This one is quite simple. By Radical Axis, $AG,CH,EF$ are concurrent so $AG\cap CH$ lies on the polar of $AC\cap BD$. Thus $AC\cap BD$ lies on the polar of $AG\cap CH$. It is also known that $AC\cap GH$ lies on the polar of $AG\cap CH$. If $AC\cap GH = AC\cap BD$, then we're done. Otherwise, the polar of $AG\cap CH$ is $AC$, and so $A=G, C=H$ and $AC$ and $GH$ become the same line. In which case we're done as well.

By Desargues' theorem we just have to prove $CH\cap GA,BH\cap DG,F$ are collinear. By Radical axis theorem $CH\cap GA$ lies on $EF$ and by Pascal's theorem $BH\cap DG$ lies on $EF$ as well and we done.

Nice and simple problem.By the radical axes concurrence theorem we have $AG,CH,EF$ are concurrent at $X$(let).It is well known that the polar of $J$ is $EF$ where $J=AC \cap BD$.From this it follows that the polar of $X$ passes through $J$.Let $AC \cap GH=J'$.Then the polar of $X$ passes through $J'$.Now if $AC$ is not the polar of $X$ then $J \equiv J'$ and we are done.If $AC$ is the polar of $X$ then certainly $XA,XC$ are tangents to $\omega$.Thus $G \equiv A,C \equiv H$ and thus $AC$ and $GH$ are same.We are done in both the cases.

We use the following well known lemma: Lemma (Well known): If for four concyclic points $A,G,C,H$ we have for some $R \in AC$, $Pole(AG), R, Pole(CH)$ colinear, then $R \in GH$. Returning to the main problem, by applying radical axis theorem to $\omega, \omega_1, \omega_2$, we see the lines $AG, CH, EF$ are concurrent at (say) the point $Q$. Let $R = AC \cap BD$. By Brocard, We have $R = Pole(EF)$. Now, applying La Hire's theorem and the lemma: $$ \text{AG, CH, EF concurrent} \Rightarrow \text{ Pole(AG), Pole(CH), Pole(EH) Colinear} \Rightarrow \text{ Pole(AG), Pole(CH), R colinear}$$$$\Rightarrow R \in GH$$, hence QED. EDIT: Well there are some minor mistakes and the solution need clarification. I will correct it later.

Let $K$ be the radical center of $\omega, \omega_1, \omega_2$, $P=AC \cap GH$ and $P'=AC \cap BD$. By Brocard's theorem on $ACGH$, we know that $P$ lies on polar of $K$. By Brocard's theorem on $ABCD$, we also know that $P'$ lies on polar of $K$. (As $E,F$ and $K$ are collinear) But as polar of $K$ is a line, it intersects $AC$ at only one point; therefore, $P=P'$. $\square$

By radical center we have $EF,AG,CH$ are concurrent consider the projective transformation that preserves the circle $(ABCD)$ and send $AC\cap BD$ to the center of the circle then $ABCD$ becomes a rectangle and line $EF$ becomes the line at infinity so $AG\parallel{CH}$ So $AGCH$ is rectangle and so $GH$ passes through $AC\cap BD$

Here is my solution for this problem Solution Since: $AG$, $CH$, $EF$ are radical axises of (($\omega$); ($\omega_1$)), (($\omega_2$); ($\omega$)), (($\omega_1$); ($\omega_2$)), we have: $AG$, $CH$, $EF$ concurrent Let $M$ $\equiv$ $CG$ $\cap$ $EF$ We have: $\widehat{MCE}$ = $\widehat{GCD}$ = $\widehat{DAG}$ = $\widehat{MEG}$ So: $\triangle$ $MCE$ $\sim$ $\triangle$ $MEG$ or $ME^2$ = $MC$ . $MG$ Similarly: $MF^2$ = $MC$ . $MG$ Then: $ME^2$ = $MF^2$ = $MC$ . $MG$ or $M$ is midpoint of $EF$ Similarly, if we let $M'$ $\equiv$ $AH$ $\cap$ $EF$ then $M'$ is also midpoint of $EF$ or $M'$ $\equiv$ $M$ Let $O$ be center of ($\omega$), $N$ $\equiv$ $AC$ $\cap$ $BD$, $N'$ $\equiv$ $AC$ $\cap$ $GH$ Applying Brocard theorem for cyclic quadrilateral $ABCD$, we have: $O$ is orthocenter of $\triangle$ $NEF$ or $ON$ $\perp$ $EF$ Similarly: $ON'$ $\perp$ $EF$ Then: $N'$ $\equiv$ $N$ or $AC$, $BD$, $GH$ concurrent at point $N$

