$\angle DAM=\angle DEM=\angle FAA'$ where $AA'$ is diameter of cirumcircle. So, $AF$ is symmedian...

Note that \[\frac{BF}{FC} = \frac{\sin \angle BCF}{\sin \angle CBF} = \frac{\sin \angle BED}{\sin \angle CED} = \frac{BD \cdot CE}{CD \cdot BE} \] Clearly we have $\frac{BD}{DC} = \frac{ c \cos B}{b \cos C}$. On the other side, if $E'$ is the second intersection of $HM$ with the circumcircle of $ABC$ by an easy law of sines we can get that cuadrilateral $BECE'$ is harmonic and therefore \[ \frac{CE}{BE} = \frac {BE'}{CE} = \frac{AE \cdot cos C}{ AE \cdot cos B} \] and therefore we get \[\frac{BF}{FC} = \frac{ c \cos B}{b \cos C} \cdot \frac{AE \cdot cos C}{ AE \cdot cos B} = \frac{AB}{AC} \] and we are done.

Well, this question is like the picture shows and i used butterfly theorem during the test without proving it...

Extend $FC$ beyond $C$ such that $CF=CG$. This is one of many threads that prove the well-known result that if $A'$ is the antipode of $A$ on $(ABC)$ then $H,M,A'$ are collinear. Therefore $\angle AEM=\angle AEA'=90^{\circ}=\angle ADM$ so $AEDM$ is cyclic. Therefore $\angle AMB=\angle AED=\angle AEF=\angle ACF$. Also clearly $\angle ABM=\angle AFC$. It follows that $\triangle ABM\sim AFC$. It follows by SAS that $\triangle AMC\sim\triangle ACG$. Therefore $\angle ACM=\angle AGC$ which implies $\angle AFB=\angle AGC$. Also $\angle ABF=\angle ACG$ so we have $\triangle ABF\sim\triangle ACG$. So $\frac{AB}{BF}=\frac{AC}{CG}=\frac{AC}{CF}$ and the results follows.

Extend $EH$ to intersect $\Gamma$ again $Q$. Then it's well known that $AQ$ is a diameter, so $\angle AEQ = 90$. Thus, $AEDM$ is cyclic, so $\angle MAD = \angle MED = \angle QBF$. Furthermore, $\angle QBC = \angle QAC = \angle DAB$. Thus, $\angle CBF = \angle QBC + \angle QBF = \angle MAD + \angle DAB = \angle MAB$, so $AF$ is the symmedian and the conclusion easily follows.

Here is my ugly non-finished solution : First we will show that: $\angle{AEM}=\frac{\pi}{2}$. Consider point $X$ the second intersection of line $(AD)$ with $(ABC)$, and point $Y$ such that: $\overrightarrow{BY}=\overrightarrow{HC} $. Clearly, $BHCY$ is a parallelogram, so: $\angle{BYC}=\angle{BHC}=\pi-\angle{A}$, therefore: $Y \in (ABC)$. And, it's a very well-known fact that $D$ is the midpoint of $[HX]$, consequently, by the converse of Thales' theorem in $\triangle HXY $ : $(MD) \parallel (XY)$. Thus: $\angle{AEM}=\angle{AEY}=\angle{AXY}=\angle{HXY}=\angle{HDM}=\angle{ADC}=\frac{\pi}{2}$. And since $\angle{ADM}=\frac{\pi}{2}=\angle{AEM}$, the quadrilateral $AEDM$ is cyclic. Now, by simple angle chasing, we get: $\angle{BCF}=\angle{BEF}=\angle{AEB}-\angle{AED}=\pi-\angle{C}-\pi+\angle{AMB}=\angle{AMB}-\angle{C}$. And: $\angle{CBF}=\angle{CEF}=\angle{CED}=\angle{AED}-\angle{AEC}=\pi-\angle{AMB}-\angle{B}$. Therefore: $\frac{BF}{FC}=\frac{\sin(\angle{BCF})}{\sin(\angle{FBC})}=\frac{\sin(\angle{AMB}-\angle{C})}{\sin(\angle{AMB}+\angle{B})}$. So it's enough to show: $\frac{\sin(\angle{AMB}-\angle{C})}{\sin(\angle{AMB}+\angle{B})}=\frac{\sin(\angle{C})}{\sin(\angle{B})}$ And this can be proved using metric relations...

