Hello. My solution. [asy][asy]import graph; size(8cm); 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 = -9.796433468422672, xmax = 18.1217326307628, ymin = -10.033513250768388, ymax = 5.762291252718078; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); /* draw figures */ draw((0.,0.)--(6.16558160759243,0.)); draw((0.,0.)--(4.362362485062396,1.8159337945353193)); draw((4.362362485062396,1.8159337945353193)--(6.16558160759243,0.)); draw(circle((3.0827908037962155,-1.2579423104755298), 3.3295672386145063)); draw((6.16558160759243,0.)--(8.72472497012479,0.)); draw((8.72472497012479,0.)--(4.362362485062396,1.8159337945353193)); draw(circle((3.0827908037962146,-3.0612059453343425), 4.344488575162886)); draw((8.72472497012479,0.)--(7.1925457277099625,-4.469927154476716)); draw((4.362362485062396,1.8159337945353193)--(3.0827908037962146,-4.587509549090036)); draw((3.082790803796218,1.283282629828544)--(3.0827908037962146,-4.587509549090036)); draw((4.362362485062396,1.8159337945353193)--(7.1925457277099625,-4.469927154476716)); draw((7.1925457277099625,-4.469927154476716)--(3.0827908037962146,-4.587509549090036)); draw((6.16558160759243,0.)--(3.082790803796218,1.283282629828544)); draw((0.,0.)--(3.0827908037962146,-4.587509549090036)); draw((3.0827908037962146,-4.587509549090036)--(6.16558160759243,0.)); draw((3.082790803796218,1.283282629828544)--(3.9994924893662827,0.), linewidth(1.2) + linetype("2 2") + qqwuqq); draw((6.16558160759243,0.)--(7.1925457277099625,-4.469927154476716)); draw((7.1925457277099625,-4.469927154476716)--(0.,0.)); draw((3.9994924893662827,0.)--(7.1925457277099625,-4.469927154476716), linewidth(1.2) + linetype("2 2") + red); /* dots and labels */ dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.5865873210409993,0.17341025729844783), NE * labelscalefactor); dot((6.16558160759243,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.261759814191527,0.14717137938567962), NE * labelscalefactor); dot((4.362362485062396,1.8159337945353193),linewidth(3.pt) + dotstyle); label("$A$", (4.477516116123282,1.9838928332794548), NE * labelscalefactor); label("$\omega$", (2.35216700518905,2.2462816124071368), NE * labelscalefactor); dot((8.72472497012479,0.),linewidth(3.pt) + dotstyle); label("$E$", (8.83316984964282,0.14717137938567962), NE * labelscalefactor); label("$\Omega$", (-0.9801704897325239,-0.7974282254739762), NE * labelscalefactor); dot((3.082790803796218,1.283282629828544),linewidth(3.pt) + dotstyle); label("$F$", (3.034377830921026,1.5115930308496268), NE * labelscalefactor); dot((7.1925457277099625,-4.469927154476716),linewidth(3.pt) + dotstyle); label("$K$", (7.337553808615027,-5.100604203167964), NE * labelscalefactor); dot((3.0827908037962146,-4.587509549090036),linewidth(3.pt) + dotstyle); label("$M$", (3.113094464659331,-5.258037470644573), NE * labelscalefactor); dot((3.9994924893662827,0.),linewidth(3.pt) + dotstyle); label("$D$", (4.267605092821136,0.17341025729844783), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] The alternate chord theorem gives $\angle{ABC}=\angle{ACF}$.Since $\angle{ABC}=\frac{1}{2}\angle{ACB}$,it follows that $CF$ bisects $\angle{ACB}$. Also,$\angle{FBC}=\angle{FCB}$,hence $F$ lies on the perpendicular bisector of $BC$.Thus,$FM$ is the perpendicular bisector of $BC$. Let $D\equiv AM\cap BC$. We shall show that $F,D,K$ are collinear.Since $FB=FC$ and $F,B,C,K$ are concyclic,$KF$ bisects $\angle{BKC}$. Also $AD$ bisects $\angle{BAC}$.Thus,due to the angle bisetor theorem,it suffices to show that $\frac{KC}{KB}=\frac{AC}{AB}$. We have $\triangle{ACE}\simeq \triangle{ABE}\Rightarrow \frac{AE}{BE}=\frac{CE}{AE}=\frac{AC}{AB}\Rightarrow \frac{CE}{BE}=\frac{AC^2}{AB^2}$. Similarly,$\frac{CE}{BE}=\frac{KC^2}{KB^2}$,whence it follows that $\frac{AC}{AB}=\frac{KC}{KB}$,q.e.d. Now,we have $\angle{DMC}=\angle{AMC}=\angle{ABC}=\angle{FBC}=\angle{FKC}=\angle{DKC}$,thus $CDMK$ is cyclic. It follows that $\angle{FKM}=\angle{DKM}=\angle{DCM}=\angle{BCM}=\angle{BAM}=\angle{FAM}$ and we are done!

