$D$ lies at $(BCF)$, $B,E$ and $F$ are collinear, $X$ lies at $(BCF)$, $BX$ is parallel to $FD$. Then $ME$ passes through intersection point of diogonals of cyclic trapezoid $BXDF$.

This problem was proposed by Belgium.

Let $\omega$ be the circumcircle of $\triangle BCF$. First of all, we show that $BA, MX, \omega$, and the perpendicular bisector of $AC$ meet at one point $Y$. First, we denote $MX \cap \omega = Y$. Then, $\angle YMF = \angle MXD = \angle MAE = 2 \angle BAF = \angle BFC$. Also, if we let $BA \cap \omega = Y'$, then $\angle Y'BF = \frac{1}{2} \angle BFC$. This shows that $Y=Y'$. Finally, this gives us $\angle YCA = \angle YAC = \frac{1}{2} \angle BFC$, so $Y$ also lies on the perpendicular bisector of $AC$. Now $D$ is just the reflection of $Y$ wrt $CA$. Therefore, $D$ lies on $(BCF)$. Now $\angle YXD = 2 \angle YCF = \angle YCD$, so $X$ also lies on that circle. Also, $\angle FXD = \angle FBA = \angle FAD$, so $FAXD$ is a parallelogram as well. This also gives us that $MFED$ is a parallelogram since $MF=DE$ as well. Note that $ME$ passes through the midpoint of $FD$. Since $BF=FA=XD$, we get $BFDX$ as an isosceles trapezoid. Finally, let us prove that $B, F, E$ are colinear. It suffices to show that $\angle BFC = \angle FED$, but we have $\angle FED = \angle FMD = 2 \angle YBF = \angle BFC$. Therefore, $B, F, E$ are colinear. This gives us that $BD, FX, ME$ are concurrent, as required. $\blacksquare$

Too easy for 1)

Yep, too easy for a imo-1. I solved it in 30 minutes

Easy one. Just angle chasing. Let $D' $ be a point on $(M)$ such that $D'A=D'C$. Then $\angle D'AC=\angle D'CA=\angle D'BF$. Moreover, $\angle FAB=\angle FBA$ then $\angle D'AB=\angle D'BA$. We get $D'A=D'B=D'C$. Hence $D'$ is the circumcenter of $(ABC)$. We get $ \angle D'AC=90^\circ-\frac{1}{2}\angle CD'A=\angle CBA-90^\circ=\angle FBA=\angle FAB$. So $D'\equiv D.$ Since $\angle EDA=\angle EAD=\angle DAF$ then $DE\parallel AC$. Let $X'$ be the intersection of $DE $ and $ (M)$. We have $\angle MFX'=\angle CBX'=\angle DBF=\angle DAF$ hence $X'F\parallel DA$ or $X'FAD$ is a parallelogram. But $MX'=MF$ and $ED=EA$ hence $FX'DA$ is a parallelogram. We get $X'\equiv X. $ Finally, since $XF=DA=DB$ then $BXDF$ is an isosceles trapezoid. We are done.

