My solution: Since $\{F,G\}=\Gamma \cap \Omega$, we get $AO$ is the perpendicular bisector of $FG$. Let $AC\cap FG=U$. We have $\angle AGU= \angle ACF =\angle ACG \Longrightarrow \triangle AUG \sim \triangle AGC$ $\Longrightarrow \angle GUC=180^{\circ}-\angle AGC=\angle ABC=\angle KFD$.$(1)$ On the other hand, $\angle DFG=\angle GEC=\angle GLU$$(2)$ From $(1),(2)$, we get $\angle XFG=\angle XGF$. Thus, $X$ lies on the perpendicular bisector of $FG$. So $X\in AO$. Q.E.D

[asy][asy] unitsize(1.cm); pair A=(3.84,4.58); pair B=(0.76,0.6); pair C=(8,0.6); pair D=(2.63,0.6); pair E=(5.05,0.6); pair F=(0.82,1.72); pair G=(7.57,2.74); pair K=(1.96,2.15); pair L=(5.65,2.85); pair O=circumcenter(A,B,C); pair R=(5.14,2.37); pair S=(1.74,1.86); pair X=(4.09,2.94); draw(A--B--C--A);draw(circumcircle(A,B,C)); draw(circumcircle(D,B,F),blue); draw(circumcircle(E,C,G),blue); draw(F--G^^A--O^^K--D--F--X--G^^R--E--L--R--C); draw(F--G--X--F,red); dot("$A$",A,dir(A)); dot("$B$",B,dir(220)); dot("$C$",C,dir(320)); dot("$D$",D,dir(280)); dot("$E$",E,dir(260)); dot("$F$",F,dir(190)); dot("$G$",G,dir(G)); dot("$R$",R,dir(100)); dot("$S$",S,dir(270)); dot("$X$",X,dir(60)); dot("$O$",O,dir(180)); dot("$K$",K,dir(90)); dot("$L$",L,dir(90)); [/asy][/asy] Let $R=FG\cap{}\odot(CGE)$. Then $$\angle{RGL}=\angle{RCL}=\angle{REL}.$$ By the construction of $R$ we have $FD\parallel{RC}\implies{}\angle{FDB}=\angle{RCE}$. Since $BDKF$ is cyclic we have $\angle{FDB}=\angle{FKB}$. Note that $AO$ is the perpendicular bisector of $FG$ which means that $FG$ is antiparallel to $BC$. So if $S=FG\cap{}AB$ then $$\angle{ASG}=\angle{C}=\angle{XFG}+\angle{FKB}.$$ But $\angle{FKB}=\angle{RCE}=\angle{C}-\angle{XGF}\implies{}\angle{XFG}=\angle{XGF}$. This means that $X$ lies on $AO$, as desired. $\square$

This is a Greek proposal by Vaggelis Psychas and Silouanos Brazitikos (me) Prove also that $KD\perp DG .$ That was a part of our initial solution.

What a nice one, the main idea is to prove $XF=XG$, which can be reached by angle chasing. My solution [http://www.artofproblemsolving.com/community/c6t79529f6h1113163_problem_4] Dear silouan, can you show us more details on your approach?

Here's a solution by pure angle chasing, without introducing any further points: Clearly, $AF=AG$ (since $FG$ is the radical axis of two circles with centres $A$ and $O$). It will therefore suffice to show that $\angle AFK=\angle AGL$, or, equivalently, that $\angle FAB-\angle CAG=\angle FKB-\angle GLC=\angle FDB-\angle GEC\,(\dag)$, as $BDKF$ and $CELG$ are cyclic. Let $\angle DAE=2\alpha$, $\angle FAB=2\beta$, $\angle EAG=2\gamma$. Then $\angle FDB=\alpha+\beta$, and $\angle GEC=\alpha+\gamma$, and so $\angle FDB-\angle GEC=\beta-\gamma$. Also notice that, on the one hand, $\angle FAB-\angle CAG=\angle BCF-\angle CBG=\angle C-\angle B$ as the angles $\angle GBA=\angle FCA(=\angle ABF=\angle GCA)=90^\circ-\alpha-\beta-\gamma$ intercept chords of the same length on the circumcircle. On the other hand, $\angle FAB-\angle CAG=2(\beta-\gamma)-(\angle BAD-\angle EAC)$. But, in triangle $BAD$, $\angle BAD=90^\circ-\alpha-\angle B$, similarly, in triangle $CEA$, $\angle EAC=90^\circ-\alpha-\angle C$. Hence $\angle FAB-\angle CAG=2(\beta-\gamma)-(\angle C-\angle B)$. It follows that $\angle C-\angle B=\gamma-\beta$, from which $(\dag)$ follows.