Sorry if my proof is basically equivalent to some above proof. Note that whenever I say similarity suffices, I mean similarity in the way which would imply the problem (and I do prove that similarity each time). Define $A'$ to be the point where $OA$ hits $\Gamma$ that is not $A$. I claim $H,M,A'$ are collinear. To prove this, we use vectors. Denote $O$ as the origin and we have $M = \frac{B + C}{2}$ and $H = A + B + C$ by well-known results. Then remark that $2M + (1-2)H = -A = A'$, therefore $H,M,A'$ are collinear and we are done with this proof. Now, remark that $\angle AEM = \angle AEA' = 90$ because that $AA'$ is a diameter. Then as $\angle ADM = 90$, we have $AEDM$ is cyclic. Rearrange what we want to show as $BF/AB = CF/AC$. Clearly by the diagram $AFB \not \sim AFC$ so that approach would be hopeless. However, remark that if we extend $CF$ to a point $X$ such that $CX = CF$, then the condition is $BF/AB = CX/AC$ and then since $ABFC$ is cyclic we have $\angle ACX = \angle ABF$, therefore the problem statement holds iff $\triangle ACX \sim \triangle ABF$. To show this, remark that $\angle BFA = \angle BCA = \angle MCA$ and that $\angle AMC = 180 - \angle AMD = \angle AEF = \angle ABF$ in the case of $D$ lies on the left of $M$. The case of $D$ lies on the right we have $\angle AMC = \angle AMD = 180 - \angle AED = 180 - \angle AEF = \angle ABF$. In either case we find $\triangle BFA \sim \triangle AMC$. I claim that $\triangle AMC \sim \triangle ACX$. To prove this assertion, first remark that $MC/CX = MC/CF$. If we can show $MC/CF = AM/AC$ then we would be done because $\angle ACG = \angle AMC$. Re-arranging we have and noting $MC = MB$ we have $MB/AC = AM/CF$. Then remark that $\angle AMB = \angle AMD = 180 - \angle AED = 180 - \angle AEF = \angle ACF$ when $D$ is on the left of $M$ and $\angle AMB = 180 - \angle AMD = \angle AED = \angle AEF = \angle ACF$ so it suffices to show $\triangle ACF \sim \triangle AMB$. To show this we only need to show one more angle relation. But $\angle ABM = \angle ABC = \angle AFC$ and therefore we are done.

Nice problem!

If $AA'$ is a diameter of the circle $\odot (ABC)$, then $BA'CH$ is a parallelogram, $E, H, M, A'$ collinear, hence $\angle AEA'=90^\circ$, or $AEDM$ cyclic, with $\angle ADE=\angle AME\ (\ *\ )$. If $AD$ intersects circle $(ABC)$ at $Q$, from $\angle ABD=\angle CAA'$, we get $Q$ the reflection of $A'$ w.r.t. midpoint of the arc $BC$. If $AM$ intersects the circle $(ABC)$ at $P$, from $(*)$ we get that the arcs $PA', QF$ are congruent, hence $P$ and $F$ are symmetrical about the midpoint of the arc $BC$, i.e. $AF$ is the symmedian, done. Best regards, sunken rock

I prove it with a complex method which has inversion and butterfly theorem ....

kakkokari wrote: Well, this question is like the picture shows and i used butterfly theorem during the test without proving it... you just need to prove AFOM is cyclic.your method is similar to mine.

You can also show that AEDM is cyclic using Power of a Point. As was proven earlier in this thread, HD extended meets the circumcircle again at a point X such that HD=DX and HM extended meets the circumcircle at a point Y such that HM=MY. But A, E, X and Y all lie on the circumcircle, so by Power of a Point: $HA \cdot HX = HE \cdot HY$ $\Rightarrow HA \cdot 2HD = HE \cdot 2HM$ $\Rightarrow HA \cdot HD = HE \cdot HM$ Therefore, by the converse of Power of a Point, AEDM is cyclic. From here we can angle chase out the result - firstly a simple angle chase yields $\angle CAM=\angle BAF$ and hence AF is a symmedian. It follows that ABFC is harmonic and therefore $AB \cdot CF=AC \cdot BF$, from which the result immediately follows. Cheers, Hyperspace Rulz!