Note that $\angle{ADC}=180^0-2\angle{BAC}=180^0-2\angle{ABF}=2(180^0-\angle{ABC})$ and since $DA=DC$ it follows that $D$ is the circumcenter of $\triangle{ABC}$.Hence $\angle{BDC}=2\angle{BAC}=\angle{BFC}$,so $BCDF$ is a cyclic quadrilateral. Hence $\angle{FBD}=\angle{ACD}=\angle{ABF}~(\bigstar)$ and so $\angle{ABD}=2\angle{ABF}=2\angle{BAC}=2\angle{DAE}=180^0-\angle{AED}$ and so $BAED$ is a cyclic quadrilateral. Now notice that this combined with $EA=ED$ implies that $BE$ is the angle-bisector of $\angle{ABD}$,so from $(\bigstar)$ we deduce that points $B,F,E$ are collinear.Hence $\angle{AEF}=\angle{AEB}=\angle{ADB}$ and from $\angle{EAF}=\angle{DAB}$ it follows that $\triangle{EAF}\sim\triangle{DAB}$.But $AD=DB$ and so $EA=EF$. Therefore $EF=EA=XM$ and so $EXMF$ is an isosceles trapezoid and so $EXMF$ is cyclic. Now in order to finish the problem we will just prove that $M\in\odot(ABDE)$ and $X\in\odot(BCDF)$. We have $\angle{AMB}=180^0-2\angle{BFC}=\angle{ADB}$ and so $M\in\odot(ABDE)$ and so $\angle{AME}=\angle{ABE}=\angle{BAF}=\angle{BEM}$,meaning that $EF=FM$. Therefore $XM=EA=EF=FM$ and since $M$ is the center of $\odot(BCDF)$(because $\angle{CBF}=90^0$) it follows that $X\in\odot(BCDF)$. Now we are done,just look at circles $\odot(BMDEA),\odot(BCXDF)$ and $\odot(FMXE)$,their radical axis are concurrent.

Note that $DA=DC$ and $\widehat{CDA}=360^\circ-2\widehat{CBA}$ so $D$ is the circumcenter of $\triangle{BCA}$. Thus, $\widehat{DEA}=360^\circ-2\widehat{DFA}$ and $EA=EC$, whence $E$ is the circumcenter of $\triangle{EAD}$. As $MD=MF$ (so $ME$ is the perpendicular bisector of $DF$), angle chasing yields $MFED$ parallelogram, which in turn implies $FAXD$ parallelgram. It follows that $ME\cap FX$ is the midpoint of $\overarc{MF}$ in $(BMF)$, but $BD$ is the bisector of $\widehat{MBF}$, implying the concurrency.

My solution. From $DA=DC$ so $\angle ADC=180^\circ-2\angle FAD= 180^\circ-2\angle FBA=180^\circ-2(ABC-90^\circ)=360^\circ-2\angle ABC$. Thus, $D$ is circumcenter of triangle $ABC$. Now $\angle AFE=2\angle FAB=\angle FAE$, hence $EF=EA=ED$. From $DA=DB$ and $FA=FB$ so $A,B$ are reflections through $FD$. Therefore $\angle FBD=\angle FAD=\angle DCF$ so $FBCD$ is inscribed in circle $(M)$. Circles $(BFDC)$ and $(AFD)$ have the same chord $FD$ and $\angle FBD=\angle FAD$ so its radius are equal. We get $MFED$ is rhombus. Thus, $E,M$ are reflections through $FD$, we get $ABME$ is isocesles trapezoid, but $AMXE$ is parallelogram. So $B,X$ are reflections through $EM$. From $MFED$ is rhombus, we get $F,D$ are reflections through $EM$, so $FX,EM,DB$ are concurrent. The same solution is for this extension by reflections. Extension. Let $ABC$ be a triangle with $\angle ABC>90^\circ$ and circumcenter $D$. Perpendicular bisector of $AB$ cuts $AC$ at $F$. $E,M$ are circumcenter of triangles $ADF,BFC$. Construct parallelogram $AMXE$. Prove that $EM,FX,BD$ are concurrent.

Was this problem actually easy? I thought it was really difficult for a problem 1.

Definitely harder then IMO 2014 4th

Got a nice and easy extension: Given triangle $ABC$ inscribed in $(O).$ An arbitrary line through $O$ intersects the circumcircles of $AOB$ and $AOC$ at $Q$ and $P$, respectively. $(AOB), (AOC)$ intersects $BC$ at $E, F$, respectively. $QE$ meets $PF$ at $J$. Then $J$ is the circumcenter of triangle $APQ.$

What about D2 predictions?