It's easy to see that $\overline{CF}$ bisects $\angle ACB$ and $\triangle CFB$ is isosceles, hence $KF$ bisects $\angle CKB.$ Consequently we get $\triangle AEB$ and $\triangle ACE$ are isosceles. Let $X$ denote the intersection of $\overline{AM}$ and $\overline{BC},$ it suffices to prove that $X$ lies on $\overline{FK}\iff \frac{AC}{AB}=\frac{KC}{KB}.$ Notice that \[\frac{KC}{KB}=\frac{\sin{\angle CFK}}{\sin{\angle BCK}}=\frac{\sin{\angle CKE}}{\sin{\angle ECK}}=\frac{EC}{EK}=\frac{AC}{AE}=\frac{AC}{AB}.\]Hence we conclude that $A,F,M,K$ are concyclic. $\square$ [asy][asy] size(8cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt); void b(){ pair B=D("B",dir(15),dir(15)); pair C=D("C",dir(165),dir(135)); pair A=D("A",dir(115),dir(115)); pair E=D("E",2*foot(A,B,C)-B,dir(180)); pair F=D("F",IP(L(bisectorpoint(C,B),midpoint(C--B),5,5),A--B),dir(35)); path c1=circumcircle(C,F,B); path c2=CP(E,A); pair K=D("K",OP(c1,c2),dir(-115)); pair M=D("M",dir(-90),dir(-45)); pair X=D("X",IP(K--F,A--M),dir(42)*1.5); D(unitcircle); D(A--E--C--A--B); D(E--K--C); D(K--B); D(c1); D(circumcircle(A,F,M),red+linetype("4 4")); D(B--C,magenta+dashed); D(A--M,magenta+dashed); D(K--F,magenta+dashed); D(C--F); } b(); pathflag=false; b(); [/asy][/asy]

Take $AM\cap BC=G$, Observe that $EA^2=EB\cdot EC=EK^2=EG^2$, so $K$ also lie on the Appolonous circle passing through $A$ with foci $BC$. Thus $KG$ is angle bisector of $\angle BGC$. Clearly from the angle condition $\angle ACF=\angle FBC$, so $\angle FCB=\angle FBC$ because $2\angle B=\angle C$, so $FC=FB$. Then $K, G, F$ are colinear, then $AG\cdot GM=BG\cdot GC=KG\cdot GF$, so $AKMF$ is cyclic. QED

My solution is the same as gavrilo's except a different end. Suppose that the circumcircle of triangle $AFK$ cuts that of triangle for second time $ABC$ in $M'$. Since the radical axes of these 3 (and the circumcircle of $BFC$) concur, $AM'$ passes through the intersection of $FK$ and $BC$. But we have shown that $MA$ passes through this point so $A,M,M'$ are collinear which means that (since both $M$ and $M'$ are on the circumcircle of $ABC$) $M\equiv M'$ and the conclusion follows.