Since $\overline{AO} \perp \overline{FG}$ for obvious reasons, we will only need to show that $XF = XG$, or that $\measuredangle KFG = \measuredangle LGF$. Let line $FG$ meet $(BDF)$ and $(CGE)$ again at $F_2$ and $G_2$. [asy][asy] size(10cm); pair A = dir(105); pair B = dir(200); pair C = dir(340); real r = 1.337; pair F = IP(CR(A, r), unitcircle); pair G = OP(CR(A, r), unitcircle); pair D = IP(CR(A, r), B--C); pair E = OP(CR(A, r), B--C); pair K = IP(circumcircle(B, D, F), A--B); pair L = IP(circumcircle(C, E, G), A--C); draw(unitcircle, orange); draw(arc(A,r,160,380), orange); filldraw(A--B--C--cycle, invisible, red); pair X = extension(F, K, G, L); draw(F--G, red); draw(F--X--G, lightblue); pair F_2 = IP(circumcircle(B, F, D), F--G); pair G_2 = IP(F--G, circumcircle(C, E, G)); filldraw(circumcircle(B, F, D), invisible, lightblue); filldraw(circumcircle(C, E, G), invisible, lightblue); draw(B--F_2--D--F--cycle, lightcyan); draw(C--G_2--E--G--cycle, lightcyan); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$F$", F, dir(F)); dot("$G$", G, dir(G)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$K$", K, dir(100)); dot("$L$", L, dir(L)); dot("$X$", X, dir(X)); dot("$F_2$", F_2, dir(70)); dot("$G_2$", G_2, dir(G_2)); /* TSQ Source: A = dir 105 B = dir 200 C = dir 340 ! real r = 1.337; F = IP CR A r unitcircle G = OP CR A r unitcircle D = IP CR A r B--C E = OP CR A r B--C K = IP circumcircle B D F A--B R100 L = IP circumcircle C E G A--C unitcircle orange !draw(arc(A,r,160,380), orange); A--B--C--cycle 0.1 lightred / red X = extension F K G L F--G red F--X--G lightblue F_2 = IP circumcircle B F D F--G R70 G_2 = IP F--G circumcircle C E G circumcircle B F D 0.1 yellow / lightblue circumcircle C E G 0.1 yellow / lightblue B--F_2--D--F--cycle lightcyan C--G_2--E--G--cycle lightcyan */ [/asy][/asy] Claim: Quadrilaterals $FBDF_2$ and $G_2ECG$ are similar, actually homothetic through $\overline{FG} \cap \overline{BC}$. Proof. This is essentially a repeated application of being ``anti-parallel'' through $\angle(FG, BC)$. Note the four angle relations \begin{align*} \measuredangle(FD, FG) = \measuredangle(BC,GE) = \measuredangle(G_2C,FG) &\implies \overline{FD} \parallel \overline{G_2C} \\ \measuredangle(F_2B, FG) = \measuredangle(BC,FD) = \measuredangle(GE,FG) &\implies \overline{F_2B} \parallel \overline{GE} \\ \measuredangle(FB, FG) = \measuredangle(BC,GC) = \measuredangle(G_2E,FG) &\implies \overline{FB} \parallel \overline{G_2E} \\ \measuredangle(F_2D, FG) = \measuredangle(BC,FB) = \measuredangle(GC,FG). &\implies \overline{F_2D} \parallel \overline{GC}. \end{align*}This gives the desired homotheties. $\blacksquare$ To finish the angle chase, \begin{align*} \measuredangle GFK = \measuredangle F_2BK &= \measuredangle F_2BF - \measuredangle ABF = \measuredangle F_2DF - \measuredangle ABF \\ &= \measuredangle F_2DF - \measuredangle GCA = \measuredangle GCG_2 - \measuredangle GCA \\ &= \measuredangle LCG_2 = \measuredangle LGF \end{align*}as needed. (Here $\measuredangle ABF = \measuredangle GCA$ since $AF = AG$.)