Proof. Let line $AD$ intersect the circumcircle at $H'$, and let line $HM$ intersect the circumcircle at $N$. Since $H'N$ is parallel to $BC$, we have that the pencil $(NH', NM; NB, NC)$ is harmonic. Intersecting this pencil at the circumcircle we get that the quadrilateral $EBH'C$ is harnomic. then the pencil $D(E, H'; B, N)$ is harmonic. Intersecting this pencil at the circumcircle we get that the quadrilateral ABFC is harmonic. So $\frac{AB}{AC}=\frac{BF}{CF}$. Thus we are done. Solution collaborated with applepi2000.

Let $O$ be the circumcenter of $ABC, AO$ cut the circumcircle again at $G$. As $GC\perp AC, GC\parallel BH$ and it is easy to see well known facts that $GC = BH$ (see the diagram) and $\angle HBC = \angle BCG = 90^{\circ} - \angle ACB$, moreover, $BM=CM$, hence $H, M, G$ are collinear. Notice that $\angle FCG = \angle FEG = \angle DAM$, hence $\angle AMB =\angle ACF$ $\Delta ABM\sim \Delta AFC$, thus $\frac{AB}{AF} = \frac{BM}{FC}$ (*) Since $\angle ABF =\angle AMC = 180^{\circ}-\angle ACF$ (or $\angle AMB$), and $\angle ABC =\angle AFC,\Delta ABF\sim \Delta AMC$, thus $\frac{AC}{AF} = \frac{CM}{BF}$ (**) As $BM = CM$, (*) and (**) give $\frac{AB}{AC} =\frac{BF}{CF}$

Let $HM$ intersect $\omega$ again at $G$, and $BH\cap AC=X$ and $CH\cap AB=Y$. then $AG$ is a diameter and $EAXY$ is cyclic. Taking the radical axis of $AEXY,XYBC,AEBC$, we get that $BC$,$AE$, $XY$ are concurrent. But this implies $BC\cap AE,B,D,C$ form a harmonic division, and using pencil $E$ with the line $BC$, we obtain $ABFC$ is harmonic, and we are done. Alternatively you we could just find $\angle FAB=\angle MAB$ as symmedian... but didn't see this until after