Nice problem [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); 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 = -98.95634688454567, xmax = 46.91295638650965, ymin = -41.65104880544451, ymax = 54.67952354953124; /* image dimensions */ pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); /* draw figures */ draw(circle((10.4,-3.06), 8), linewidth(1.2) + wvvxds); draw((5.297234304678813,3.1013133550123295)--(2.7848174957253233,-5.511325238843035), linewidth(1.2) + rvwvcq); 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draw(circle((-11.861745464607303,0.4638453701541232), 25.54594539978864), linewidth(1.2) + sexdts); draw(circle((-16.356710382141106,11.406039711699496), 25.545945399788643), linewidth(1.2) + sexdts); draw((8.28082556935351,4.654214135768732)--(10.946741233431107,-11.041295259772454), linewidth(1.2) + rvwvcq); draw((5.297234304678813,3.1013133550123295)--(17.73124135962027,-6.262015010425956), linewidth(1.2) + rvwvcq); draw((2.7848174957253233,-5.511325238843035)--(13.38499201175277,4.362251861111435), linewidth(1.2) + rvwvcq); draw((xmin, 0.27471034551924733*xmin + 2.379385682427012)--(xmax, 0.27471034551924733*xmax + 2.379385682427012), linewidth(1.2) + rvwvcq); /* line */ draw((xmin, 0.06850282011076657*xmin-11.791177905285291)--(xmax, 0.06850282011076657*xmax-11.791177905285291), linewidth(1.2) + rvwvcq); /* line */ draw((xmin, 0.4107919104002106*xmin + 11.730899233841692)--(xmax, 0.4107919104002106*xmax + 11.730899233841692), linewidth(1.2) + dbwrru); /* line */ draw((8.878539073523909,15.378131261417495)--(8.28082556935351,4.654214135768732), linewidth(1.2) + dtsfsf); draw((xmin, 6.317458096532102*xmin-80.1968742955536)--(xmax, 6.317458096532102*xmax-80.1968742955536), linewidth(1.2) + rvwvcq); /* line */ draw((xmin, 1.8496223510641594*xmin-10.662185922571112)--(xmax, 1.8496223510641594*xmax-10.662185922571112), linewidth(1.2) + rvwvcq); /* line */ draw((xmin, -0.0572007746910497*xmin + 5.127883773417206)--(xmax, -0.0572007746910497*xmax + 5.127883773417206), linewidth(1.2) + dtsfsf); /* line */ draw((xmin, -0.6775326747274687*xmin-3.624520392336405)--(xmax, -0.6775326747274687*xmax-3.624520392336405), linewidth(1.2) + dbwrru); /* line */ draw((xmin, -2.503335033152745*xmin + 16.362065568733325)--(xmax, -2.503335033152745*xmax + 16.362065568733325), linewidth(1.2) + rvwvcq); /* line */ draw((xmin, -1.1551056999457747*xmin + 14.219442951185687)--(xmax, -1.1551056999457747*xmax + 14.219442951185687), linewidth(1.2) + dbwrru); /* line */ /* dots and labels */ dot((10.4,-3.06),dotstyle); label("$O$", (10.824511875943365,-2.15267253010046), NE * labelscalefactor); dot((5.297234304678813,3.1013133550123295),dotstyle); label("$B$", (5.70961422877649,4.004148711859641), NE * labelscalefactor); dot((2.7848174957253233,-5.511325238843035),dotstyle); label("$C$", (3.1521654051930534,-4.520680700085115), NE * labelscalefactor); dot((17.73124135962027,-6.262015010425956),dotstyle); label("$D$", (18.11797703949613,-5.278443314480204), NE * labelscalefactor); dot((13.38499201175277,4.362251861111435),dotstyle); label("$A$", (13.76084200672435,5.330233287051048), NE * labelscalefactor); dot((-37.096994920272316,-3.5082461795638835),linewidth(4pt) + dotstyle); label("$E$", (-36.72509217734869,-2.720994490896777), NE * labelscalefactor); dot((8.878539073523909,15.378131261417495),linewidth(4pt) + dotstyle); label("$F$", (9.214266320353794,16.12835054218107), NE * labelscalefactor); dot((10.946741233431107,-11.041295259772454),linewidth(4pt) + dotstyle); label("$G$", (11.298113509940299,-10.298620634847671), NE * labelscalefactor); dot((8.28082556935351,4.654214135768732),linewidth(4pt) + dotstyle); label("$H$", (8.645944359557474,5.424953613850434), NE * labelscalefactor); dot((9.020862616514224,0.29726519575851285),linewidth(4pt) + dotstyle); label("$I$", (9.403706973952566,1.06781858107867), NE * labelscalefactor); dot((-68.71991485102953,-16.498685870352563),linewidth(4pt) + dotstyle); label("$J$", (-68.3616813283438,-15.697679262412683), NE * labelscalefactor); dot((15.563394075871452,18.124215618580248),linewidth(4pt) + dotstyle); label("$M$", (15.939409523110239,18.87524001936327), NE * labelscalefactor); dot((-14.109227923374217,5.934942540926801),linewidth(4pt) + dotstyle); label("$N$", (-13.708052765097754,6.656317862242454), NE * labelscalefactor); dot((1.5892122836780889,12.383734783885288),linewidth(4pt) + dotstyle); label("$P$", (1.920801156801028,13.097300084600715), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution:We add some more points. Let $AG\cap HC=M$,$AH \cap GC=N$ and $AB \cap CD=I$. We have by radical axis theorem that $M\in EF$ . Now by pascals on $CCHAAG$ we get that $CC \cap AA$, $M$ and $N$ are collinear. Now by pascals on $ABCCDA$ we get that $E-F-CC\cap AA$ are collinear $\implies N \in EF$. Now by brocards theorem we know that $MN$ is the polar of $HG \cap AC$ and since $MN \equiv EF$ we get that $I\in HG$ and we are done. $\blacksquare$.