silouan wrote: This is a Greek proposal by Vaggelis Psychas and Silouanos Brazitikos (me) Congratulations This is a very rich configuration, and I proved plenty more things than I actually used in my solution. Just for starters, in my diagram above the quadrilateral $DEG_2F_2$ is also cyclic, and also the circumcircle of $\triangle ADE$ passes through $FG \cap AB$, $FG \cap AC$.

v_Enhance wrote: silouan wrote: This is a Greek proposal by Vaggelis Psychas and Silouanos Brazitikos (me) Congratulations This is a very rich configuration. Indeed! I already proposed one more property above and for the collection, here is another one Let $GL$ and $FK$ intersect $\Omega$ at $S,T.$ Then $AS=AT.$

Just finished an angle chase solution and I must comment that the richness of the configuration truly appears to come from the magical interdependence of all of these circles. Simply stunning.

silouan wrote: This is a Greek proposal by Vaggelis Psychas and Silouanos Brazitikos (me) Prove also that $KD\perp DG .$ That was a part of our initial solution. Really very interesting property.More interesting and difficult than the problem posed in the competition. Congratulations once more Silouan to you and Vaggelis Psychas! We consider the line $AM$ which is perpndicular to $DG$,with $M$ the being the midpoint of $DG$. This line is obviously the perpendicular bisector of $DG$. Suppose that $H\equiv AM\cap FD$.We have $\angle{FDG}=180^{\circ}-\frac{\angle{FAG}}{2}$. Thus $\angle{HDG}=\frac{\angle{FAG}}{2}$.Thus $\angle{MHD}=90^{\circ}-\frac{\angle{FAG}}{2}$. Since $HM$ is the perpendicular bisector of $DG$,we get $\angle{DHG}=2\cdot \left(90^{\circ}-\frac{\angle{FAG}}{2}\right)=180^{\circ}-\angle{FAG}$. Thus $FAGH$ is cyclic.Thus $H$ belongs to $\Omega$. Thus $\angle{BAH}=\angle{BFH}=\angle{BKD}$ which gives $AH\parallel KD\Leftrightarrow KD\perp DG$ which is the desired result. Lets hope it's correct. Edit:Well,this isn't as difficult as I initially thought.I just noticed thet it can be proved using angles only,and not considering additional lines.

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Let $FD$ $\cup$ $\omega$ $=$ $X$ and $\omega$ $\cup$ $GE$ $=$$Y$. Then, $\angle$$FXY $ $=$ $\angle$$FGE$ $=$ $\angle$$FDB$ ($GEDF$ cyclic.) $\implies$ arc $BX$ $=$ arc $CY $. $\implies$$BFD$ $=$ $\angle$$CGE$. Since $AF$ $=$ $AG$ $\implies$ $\angle$$FBA$ $=$ $\angle$$GCA$. From the above resuts we get that $\angle$$KBD$ $-$ $\angle$$GEC$ $=$ $\angle$$LCE$ $-$ $\angle$$FDB$ Since $BFKD$ , $CGLE$ and $FDEG$ are cyclic this is equivelant to, $\angle$$KFG$ $=$ $\angle$$LGF$. Rest is very similar to the THVSHs solution.

Because $GA=FA$ and $GO=FO$, it is enough to show that $GX=FX$. $\angle XGF = $ $=\angle XGE - \angle FGE =$ $=\angle LGE - \angle FGE =$ $\qquad$ ($FDEG$ cyclic) $= \angle LGE - (180 - \angle FDE)=$ $\qquad$ ($LECG$ cyclic) $= \angle LCE - 180 + \angle FDE=$ $= \angle ACD - 180 + \angle FDE=$ $= \angle ACD - 180 + \angle FDA + \angle ADE=$ $\qquad$ (sum of angles in the isosceles $\triangle AFD$) $= \angle ACD - 180 + 90 - \frac{\angle FAD}{2} + \angle ADE=$ $= \angle ACD - \frac{\angle FAD}{2} + \angle ADE -90=$ $\qquad$ (central angle / inscribed angle) $= \angle ACD - \angle FCD + \angle ADE -90=$ $= \angle ACF + \angle ADE -90$ Similarly, we can get $\angle XFG = \angle ABG + \angle AED -90$. $ \angle ACF = \angle ABG$ because they are inscribed angles in the same circle over equal arcs ($AF=AG$) $ \angle ADE = \angle AED$ because the triangle ADE is isosceles ($AD=AE$) So, $\angle XGF = \angle XFG$, i.e $GX=FX$ $\blacksquare$

