Inversion with center $A$ and power $\overline{AH} \cdot \overline{AD}$ takes $\odot(AEF)$ into the sideline $BC$ and $\odot(BHC)$ into the 9-point circle $\odot(DEF)$ $\Longrightarrow$ $M$ is the inverse of the second intersection of $\odot(AEF)$ and $\odot(BHC),$ i.e. $N \in \odot(AEF)$ $\Longrightarrow$ $\angle HNA$ is right. From the harmonic cross ratio $(G,D,H,A),$ it follows that $NA,NH$ bisect $\angle DNG$ externally and internally $\Longrightarrow$ $\odot(AEF)$ is N-Apollonius circle of $\triangle GND$ $\Longrightarrow$ $\odot(AEF) \perp \odot(GND).$ Hence, $\angle GNS=\angle HDN=\angle HMA.$

Denote $L\in EF\cap BC$ . From an well-known property obtain that $LH\perp AM$ . Denote $N'\in LH\cap AM$ . Therefore, $AF\cdot AB=AH\cdot AD=AN'\cdot AM\implies $ $AF\cdot AB=AN'\cdot AM\implies BFN'M$ is cyclically $\implies m\left(\widehat{BN'M}\right)=$ $m\left(\widehat{BFM}\right)=B$ . Also, $AE\cdot AC=AH\cdot AD=AN'\cdot AM\implies $ $AE\cdot AC=AN'\cdot AM\implies CEN'M$ is cyclically $\implies m\left(\widehat{CN'M}\right)=$ $m\left(\widehat{CEM}\right)=C$ . Therefore, $m\left(\widehat{BN'C}\right)=$ $m\left(\widehat{BN'M}\right)+m\left(\widehat{CN'M}\right)=$ $B+C=180^{\circ}-A$ $\implies$ $BHN'C$ is cyclically $\implies $ $N'\equiv N$ . Since the division $(A,G,H,D)$ is harmonically and $NH\perp NA$ obtain that $\widehat{HNG}\equiv$ $\widehat{HND}$ . Since the quadrilateral $DHNM$ is cyclically obtain that $\widehat{HND}\equiv \widehat{HMD}$ . Thus, $\widehat{HNG}\equiv\widehat{HMD}$ . In conclusion, $m\left(\widehat{HMA}\right) =$ $90^{\circ}-m\left(\widehat{DAM}\right)-$ $m\left(\widehat{DMH}\right)=$ $90^{\circ}-m\left(\widehat{ANS}\right)-$ $m\left(\widehat{HND}\right)=$ $90^{\circ}-m\left(\widehat{ANS}\right)-$ $m\left(\widehat{HNG}\right)=$ $m\left(\widehat{GNS}\right)$ $\implies$ $\widehat{HMA}\equiv $ $\widehat{GNS}$ . Nice problem ! Thank you. See the problem PP6 from here.

$\square BCEF$ is cyclic. $M$ is the orthocenter of $\triangle AHK$ (Brocard). $HK\cap AM=N$ Since ($K$,$D/C$,$B$)$=-1$ and $\square NMDH$ is cyclic , $KH.KN=KD.KM=KC.KB$ Then $\square BCHN$ is cyclic. $N$ is that of the problem. $\angle GNH=\angle HND$ since ($D$,$G/H$,$A$)$=-1$ and $\angle HNA=\pi$/2 Since $SN=SH$, $\angle SNH=\angle SHN$ . $\angle SNH=\angle SNG+\angle GNH$ and $\angle SHN=\angle HND+\angle HDN=\angle HND+\angle HMA$ . Thus, $\angle SNG=\angle HMA$

let $A'$ be the reflection of A in respect of M. then B,H,C,A' cyclic and HA' is the diameter. hence $\angle HNM=90$,obtaining B,H,N,C;F,H,N,E cyclic. hence FE,HN,BC concurrent(at T). and $\angle HNS=90-\angle SNA=90-\angle SAN=\angle AMD$ so it siffices to prove $\angle GNH=\angle HMB$. let Q be midpoint of EF by MF=ME,we get $\angle MQG=90$ yielding G,D,M,Q cyclic then$TG*TQ=TD*TM=TH*TM$ G,Q,N,H cyclic hence $\angle GNH=\angle HQF=\angle HMB$ QED

Anyone give reason for $ \angle HQF = \angle HMB $ ????

any help for little tush solution???????