Oops darn did not notice symmedians. [asy][asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=1.2000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-2.5, 2.5); pair B = (-3.5, -0.5); pair C = (0.5, -0.5); pair M = midpoint(B--C); pair X = foot(B,A,C); pair Y = foot(C,A,B); pair K = IntersectionPoint(Line(X,Y,lisf),Line(B,C,lisf)); pair H = orthocenter(A,B,C); pair D = foot(A,B,C); pair E = foot(H,A,K); pair F = (2)*(foot(circumcenter(A,B,C),D,E))-E; pair A_prime = (2)*(M)-H; /* Draw objects */ draw(A--B, rgb(0.6,0.2,0.0)); draw(B--C, rgb(0.6,0.2,0.0)); draw(C--A, rgb(0.6,0.2,0.0)); draw(circumcircle(A,B,C), rgb(1.0,0.6,0.0)); draw(CirclebyPoint(M,B), rgb(1.0,0.6,0.0) + dashed); draw(K--B, rgb(0.8,0.8,0.0) + linewidth(1.0) + linetype("4 4")); draw(X--K, rgb(0.8,0.8,0.0) + linewidth(1.0) + linetype("4 4")); draw(A--D, rgb(0.6,0.6,0.0) + dotted); draw(B--X, rgb(0.6,0.6,0.0) + dotted); draw(C--Y, rgb(0.6,0.6,0.0) + dotted); draw(A--K, rgb(0.8,0.8,0.0) + linewidth(1.0) + linetype("4 4")); draw(E--F, rgb(0.8,0.0,0.2) + linetype("4 4")); draw(A--A_prime, rgb(0.4,0.0,0.0)); draw(E--A_prime, rgb(0.4,0.0,0.0)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(M); dot(X); dot(Y); dot(K); dot(H); dot(D); dot(E); dot(F); dot(A_prime); /* Label points */ label("$A$", A, lsf * dir(120)); label("$B$", B, lsf * dir(220)); label("$C$", C, lsf * dir(-40)); label("$M$", M, lsf * dir(225)); label("$X$", X, lsf * dir(60)); label("$Y$", Y, lsf * dir(120)); label("$K$", K, lsf * dir(225)); label("$H$", H, lsf * dir(45)); label("$D$", D, lsf * dir(225)); label("$E$", E, lsf * dir(120)); label("$F$", F, lsf * dir(45)); label("$A'$", A_prime, lsf * dir(65)); [/asy][/asy] Let $A'$ be the reflection of $H$ over $M$ and note that $AA'$ is a diameter. Set $X = BH \cap AC$ and $Y = CH \cap AB$. Let $XY$ meet $BC$ at $K$. Let $E' = HM \cap AK$. By Brokard's Theorem on the $XCBY$ we find that $\angle A'EA = \angle MEA = 90^{\circ}$, but $AA'$ is a diameter, so $E'$ lies on $(ABC)$ and hence $E'=E$. But $(K,D;B,C)$ is a harmonic bundle. Taking perspectivity at $E$ onto $(ABC)$ gives that $ABFC$ is harmonic as desired.

Let the line through $A$ parallel to $BC$ intersect the circumcircle at $G$. Then $(P_{\infty}, M; B, C)$ is harmonic, and taking the perspective of $G \implies GBFC$ is a harmonic quadrilateral. Since $AGCB$ is a isosceles trapezium, it follows $\triangle GBC \sim \triangle ABC \implies ABFC$ is a harmonic quadrilateral. Note that $G, M, F$ is collinear. To prove this, we are going to show that if we let $GM \cap \Gamma = F' \implies \angle F'GA = \angle CFA'$. Let $A'$ be the antipode of $A$. Let the reflection of $H$ in $BC$ be $H'$. Then $GA'H' = 180 - \angle GAH' = 90$. Let $GA' \cap BC = P \implies \angle MGP = \angle A'GF = \angle DAM$ since $ADPG$ is a rectangle. Now, by power of a point $AH \cdot HH' = EH \cdot HA'$. It is well known that $A'$ is the relflection of $H$ in $M$, so $AH \cdot HD = EH \cdot EM \implies EAMD$ is cyclic. Thus, $\angle DEM = \angle FEA' = \angle DAM = \angle A'GF$ as desired $\blacksquare$.

(Using the figure of v_Enhance) $K$ is the radical center of circle $AEYX$ and the other two drawn circles. So $ (K,D;B,C) = -1$. Therefore $ (K,D;B,C) \stackrel{E}{=} (EA,EF;EB,EC)\implies ABFC$ is a harmonic quad. So the result follows.

Extend $HM$ past $M$ to intersect $\Gamma$ at point $Y$. By Lemma 1.17 (EGMO), $AY$ is a diameter. WLOG, let $AB < AC$ (AB=AC is a trivial case which we can ignore since D=M). Then let $\angle MAC = \theta$. We can see that $\angle DAM = \angle A - (\theta + 90 - \angle B) = \angle A + \angle B - \theta -90$. But $\angle DAM = \angle DEM = \angle FEY$. Since arc $AY$ is 180, then arc $YC$ is $180 - 2B$, and we know arc $FY$ is $2A + 2B - 2\theta - 180$. So then arc $BF$ is $2A - (2A+2B-2\theta - 180 + 180 - 2B) = 2\theta$, which means $\angle BAF = \theta = \angle MAC$, so $AF$ is a symmedian, and the conclusion follows.