$OA$ is perpendicular bisector of $FG$. $X$ lies on $AO$ $ \iff $ we have $XF=XG$ $ \iff $ we have $ \angle XFG= \angle XGF$ $ \iff $ we have $ \angle XFA= \angle XGA$ $ \iff $ we have $ \angle KFA= \angle LGA$ $ \iff $ we have $ \angle AFB - \angle KFB = \angle AGC - \angle LGC$ $ \iff $ we have 180° -$ \angle C$-$( \angle KFD + \angle BFD )$ =180°- $ \angle B$-$( \angle LGE + \angle CGE)$.Also $BFKD$ concyclic so $ \angle KFD = \angle B$.And $LGCE$ concyclic so $ \angle LGE = \angle C$ So it reduces to showing $ \angle BFD= \angle CGE$. Now see that $FG$ antiparallel to $DE$ $FG$ antiparallel to $BE$. This implies the equality of angles.Done.

Here's a different solution: Let $KD$ intersect $AC$ at $P$, and $LE$ intersect $AB$ at $Q$. Then $F$ is the Miquel Point of the complete quadrilateral consisting of points $P,A,C,D,B,K$, so $PAKF$ is cyclic, and thus $\angle AFK=\angle APK =\angle BAC - \angle PKA = \angle BAC - \angle BKD = \angle BAC - \angle BFD$. Similarly, $\angle AGL = \angle BAC - \angle EGC$. It suffices to prove that $\angle AFK = \angle AGL \iff \angle BFD = \angle EGC$. Now let $FD, GE$ intersect $\Omega$ again at $F', G'$, respectively. Then $\angle DF'G' = \angle FGG'=\angle F'DE \implies BC || F'G' \implies BF' = CG'$. Thus $\angle BFF'=\angle G'GC$ so $\angle BFD = \angle EGC$ and we are done. I would like to add that for some weird reason, I took over an hour to figure out why $\angle BFD = \angle EGC$....

My solution: Let $FG\cap \odot (\triangle BFD)=T,FG\cap \odot (\triangle CEG)=S$ then $AG=AF\Longrightarrow \angle GCA=\angle ABF\Longrightarrow \angle KBF=\angle GCL=\angle GEL$(1) Note that $\angle TBD=\angle DFT=\angle GEC\Longrightarrow GE\parallel BT$(2) Also $SEC=\angle 180-\angle FGC=\angle FBC\Longrightarrow SE\parallel FB$(3) From (2),(3): $\angle TBF=\angle GES$(4) From (1),(4) we get $\angle KBT=\angle LCS\Longrightarrow \angle SGL=\angle TFK\Longrightarrow XG=XF$ Q.E.D

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does inversion work?

Because I love long solutions : Notice that , since $(AO)$ is the perpendicular bisector of $[FG]$ , it is enough to prove that $MGF$ is isocele . Now , let $MF$ intesect the circumcercle at $Q$ , and let $P$ be the point on the circumcercle such that $A$ is the midpoint of arc $QP$ . We claim that $G,K,$ and $P$ are collinear . That is equivalent to $FGK=FGP$ Notice that $(PQ)$ and $(FG)$ are parallel . And let $R$ be the intersection of $(AB)$ and $(EG)$ Then : $FGP=QPG=LFG=LFR$ And $FGK=EGK-EGF=ACB-(\pi - FDE)= \pi-AFB-FDB=\pi-AFB-FLR$ So $FGK=FGP <=> \pi-AFB-FLR=LFR <=> \pi-AFB=\pi-FRL <=> AFB=FRL=FRA $ Wich is obvious since $FBA=FGA=AFG=AFR$ and $FAR=FAB$ Now since $G$, $K$ and $P$ are collinear we get $MGF=PGF=PQF=AQF-AQP=APG-APQ=QPG=MFG$

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silouan wrote: This is a Greek proposal by Vaggelis Psychas and Silouanos Brazitikos (me) My congratulations dear silouan!

