[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -25.63510486570654, xmax = 21.938403102039047, ymin = -11.541814982676225, ymax = 16.471853472198525; /* image dimensions */ /* draw figures */ draw((-11,-3)--(1.6,-3), linewidth(0.7)); draw((1.6,-3)--(-8.205564172850684,8.384802625282969), linewidth(0.7)); draw((-8.205564172850684,8.384802625282969)--(-11,-3), linewidth(0.7)); draw(circle((-9.228436136853036,2.6005169148470095), 5.874029982447796), linewidth(0.7) + blue); draw(circle((-1.8375280186455198,3.9544032343294724), 7.757597774093809), linewidth(0.7) + blue); draw((-3.3564297387454602,2.7546892057405303)--(-11,-3), linewidth(0.7)); draw((-3.3564297387454602,2.7546892057405303)--(-6.221622026829818,-2.4456001212968164), linewidth(0.7)); draw((-9.533176844337557,2.9759798207451595)--(1.6,-3), linewidth(0.7)); draw(circle((-4.7,-5.9108730159163905), 6.939969864112536), linewidth(0.7) + red); draw((-3.593883913776077,0.9403815347908624)--(-11,-3), linewidth(0.7) + linetype("4 4")); draw((-9.533176844337557,2.9759798207451595)--(-6.221622026829818,-2.4456001212968164), linewidth(0.7)); draw((-8.463528506828252,-0.0800038788939097)--(1.6,-3), linewidth(0.7) + linetype("4 4")); draw(circle((-8.886007596464973,-0.3508780216067162), 3.3892198421775697), linewidth(0.7) + linetype("4 4") + green); draw(circle((-2.207672768100939,-1.2676980623519316), 4.18320952285643), linewidth(0.7) + linetype("4 4") + green); /* dots and labels */ dot((-11,-3),dotstyle); label("$A$", (-10.840904217200295,-2.590701985260625), NE * labelscalefactor); dot((1.6,-3),dotstyle); label("$B$", (1.756958519903066,-2.590701985260625), NE * labelscalefactor); dot((-8.205564172850684,8.384802625282969),dotstyle); label("$C$", (-8.35448394014042,9.509876696430835), NE * labelscalefactor); dot((-3.3564297387454602,2.7546892057405303),dotstyle); label("$D$", (-3.174441696265684,3.169504989928118), NE * labelscalefactor); dot((-9.533176844337557,2.9759798207451595),dotstyle); label("$E$", (-9.34905205096437,3.3767066796831084), NE * labelscalefactor); dot((-6.221622026829818,-2.4456001212968164),linewidth(4pt) + dotstyle); label("$F$", (-6.075265352835537,-2.093417929848647), NE * labelscalefactor); dot((-5.755693768621451,0.9483319220982207),linewidth(4pt) + dotstyle); label("$I$", (-5.5779812974235625,1.263249444182203), NE * labelscalefactor); dot((-3.593883913776077,0.9403815347908624),linewidth(4pt) + dotstyle); label("$N$", (-3.4230837239716716,1.263249444182203), NE * labelscalefactor); dot((-4.6721317603415615,0.36670595975665005),linewidth(4pt) + dotstyle); label("$H$", (-4.500532510697617,0.6830847128682289), NE * labelscalefactor); dot((-8.463528506828252,-0.0800038788939097),linewidth(4pt) + dotstyle); label("$M$", (-8.313043602189424,0.2686813333582474), NE * labelscalefactor); dot((-7.49487729580251,-0.36106412302008367),linewidth(4pt) + dotstyle); label("$G$", (-7.318475491365474,-0.02140103229873962), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $I$ be intersections of $AD$ and $BE$ , $\omega_1$ as a circumcircle of $\triangle ACD$ , $\omega_2$ as a circumcircle of $\triangle BCE$ , $M$ and $N$ be intersections of the circumcircle of $\triangle ABI$ with respect to $\omega_2$ and $\omega_1$ respectively. Claim 1: $A,E,I,F$ are concyclic ( $B,D,I,F$ as well) Proof. $$\angle FEI = \angle FEB = \angle FCB = \angle FCD = \angle FAD = \angle FAI$$done. $\square$ Claim 2: $H$ lies on $AN$ (Similarly, $G$ lies on $BM$) Proof. It suffices to show that $H$ lies on radical axis of $\omega_1$ and $(ABI)$ which is clearly true since, $HI \times HB = HD \times HF$ (by Claim 1) $\square$ Finally, applied radical axis theorem to $\omega_1$ , $\omega_2$ and $(ABI)$ implies our results. $\blacksquare$

From the cyclic quadrilaterals $AFDC$ and $BFEC$ we get $\angle AFD = 180 - \angle ACD=180-\gamma = 180-\angle BCE=\angle BFE$, so $(FD, FE)$ are isogonal w.r.t. $\angle AFB$. Also, $(FA, FB)$ are isogonal w.r.t. $\angle AFB$, so by the Isogonal Lines Lemma, if $DA\cap EB=X$ and $DB\cap EA=C$, we get that $(FX, FC)$ are isogonal w.r.t. $\angle AFB$. $...(1)$ Since $FD\equiv FH$ and $FE\equiv FG$, $(FH, FG)$ and $(FA, FB)$ are pairs of isogonal lines w.r.t. $\angle AFB$. So, by using the Isogonal Lines Lemma again, if $HA\cap GB=Y$ and $HB\cap GA=X$, we get that $(FY, FX)$ are also isogonal w.r.t. $\angle AFB$. $...(2)$ From $(1)$ and $(2)$ we get that $FC\equiv FY$, i.e. $Y\in FC \,\blacksquare$

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The circles $ACDF$ and $BCEF$ are totally not needed, the point $F$ could be arbitrary. Applying Pappus' theorem to the lines $ADG$ and $BEH$ gives that $AE \cap BD = C$, $DH \cap EG = F$ and $AH \cap BG$ are collinear, done.