Let $AH \cap BC = D$, $MH \cap (ABC) = L$, $EF \cap AH =T$, $EF \cap AL \cap BC = N$, $TL \cap AM = R$, then $LRMN$ are concyclic. Proof: Let $AH \cap (ABC) = Q$ , inverting wrt $A$ then $T->Q, L->N$ so $TLQN$ are concyclic. $(A,C;D,N)=-1 \implies DM \times DN = DB \times DC = DA \times DQ $ implies $ANQM$ are concyclic, so $$ \angle ALT = \angle NQT = \angle NMA$$Let $MX \cap AH =S$, $LT \cap BC =Z$, $$ -1=(A,H;T,D)=L(N,M;Z,D)=T(X,M;Z,S)$$so $RL$ is parallel to $MX$, then apply angle chasing.

The intersection of $AH$ and $MX$ is radical center of $\odot (M,0),\odot AEF$ and $\odot AHM$ so it is the midpoint of $MX$.

Let $T$ be the projection of $H$ on to $AM$, $AH\cap BC=D$ , $EF\cap BC=S$ and $MH\cap AS=L$ , since $H$ is orthocenter of $\triangle ASM$ (well known) $\Longrightarrow EF,LT$ intersect $AH$ at harmonic conjugate of $D$ WRT $(A,H)$ (call it $Y$). $$\angle AMX=\angle MHD=\angle ATL\Longrightarrow TL\parallel MX$$Hence because $Y(T,H;M,S)=-1,YT\parallel MX\Longrightarrow AH$ bisects $MX$. 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Dadgarnia wrote: Let $ABC$ be an acute-angle triangle. Suppose that $M$ be the midpoint of $BC$ and $H$ be the orthocenter of $ABC$. Let $F\equiv BH\cap AC$ and $E\equiv CH\cap AB$. Suppose that $X$ be a point on $EF$ such that $\angle XMH=\angle HAM$ and $A,X$ are in the distinct side of $MH$. Prove that $AH$ bisects $MX$. Let $Q$ be on $\odot(ABC)$ such that $\measuredangle AQH=90^{\circ}$, $T=\overline{AH} \cap \overline{XM}$, $N=\overline{AH} \cap \overline{EF}$, $Y=\overline{AM} \cap \odot(AH)$ and $Z=\overline{AM} \cap \overline{EF}$. Note that $Q, H, M$ are collinear. By Brokard's Theorem, the polar $\overline{EF}$ of point $M$ in $\odot(AH)$ passes through $\overline{AH} \cap \overline{QY}$, hence, $N$ lies on $\overline{QY}$. Observe that $$\measuredangle XMH=\measuredangle MAH=\measuredangle YQM \implies \overline{QY} \parallel \overline{XM}.$$Finally, consider $$(XM, T\infty) \overset{N}{=} (ZM, AY)=-1,$$so $T$ bisects $\overline{XM}$ as desired. $\blacksquare$

Generalization: Let $ABC$ be an acute-angle triangle and $BE,CF$ are altitudes. Suppose that $M$ be the midpoint of $BC$ and $K$ is an arbitary point such that $A,E,F$ and $K$ lie on a circle and $X$ is a point such that $X\in EF$, $\angle XMK=\angle KAM$ and $A,X$ are in the distinct side of $MK$. Prove that $AK$ bisects $MX$.

Radical axis on the point circle $M$, $(AH)$ and $(AMH)$ does the original problem instantly.

WizardMath wrote: Radical axis on the point circle $M$, $(AH)$ and $(EMH)$ does the original problem instantly. I don't think so. I believe the solution in #3 is the best one.

Dadgarnia wrote: Generalization: Let $ABC$ be an acute-angle triangle and $BE,CF$ are altitudes. Suppose that $M$ be t he midpoint of $BC$ and $K$ is an arbitary point such that $A,E,F$ and $K$ lie on a circle and $X$ is a point such that $X\in EF$, $\angle XMK=\angle KAM$ and $A,X$ are in the distinct side of $MK$. Prove that $AK$ bisects $MX$. I will post my solution for this generalization, which is basically the same idea as above solution.

@WizardMath and @Dadgarnia It seems that I am missing something obvious, but i don't see that why the intersection of $AH$ dan $MX$ is the radical center of $(M,0), (AEF), (AHM)$ implies that it is the midpoint of $MX$. Could you explain further?

@above, the radical center is on the $M$-midline of $MEF$, as $ME, MF$ are tangents to $(AH)$. So done.