Let $\{P\}=FG\cap\odot (\triangle{BFD})$ and $\{Q\}=FG\cap\odot (\triangle{EGC})$.We will also consider the points $\{M\}=B=CQ\cap AB$ and $\{N\}=BP\cap AC$.We will prove that $MN\parallel FG$.Let $N'\in AC$ such that $MN'\parallel FG$ and let $\{N''\}=MN'\cap BP$. Since $MN''\parallel FG$ it follows that $$\widehat{MN''B}\equiv\widehat{N''PG}\equiv\widehat{FPB}\equiv\widehat{FDB}\equiv\widehat{FGE}\equiv\widehat{QGE}\equiv\widehat{QCE}\equiv\widehat{MCB}$$,so the quadrilateral $BMN''C$ is cyclic$(1)$. The line $FG$ is anti-parallel to line $BC$ and since $MN'\parallel FG$,it follows that the quadrilateral $BMN'C$ is cyclic$(2)$. From $(1)$ and $(2)$ it follows that $N',N''\in\odot (\triangle{BMC})$,so $N\equiv N'\equiv N''$. Thus $MN\parallel FG$,from which it results that the quadrilateral $BMNC$ is cyclic,so $\widehat{MBN}\equiv\widehat{MCN}$.But $\widehat{MBN}\equiv\widehat{KBP}\equiv\widehat{KFP}\equiv\widehat{XFG}$ and $\widehat{MCN}\equiv\widehat{QCL}\equiv\widehat{LGQ}\equiv\widehat{XGF}$.Therefore $\widehat{XFG}\equiv\widehat{XGF}$,which implies $XF=XG$,so $X$ is situated on the perpendicular bisector of $FG$.Hence $X\in AO$,which ends our proof.

Note that $AF=AG$, and $AO$ is the perpendicular bisector of $FG$ so it suffices to show $\angle AFL=\angle AGK.$ We claim that $\angle BFD=\angle EGC.$ Let $FD$ intersect $\Omega$ in $I$ and $GE$ intersect $\Omega$ in $H.$ We have $\angle FIH=\angle FGH=\angle EDI$ so $DE\parallel HI.$ Thus, $\angle BFD=\angle EGC$ Now, \[0=\angle BFD-\angle EGC\]\[=\angle BKD-\angle CLE\]\[=\angle ACB+\angle KBD+\angle BKD-\angle ELC-\angle LCE-\angle ABC\]\[=\angle ACB-\angle KDB+\angle LEC-\angle ABC\]\[=\angle AFB-\angle KFB-\angle AGC+\angle LGC\]\[=\angle AFL-\angle AGK\]so we're done.

There is no need to introduce any points. Notice that $AO$ is the perpendicular bisector of $FG$. Thus, \begin{align*} \measuredangle XFG + \measuredangle XGF &= (\measuredangle KFD + \measuredangle DFG) + (\measuredangle LGE + \measuredangle EGF) \\ &= \measuredangle KBD + \measuredangle DEG + \measuredangle LCE + \measuredangle EGF \\ &= \measuredangle ABC + \measuredangle(DE, FG) + \measuredangle ACB \\ &= \measuredangle (AB, FG) + \measuredangle ACB \\ &= \measuredangle (AB, AA) + \measuredangle ACB \\ &= 0, \end{align*}so $XF=XG$, done.

Let $BG$ and $CF$ meet $\Gamma$ again at $P$ and $Q$ respectively. Reim's yields $PQ \parallel \overline{BDEC}$. It's clear that $AO$ is the perpendicular bisector of $FG$, so it suffices to show $XF = XG$. By the Spiral Center Lemma, we know $FAK \overset{+}{\sim} FCD$ and $GAL \overset{+}{\sim} GBE$. Thus, $$\measuredangle AFX = \measuredangle AFK = \measuredangle CFD = \measuredangle QFD = \measuredangle QPD$$$$= \measuredangle EDP = \measuredangle EGP = \measuredangle EGB = \measuredangle LGA = \measuredangle XGA.$$This implies $$\measuredangle XFG = \measuredangle AFG - \measuredangle AFX = \measuredangle FGA - \measuredangle XGA = \measuredangle FGX$$which finishes. $\blacksquare$

Nice configuration. Set $P = \overline{FG} \cap (BDF)$ and $Q = \overline{FG} \cap (CEG)$. Note that $\{\overline{BP}, \overline{FD}\}, \{\overline{EG}, \overline{CQ}\}$, and $\{\overline{DF}, \overline{EG}\}$ are all antiparallel with respect to $\overline{FG}$ and $\overline{BC}$. Thus by many applications of Reim's Theorem, the two quadrilaterals $FBDP$ and $QECG$ are homothetic. Thus $$\angle XFG = \angle XGF \iff \angle FBA = \angle ACG,$$which is clearly true. As $\overline{AO}$ is the perpendicular bisector of $\overline{FG}$ the result immediately follows.