I'm guessing NAG

Juraev wrote: What about D2 predictions? The topics of Day 2 will be 4. Algebra 5. Number theory 6. Geoff Smith's geometry problem See also: http://www.artofproblemsolving.com/community/c6h1270207_question_number_6

Wait, wasn't number 3 number theory + geometry? I really think #6 is combinatorics. Don't get me wrong, I really hope it is geometry though.

IMO, this is the hardest easy geometry problem that has been in the IMO in the last few years. My solution is ugly and kind of long and probably similar to those already written here, so I won't bother to write it.

In that video he indicated that problem 6 may be geometry :-o Waiting for something like P6 2011. It was really nice and challenging.

MathPanda1 wrote: I really think #6 is combinatorics. Maybe it's an inequality at last?

I think we can use Seva’s Theorem

i immediately saw this was easily coordbashable (only linear equations) Let $A = (0, 0)$ and $B = (2, 0)$. Then, let $F = (1, a)$. We will express every point or line in terms of $a$. First, $FB$ is $y=-a\left(x-2\right)$ and $FA$ is clearly $y=ax$. Then, we can get that $C = \left(-\frac{2}{a^{2}-1},-\frac{2a}{a^{2}-1}\right)$. Next, the perpendicular bisector of $AC$ is $y+\frac{a}{a^{2}-1}=-\frac{1}{a}\left(x+\frac{1}{a^{2}-1}\right)$, and since $\angle DAB = 2 \angle CAB$, we have that $\tan(\angle DAB) = \frac{2a}{1-a^2}$, so $AD$ is $y=\frac{2a}{1-a^{2}}x$. We can solve for $D = \left(1,\frac{2a}{1-a^{2}}\right)$. Next, the perpendicular bisector of $AD$ is $y-\frac{a}{1-a^{2}}=\frac{a^{2}-1}{2a}\left(x-\frac{1}{2}\right)$, and since $\angle EAB = 3 \angle CAB$, we have $\tan(\angle EAB) = \frac{3a-a^3}{1-3a^2}$. Thus, $AE$ is $y=\frac{3a-a^{3}}{1-3a^{2}}x$. We can solve for $E = \left(\frac{3a^{2}-1}{2\left(a^{2}-1\right)},\frac{a\left(a^{2}-3\right)}{2\left(a^{2}-1\right)}\right)$. Since $AEXM$ is a parallelogram and $A$ is the origin, we have $X = E+M = \left(2,\frac{a\left(a^{2}-3\right)}{a^{2}-1}\right)$. Now, we have $FX$ is $y-a=\frac{-2a}{a^{2}-1}\left(x-1\right)$, $EM$ is $y=\frac{a\left(a^{2}-3\right)}{2\left(a^{2}-1\right)}$, and $BD$ is $y=-\frac{2a}{1-a^{2}}\left(x-2\right)$. We can verify that the point $\left(\frac{a^{2}+5}{4},\frac{a\left(a^{2}-3\right)}{2\left(a^{2}-1\right)}\right)$ lies on all three lines. Thus, we are done.

On the difficult side for problem 1... Notice that $X$ lies on $\overline{DE}$. The key claim is the following: Claim. $BFDXC$ is cyclic with center $M$. Proof. Observe that side ratios imply $\triangle ABC \sim \triangle AFD$. So $\angle FDC = \angle ADC - \angle ADF = 90^\circ$, hence $D$ lies on the circle. On the other hand, note that as $\triangle AED \sim \triangle AFB$, we have $\triangle AFE \sim \triangle ABD$. Because $F$ is the incenter of triangle $ABD$ by Fact 5 and $AF=BF$, it follows that $AD=BD$ and $AE=EF$. Then, set $\angle EAD = \theta$. It follows that $\angle DMA = \angle DBA = 2\theta$ and $\angle FED = (180^\circ - 2\theta) - (180^\circ-4\theta) = 2\theta$. Hence $FMDE$ is a parallelogram, and $MX=AE=EF=MD$. Thus $X$ lies on $(BFDC)$. $\blacksquare$ To finish, we have $ED=EA=EF$, so the result follows by symmetry about $\overline{EM}$.