My soluttion: LEMMA 1 (well-known): Intriangle $ABC$, $H$ is orthocenter and $F(A)$ is the foot of prependicular from $H$ to $A$-median. Then $\angle F(A)AC = \angle F(A)BC, \angle F(A)AC=\angle F(A)CB$ LEMMA 2: $w$ is an arbitrary circle passing through $A,F(A)$. Then the polar of $B$ wrt $w$ passes through $C$ and the polar of $C$ wrt $w$ passes through $B$. PROOF: Let $w$ cut $AB,AC$ at $F,E$ respectively. We know that the intersection of $BE,CF$ lies on $w$. So we're done! BACK TO THE PROBLEM: According to LEMMA 1 we wish to prove that $H$ is $F(A)$ in triangle $AMX$. Let $w$ be the circle with diameter $AH$. By angle chasing we get that $ME,MF$ are tangent to $w$. So $X$ lies on the polar of $C$ wrt $w$ so we're done!! (LEMMA 2)

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.910442717432293cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(9); defaultpen(dps); /* default pen style */ real xmin = -1.7467201356653412, xmax = 2.708501223050805, ymin = -1.062976861604131, ymax = 1.1942634640331873; /* image dimensions */ pen ffdxqq = rgb(1.,0.8431372549019608,0.); pair A = (-0.5171967708187546,0.8558665201155217), B = (-0.855518361466782,-0.5177724724173665), C = (0.8517317356916332,-0.5239781010841177), M = (-0.0018933128875743832,-0.520875286750742), H = (-0.5209833965939031,-0.18588405338596203), F = (0.0017838113556972077,0.33274748108991364), X = (-1.0416980178789688,-0.29782448288336055), D = (-0.5217956653832717,-0.4093498848170513), G = (-5.475076593858773e-5,-0.09406390283041419), J = (-0.3806055854989826,-0.324029275035587); draw(A--B--C--cycle, linewidth(2.) + ffdxqq); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(2.) + linetype("0 3 4 3") + blue); draw(A--B, linewidth(1.2) + ffdxqq); draw(B--C, linewidth(1.2) + ffdxqq); draw(C--A, linewidth(1.2) + ffdxqq); draw((-0.7593178581103909,-0.12718326332043192)--F, linewidth(1.2)); draw(B--F, linewidth(1.2)); draw(C--(-0.7593178581103909,-0.12718326332043192), linewidth(1.2)); draw(H--M, linewidth(1.2)); draw(A--M, linewidth(1.2)); draw(H--A, linewidth(1.2)); draw(M--X, linewidth(1.2) + linetype("2 2")); draw(X--H, linewidth(1.2)); draw(X--(-0.7593178581103909,-0.12718326332043192), linewidth(1.2)); draw(H--D, linewidth(1.2)); draw(A--X, linewidth(1.2)); draw(M--(-0.7593178581103909,-0.12718326332043192), linewidth(1.2)); draw(M--F, linewidth(1.2)); draw(B--X, linewidth(1.2)); draw(G--D, linewidth(1.2)); /* dots and labels */ dot(A,blue); label("$A$", (-0.5642278887025913,0.9302902517271522), NE * labelscalefactor,blue); dot(B,blue); label("$B$", (-0.9467196861256221,-0.6050641463793771), NE * labelscalefactor,blue); dot(C,blue); label("$C$", (0.9037864042238293,-0.5404176454064705), NE * labelscalefactor,blue); dot(M,blue); label("$M$", (0.014897015846363593,-0.6373873968658303), NE * labelscalefactor,blue); dot(H,blue); label("$H$", (-0.5776959097386135,-0.13368341011860055), NE * labelscalefactor,blue); dot(F,blue); label("$F$", (0.04722026633281689,0.356552555592607), NE * labelscalefactor,blue); dot((-0.7593178581103909,-0.12718326332043192),blue); label("$E$", (-0.8443627262518533,-0.1229089932897828), NE * labelscalefactor,blue); dot(X,blue); label("$X$", (-1.1379655848371375,-0.26567001627161796), NE * labelscalefactor,blue); dot(D,blue); label("$D$", (-0.5507598676665691,-0.4892391654695863), NE * labelscalefactor,blue); dot(G,blue); label("$G$", (0.06338189157604354,-0.1148281806681695), NE * labelscalefactor,blue); dot(J,blue); label("$J$", (-0.38375640681989376,-0.40035022663183983), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] solution(with sina feizi): define $G$ and $J$ as the midpoints of $MF$ and $ME$ it is trivial that $GJ$ is the radical axis of $(AEF)$ and $M$(consider a point as a circle with radius zero) on the other hands $\triangle DHM \sim \triangle AMD$ so $DH \times DA=DM^2$ and it is obviously on the line $GJ$ and simply we conclude that D is the midpoint of $MX$.$\boxed{Q.E.D}$

