Hello. I suppose you mean that $M$ is the midpoint of $\overarc{ABF}$.If so,then we have the following: [asy][asy] import graph; size(7cm); 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 = -7.192810661043649, xmax = 15.550902660456341, ymin = -5.295909126808232, ymax = 7.572244462987871; /* image dimensions */ pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); pen ffxfqq = rgb(1.,0.4980392156862745,0.); draw((1.6,4.96)--(0.,0.)--(7.,0.)--cycle, cqcqcq); draw(arc((5.099157282690777,-1.895490647083808),0.42751340829887197,32.08192406887382,164.5612259724581)--(5.099157282690777,-1.895490647083808)--cycle, linewidth(1.6) + red); draw(arc((1.6,4.96),0.42751340829887197,-102.92604855350807,-55.40535045709237)--(1.6,4.96)--cycle, linewidth(1.6) + blue); draw(arc((5.957313443436363,-1.3575474978495434),0.42751340829887197,124.59464954290763,172.11534763932332)--(5.957313443436363,-1.3575474978495434)--cycle, linewidth(1.6) + blue); draw(arc((5.957313443436363,-1.3575474978495434),0.42751340829887197,172.11534763932335,304.5946495429077)--(5.957313443436363,-1.3575474978495434)--cycle, linewidth(1.6) + red); draw(arc((0.3289589776455144,-0.578085398129868),0.6412701124483079,-15.438774027541903,-7.884652360676656)--(0.3289589776455144,-0.578085398129868)--cycle, linewidth(1.6) + ffxfqq); /* draw figures */ draw((1.6,4.96)--(0.,0.), sqsqsq); draw((0.,0.)--(7.,0.), sqsqsq); draw((7.,0.)--(1.6,4.96), sqsqsq); draw(circle((3.5,1.6090322580645162), 3.8521402891759013)); draw((0.3289589776455144,-0.578085398129868)--(4.805180000978146,5.2333239092066925)); draw((4.805180000978146,5.2333239092066925)--(5.099157282690777,-1.895490647083808)); draw((0.3289589776455144,-0.578085398129868)--(6.617571547621305,-2.3148373481612667)); draw((1.6,4.96)--(6.617571547621305,-2.3148373481612667)); draw((3.014734473399881,2.908815858998591)--(5.099157282690777,-1.895490647083808)); draw((4.805180000978146,5.2333239092066925)--(6.617571547621305,-2.3148373481612667)); draw((0.3289589776455144,-0.578085398129868)--(5.957313443436363,-1.3575474978495434)); draw((0.3289589776455144,-0.578085398129868)--(1.6,4.96)); draw((1.6,4.96)--(4.805180000978146,5.2333239092066925)); draw((5.099157282690777,-1.895490647083808)--(5.957313443436363,-1.3575474978495434)); /* dots and labels */ dot((1.6,4.96),linewidth(3.pt) + dotstyle); label("$A$", (1.507087197838396,5.156793706099234), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.5663528324111333,-0.03749420473208387), NE * labelscalefactor); dot((7.,0.),linewidth(3.pt) + dotstyle); label("$C$", (7.235766869043281,-0.03749420473208387), NE * labelscalefactor); label("$\omega$", (0.053541609622231104,4.066634514937105), NE * labelscalefactor); dot((5.020991244909683,0.),linewidth(3.pt) + dotstyle); label("$D$", (5.09819982754892,0.13351115858746568), NE * labelscalefactor); dot((3.014734473399881,2.908815858998591),linewidth(3.pt) + dotstyle); label("$E$", (2.575870718585576,2.805469960455427), NE * labelscalefactor); dot((5.957313443436363,-1.3575474978495434),linewidth(3.pt) + dotstyle); label("$F$", (5.910475303316778,-1.1062777254792686), NE * labelscalefactor); dot((0.3289589776455144,-0.578085398129868),linewidth(3.pt) + dotstyle); label("$M$", (0.13904429128200552,-1.0635263846493812), NE * labelscalefactor); dot((4.805180000978146,5.2333239092066925),linewidth(3.pt) + dotstyle); label("$G$", (4.8844431233994845,5.37055041024867), NE * labelscalefactor); dot((5.099157282690777,-1.895490647083808),linewidth(3.pt) + dotstyle); label("$H$", (5.012697145889146,-2.3033152687161156), NE * labelscalefactor); dot((6.617571547621305,-2.3148373481612667),linewidth(3.pt) + dotstyle); label("$K$", (6.530369745350142,-2.7094530066000457), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $\angle{MHF}+\angle{MAF}=180^{\circ}\overset{MF=MA}\Rightarrow \angle{MHF}+\angle{MFA}=180^{\circ}\Rightarrow \angle{MHF}+180^{\circ}-\angle{MFK}=180^{\circ}\Rightarrow $ $\Rightarrow \angle{MHF}=\angle{MFK}$.Thus,$\triangle{MHF}\simeq \triangle{MFK}$ which gives $\angle{MFH}=\angle{MKF}\Rightarrow $ $\Rightarrow \angle{MGH}=\angle{HKF}\Rightarrow \angle{EGH}=\angle{EKH}$ whence we get that $EHKG$ is cyclic. Thus $HD\cdot DG=ED\cdot DK$ and since $HD\cdot DG=BD\cdot DC$ we deduce that $ED\cdot DK=BD\cdot DC$ yielding that $EBKC$ is cyclic,and we are done.