Let $\angle ABF = \angle BAF = \theta.$ We first see that $D,E,$ and $X$ are collinear since $\angle AED = 180^\circ - 2\theta$ and $\angle EAC = 2\theta.$ Next, we note that $\angle ADC = 180^\circ - 2\theta$ and $\angle ABC = 90 + \theta$, so combining this with $AD = DC$ gives that $D$ is the circumcenter of $\triangle ABC.$ Thus $AD = DB.$ On top of that, we have that $\triangle AFB \sim \triangle AED,$ so spiral similarity implies $\triangle ADB \sim \triangle AEF,$ implying $AE = EF = MX.$ Now, we have that $\angle FBD = \angle ABD - \angle ABF = \angle BAD - \theta = 2\theta - \theta = \theta$ and $\angle FCD = \angle CAD = \angle BAC = \theta,$ so $FBCD$ is cyclic. Moreover, $M$ is the cirucmcenter of $\triangle FBC,$ so $MC = FM = MD = MB.$ This means that $\angle CMD = 180^\circ - 2 \angle MCD = 180^\circ - 2\theta,$ so $\angle AMD = \angle EAM = 2 \theta.$ Thus $AEDM$ is an isosceles trapezoid, implying that $AE = DE = DM = FM = CM.$ Thereore, $EM \parallel CD.$ We also know that $FM = DE$ and $EX = AM,$ so $AF = DX,$ implying that $FX \parallel AD.$ This means that we can define $O = EM \cap FX$ so that there is a homothety centered at $I = OD \cap AC$ sending $\triangle FOM$ to $\triangle ADC,$ implying that $\frac{FI}{IM} = \frac{AI}{IC}.$ To finish off the problem, we now let $I' = BD \cap AC.$ Then $\angle FBI' = \angle I'BM = \theta,$ so the Angle Bisector Theorem implies that $\frac{FI'}{I'M} = \frac{FB}{BM}.$ By looking at $\triangle ABF,$ we see that $BF = \frac{AB}{2 \cos \theta},$ and by looking at $\triangle BMC,$ we see that $BM = \frac{BC}{2 \cos (90 - 2\theta)}.$ Therefore, plugging in these two quantities gives $$\frac{FI'}{I'M} = \frac{AB}{BC} \cdot \frac{\cos(90 - 2\theta)}{\cos \theta} = \frac{AB}{BC} \cdot \frac{\sin 2\theta}{\sin (90 - \theta)}.$$By the Ratio Lemma this is equal to $\frac{AI'}{I'C}.$ Therefore, $I'$ is the center of a homothety taking segment $FM$ to segment $AC,$ but we know that this center of homothety is just $I.$ Thus $I = I',$ and so $B,I,O,D$ are collinear, as desired.