WLOG, let $ABC$ be an acute triangle. Note that both $A$ and $O$ lie on the perpendicular bisector of $FG$, so we only have to prove that $H$ also lies on $FG$ or that $HF=HG$. Perform an inversion around $\Gamma=(A,AD)=(FDEG)$. Let $B',C'$ be the image of $B,C$, respectively. Note that $B=(AFG)\cap AB$ implies that $B'=FG\cap AB$. Similarly, we have $C'=FG\cap AC$. Since $B,B',C',C$ are concyclic, we can get the following angle chase $$\angle KB'C'=180^{\circ}-\angle BB'C'=\angle BCC'=\angle ECL=\angle EGL$$However, we also have that $$\angle FKB'=\angle FKB=\angle FDB=180^{\circ}-\angle FDE=\angle FGE=\angle C'GE.$$Hence, we have $$\angle KB'C'-\angle FKB'=\angle EGL-\angle C'GE\implies \angle KFB'=\angle LGC'\implies \angle HFG=\angle HGF$$or that $HF=HG$, as desired.

Extend $\overline{DF},\overline{EG}$ to hit $(ABC)$ again at $P,Q$ respectively. Let $Y=\overline{AB} \cap \overline{DF}$ and $Z=\overline{AC} \cap \overline{EG}$. Also let $\overline{FK} \cap \overline{AO}=X_1$ and $\overline{GL} \cap \overline{AO}=X_2$. By Reim's, we have $\overline{DE} \parallel \overline{PQ}$, so $BCPQ$ is an isosceles trapezoid. Because $AF=AG$, we have $\measuredangle FBA=\measuredangle ACG$, hence $\measuredangle FDK=\measuredangle FBK=\measuredangle LCG=\measuredangle LEG$. Also, since $AD=AE$, we have $\measuredangle ADE=\measuredangle EDA$. Combining this with the isosceles trapezoid, \begin{align*} \measuredangle BFP+\measuredangle ABF&=\measuredangle QGC+\measuredangle GCA &\implies\\ \measuredangle FYB&=\measuredangle CZG&\implies\\ \measuredangle DYB&=\measuredangle CZE&\implies\\ \measuredangle KBD+\measuredangle BDF&=\measuredangle ECL+\measuredangle GEC&\implies\\ \measuredangle KBD+\measuredangle EDA+\measuredangle ADF&=\measuredangle ECL+\measuredangle GEA+\measuredangle AED&\implies\\ \measuredangle KFD+\measuredangle DFA&=\measuredangle EGL+\measuredangle AGE&\implies\\ \measuredangle KFA&=\measuredangle AGL&\implies\\ \measuredangle X_1FA&=\measuredangle AGX_2. \end{align*}Furthermore, since $AF=AG$ and $\measuredangle FAX_1=\measuredangle FAO=\measuredangle OAG=\measuredangle X_2AG$ by symmetry, it follows that $\triangle AFX_1 \stackrel{-}{\cong} \triangle AGX_2$, hence $X_1=X_2$, so $\overline{FK}$ and $\overline{GL}$ concur on $\overline{AO}$ as desired. $\blacksquare$ Remark: what mindlessly angle chasing does to a mf Edit: After writing up this solution and looking back at the diagram I have realized that the angle chase can be made much better if we use our minds. We still want to prove that $\measuredangle KFA=\measuredangle AGL$, which after $\measuredangle BFP=\measuredangle QGC$ is equivalent to $\measuredangle BFA-\measuredangle DFK=\measuredangle AGC-\measuredangle LGE$. But we have $$\measuredangle BFA-\measuredangle DFK=\measuredangle BCA-\measuredangle DBK=\measuredangle BCA+\measuredangle ABC,$$and it's not hard to see that $\measuredangle AGC-\measuredangle LGE$ is also equal to this, so we're done.