It suffices to show that $AM, BC, FK$ concur, and the rest is PoP. Now let $AM \cap BC = X$. It suffices to show that $KX$ is a bisector of $\angle CKB$. Therefore, it suffices to show that $\frac{AC}{AB} = \frac{CX}{CB} = \frac{CK}{KB}$, or just $\frac{AC}{AB} = \frac{CK}{KB}$. Since $EA^2 = EK^2 = EC \cdot EB$ by PoP, we have $ACKB$ is a harmonic quadrilateral, so we are done. $\blacksquare$

Nice problem! We will show that $AM, FK$ intersect on $BC$ at $D$, where $D=AM \cap BC$. Then $DF \cdot DK \overset{\Omega}{=} DB \cdot DC \overset{\omega}{=} DM \cdot DA$, which would imply that $AFMK$ is cyclic. Firstly, since $AC$ is tangent to $\Omega$, hence $\angle ACF=\angle CBF=\angle B$ and since $\angle C=2\angle B$, we obtain that $F$ is the midpoint of arc $BC$ not containing $K$ in circle $\Omega$. It thus suffices to show that $KD$ bisects $\angle BKC$, which is equivalent to showing $\frac{KB}{KC}=\frac{BD}{DC}=\frac{AB}{AC}$. To see this, note that $\triangle EKB \sim \triangle ECK \implies \frac{KB}{KC}=\frac{EB}{EK}$. Also, $\triangle EBA \sim \triangle EAC \implies \frac{AB}{AC}=\frac{EB}{EA}$. Hence, it suffices to show $EK=EA$, but this follows as $EA^2 \overset{\omega}{=} EC \cdot EB \overset{\Omega}{=} EK^2$, proving the result. $\blacksquare$

Let $D=\overline{AM}\cap\overline{BC}$ and $D'=\overline{KF}\cap\overline{BC}$. By radical axes, it suffices to show that $D=D'$. [asy][asy] size(8cm); defaultpen(fontsize(9pt)); pair A=(53.6,105.9), B=(-70,0), C=(70,0), D=(14.4,0), E=(177.2,0), F=(0,60), M=(0,-39), K=(32.2,-73.9); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(M); dot(K); draw(A--B--C--cycle, linewidth(0.5)); draw(circumcircle(A,B,C), linewidth(0.5)); draw(circumcircle(B,F,C), linewidth(0.5)); draw(A--E--K, linewidth(0.5)); draw(F--K, linewidth(0.4)+dashed); draw(A--M, linewidth(0.4)+dashed); draw(C--D, linewidth(0.4)+dashed); draw(C--E, linewidth(0.5)); label("$A$", A, NE); label("$B$", B, (-1,-1)); label("$C$", C, (1,-1)); label("$M$", M, (0,-1)); label("$K$", K, (0.5,-1)); label("$F$", F, (0,1)); label("$D$", D, (-1,-1)); label("$E$", E, (0.5,-1)); [/asy][/asy] Since $\overline{AM}$ is, by fact 5, the bisector of $\angle A$, we must have that that $\angle DAE=\angle ADE$; thus, $AE=DE$. Furthermore, it's not difficult to see that $$\angle FCB=\frac{\angle C}{2}=\angle B,$$so $BF=CF$ and $\overline{KF}$ is the bisector of $\angle BKC$, from which it follows that $KE=KD'$ using similar logic to above. Finally, $E$ lying on the radical axis of $(ABC)$ and $(FBC)$ tells us that $AE=KE$, so $ED=ED'$ and we're done.

Iran Round 2 2016 P2 wrote: Let $ABC$ be a triangle such that $\angle C=2\angle B$ and $\omega$ be its circumcircle. a tangent from $A$ to $\omega$ intersect $BC$ at $E$. $\Omega$ is a circle passing throw $B$ that is tangent to $AC$ at $C$. Let $\Omega\cap AB=F$. $K$ is a point on $\Omega$ such that $EK$ is tangent to $\Omega$ ($A,K$ aren't in one side of $BC$). Let $M$ be the midpoint of arc $BC$ of $\omega$ (not containing $A$). Prove that $AFMK$ is a cyclic quadrilateral. Solution: Let $M_{AB}$ be midpoint of arc $AB$ not containing $C$. Then, $CM_{AB}||AE$ $\implies$ $\angle ACF=\angle ABC=\angle EAC $$=$ $\angle ACM_{AB}$ $\implies$ $F$ $\in$ $CM_{AB}$. Clearly, $FM \perp BC$. Note: $\Delta EAK $ & $\Delta ECA$ are isosceles. Let $AM$ $\cap$ $BC$ $=$ $D$ $$\frac{AC}{AB}=\frac{EC}{CM_{AB}}=\frac{EC}{EA}=\frac{EC}{EK}=\frac{CK}{BK} \implies D \in FK$$So, we're done by POP/Radical Axes $\qquad \blacksquare$