Let the perpendicular bisector of $\overline{AC}$ meet $(BCF)$ at $D'.$ We show that $D' \equiv D.$ Indeed, as \[ \measuredangle D'AC = \measuredangle D'CA = \measuredangle D'CF = \measuredangle D'BF.\] Furthermore, \[\measuredangle FAB = \measuredangle FBA.\] Hence, $$\measuredangle D'AB = \measuredangle D'AF + \measuredangle FAB = \measuredangle D'BF + \measuredangle FBA = \measuredangle D'BA.$$ Thus $D'A = D'B = D'C.$ Now, as $$\measuredangle D'BM = \measuredangle D'BC - \measuredangle MBC = \measuredangle D'CB - \measuredangle MCB = \measuredangle D'CM = \measuredangle D'AM,$$ we see that $D'ABM$ is cylic. Thus, \[\measuredangle D'AM = \measuredangle D'BM = \measuredangle BD'M = \measuredangle BAM\] which implies $CA$ bisects $\angle D'AB.$ This implies $D' \equiv D$ as claimed. Now note that as $\angle EDA = \angle EAD = \angle CAD$ we see that $AC \parallel ED.$ Furthermore we know that $AC \parallel EX.$ So we get that $E-D-X$ collinear. Let $$\angle EAD = \angle DAC = \angle DCA = \angle DBF = \angle FBA = \alpha.$$ Then $\angle DMA = \angle DBA = \angle DBF + \angle FBA = 2\alpha.$ And also, $\angle AED = 180 - 2\alpha.$ Thus, $AMDE$ is cyclic. As a whole, $(ABMDE)$ is cyclic. This means that $$\measuredangle EBA = \measuredangle EDA = \alpha = \measuredangle FBA .$$ Thus, $B-F-E$ collinear as well. Now, as $\angle MXE = \angle EAM = 2\alpha$ and $\angle EFM = 180 - \angle BFC = 180 - 2\alpha$ we get $EFMX$ cyclic as well. Furthermore as, $$\angle DXF = \angle EXF = \angle EMF = \angle EMA = \angle EBA = \alpha = \angle DBF$$ we get that $X$ lies on $(BFDC)$ too. Now consider the radical axes of $(BMDEA), (EFMX)$ and $(BCXDF)$ we get that $\overline{BD} \cap \overline{ME} \cap \overline{FX}$ are concurrent, as needed.

Let $\alpha$ be $\angle BAC = \angle CAD = \angle DAE$. Since \[\angle CBA = 90^{\circ} + \alpha = 180^{\circ} - \frac{180^{\circ} - 2 \alpha}{2} = 180^{\circ} - \frac{\angle CDA}{2},\]it follows that $D$ is the circumcenter of $\triangle ABC$; in particular, $\overline{MD} \perp \overline{BC}$, so $\angle DMA = 2 \alpha$. We also have $\angle EAM = 2 \alpha$ and $\angle MAD = \alpha = \angle ADE$, so $MDEA$ is an isosceles trapezoid with $MD = DE = EA$. Furthermore, by definition, $MX$ is equal to all these lengths, and even further more, since $\angle MCD = \alpha$ and $\angle CMD = 180^{\circ} - 2 \alpha$, it follows that $CM$ and $MF$ are equal to all these lengths. In full: \[ CM = MX = MD = MF = MB = DE = EA = EF.\]It's easy to see from this that $BFDX$ is an isosceles trapezoid and line $ME$ is its line of symmetry; this finishes.