gavrilos wrote: Hello. I suppose you mean that $M$ is the midpoint of $\overarc{ABF}$. Indeed! That was the official wording, so I had to type it like that. Sorry for the confusion there. My solution was pretty much the same (and everybody solved it like that) From $\angle MAE = \overarc{MF} = \overarc{AM} = \angle AGM$, so $\triangle AME \sim \triangle GMA$. Therefore, we have $\angle MAG = \angle MEA$. Now we have $\angle GHK = \angle MAG = \angle MEA = \angle GEK$, so $E, G, H, K$ are cyclic. From Power of a Point, we have $DB \cdot DC = DH \cdot DG = DE \cdot DK$, so $B, C, K, D$ are cyclic as desired. $\blacksquare$

Let $t$ tangent for $\omega$ into M, then $\angle (t,KM)=\angle (AK,KM)=\angle (GM,GH) \Rightarrow EGKH$ is cyclic $\Rightarrow ED\cdot DK=DG\cdot DH = DC\cdot DB \Rightarrow BECK$ is cyclic

Since $M$ is a midpoint of arc $\widehat{AF}$, it is well-known that $ME \cdot MG = MK \cdot MH = MA^2.$ Therefore, $E, G, K, H$ are concyclic, so from Power of a Point, $DE \cdot DK = DG \cdot DH = DB \cdot DC \implies E, K, B, C$ are concyclic. $\square$

It is quite obvious that $\angle KHG = \widehat {MHG} / 2 = \widehat {MA}/2 + \widehat {FG}/2 = \angle KEG$, so $HEGK$ is cyclic and so Power of a Point gives us $BD \cdot CD = HD \cdot DG = KD \cdot DE$, meaning that $BECK$ is cyclic.

let MP is tangent to w where P is at the side of K to ME then it is easy to show that MP || AK. <PMH = <MGH = <EGH by rule of tangecy. <PMH = <PMK = <AKM = <EKH by the fact MP || AK therefore <EGH = <EKH then E,G,K,H are cyclic; ED*DK = GD*DH by power of point. B,C,G,H cyclic by figure. by power of point, BD*CD = GD*DH = ED*DK therefore B,C,E,K is cyclic problem is really hard for me as I am a bigginer to olympiad level any other nice solution because it is ugly to me

please a nicer solution

$\angle GEK = \angle EAG + \angle AGE = \angle MAG = \angle GMK$ so $GEHK$ is cyclic Then we have : $DE.DK = DH.DG = DB.DC$ so $BECK$ is cyclic

I'm not too sure if it works but use pure projective to send D to the circumcenter and the problem becomes really nice!

Good problem for training Desargues's involution theorem. We want to show that $ BD*DC = DE*DK $ But we known that $BD*DC=DF*DA$ So we want to show that $\boxed{DF*DA=DE*DK}$ Or we want to show that there is an involution with center $D$ which sends $E$ to $K$ and $A$ to $F$. But it is obvious after using Desargues's Involution theorem on quadrilateral $BMMG$ and line $AF$