Let AM meet BC at S. Claim1 : K,S,F are collinear. Proof: First Note that ∠FCA = ∠FBC and ∠ACB = 2∠FBC so F is midpoint of arc BC in $\Omega$. Let KF meet BC at S'. KF is angle bisector of ∠CKB so CS'/S'B = KC/KB = EK/EB = EA/EB = AC/AB = CS/SB so S is S'. Claim2 : CSMK is cyclic. Proof: ∠SKC = ∠FKC = ∠FBC = ∠ABC = ∠AMC = ∠SMC. Now we have ∠MKF = ∠MKS = ∠MCS = ∠MCB = ∠MAB = ∠MAF so AFMK is cyclic. we're Done.

$\angle ACF= \angle FCB = \angle FBC$. So, F is CF is angle bisector of $ \angle ACB$. Let $\ \angle ABC=x$. Let circumcircle of $\Omega$ be O. As MF is the perpendicular bisector of CB, O must lie on this line as well. $ BC \cap FM= T$, $ \angle ETO= \angle EKO=90$, so $ETOK$ is cyclic. $ \angle COF= 2* \angle CBF=2x$ , $ \angle KCO=a$. So, $ \angle KOT=180-2a+2x \implies \angle KET=2a-2x$. We know that $ \angle CEA=x$. By power of point we get $AE^2=EC*EB=EK^2 \implies AE=KE$. Hence, $ \angle EKA=EAK=90-a+\frac{x}{2}$. $\angle CBK=\frac{\angle COK}{2}=90-a$. Hence, $ \angle CKA=\frac{x}{2}$ $ AM \cap BC=L$. $ \angle ESA= \angle EAS=90-\frac{x}{2}$, so $EA=ES=EK$. Additionally, as $ \angle ACK=90+a$, $ \angle CAK=90-a-\frac{x}{2} \implies \angle KAI=a-x= \frac{ \angle KEI}{2}$.So, E is the center of $(AIK)$. Hence, $ \angle CKL= \frac{\angle AEL}{2}=\frac{x}{2} \implies \angle CKL=x= \angle CBF$. By this result, we get $K, L, F$ are collinear. $\angle AMF=\angle MFB- \angle MAF=\frac{x}{2} \implies \angle AMF=\angle AKF \implies AKFM$ is cyclic.

Let $L$ be the intersection of $(AFM)$ and $(CFB)$. We will show that $EL$ is tangent to $(CFBL)$. Take $\sqrt{bc}$ inversion. $AE^*CB$ is an isosceles trapezoid. $AM^*$ is the angle bisector of $\angle BAC$ where $M^*$ lies on $BC$. $F^*$ is on $AB$ such that $\angle BF^*C=\angle E^*AC$ and $L^*$ is the intersection of $F^*M^*$ with $(F^*BC)$. We want to prove that $(E^*AL^*)$ and $(BL^*CF^*)$ are tangent to each other. Note that $M^*$ is the incenter of $ACF^*$ hence $\angle L^*BC=\angle M^*F^*C=\angle BF^*M^*=\angle BCL^*$ which yields $L^*B=L^*C$. \[\angle E^*AL^*+\angle L^*F^*C=\angle E^*AL^*+\frac{\angle C}{2}=\angle CAL^*+\frac{3\angle C}{2}=180-\angle AL^*C+\angle C=180-\angle BL^*E+\angle C=\angle E^*L^*C\]As desired.$\blacksquare$