Denote $\angle BAC = \angle CAD = \angle DAE = \alpha$. We notice that $\triangle AED \sim \triangle ADC$. From that similarity we get that $\frac{AD}{AB} = \frac{AE}{AF}$, which means $\triangle BAD \sim \triangle FAE$ and we also get $\frac{AC}{AB} = \frac{AD}{AF}$, which means $\triangle CAB \sim \triangle DAF$. From AF=FB $\angle FBA = \angle FAB = \alpha$, $\angle FBC = 90^{\circ}$ $\Rightarrow$ $\angle BCA = 90 - 2\alpha$, by $\triangle CAB \sim \triangle DAF$, we have that $\angle BCA = \angle ADF = 90 - 2\alpha$, also $\angle DFC = \angle DAF + \angle ADF = \alpha + 90 - 2\alpha = 90 - \alpha$, DA = DC $\Rightarrow$ $\angle DAC = \angle DCA = \alpha$, $\angle FDC = 180 - \angle DFC - \angle DCF = 180 - (90 - \alpha) - \alpha = 90^{\circ}$ $\Rightarrow$ $\angle FDC + \angle FBC = 90 + 90 = 180^{\circ}$ $\Rightarrow$ FDCB is cyclic, and since M is midpoint of FC, we have that FM=MC=MB=MD $\Rightarrow$ M is the center of the circle around FDCB. Also $\angle FDB = \angle FCB = 90 - 2\alpha$. Now from $\triangle BAD \sim \triangle FAE$ we get that $\angle AFE = \angle ABD = \angle ABF + \angle FBD = \alpha + \angle FCD = 2\alpha$ $\Rightarrow$ $\angle EAF = \angle AFE = 2\alpha$ $\Rightarrow$ EA = EF. Also $\angle AFE + \angle AFB = 2\alpha + 180 - 2 \alpha = 180^{\circ}$ $\Rightarrow$ E, F and B lie on one line. Now let $EM \cap BD$ = O. So our new goal is for $X \in FO$. From $\triangle FMB \sim \triangle AEF$ we get that $\frac{FM}{EF} = \frac{FB}{AF}$ and from $\angle AFB = \angle EFM = 180 - 2\alpha$ $\Rightarrow$ FE = FM, $\angle FEM = \angle FME = \alpha$ $\Rightarrow$ $\angle OMF = \angle OBF = \alpha$ $\Rightarrow$ FOMB is cyclic. We know that AMXE is a parallelogram also $\angle FAE + \angle AED = 2\alpha + 180 - 2\alpha = 180^{\circ}$ $\Rightarrow$ $ED \parallel AF$ $\Rightarrow$ $X \in ED$. Also EX = AM $\Rightarrow$ ED + DX = AF + FM, but ED = EF = FM $\Rightarrow$ DX = AF $\Rightarrow$ $X \in ED$ and DX = AF. Now we have that AE = XM, AF = DX and $\angle FAE = \angle MXD = 2\alpha$ all from AMXE being a parallelogram $\Rightarrow$ $\triangle FAE \cong \triangle DXM$ $\Rightarrow$ DM = MX, $\angle MDX = \angle MXD = 2\alpha$. Also $\angle ODX = 180 - \angle ADE - \angle ADO = 180 - \alpha - (180 - 4\alpha) = 3\alpha$. Now $\angle OMX = 180 - \angle AMO - \angle XMC = 180 - \angle FME - \angle MXD = 180 - \alpha - 2\alpha = 180 - 3\alpha$ $\Rightarrow$ $\angle ODX + \angle OMX = 3\alpha + 180 - 3\alpha = 180^{\circ}$ $\Rightarrow$ ODXM is cyclic. Now $\angle FOM = 180 - \angle FBM = 180 - 2\alpha$, $\angle MOX = \angle MDX = 2\alpha$ $\Rightarrow$ $\angle FOM + \angle MOX = 180 - 2\alpha + 2\alpha = 180^{\circ}$ $\Rightarrow$ $X \in FO$ $\Rightarrow$ EM, BD and FX are concurrent at O. We are ready.

Suppose $\angle CAB = \theta$ and $CF = 2$. Some trig gives us $AE = ED = 1$, and further angle chase gives: $X \in ED$, as $\measuredangle AED = \measuredangle AEX = 2\theta$. This induces $EF \parallel DM$, $EM \parallel DC$, and $AD \parallel FX$. $EAMD$ is an isosceles trapezoid, as $\measuredangle ADE = \measuredangle XDC = \measuredangle ACD = \measuredangle AME$, so \[EA = ED = EF = MB = MC = MD = MF = MX = 1.\] This mess of equal angles and lengths eventually outputs the isosceles trapezoids $FBXD$, $EFMX$, and $BEDM$, from which we finish with radical axis. $\blacksquare$