Korea MO 2015 P2 wrote: Let the circumcircle of $\triangle ABC$ be $\omega$. A point $D$ lies on segment $BC$, and $E$ lies on segment $AD$. Let ray $AD \cap \omega = F$. A point $M$, which lies on $\omega$, bisects $AF$ and it is on the other side of $C$ with respect to $AF$. Ray $ME \cap \omega = G$, ray $GD \cap \omega = H$, and $MH \cap AD = K$. Prove that $B, E, C, K$ are cyclic. Solution: Obviously, $\angle MFH=\angle AKH$ $\implies$ $EGKH$ cyclic $\implies $ $ED \cdot DK=GD \cdot DH=BD \cdot DC$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -12.650996203483539, xmax = 11.4646320552933, ymin = -5.259993148839092, ymax = 6.722459642240697; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen ffcqcb = rgb(1,0.7529411764705882,0.796078431372549); pen afeeee = rgb(0.6862745098039216,0.9333333333333333,0.9333333333333333); pen cczzff = rgb(0.8,0.6,1); draw((-1,5)--(-3,-1)--(4,-1)--cycle, linewidth(0.8) + zzttqq); draw((2.2930750854750785,4.871977204569392)--(-0.47748111566373863,2.6744806393444374)--(-0.2733722531386903,-2.8763940306911686)--(1.059862274657048,-4.16761047992336)--cycle, linewidth(0.8) + afeeee); /* draw figures */ draw((-1,5)--(-3,-1), linewidth(0.8) + zzttqq); draw((-3,-1)--(4,-1), linewidth(0.8) + zzttqq); draw((4,-1)--(-1,5), linewidth(0.8) + zzttqq); draw((xmin, -4.450593902667449*xmin + 0.5494060973325511)--(xmax, -4.450593902667449*xmax + 0.5494060973325511), linewidth(0.8)); /* line */ draw(circle((0.5,1.1666666666666667), 4.1163630117428225), linewidth(0.8)); draw((xmin, 0.7931607972152628*xmin + 3.0531999416995212)--(xmax, 0.7931607972152628*xmax + 3.0531999416995212), linewidth(0.8)); /* line */ draw((xmin, 3.0191039257582184*xmin-2.0510547880467795)--(xmax, 3.0191039257582184*xmax-2.0510547880467795), linewidth(0.8)); /* line */ draw((xmin, -0.9684841056186747*xmin-3.141150712773155)--(xmax, -0.9684841056186747*xmax-3.141150712773155), linewidth(0.8)); /* line */ draw(circle((0.5,-0.6996477166375639), 3.512863716986617), linewidth(0.8) + linetype("4 4")); draw(circle((3.8572045335939733,0.0546795949560292), 5.064864972539677), linewidth(0.8) + linetype("0 3 4 3") + ffcqcb); draw((2.2930750854750785,4.871977204569392)--(-0.47748111566373863,2.6744806393444374), linewidth(0.8) + afeeee); draw((-0.47748111566373863,2.6744806393444374)--(-0.2733722531386903,-2.8763940306911686), linewidth(0.8) + afeeee); draw((-0.2733722531386903,-2.8763940306911686)--(1.059862274657048,-4.16761047992336), linewidth(0.8) + afeeee); draw((1.059862274657048,-4.16761047992336)--(2.2930750854750785,4.871977204569392), linewidth(0.8) + afeeee); draw((-0.2733722531386903,-2.8763940306911686)--(2.2930750854750785,4.871977204569392), linewidth(0.8) + afeeee); draw((-0.47748111566373863,2.6744806393444374)--(1.059862274657048,-4.16761047992336), linewidth(0.8) + afeeee); draw((-3.5162311340429033,0.26426325222892455)--(-1,5), linewidth(0.8) + cczzff); draw((-3.5162311340429033,0.26426325222892455)--(0.7840062256601475,-2.9398872302438215), linewidth(0.8) + cczzff); draw((xmin, -4.45059390266745*xmin-15.385053593311873)--(xmax, -4.45059390266745*xmax-15.385053593311873), linewidth(0.8)); /* line */ draw((-0.2733722531386903,-2.8763940306911686)--(0.7840062256601475,-2.9398872302438215), linewidth(0.8)); draw((-3,-1)--(0.7840062256601475,-2.9398872302438215), linewidth(0.8)); draw((-1,5)--(2.2930750854750785,4.871977204569392), linewidth(0.8)); /* dots and labels */ dot((-1,5),dotstyle); label("$A$", (-1.6482408104166058,5.139871537758461), NE * labelscalefactor); dot((-3,-1),dotstyle); label("$B$", (-3.607635606442224,-1.3412035567878393), NE * labelscalefactor); dot((4,-1),dotstyle); label("$C$", (4.3618259197004345,-1.209321214747653), NE * labelscalefactor); dot((0.3481346829698211,-1),dotstyle); label("$D$", (0.6502800079980617,-0.775993519472755), NE * labelscalefactor); dot((-0.47748111566373863,2.6744806393444374),dotstyle); label("$E$", (-1.0830307731015236,2.9355523922296323), NE * labelscalefactor); dot((0.7840062256601475,-2.9398872302438215),linewidth(4pt) + dotstyle); label("$F$", (0.9517253612327723,-2.584665638881025), NE * labelscalefactor); dot((-3.5162311340429033,0.26426325222892455),linewidth(4pt) + dotstyle); label("$M$", (-4.229366647488814,0.26022488227156637), NE * labelscalefactor); dot((2.2930750854750785,4.871977204569392),linewidth(4pt) + dotstyle); label("$G$", (2.440111792829155,4.179014474322818), NE * labelscalefactor); dot((-0.2733722531386903,-2.8763940306911686),linewidth(4pt) + dotstyle); label("$H$", (-0.7815854198668131,-3.2817580182362955), NE * labelscalefactor); dot((1.059862274657048,-4.16761047992336),linewidth(4pt) + dotstyle); label("$K$", (0.6502800079980617,-4.694783111524007), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Notice that $\measuredangle AGM=\measuredangle AFM=\measuredangle MAE,$ so $\triangle AME\sim\triangle GMA$ and $$\measuredangle KHG=\measuredangle MHG=\measuredangle MAG=\measuredangle AEM=\measuredangle KEG.$$Hence, $KHEG$ is cyclic and $$ED\cdot DK=HD\cdot DG=BD\cdot DC$$meaning $BECK$ is cyclic. $\square$