Here's another solution: Let $MH \cap (ABC) = Y$ and $BC \cap EF = T$ .It's well-known that $\angle AYM=90$ and $A ,Y , T$ are collinear so $H$ is the orthocenter of $\triangle AMT$ so $\angle XMH=\angle HAM = \angle HTM$ . so we have $HT.HX = HM^2$ $*$.Now , by Ratio Lemma we should prove that $\frac{HX}{HM} = \frac{sin HMT}{sin HMT}$ which is equivalent to $\frac{HM}{HT} = \frac{sin HMT}{sin HMT}$(by $*$) which is clear by sines law in $\triangle HMT$

iman007 wrote: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.910442717432293cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(9); defaultpen(dps); /* default pen style */ real xmin = -1.7467201356653412, xmax = 2.708501223050805, ymin = -1.062976861604131, ymax = 1.1942634640331873; /* image dimensions */ pen ffdxqq = rgb(1.,0.8431372549019608,0.); pair A = (-0.5171967708187546,0.8558665201155217), B = (-0.855518361466782,-0.5177724724173665), C = (0.8517317356916332,-0.5239781010841177), M = (-0.0018933128875743832,-0.520875286750742), H = (-0.5209833965939031,-0.18588405338596203), F = (0.0017838113556972077,0.33274748108991364), X = (-1.0416980178789688,-0.29782448288336055), D = (-0.5217956653832717,-0.4093498848170513), G = (-5.475076593858773e-5,-0.09406390283041419), J = (-0.3806055854989826,-0.324029275035587); draw(A--B--C--cycle, linewidth(2.) + ffdxqq); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(2.) + linetype("0 3 4 3") + blue); draw(A--B, linewidth(1.2) + ffdxqq); draw(B--C, linewidth(1.2) + ffdxqq); draw(C--A, linewidth(1.2) + ffdxqq); draw((-0.7593178581103909,-0.12718326332043192)--F, linewidth(1.2)); draw(B--F, linewidth(1.2)); draw(C--(-0.7593178581103909,-0.12718326332043192), linewidth(1.2)); draw(H--M, linewidth(1.2)); draw(A--M, linewidth(1.2)); draw(H--A, linewidth(1.2)); draw(M--X, linewidth(1.2) + linetype("2 2")); draw(X--H, linewidth(1.2)); draw(X--(-0.7593178581103909,-0.12718326332043192), linewidth(1.2)); draw(H--D, linewidth(1.2)); draw(A--X, linewidth(1.2)); draw(M--(-0.7593178581103909,-0.12718326332043192), linewidth(1.2)); draw(M--F, linewidth(1.2)); draw(B--X, linewidth(1.2)); draw(G--D, linewidth(1.2)); /* dots and labels */ dot(A,blue); label("$A$", (-0.5642278887025913,0.9302902517271522), NE * labelscalefactor,blue); dot(B,blue); label("$B$", (-0.9467196861256221,-0.6050641463793771), NE * labelscalefactor,blue); dot(C,blue); label("$C$", (0.9037864042238293,-0.5404176454064705), NE * labelscalefactor,blue); dot(M,blue); label("$M$", (0.014897015846363593,-0.6373873968658303), NE * labelscalefactor,blue); dot(H,blue); label("$H$", (-0.5776959097386135,-0.13368341011860055), NE * labelscalefactor,blue); dot(F,blue); label("$F$", (0.04722026633281689,0.356552555592607), NE * labelscalefactor,blue); dot((-0.7593178581103909,-0.12718326332043192),blue); label("$E$", (-0.8443627262518533,-0.1229089932897828), NE * labelscalefactor,blue); dot(X,blue); label("$X$", (-1.1379655848371375,-0.26567001627161796), NE * labelscalefactor,blue); dot(D,blue); label("$D$", (-0.5507598676665691,-0.4892391654695863), NE * labelscalefactor,blue); dot(G,blue); label("$G$", (0.06338189157604354,-0.1148281806681695), NE * labelscalefactor,blue); dot(J,blue); label("$J$", (-0.38375640681989376,-0.40035022663183983), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] solution(with sina feizi): define $G$ and $J$ as the midpoints of $MF$ and $ME$ it is trivial that $GJ$ is the radical axis of $(AEF)$ and $M$(consider a point as a circle with radius zero) on the other hands $\triangle DHM \sim \triangle AMD$ so $DH \times DA=DM^2$ and it is obviously on the line $GJ$ and simply we conclude that D is the midpoint of $MX$.$\boxed{Q.E.D}$ what a synthethic solution

My sol:- Construct a circle omega of radius 0 with M as center. Let Applying radax theorem on omega, (AH), (AHM) implies that the intersection of AH& MX belongs to M- midline in ∆ MEF which implies the result.

Claim : $ME,MF$ are tangent to $AEHF$. Proof : $\angle HEF = \angle FBC = \angle HFM$ and $HFE = \angle ECB = \angle HEM$. Now let $P,Q$ be midpoints of $MF,ME$. $P,Q$ have same power w.r.t $M$ and $AEHF$ so $PQ$ is Radical Axis of $M$ and $AEHF$. Now let $AH$ meet $MX$ at $S$. $\angle HAM = \angle HMX$ so $SM^2 = SH.SA$ or $S$ has same power w.r.t $M$ and $AEHF$ so $S$ lies on $PQ$ so obviously $S$ is midpoint of $MX$.

Radical axis theorem in $(AEF),(AHM),(M)$ gives $AH$ bisects $MX$