The conclusion is a contrived way of saying: Claim: $\overline{AF}$ is the $A$-symmedian of $\triangle ABC$. Proof. [Proof of main claim] It's well known that $\angle AEM = 90^{\circ}$, since the second internsection of $\overline{EHM}$ is the $A$-antipode. That means $MDEA$ is cyclic. [asy][asy] size(7cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A, B, C); pair T = 2*B*C/(B+C); pair M = midpoint(B--C); pair H = orthocenter(A, B, C); draw(A--B--C--cycle, red); pair O = origin; pair L = 2*M-H; pair E = -L+2*foot(O, M, H); pair F = -A+2*foot(O, A, T); filldraw(unitcircle, invisible, red); draw(E--M--A--F, red); draw(A--E, orange); draw(A--D, red); draw(E--F, red); draw(circumcircle(A, D, M), dashed+brown); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$H$", H, dir(H)); dot(L, dir(L)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); /* Source generated by TSQ */ [/asy][/asy] Now, \begin{align*} \measuredangle BAF = \measuredangle BEF &= \measuredangle EBC + \measuredangle BDE = \measuredangle EBC - \measuredangle EDM \\ &= \measuredangle EAC - \measuredangle EAM = \measuredangle MAC. \end{align*}$\blacksquare$

Literally one application of Pascal's theorem. Let $K=(AM) \cap (ABC), G=(AD) \cap (ABC), A'=(MH) \cap (ABC) \neq E$. We apply Pascal's theorem on $AGA'EFK$. Then $(AG) \cap (EF)=D, (GA')\cap (FK), (A'E)\cap(KA)=M$ all lie on a line. But $(AG)$ and $(AA')$ are isogonal, which implies $(GA')\parallel (BC) \equiv (DM)$. Hence, $(FK)$ is parallel to $(BC)$, which implies the desired equality.

Note $\angle{AEM}= \angle{ADM}=90$ So, we have $A,E,D,M$ cyclic $\angle{ABF}=\angle{AEF}=90+\angle{FEM}=90+\angle{DEM}=90+\angle{DAM}=\angle{AMC}$ So, $\triangle{ABF}=\triangle{AMC}$ which implies $F$ is the $A$ symmedian. Q.E.D

Note that the result is equivalent to $AF$ being the $A$-Symmedian. We claim that quadrilateral $MDEA$ is cyclic. This follows since $HM \cdot HE = HD \cdot HA$ follows from reflecting the orthocenter. Then, \[ \measuredangle MAC = \measuredangle MAE + \measuredangle EAC = \measuredangle MDE + \measuredangle EFC = \measuredangle CDF + \measuredangle DFC = \measuredangle DCF = \measuredangle BAF \]gives the result.

Project through $D$, then through $M$... resulting quadrilateral is obviously harmonic

It is well known that $E$ is the A-Queue point $(ABC)$ intersect $(AEF)$. Furthermore, by radical axis theorem on $(AEF),(ABC),(BCEF)$, we have that $AE,FE,BC$ concur at a point, say $P$. Thus, we have $$(AF;BC)^E=(PD;BC)=-1$$by projecting from $E$ onto line $BC$, hence done.

Due to monotonicity, there exists a unique point $F'$ on minor arc $AB$ such that $\frac{BF'}{CF'} = \frac{AB}{AC}$, which is well known to be the intersection of the $A$-symmedian with $(ABC)$. We want to show $F$ is this point, or \[\angle BAF = \angle MAC.\]

