$ cyclic. We know that $EFBC$ is parallelogram,let's call $\angle DFC=\angle DBH=\angle FCB=\angle HGB=\alpha$ and $\angle DCF=\angle HBM=\beta$, $\angle GBC=x;\angle DCA=y$ from these angles $\angle BAC=180-2\alpha-2\beta-x-y$ and $\angle BFC=\beta+y$ so $\angle FBA=180-2\alpha-2\beta-x-y=\angle FHD$ which quickly implies that $ADHC$ is cyclic.Thus,$\angle DAH=\beta=\angle HBG$ and $\angle HGB=\angle ABH=\alpha$ this result immediately tells us $H$ is $B-Dumpty$ so we know the dumpty property that $BDHM$ is cyclic. $\blacksquare$