Let the perpendicular from $A$ to $BC$ meet $\Gamma$ at $H$. Let $P \equiv AD \cap \Gamma$ and let $A’ \equiv AO \cap \Gamma$. Since $ \angle EMF = \angle EAF = \frac{ \overarc{CF} + \overarc{A’C}}{2} = \frac{ \overarc{CF} + \overarc{BH}}{2}$ we get that $F$, $M$ and $H$ are collinear. On the other hand $\angle FEC = \angle MAF= \frac{ \overarc{CF} + \overarc{CN}}{2} = \frac{ \overarc{CF} + \overarc{BP}}{2}$, thus $P$, $E$ and $F$ are collinear. $\angle BFM= \angle CFA’$, thus $FA’$ is the symmedian of $\angle BFC$ in triangle $BFC$, thus $F$, $A’$ and $D$ are collinear. On the other hand $\angle A’FH= \angle A’AH= \angle A’OM$, furthermore $FOMA’$ is cyclic. Notice that if we prove that $\tfrac{MF}{FD} = \tfrac{NA’}{NH}$ we are done by the general angle bisector theorem in triangle $MFD$. Also $\tfrac{MF}{FD} = \tfrac{OA’}{OD}$. Thus if we prove that $\tfrac{OA’}{NA’} = \tfrac{OD}{NH}$ we are done. $\frac{OA’}{NA’} = \frac{ \cos \angle NAA’}{ \sin 2 \angle NAA’} = \frac{1}{2 \sin \angle NAA’} = \frac{1}{2 \sin \angle ADO} = \frac{ AO }{AA’ \sin \angle ADO} = \frac{ OD \sin \angle DAO}{ A’P \sin \angle DAO} = \frac{OD}{A’P} = \frac{OD}{NH}$

Let $G=AD\cap \Gamma$, $A'=AO\cap \Gamma$, $F'=DA'\cap \Gamma$, $P=AF'\cap GA'$ 1) Prove that $GMEA'$ is cyclic. So from $DG$ is the G-symmedian of $\triangle BGC$ $\Longrightarrow$ $\measuredangle MEA$ $=$ $\measuredangle CAE$ $+$ $\measuredangle ECA$ $=$ $\measuredangle BGA$ $+$ $\measuredangle CGA'$ $=$ $\measuredangle CGM$ $+$ $\measuredangle CGA'$ $=$ $\measuredangle MGA'$, hence $GMEA'$ is cyclic. 2) Prove that $P\in BC$ and $E\in F'G$. By pole and polars' theorem we get $AA'\cap GF'$ and $AF'\cap GA'$ lie in $BC$, hence $P\in BC$ and $E\in F'G$. 3) Prove that $F'=F$ Since $GMEA'$ is cyclic and $AF'$, $ME$, $GA'$ are concurrent in $P$ we get $PF'.PA=PA'.PG=PE.PM$ $\Longrightarrow$ $AF'EM$ is cyclic, hence $F'=F$. 4) Prove that $GMFD$ is a harmonic quadrilateral. It's so easy prove that $OF^2=OG^2=OA^2=OM.OD$ $\Longrightarrow$ $\measuredangle MFG$ $=$ $\measuredangle OAM$ $=$ $\measuredangle MDG$, hence $GMFD$ is cyclic and $OG$ and $OF$ are tangent to $\odot (GMDF)$ and $O$, $M$, $D$ are collinear, hence $GMFD$ is a harmonic quadrilateral. Finally, since $GMFD$ is a harmonic quadrilateral we get $FG$ is F-symmedian of $\triangle FMD$, but $\measuredangle MFG$ $=$ $\measuredangle MAE$ $=$ $\measuredangle DFN$ $\Longrightarrow$ $FN$ is F-median of $\triangle FMD$, hence $FN$ bisects the segment $MD$.

abeker wrote: Let $ABC$ be an acute-angled triangle with $AB \neq AC$, circumcentre $O$ and circumcircle $\Gamma$. Let the tangents to $\Gamma$ at $B$ and $C$ meet each other at $D$, and let the line $AO$ intersect $BC$ at $E$. Denote the midpoint of $BC$ by $M$ and let $AM$ meet $\Gamma$ again at $N \neq A$. Finally, let $F \neq A$ be a point on $\Gamma$ such that $A, M, E$ and $F$ are concyclic. Prove that $FN$ bisects the segment $MD$. Let $A_0$ lie on $\odot(ABC)$ such that $\overline{AA_0} \perp \overline{BC}$ and $M_0$ be symmetric to $M$ in point $O$. Let $\Gamma$ denote inversion about circle $\odot(BC)$ followed by point-symmetry at $M$. Let $K$ be the midpoint of $\overline{MD}$. Notice that $AA_0MM_0$ is an isosceles trapezoid, hence it is cyclic; $\measuredangle AFM=\measuredangle AEM=\measuredangle AFA_0$ so $A_0$ lies on $\overline{MF}$; $\Gamma$ switches each of the pairs $\{A, N\}, \{A_0, F\}, \{M_0, K\}$. Consequently, we see $K, N,$ and $F$ are collinear. $\blacksquare$

hehe here is solution of LBT : by symmetry problem : triangle ABC is in (O). AO cut (O) = X, M is midpoint of BC,XM cut (O)=Y, Tagent of (O) at B and C meet each other at D,AD cut (O)=P. Prove YP bisects MD