WHY IS IT ANGLE CHASING. I SPENT LIKE AN HOUR. We'd like to show that the following are equal: \[\measuredangle XFG=\measuredangle KFD-\measuredangle GFD=\measuredangle ABC-\measuredangle GFD\]\[\measuredangle FGX=\measuredangle EGL-\measuredangle EGF=\measuredangle BCA-\measuredangle EGF\] Let $H\in (ABC)$ satisfy $AH\perp BC$. Ignoring configuration issues (yay! also this is not allowed, so this solution is a 6/7) \[\measuredangle GFD-\measuredangle EGF=\frac{1}{2}\measuredangle GAD-\frac{1}{2}\measuredangle EAF=\frac{1}{2}\measuredangle GAH-\frac{1}{2}\measuredangle HAF\]which is now \begin{align*} \frac{1}{2}\measuredangle ABH-\frac{1}{2}\measuredangle HCA &= \frac{1}{2}\left(\measuredangle ABC+90^{\circ}-\measuredangle BCA\right))-\frac{1}{2}\left(\measuredangle BCA+90^{\circ}-\measuredangle ABC\right) \\ &= \measuredangle ABC-\measuredangle BCA \end{align*}which is what we aimed to prove. $\blacksquare$ gosh i hate my life. what even is this question (though you can get stuck pretty easily if you don't realize how easy it is. the point is to focus on cyclic FDEG)

Let $\angle FDB = \alpha, \angle GEC = \beta$. Let $x = \angle AFG$. Then we have $\angle AFG = \angle AGF = \angle ABF = \angle FDK = x$ and $\angle AFG = \angle ACG = \angle LEG = x$. Note that $\angle BFG + \angle GCB = 180^{\circ}$ and $\angle DFG + \angle DEG = 180^{\circ}$, so we get $\angle DFB = EGC$, which is equivalent to $\angle BKD = \angle CLE$. Now consider the following claim: Claim: $\beta - \alpha = \angle ABC - \angle ACB$. Proof.$\beta - \alpha = \beta + \angle LEG - \alpha - \angle FDK = \angle LEC - \angle KDB = \angle ALE - \angle ACB - (\angle AKD - \angle ABC) = \angle ABC - \angle ACB$, so we're done. Claim: $FX = GX$. Proof. Note that $\angle XFG = \angle KFB - \angle GFB = 180^{\circ} - x - \alpha - (180^{\circ} - \angle ACB - x) = \angle ACB - \alpha$. Similarly $\angle XGF = \angle ABC - \beta$. Thus $\angle XGF = \angle XFG$ is equivalent to $\angle ABC - \angle ACB = \beta - \alpha$, which is above claim. Note that $AF = AG$ and $FO = GO$, thus $AO$ is the perpendicular bisector of $FG$ and since $FX = GX$, thus $X$ lies on $AO$, as required. $\blacksquare$

Sketch

The key observation is to see that $AO$ is the perpendicular bisector of $FG$. So, to get the desired result, we must show that $XF=XG$, or equivalently, $\angle XFA=\angle XGA$. Let $\angle XFA$ be $x$ and $\angle XFD$ be $y$. Then, $\angle LGE= \angle ACD=180-\angle AFD=180-x-y$ and $\angle KBD=\angle KFD=x$. However, $\angle AGE=180- \angle ABD=180-x$, and so $\angle XGA=x$, as desired.

Extend $FK, GL$ to meet $\Omega$ again at points $P,Q,$ respectively. It suffices to show that $AP = AQ,$ since then $X$ would be on the perpendicular bisector of $FG$ along with $A$ and $O.$ By the angle between chords formula, $$\angle FKB = \angle FCB + \angle ABP$$and $$\angle GLC = \angle GBC + \angle ACQ.$$Subtracting the two, we see that it suffices to show that $$\angle FKB - \angle GLC = \angle FCB - \angle GBC,$$or equivalently $$\angle FDB - \angle FCB = \angle GEC - \angle GBC.$$However, by sum of angles in $\triangle FEC,$ we see that $\angle FEB - \angle FCB = \angle EFC$ and similarly $\angle GDC - \angle GBD = \angle BGD.$ Thus we would like to show that $\angle BGD = \angle EFC.$ However, $\angle BFC = \angle BGC,$ so if we can prove that $\angle BFD = \angle EGC,$ then we are done. Extend $BF$ and $CG$ to hit $\Gamma$ again at points $Y$ and $Z,$ respectively; our desired angle condition translates into proving $YZ \parallel BC,$ which follows since $$\measuredangle YZC = \measuredangle YZG = \measuredangle YFG = \measuredangle BFG = \measuredangle BCG = \measuredangle BCZ.$$Thus we are done.