Claim: $\triangle ABD \sim \triangle AFD$ and $\triangle ABC \sim \triangle AFD$. Proof. $\triangle AED \sim \triangle ADE \sim \triangle AFB$, so $\frac{AB}{AF} = \frac{AD}{AE} = \frac{AC}{AD}$, so $\triangle ABD \sim \triangle AFD$ and $\triangle ABC \sim \triangle AFD. \ \square$ Claim: $D$ is the circumcenter of $\triangle ABC$. Proof. Let $\angle BAC = \alpha$. We have $\angle ABC = 90^\circ + \alpha$ and $\angle ADC = 180^\circ - 2 \alpha$, and since $AD = CD$, $D$ is the circumcenter of $\triangle ABC$. $\square$ Claim: $FMXE$ is an isosceles trapezoid. Proof. We have $AD = BD = CD$, so $\angle AFE = \angle ABD = \angle BFC = 2 \alpha$, so $B, F, E$ are collinear. Note that $\angle ADE = \angle DAC = \alpha$, so $ED \parallel AC$ and $X$ lies on line $ED$. Then $\angle MXE = \angle MAE = \angle FEX = 2 \alpha$, so $FMXE$ is an isosceles trapezoid. $\square$ Claim: $BCXDF$ is cyclic. Proof. We have $\angle FDC = \angle ADC - \angle ADF = 90^\circ$, so $FBCD$ is cyclic with circumcenter $M$. Then $\angle CFD = 90^\circ - \alpha$, so $\angle MDX = \angle DMA = \angle MXD = 2 \alpha$, so $\triangle MDX$ is isosceles so $FDXC$ is an isosceles trapezoid, therefore $BCXDF$ is cyclic. $\square$ Claim: $\triangle EFD$ is isosceles. Proof. $\angle FED = 2\alpha$ and $\angle EFD = 180^\circ - \angle MXE - \angle CFD = 90^\circ - \alpha$, so $\angle EFD = \angle EDF = 90^\circ - \alpha$. Therefore, $EF = ED$ and $\triangle EFX$ is isosceles. $\square$ Now note that $\angle MFD = \angle MDF = 90^\circ - \alpha$ and $\angle MFB = \angle MDX = 2\alpha$, so $BFDX$ is an isosceles trapezoid, and $EM$ is the perpendicular bisector of $DF$, so $FX$ and $BD$ concur on $EM$.

almost pure angle chasing first we just angle chase and easily prove that $D$ lies on $EX$ next, even more angle chasing yields that $D$ is the circumcenter of $ABC$, which yields that $BFDC$ and $BAED$ are cyclic then, $BED=BAD=BFC$, so $EFB$ are collinear then, still more angle chasing yields that $EM$ and $DC$ are parallel, $AD$ and $FX$ are parallel, and $EF$ and $DM$ are parallel let $EM$ intersect $AD$ at $O$, and let $EM$ intersect $FX$ at $P$ we easily prove $EO=MP$ and $FE=FM$, so we get that $EFO=PFM=DXP$ and more angle chasing yields that $OF$ is parallel to $DB$ let $DB$ intersect $AC$ at $Q$, and we do length ratio chasing to get that $ED:DX=AF:FM=FQ:QM$, so $DPQ$ are collinear and we are done

Claim: $B,F,C,D,X$ are cyclic with center $M$. Proof: Reflect $B$ over $FC$ to get $B'$ on $AD$. Since $\angle FB'C = 90$, proving $AF \cdot AC = AB' \cdot AD$ will show that $D$ lies on the circle with diameter $FC$, however this is trivial since $\triangle ADC \sim \triangle AFB \sim \triangle AFB' $. For showing $X$ lies on the circle, note at $ED$ is parallel to $AC$, it suffices to prove $MD$, $MX$, are reflections over the perpendicular from $M$ to $FC$. Indeed, this is trivial since $\angle XMA = 180 - \angle 2 \angle CAD, \angle AMD = \angle FMD = 2 \angle FCD = 2 \angle CAD$. Finish: Note that arc $DX$ has measure $180 - 4 \angle CAD$, and arc $BF$ has measure $2 \angle BCF = 180 - 4 \angle CAD$, so $BFDX$ is an isosceles trapezoid. We then show $EM$ is the perpendicular bisector of this trapezoid, finishing. It is sufficient to prove $EF = ED$, equivalently showing that $E$ is the center of $(ADF)$. Indeed, we find $\angle AFD = 180 - \angle DFC = 180 - \frac 12 \angle DMC = 180 - \frac 12 (180 - \angle AMD) = 90 + \frac 12 AMD = 90 + \angle CAD$, thus letting $Y$ be the center of $(ADF)$ we have $\angle AYD = 180 - 2 \angle CAD$ and $Y$ lies on the perpendicular bisector of $AD$ on the opposite side of $C$, forcing $Y = E$ as desired.