H is othorcentre of ABC , X,Y,M,H collinear , AH cut BC at G,cut (O) at S. BH ,CH cut AC,AB at M,N. YA,MN,BC concurrent at T (radical ). Y(AGCB)=-1 ==> Y,G,P collinear , A(YSBC)=-1 , Y,S,D collinear. HS//MD , GH=GS... .WED

Solved with jclash and nukelauncher. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair A,B,C,O,Ap,EE,M,D,NN,F,X,Fp,HA,K,SS; A=dir(120); B=dir(215); C=dir(325); O=(0,0); Ap=2O-A; EE=extension(A,O,B,C); M=(B+C)/2; D=-1/M; NN=2*foot(O,A,M)-A; F=2*foot(O,D,Ap)-Ap; X=(D+M)/2; Fp=reflect(O,D)*F; HA=-B*C/A; K=extension(A,D,X,Fp); SS=foot(A,B,C); draw(X--Fp--D,gray); draw(X--F--D,gray); draw(O--D,Dotted); draw(circumcircle(A,EE,F),dashed); draw(A--M,dashed); draw(circumcircle(A,B,C)); draw(Ap--A--HA); draw(A--B--C--A--D); draw(B--D--C); dot("$A$",A,NW); dot("$B$",B,B); dot("$C$",C,C); dot("$O$",O,NE); dot("$A'$",Ap,dir(165)); dot("$E$",EE,dir(240)); dot("$M$",M,NE); dot("$D$",D,S); dot("$F$",F,E); dot("$X$",X,SE); dot("$F'$",Fp,W); dot("$H_A$",HA,NE); dot("$K$",K,SW); dot("$S$",SS,NE); [/asy][/asy] Let $A'$ be the antipode of $A$, let $S$ be the foot from $A$ to $\overline{BC}$, let $H_A=\overline{AS}\cap(ABC)$, let $K=\overline{AD}\cap(ABC)$, let $F'=\overline{DH_A}\cap(ABC)$. Claim: $\overline{FF'}\parallel\overline{BC}$. Proof. Let $T=\overline{AF}\cap\overline{BC}$. Then $TB\cdot TC=TA\cdot TF=TM\cdot TE$, so \[-1=(BC;ET)\stackrel A=(BC;A'F).\]But $-1=(BC;H_AF')$, so the claim is true. $\blacksquare$ Claim: $S$, $K$, $X$ collinear. Proof. Let $L$ be the midpoint of $\overline{AK}$, let $Y$ be the point so that $ABCY$ is an isosceles trapezoid, and let $Z$ be the reflection of $A$ over $\overline{BC}$. Note that $K$, $M$, $Y$ collinear, and since $BYCZ$ is a parallelogram, $K$, $M$, $Z$ collinear. But by homothety $(A,2)$, we have $\overline{SL}\parallel\overline{ZK}=\overline{KM}$, so $\triangle ASL\sim\triangle DKM$. In particular, \[\frac{AK}{KD}=\frac{2AL}{KD}=\frac{2AS}{MD}=\frac{AS}{XD},\]so $S$, $K$, $X$ collinear, as claimed. $\blacksquare$ By reflection over $\overline{OD}$, it suffices to prove $X$, $K$, $F'$ collinear. By Brokard's theorem, $\overline{AH_A}\cap\overline{KF'}$ lies on the polar of $\overline{AK}\cap\overline{H_AF'}=D$, or $\overline{BC}$. Thus $X$, $K$, $S$, $F'$ are all collinear, and we are done.

I'm surprised nobody complex-bash it because it's fairly easy. First let's make some synthetic observations. Let $\{T\}=AF\cap BC$ and $A'$ the antipode of $A$.By the power of point we have $TB\cdot TC=TA\cdot TF=TM\cdot TE$.So there is a inversion of pole $T$ which swaps $\{B,C\}$,$\{M,E\}$,$\{T,\infty\}$. By the famous fact that inversion preserves cross-ratio we have $-1=(B;C;M;\infty)=(C;B;E;T)\overset{A}{=}(C,B,A',F)$ so $FBA'C$ is harmonic which means $F,A',D$ are collinear. Now we are ready to bash!!! As usual let $(ABC)$ be the unit circle and we denote by $x$ the affixe of $X$.Also set $OD$ to be the real axis,so $b=\overset{-}{c}$ ,$d=\frac{2}{b+c}=\frac{1}{m}$ and $t=\frac{d+\frac{1}{d}}{2}=\frac{d^2+1}{2d}$,where $T$ is the midpoint of $MD$. We use the following lemma. Lemma:Let P be a point on the unit circle and X be other point.Then the affixe of $P'$, the second intersection of $PX$ with the unit circle is $\frac{p-x}{p\overset {-}{x}-1}$