We can simply show that \( GX = FX \): \[ \angle XGF = \angle XGE - \angle FGE = \angle LGE - \angle FGE \quad (\text{since } FDEG \text{ is cyclic}) \]\[ = \angle LGE - (180^\circ - \angle FDE) \quad (\text{as } LECG \text{ is cyclic}) \]\[ = \angle LCE - 180^\circ + \angle FDE = \angle ACD - 180^\circ + \angle FDE \]\[ = \angle ACD - 180^\circ + \angle FDA + \angle ADE \quad (\text{in } \triangle AFD) \]\[ = \angle ACD - 180^\circ + 90^\circ - \frac{\angle FAD}{2} + \angle ADE \]\[ = \angle ACD - \frac{\angle FAD}{2} + \angle ADE - 90^\circ \]\[ = \angle ACD - \angle FCD + \angle ADE - 90^\circ \]\[ = \angle ACF + \angle ADE - 90^\circ \] Similarly, \[ \angle XFG = \angle ABG + \angle AED - 90^\circ \] Since \[ \angle ACF = \angle ABG \quad (\text{as } AF = AG) \]\[ \angle ADE = \angle AED \quad (\text{since } AD = AE) \] Thus, \[ \angle XGF = \angle XFG \implies GX = FX \quad \blacksquare \]

We first note that since $AF=AG$ and $OF=OG$, line $\overline{AO}$ must be the perpendicular bisector of segment $FG$. Now, the entirety of the solution is the following key claim. Claim : Triangle $\triangle XFG$ is isoceles. Proof : We simply note that, \begin{align*} \measuredangle KFG &= \measuredangle KFB + \measuredangle BFG\\ &= \measuredangle FKB + \measuredangle KBF + \measuredangle BFG\\ &= \measuredangle FDB + \measuredangle ABF + \measuredangle BFG\\ &= \measuredangle FGE + \measuredangle AGF + \measuredangle BCG \\ &= \measuredangle FGE + \measuredangle GFA + \measuredangle ECG \\ &= \measuredangle FGE + \measuredangle GCA + \measuredangle ECA + \measuredangle ACG\\ &= \measuredangle FGE + \measuredangle ECA \\ &= \measuredangle FGE + \measuredangle EGL\\ &= \measuredangle FGL \end{align*}from which it is quite clear that the claim follows. But, now if $\triangle XFG$ is isosceles, $X$ must lie on the perpendicular bisector of segment $FG$, which is indeed $\overline{AO}$, as we mentioned before. Thus, point $X$ lies on $AO$ as desired.

This took about 2 hours, but here it is: Let $FD \cap \Omega=M$ and $GE \cap \Omega=N$ Claim 1: $\angle FEL=\angle GDK=90$ Proof:Indeed, note that $\theta=\angle LEG=\angle LCG=\angle ACG=\angle ABF=\angle KBF=\angle KDF$ and $\angle FAG=180-2 \theta$ give us $\angle FDG=\angle FEG=90+\theta$, proving the claim. Claim 2: $\angle AGX=\angle AFX$ Proof: Observe that $$\angle AGX=\angle AGC-\angle LGC$$$$=180-\angle B-\angle LED=180-\angle B- 90-\angle FED$$$$=90-\frac{\angle FAD}{2}-\angle B=\angle AFD-\angle AFC=\angle CFM=\angle CBM$$Similarly, $\angle AFX=\angle BCN$ Also note that $\angle DMN=\angle FMN=\angle FGN=\angle FGE=\angle EDM$, therefore $ BC \parallel MN$ $\implies \angle CBM=\angle BCN$, proving the claim Now, to finish the problem, observe that $\angle XFG=\angle AFG-\angle AFX=\angle AGF-\angle AGX=\angle XGF$ $\implies XF=XG$, so $AX \perp FG$ But $AO \perp FG$, hence $A$, $X$ and $O$ are collinear.