that $H$, $M$, and $A'$ are collinear, so $\angle AEM = \angle AEA' = 90$. Claim 1: $\angle BAD = \angle A'AC$. We may simply state $H$ and $O$ are isogonal conjugates, or \[\angle BAD = 90 - \angle ABC = 90 - \angle AA'C = \angle A'AC.~\Box\] Claim 2: $\angle ADE = \angle AME$ and $\angle AFE = \angle AA'E$. This follows directly from cyclic quadrilaterals $DEAM$ and $AEFA'$. $\Box$ From these results, we find \begin{align*} \angle BAF &= \angle BAD + \angle DAF \\ &= \angle A'AC + \left(\angle ADE - \angle AFE\right) \\ &= \angle A'AC + \left(\angle AME - \angle AA'E\right) \\ &= \angle A'AC + \angle MAA' \\ &= \angle MAC.~\blacksquare \end{align*} [asy][asy] size(225); defaultpen(linewidth(0.4)+fontsize(10)); pair A, B, C, H, D, M, E, F, A1; A = dir(120); B = dir(210); C = dir(330); H = orthocenter(A, B, C); D = foot(A, B, C); M = .5B + .5C; E = IP(circumcircle(A, B, C), H--5H-4M); F = IP(circumcircle(A, B, C), D--3D-2E); A1 = -A; draw(E--A--B--C--A--F--E--A1--A--D--A--M); draw(circumcircle(A, B, C)); markscalefactor = .02; filldraw(anglemark(B, A, D)^^anglemark(A1, A, C), cyan); markscalefactor = .015; filldraw(anglemark(A, D, E)^^anglemark(A, M, E), blue); filldraw(anglemark(A, F, E)^^anglemark(A, A1, E), green); markscalefactor = .008; draw(rightanglemark(A1, E, A)^^rightanglemark(C, D, A)); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$H$", H, NE); label("$D$", D, S); label("$M$", M, S); label("$E$", E, W); label("$F$", F, S); label("$A'$", A1, SE); dot(A); dot(B); dot(C); dot(H); dot(D); dot(M); dot(E); dot(F); dot(A1); [/asy][/asy]

[asy][asy] unitsize(4cm); pair A = dir(110); pair B = dir(205); pair C = dir(335); pair D=foot(A, B, C); pair BB=foot(B, C, A); pair CC=foot(C, A, B); pair H=A+B+C; pair M=(B+C)/2; pair G=IP(Line(M, H, 10),unitcircle,1); pair T=extension(A, G, B, C); pair F=IP(Line(D, G, 10), unitcircle, 0); pair I=IP(Line(D, A, 10), unitcircle, 0); draw(A--B--C--cycle, heavygreen); draw(A--T--BB, red); draw(T--C, red); draw(G--F, heavygreen); draw(A--I, heavygreen); draw(G--M, heavygreen); draw(circumcircle(A, G, H), red); draw(circumcircle(A, B, C), heavygreen); draw(circumcircle(B, CC, C), red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B'$", BB, NE); dot("$C'$", CC, W); dot("$H$", H, NE); dot("$M$", M, dir(M)); dot("$G$", G, dir(G)); dot("$T$", T, dir(T)); dot("$F$", F, SE); dot("$$", I, SW); [/asy][/asy] oops pretend $E$ was renamed to $G$ Let $B', C',$ be the feet of the altitudes in $\triangle ABC$ from $B$ and $C$. Since $MH$ passes through the $A$-antipode of $(ABC)$ we find that $G$ lies on $(AB'HC')$. Also, by radical center on $(AB'HC'), (BCB'C')$, and $(ABC)$ we find that $B'C'$, $AG$, and $BC$ concur at some point $T$. Then by Ceva-Menelaus, we have $$-1 = (T, D; B, C) \overset{A}{=} (G, \overline{AD} \cap (ABC); B, C) \overset{D}{=} (F, A; C, B),$$which implies that $ABFC$ is harmonic.

Not a bad problem. Perhaps when it was first proposed, this config was still fresh. Nowadays, it is textbook. Let $A'$ be the antipode of $A$, then it follows from $\overline{HMA'}$ that $\angle AEA' = \angle AEH = 90^{\circ}$, so let $V, W$ be the feet of the altitudes from $B, C$ in $\triangle ABC$, then we have $(AEHVW)$. Let $R = VW \cap BC$, then by radical axis theorem it follows that $AE, VW, BC$ concur at $R$. Now by Ceva-Menelaus theorem it follows that $-1 = (R, D; B, C) \stackrel{E}= (A, F; B, C)$, so we are done.

Ok. Note that $E$ is the $A$-queue point. Projecting through $E$ we find that, \begin{align*} (AF, BC) \overset{E}{=} (\overline{AE} \cap \overline{BC}, D; BC) = -1 \end{align*}Thus $ABFC$ is harmonic which implies the conclusion.