Proof:Click to reveal hidden text

With the lemma we easily obtain $f=\frac{a+d}{ad+1}$,and $n=\frac{a-m}{\overset{-}{m}a-1}=\frac{ad-1}{a-d}$. Now,we have $\frac{f-n}{f-t}=\frac{\frac{a+d}{ad+1}-\frac{ad-1}{a-d}}{\frac{a+d}{ad+1}-\frac{d^2+1}{2d}}=\frac{\frac{a^2-d^2-a^2d^2+1}{(ad+1)(a-d)}}{\frac{2ad+2d^2-ad^3-ad-d^2-1}{2d(ad+1)}}=2d\cdot \frac{a^2(1-d^2)+1-d^2}{(a-d)(d^2-1+ad(1-d^2))}=2d\cdot \frac{a^2+1}{(a-d)(ad-1)}=2d\cdot \frac{\frac{1}{a^2}+1}{(\frac{1}{a}-d)(\frac{d}{a}-1)}=2d\cdot \frac{\overline{a^{2}}+1}{ \overline{(a-d)}\overline{(ad-1)}}=\overline{2d\cdot \frac{a^2+1}{(a-d)(ad-1)}}$ so $\frac{f-n}{f-t} \in \mathbb{R}$ which means that $FN$ bisects $MD$. $\square$

Let $A'$ is antipode of $A$ with respect to $\odot(ABC)$. Assume that $FN \cap MD = G$. Claim 1: Quadrilateral $AOND$ is cyclic Proof: Clearly quadrilateral $OBDC$ is cyclic. From Pop we deduce that: $$ OM \cdot MD = BM \cdot MC = AM \cdot MN $$This proves that $OBDC$ is cyclic as desired. Claim 2: Points $F, A',D$ are collinear Proof: Consider $\sqrt{bc}$ inversion followed by reflection over angle bisector of $\angle BAC$. The claim transforms into: Quote: Given triangle $\triangle ABC$ with orthocenter $H$ and Humpty point $H_A$. Let $M$ be a intersection of $\odot(ABC)$ and $A$ symmedian. Assume that altitude from $A$ intersects $BC$ at point $D$ and $\odot(ABC)$ at point $E$. Let $F$ be intersection of $ME$ and $BC$. Prove that quadrilateral $AFDH_A$ is cyclic. Clearly $-1 =(M,A;B,C) \overset {E}{=} (F,D;B,C) $ , therefore $F$ is $A$ Ex - point. Consequently points $F,H,H_A$ are collinear. But note that $\angle FH_AA = \angle FDA = 90$, which implies that $AFDH_A$ is cyclic as desired. Now we go back to the main problem. Note that: $$ \angle ADO = \angle ODN = \angle OAN =\angle ONA = \angle NFA'= \angle NFD = x $$This implies that $GD$ is tangent to $\odot(FND)$, therefore $GD^ 2 = GN \cdot GF $. Claim 3: Quadrilateral $OMA'F$ is cyclic Proof: From PoP we deduce that: $$ DA' \cdot DF = DB^ 2 = DM \cdot DO $$This implies that $OMA'F$ is cyclic as desired. Claim 4: $ \angle GMN = \angle MFN $ Sketch of Proof: Note that: $$ \angle MFN = \angle MFA'- \angle NFD = \angle MOA'-x =\angle B - \angle C - x $$On another hand: $$ \angle GMN = \angle DAA'= \angle A - 2 \angle BAD + \angle OAN = \angle B - \angle C -x$$ Now we deduce that $MG$ is tangent to $\odot(MFN)$, therefore: $$ GM^ 2 = GF \cdot GN = GD^2 $$This implies that $FN$ bisects $MD$ as desired.

Interestingly the first solution that is neither a bash nor involves any projective geometry. Easy to see $O,M,D$ are collinear. Also, $AOND$ is cyclic by radax. Let $FE \cap \Gamma=K$. By Reim, $KN//ME \implies BK=NC \implies AK$ is a symmedian so $AKD$ is collinear. Let $A'$ be the $A$ antipode. $\angle FKD = 180-\angle AKF = 180- \angle AA'F = 90+\angle FAA' = 90+\angle FAE = 90+\angle FME = \angle DMF$ so $DKMF$ is cyclic. Now $\angle KFD = \angle KMD = \angle NMT = \angle OMA = 90-\angle AMB = 90-\angle ACK = 90-\angle AA'K = \angle KAA' = \angle KFA'$, which means $FA'D$ collinear. Let $T$ be the intersection of $FN$ with $MD$. Lastly, $\angle TDN = \angle MAO = \angle NAA' = \angle NFA' = \angle TFD$ and $\angle MFT = \angle MFE + \angle KFN = \angle NAA'+\angle KAN=\angle KAA'=\angle KFD=\angle KMD=\angle TMN$. Thus, $TM^2=TN \times TF=TD^2$

$N$ is humpty point in $\bigtriangleup FMD$ and done

The diametrical opposite of $A$ lies at $DF,FE,AD$ and $(ABC)$ intersect at one point.