N is a point on the median AM whoch (BC/2)^2=MN.MA which implies that HN is perpendicular to AM ( not hard to prove) . then HNMD is inscribed quad. so HMA is HDN . so all we have to prove is that SN^2=SG.SD and since G and D are inverses about circle EFH and N is on the very circle the angle equality is an instant resolve. done

Dear Mathlinkers, the kern of this problem is to prove that NS is tangent to circumcircle of the triangle GND which is orthogonal to the circle with diameter AH (Apollonius circle)...and we finsh with a small angle chasing... Sincerely Jean-Louis

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -18.910594552715125, xmax = 22.826615668950918, ymin = -16.282277070149565, ymax = 9.796659512799902; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((1,6)--(-5,-1)--(6,-5)--cycle, linewidth(0.4) + zzttqq); /* draw figures */ draw((1,6)--(-5,-1), linewidth(0.4) + zzttqq); draw((-5,-1)--(6,-5), linewidth(0.4) + zzttqq); draw((6,-5)--(1,6), linewidth(0.4) + zzttqq); draw(circle((1.4306930693069306,-0.4405940594059406), 6.454978571459716), linewidth(0.4) + blue); draw(circle((-0.4306930693069312,-5.559405940594061), 6.4549785714597165), linewidth(0.4) + blue); draw((1,6)--(0.5,-3), linewidth(0.4) + red); draw((6,-5)--(-2.317647058823531,2.129411764705881), linewidth(0.4) + red); draw((-5,-1)--(2.609589041095891,2.458904109589041), linewidth(0.4) + red); draw((1,6)--(-1.9489051094890524,-2.10948905109489), linewidth(0.4) + red); draw((0.7107692307692306,0.7938461538461522)--(-0.3598796820167596,2.260330874453913), linewidth(0.4) + red); draw((0.06930693069306859,3.44059405940594)--(0.7107692307692306,0.7938461538461522), linewidth(0.4) + red); draw(circle((0.06930693069306859,3.44059405940594), 2.723370771306052), linewidth(0.4) + blue); draw((-0.8613861386138628,0.8811881188118803)--(0.5,-3), linewidth(0.4) + red); draw((0.7107692307692306,0.7938461538461522)--(-1.9489051094890524,-2.10948905109489), linewidth(0.4) + red); draw((-0.8613861386138628,0.8811881188118803)--(0.7107692307692306,0.7938461538461522), linewidth(0.4) + red); /* dots and labels */ dot((1,6),dotstyle); label("$A$", (1.112354664737735,6.27763982744375), NE * labelscalefactor); dot((-5,-1),dotstyle); label("$B$", (-4.889074256024703,-0.7331203209014516), NW * labelscalefactor); dot((6,-5),dotstyle); label("$C$", (6.104452357917399,-4.715886786498337), SE * labelscalefactor); dot((0.5,-3),linewidth(4pt) + dotstyle); label("$M$", (0.6213286621298992,-2.779061998434098), NE * labelscalefactor); dot((-0.8613861386138628,0.8811881188118803),linewidth(4pt) + dotstyle); label("$H$", (-0.7426324562252004,1.0945875776943792), NE * labelscalefactor); dot((0.7107692307692306,0.7938461538461522),linewidth(4pt) + dotstyle); label("$N$", (0.8122832186996131,1.0127499105930733), NE * labelscalefactor); dot((-1.9489051094890524,-2.10948905109489),linewidth(4pt) + dotstyle); label("$D$", (-1.8338013509092799,-1.8788476603197335), NW * labelscalefactor); dot((2.609589041095891,2.458904109589041),linewidth(4pt) + dotstyle); label("$E$", (2.7218287843967524,2.6767824749862923), NE * labelscalefactor); dot((-2.317647058823531,2.129411764705881),linewidth(4pt) + dotstyle); label("$F$", (-2.215710464048708,2.349431806581069), NW * labelscalefactor); dot((0.06930693069306859,3.44059405940594),linewidth(4pt) + dotstyle); label("$S$", (0.18486110425626734,3.658834480201963), NE * labelscalefactor); dot((-0.3598796820167596,2.260330874453913),linewidth(4pt) + dotstyle); label("$G$", (-0.2516064536173645,2.485827918416579), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice exercise for projective geo and inversion Let's do an inversion around $(AFE)$, we shall denote this by $\psi$. Since we have that $-1=(A,H;G,D)=(G,D;A,H)$ this implies that $SG.SD=\left(\frac{1}{2}AH\right)^2$. Thus we have that $G\overset{\psi}{\rightarrow} D$, this implies that $\angle GNS=\angle SDN = \angle HDN$. Since we have that $HNMD$ is cyclic this implies that $\angle HDN=\angle HMN=\angle HMA$, thus we have that $\angle GNS = \angle HMA$.