Easy but a nice problem. ELMO 2014 Shortlist G2 wrote: $ABCD$ is a cyclic quadrilateral inscribed in the circle $\omega$. Let $AB \cap CD = E$, $AD \cap BC = F$. Let $\omega_1, \omega_2$ be the circumcircles of $AEF, CEF$, respectively. Let $\omega \cap \omega_1 = G$, $\omega \cap \omega_2 = H$. Show that $AC, BD, GH$ are concurrent. Proposed by Yang Liu Solution:- Radical axis on $\odot(AEF),\odot(ABCD),\odot(EFC)$ we get that $EF,AG,HC$ are concurrent at a point $T$. Also by Pascal's Theorem on $DGABHC$ we get that $DG\cap BH$, $AG\cap HC$ and $AB\cap CD$ are concurrent. If $DG\cap BH=R$, then $E-R-F-T$. Let $AC\cap BD=K$ and $GH\cap BD=K'$. So, it suffices to prove that $K\equiv K'$. Now notice that $K\in$ Polar of $E,F$. So, by La Hire's Theorem we get that $EF$ is the Polar of $K$. Now as $R\in$ Polar of $K$, then again by La Hire's Theorem we get that $K\in$ Polar of $R$, but $K'\in$ Polar of $R$. Hence, $BD$ is the polar of $R$ which is not possible. Hence, $K'\equiv K$. Hence, $AC,BD,GH$ concur.$\blacksquare$.

Let me put it concise.

1) $Q \in EF$ (Radical center of $\omega, \omega_1, \omega_2$) 2) $P \in EQ$ (Pascal on $DGABHC$)

)

Let $K = AG \cap CH$ and $J = DG \cap BH$. By Desargue's Theorem it suffices to show that $K,J,F$ are collinear. Note that by Radical Axes, $E,F,K$ are collinear. Moreover, by Pascal on $ABHCDG$ $E,J, K$ are collinear, so we're done.

By radical axis $EF,AG,CH$ concur. We take the dual wrt $\omega$ to get $K=AA\cap GG, J=CC\cap HH, I=AC\cap BD$ collinear. It's well-known that $AC,GH,KJ$ concur, and thus $G,H,I$ are collinear, as desired.