it was fun

I will just write the steps since writing up the solution will take ages. Step 1. $D \in \odot(BCF)$ (Do this by showing $\triangle DAF \sim \triangle CAB$) Step 2. $X \in \odot(BCF)$ (Do this by just angle chasing) Step 3. $\overline{B-F-E}$ collinear (Do this by showing $\triangle EAF \thicksim \triangle MFB$) Step 4. $\overline{E-D-X}$ collinear (Do this by showing $\square EDMF$ is a rhombous) Step 5. Show that qudrilaterals $\square MXEF$ and $\square EDMB$ are cyclic (Do this by showing they are isosceles trapezoid) Step 6. Finish with the Radical Axis Theorem

Let $\alpha=\angle BAC$ and $FM=x$. Then $AF=2x\cos 2\alpha$, so $AC=2x(1+\cos 2\alpha)=4x\cos^2\alpha$. So $AE=ED=x$. ALso, $D$ lies on $EX$ because $\angle EDA=\angle CAD$. Now the rest of this problem is just a lot of discoveries. In particular, eventually you get $DMC\cong EFM\cong AED\cong BMD$ and $DMX\cong BMF$. Eventually though, we get $BF=DX$, $EF=ED$, $BFE$ are collinear, and $MB=MX$. But then we can just use Ceva on triangle $BEX$ to show that $M$ lies on the $E$-median, which is true because $EB=EX$ and $MB=MX$. $\blacksquare$

[asy][asy] size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt); void b(){ pair B=D("B",origin,dir(-100)*1.5); pair C=D("C",(3,0),dir(-45)); pair F=D("F",(0,2),dir(-160)*1.5); pair A=D("A",IP(L(F,C,5,5),CP(F,B)),dir(135)); pair M=D("M",midpoint(F--C),dir(-80)*1.5); pair D=D("D",IP(CP(M,B),L(midpoint(A--C),bisectorpoint(A,C),5,5)),dir(60)); pair E=D("E",OP(L(B,F,5,5),circumcircle(A,B,M)),dir(100)*1.5); pair Y=IP(M--E,B--D); dot(Y); pair X=D("X",IP(CP(M,B),L(F,Y,5,5)),dir(-10)); D(anglemark(A,D,F,14),blue); D(anglemark(F,C,B,14),blue); D(anglemark(F,A,D,14),deepgreen); D(anglemark(B,A,F,16),deepgreen); D(anglemark(F,B,A,14),deepgreen); D(anglemark(D,C,F,16),deepgreen); D(anglemark(F,D,B,16),blue); D(B--D,dashed); D(E--M,dashed); D(F--X,dashed); D(circumcircle(A,E,D),red+dotted); D(CP(M,B),red+dotted); D(circumcircle(E,F,M),red+dotted); D(A--C--B--E); D(A--E--X--M); D(B--A--D--C); D(D--F); } b(); pathflag=false; b(); [/asy][/asy] claim:A,B,D,E are concyclic: proof: $$\angle ADE =\angle EAD= \angle DAF =\angle FAB =\angle ABF $$claim: E,F,M,X are concyclic: proof $$\angle EFM =\angle 180-BFM =180-(\angle FAB+\angle FBA )=180-\angle FAE =180 - \angle MXE$$claim:$$\angle FBA=\angle FBD$$proof: $$\angle FBD=\angle DAE=\angle DAF=\angle FAB=\angle FBA$$claimB,F,D,C are concyclic points proof: $$\angle FCD= \angle FAD =\angle FAB =\angle ABF=\angle FBD$$Hence the required result is obtained by radial axis theorem