Let $A'$ and $H'_A$ denote the intersection between $(ABC)$ and rays $HM$, $HD$ respectively. It is well-known that $BC \parallel A'H'_A$. Since $$(B,C;A,F) \stackrel{D}{=} (C,B;H'_A,E) \stackrel{A'}{=} (C,B;\infty_{BC},M) = -1,$$$ABFC$ is a harmonic quadrilateral so we are done. $\square$

It is well-known that $H$ reflected over $M$ is the antipode of $A$, call it $A'$. Hence, $\angle MEA = \angle MDA = 90^\circ$, so $AMDE$ is cyclic. This means that \[\angle MAD = \angle MED = \angle FAA'.\] It is well-known that $\overline{AD}$ and $\overline{AA'}$ are isogonal with respect to $\angle BAC$. This means that $\angle CAA' = \angle BAD$. Thus, \[\angle BAM = \angle BAD + \angle MAD = \angle CAA' + \angle FAA' = \angle CAF,\] which implies that $\overline{AM}$ and $\overline{AF}$ are isogonal with respect to $\angle BAC$. This means $\overline{AF}$ is the $A$-symmedian of $\triangle ABC$, so $ABFC$ is a harmonic quadrilateral. Thus, we have $\tfrac{BF}{CF} = \tfrac{AB}{AC}$, as desired. $\square$

Claim: $AEDM$ is cyclic. Proof. We first present the following lemma: Lemma: In $\Delta ABC$, let $H$ be the orthocenter and $M$ be the midpoint of $BC$. Let $X$ be a point on $HM$ such that $M$ is the midpoint of $HX$. Then, $AX$ is a diameter in $(ABC)$. Proof. Let $Y$ be the reflection of $H$ across $BC$, $D$ be the projection of $A$ on $BC$, and $E$ be the projection of $B$ on $AC$. Since $B$ lies on the perpendicular bisector of $HY$, we have \begin{align*} \angle BYH &= \angle BHY \\ &= \angle AHE \\ &= 90^{\circ} - \angle DAC \\ &= \angle ACB \end{align*}which implies that $ABYC$ is cyclic. However, we know that $MD$ is the $H$-midline in $\Delta HYX$, so $DM \parallel XY$, implying that $\angle AYX = 90^{\circ}$. Further, by construction, $BHCX$ is a parallelogram, so $BH \parallel CX$, or $BE \parallel CX$, implying that $\angle ACX = \angle AYX = 90^{\circ}$, so the result follows. $\blacksquare$ Returning to the original problem and applying the lemma by letting $X$ be the reflection of $H$ across $M$, we know that $AX$ is a diameter in $\Gamma$, so $$\angle AEX = 90^{\circ} = \angle ADM$$and the desired condition follows. $\blacksquare$ Then, we have \begin{align*} \angle AMB &= 180^{\circ} - \angle AED \\ &= 180^{\circ} - \angle AEF \\ &= \angle ACF \end{align*}by angle chasing in cyclic quadrilaterals. However, we know that $\angle ABM = \angle ABC = \angle AFC$, so it follows that $$\Delta ABM \sim \Delta AFC$$by angle-angle similarity, implying that $\dfrac{AM}{MB} = \dfrac{AC}{CF}$ and $\angle BAM = \angle CAF$, which in fact gives $\angle BAF = \angle CAM$. Thus, since $\angle AFB = \angle ACB$, we also obtain $\Delta ABF \sim \Delta AMC$, giving \begin{align*} \dfrac{AB}{BF} &= \dfrac{AM}{MC} \\ &= \dfrac{AM}{MB} \\ &= \dfrac{AC}{CF} \end{align*}which is also the desired conclusion. $\blacksquare$

Let $X$ be the $A$ expoint of $\Delta ABC$. Then it is well known that $E$ is the $A$ queue point and $X,E,A$ lie on a line and that $(XD;BC)=-1$. $$\implies -1=(XD;BC) \stackrel{E}{=} (AF;BC)$$And we are done.