By radical axis theorem $EF,AG,CH$ are concuurent .Let $AG\cap CH=P$ $\Longrightarrow E,F,P $ are collinear. By Pascal's theorem on hexagon $ BAGDCH$, $E,P,GD\cap BH$ are collinear.Let $GD\cap BH=R$ then $R\in EF$ Similarly we can show $BG\cap DH=K \in EF$. Let $BD\cap HG=Q$ then by Brocards theorem,$KR$ is the polar of $Q$ wrt $\omega$ But,$K,R\in EF \Longrightarrow EF$ is the polar of $Q$ wrt $\omega$ $\Longrightarrow $$Q$ is the pole of $EF$ wrt $\omega$. But by Brocards theorem on quadrilateral $ABCD$,$AC\cap BD$ is the pole of $EF$ wrt $\omega$ $\Longrightarrow Q=AC\cap BD$ (A.W.D)

define $I$ the intersection of tangents to $\omega$ at $G$ and $H$ and let $X=AC\cap BD$ Its well-known that $AG,CH$,and $EF$ concur on the radical centre of the three circles applying Pascal's Theorem on $CCHAAG$ gives that $CG\cap AH \in EF$. Pascal's theorem again on $HHCGGA$ yields $I,CH\cap AG,AH\cap CG$ are collinear so $I\in EF$ By applying La hire's Theorem we get that $X$ lies on the polar of $I$ which is the line $GH$ and we are done

By Radical Axis thm, we know $AG, CH, EF$ concur, say at point $L$. Let $BG \cap DH =X$. By pascals thm on $BGADHC$, we get that $X-L-F$ Thus $AB \cap CD=E , BG \cap DH=X, AG \cap CH=L$, are collinear. By desargues on $ABG$ and $CDH$ we get that $AC,BD,GH$ are concurrent

v_Enhance wrote: $ABCD$ is a cyclic quadrilateral inscribed in the circle $\omega$. Let $AB \cap CD = E$, $AD \cap BC = F$. Let $\omega_1, \omega_2$ be the circumcircles of $AEF, CEF$, respectively. Let $\omega \cap \omega_1 = G$, $\omega \cap \omega_2 = H$. Show that $AC, BD, GH$ are concurrent. Proposed by Yang Liu We will use polars w.r.t. $\omega$ We know by radical axis that $AG \cap HC \cap EF=P$ exists, now $P \in \mathcal{P}_{AC \cap GH}$ and $P \in \mathcal{P}_{AC \cap BD}$ hence $AC \cap BD \equiv AC \cap GH$. This means $AC, BD, GH$ concurrent thus we are done

Let $T = AC \cap BD$. It suffices to show the polar of $AC \cap GH$ is $EF$ (which is also the polar of $T$ by Brocard's). Applying the Radical Axis Theorem on $\omega$, $\omega_1$, and $\omega_2$ yields $AG, CH, EF$ are concurrent at some point $P$, so $E, F, P$ are collinear. Claim: $AA \cap CC$ lies on $EF$. Proof. Notice $T \in AC$, which is the polar of $AA \cap CC$. As a result, $AA \cap CC$ lies on the polar of $T$ (aka $EF$) by La Hire's. $\square$ Applying similar logic to $AHCG$, we conclude $AA \cap CC$, $P = AG \cap CH$, and $AH \cap GC$ all lie on the polar of $AC \cap GH$. Now, combining all our previously shown results yields $AA \cap CC$, $E$, $F$, $P$, and $AH \cap GC$ are all collinear. Thus, $P = AG \cap CH$ and $AH \cap GC$, both of which lie on the polar of $AC \cap GH$, each lie on $EF$. The result follows easily. $\blacksquare$ Remarks: It's easy to see we also want the polar of $AC \cap GH$ to be $EF$. In order to prove this, we have to relate key quadrilaterals $ABCD$ and $AHCG$, which is possible by analyzing their common diagonal, i.e. $AC$. Fortunately, a quick application of Radical Axes and recalling a well-known collinearity (the main claim of this problem) suffices.

By radical axis on $(AGBCHD),(ECHF),(AGEF)$, $AG,CH,EF$ are concurrent at $P$. By Brocard on $ABCD$, the polar of $AC\cap BD$ is $EF$ hence by La Hire, $AC\cap BD$ lies on the polar of $P$. Also, by Brocard on $AGCH$, $AC\cap GH$ lies on the polar of $P$. Therefore, $AC\cap BD$ and $AC\cap GH$ coincide, meaning $AC,BD,GH$ are concurrent. [asy][asy] size(9cm);defaultpen(fontsize(10pt));pen sec=mediummagenta;pen cho=pink;pen ges=deepmagenta; pair O,A,B,D,C,E,F,G,H,P; O=(0,0);A=dir(105);B=dir(235);D=dir(5);C=dir(275);E=extension(A,B,C,D);F=extension(A,D,B,C);path abc=circumcircle(A,B,C);path cef=circumcircle(C,E,F);path aef=circumcircle(A,E,F); H=intersectionpoints(aef,abc)[1];G=intersectionpoints(cef,abc)[0];P=extension(E,F,A,H); draw(A--F--B);draw(D--E--A);draw(abc);draw(cef);draw(aef);draw(A--C);draw(H--G);draw(B--D);draw(P--F);draw(P--H);draw(P--C); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$D$",D,dir(D)); dot("$C$",C,dir(C)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$G$",H,dir(H)); dot("$H$",G,dir(G)); dot("$P$",P,dir(P)); [/asy][/asy]

Let $K = AC \cap BD$. By Radical Axis theorem on $(FEHC) , (ABCD) , (EAGF)$ we get $AG , EF , HC$ are concurrent at a point, say $X$. Now we chase polars with respect to $(ABCD)$. Since $EF$ is the polar of $K$ and $X \in EF$, by La Hire we have $K$ lies on the polar of $X$. Let $K' = AC \cap GH$. We have that $K'$ lies on the polar of $X$. But the polar of $X$ intersects line $AC$ at only 1 point, this implies $K' = K$, or $AC , BD , GH$ are concurrent.

[asy][asy]import geometry; size(10cm); pen c1=white+white+brown; pen c2=white+white+white+red+yellow; pen c3=white+orange+red; pen c4=white+pink; pen c5= purple+white; pair A,B,C,D,E,F,G,H,X,Y; A=dir(73); B=dir(210); C=dir(330); D=dir(44); E=extension(B,A,C,D); F=extension(A,D,B,C); G=intersectionpoints(circle(A,B,C),circle(A,E,F))[0]; H=intersectionpoints(circle(C,E,F),circle(A,B,C))[1]; X=extension(B,H,D,G); Y=extension(H,C,A,G); draw(circle(A,B,C),c1); draw(circle(A,E,F),c2); draw(circle(C,E,F),c4); draw(A--B--C--D--cycle,c3); draw(E--A); draw(E--D); draw(F--C); draw(F--D); draw(X--G); draw(X--B); draw(Y--C); draw(Y--G); draw(Y--F,dotted); draw(A--C,c5); draw(B--D,c5); draw(G--H,c5); dot(A^^B^^C^^D^^E^^F^^G^^H^^X^^Y); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,2S); label("$D$",D,NE); label("$E$",E,2N); label("$F$",F,SE); label("$G$",G,2S); label("$H$",H,NE); label("$X$",X,NE); label("$Y$",Y,NW);[/asy][/asy] Let $X=\overline{DG}\cap\overline{BH}$ and $Y=\overline{AG}\cap\overline{CH}.$ By Desargues on $\triangle ADG$ and $\triangle CBH,$ it suffices to prove that $F,X,$ and $Y$ are collinear. By Pascal on $BHCDGA,$ $E,X,$ and $Y$ are collinear and by Radical Axis on $\omega,\omega_1,\omega_2,$ $E,F,$ and $Y$ are collinear. $\square$

This problem is a some generalization

Let $R = AC \cap BD$, and similarly let $R' = AC \cap GH$. By the radical axes theorem we observe that $AG, CH, EF$ all concur at the radical center $X$. Now considering Brokard's theorem on $ABCD$, this provides us with the incentive to prove that the polar of $R'$ is $EF$, effectively showing that $R = R'$. (The pole of a polar is always one point). In $AGCH$, $R'$ lies on the polar of $X$. In $ABCD$, the polar of $R$ is $EF$, and $X$ lies on $EF$. Therefore in $ABCD$, $R$ lies on the polar of $X$. As the polar of $X$ is a line and can only meet $AC$ at one point, we must have $R = R'$, as desired. $\blacksquare$

Let $X=AC \cap BD$. By radax on $\omega, \omega_1, \omega_2$, $P=AG \cap CH$ lies on $EF$, the polar of $X$. By Brocard on $ACGH$, $P$ lies on the polar of $AC \cap GH$. It follows that $X=AC \cap GH$, so we are done.

By the Radical Axes Theorem we have that $AG,BD,CH$ concur at $I$. We now consider the generalization where $I$ is an arbitrary point on $EF$, $AI$ cuts $\omega$ again at $G$ and $CI$ cuts $\omega$ again at $H$. We will prove that $AC,BD,GH$ are concurrent. Consider a projective transformation taking $\omega$ to itself and $AC\cap BD$ to the centre of $\omega$. Then $ABCD$ is a rectangle and hence $EF$ is the line at infinity. Therefore $AG\parallel CH$, so $AGCH$ is a rectangle because $AC$ is a diametre of $\omega$. Hence $AC,BD,GH$ concur at the centre of $\omega$. Done.