Easy geo from China TST as always Here's my solution: Let $H$ and $I$ denote the center of $(ABC)$ and $(AEF)$, respectively. It's obvious that $FB=FD$ and $ED=EC$. Note that we've $BF=FD=DE,BH=HA,\angle{HBF}=\angle{HAB}=\angle{HAE}\implies \triangle{HBF}\equiv \triangle{HAE}$. So, $\angle{HFB}=\angle{HEA}\implies H\in (AEF)$. Since $\angle{HKE}=\angle{HAE}=\angle{HBA}$, we get $H,K,B$ collinear. Consider $(H)$ and $(P)$, we get $HP\perp LG\implies HP\parallel KG$. And consider $(I)$ and $(Q)$, we get $IQ \perp KG\implies IQ\parallel CG$. From spiral similarity, we've $\triangle{GFB}\sim \triangle{GEC}\implies \frac{GF}{GE}=\frac{BF}{EC}=\frac{AE}{AF}\implies \triangle{GFE}\equiv \triangle{AEF}$. In other words, $AGFE$ is an isosceles trapezoid. This means $FG=AE=FD$, and similarly, $EG=AF=DE$. Hence, $EF$ is the perpendicular bisector of $FD$. This gives $P,Q\in EF$. Let $\angle{HBA}=\angle{HAB}=\angle{HAC}=x$. We easily get $\angle{ACB}=\angle{ABC}=90^{\circ} -x$. Since $FP\parallel AG$ and $HP\parallel KG$, we get $\angle{HPE}=\angle{KGA}=180^{\circ}-\angle{AHB}=2x$. While $\angle{HEP}=\angle{HAE}=x$. This gives $\angle{PFH}=\angle{PHF}\implies PF=PH$. Combine with $IF=IH$, we get that $IP$ is the perpendicular bisector of $FH$. This gives $\angle{PIF}=\frac{\angle{HIF}}{2}=\angle{HAF}$. We get $$\angle{IPF}=180^{\circ}-\angle{PIF}-\angle{IFP}=180^{\circ}- \angle{HAF} -(90^{\circ}-\angle{FAE})=180^{\circ} -x-(90^{\circ}-2x)=90^{\circ}+x.$$Since $IQ\parallel GC$ and $QE\parallel GA$, we get $$\angle{IQE}=180^{\circ}-\angle{CGA}=180^{\circ}-\angle{CBA}=180^{\circ}-(90^{\circ}-x)=90^{\circ}+x.$$Hence, $\angle{IPF}=\angle{IQE}$, combine with $IF=IE$ and $\angle{IFB}=\angle{IEQ}$ gives $\triangle{IPF}\equiv \triangle{IQE}$. In other words, $PF=QE$. This easily gives us $AGQP$ is also, as $AGFE$, an isosceles trapezoid, and hence it must be concyclic, done.

Notice AE=BF so OE=OF hence GE=FA=ED, GF=AE=FD so P, Q on EF. FKD=FAE=FDE so let FK=FD=FB and BO cuts (w2) at K' then FK'=FB so K=K'. Because OE//GM so EM=OG=OC, OM=EG=EC so CM cuts H-midpoint OE. From HQ//CG, HO=HE hence QO=QE. Thus OPE=KGA=KOA=A=2OEG=OQF so OQ=OF so AEPQ isoceles trapezoid hence cyclic

Beautiful problem . Although the proof is elementary but it is still a rich configuration. Here goes my solution. Proof: We break the problem mainly into two claims. Claim 1: $\overline{PF}=\overline{PO}$. Proof: Define $O$ as the circumcentre of $\triangle{ABC}$ . Now define $K=\overline{BO} \cap \overline{DE}$. Notice that since $$\angle OKE=\angle BKD=180^\circ-(\angle KBD+\angle KDB)=180^\circ-(90^\circ-\angle A+\angle EDF+\angle EDB) =\angle OAE .$$so we have that $(AOEK)$ cyclic or infact $(AFEKG)$ is cyclic since we have that $\overline{OA}=\overline{OG}$ and $\overline{OE}=\overline{OF}$. Now we claim that $\overline{AL} \parallel \overline{BO}$ . For this define $Z$ as the $\overline{GK} \cap \omega_1$ and then notice that $Z$ is the $L$-antipode w.r.t $\omega_1$ so $\angle LAB=90^\circ-\angle A=\angle OBA$. Now we state a subclaim. Subclaim: If $I=\overline{AL} \cap \overline{OP}$ then $I \in \odot(AFEKG)$. Proof: We'll use phantom points. Redefine $I=\overline{OP} \cap \odot(AFEKG)$. Then notice that by the converse of reims theorem we just need to show that $\overline{IK} \parallel \overline{LZ}$. Notice that $\overline{OP} \perp \overline{LG}$ since $\overline{LG}$ is the radical axis of $\odot(DLG)$ and $\odot(ABC)$. By this we have that $(IGKO)$ is a cyclic trapezoid . So we have that $$\angle IKG=\angle IOG=\angle OGK=\angle OGZ=\angle OZG=\angle LZG$$and so we are done $\square$. Now back to the claim. Notice that with this subclaim we are done since $$\angle IOF=\angle IAF=\angle LAF=\angle EAO=\angle EFO$$$\square$. Now we state the second claim. Claim 2: $\overline{OQ}=\overline{OE}$. Proof: For this part clearly notice that $E$ is the circumcentre of $\triangle{GDC}$. Now notice that $\overline{GC}$ is the radical axis of $\odot(ABC)$ and $\odot(DGC)$ $\implies$ $\overline{OE} \perp \overline{CG}$ $\implies$ $\overline{OE} \parallel \overline{MG}$. But now notice that clearly if $O_1$ is the centre of $\odot(AEF)$ then $\overline{O_1Q} \perp \overline{MG} \parallel \overline {OE}$ which implies the result. $\blacksquare$.

[asy][asy] import olympiad; size(230); // Co-ordinates from Geogebra // pair A = (-0.08 , 4.83) , B = (-2.08 , -1.17) , C = (4.92 , -1.17); draw(A--B--C--cycle , green); pair D = (2.12 , -1.17) , E = (2.92 , 1.23) , F = (-0.9 , 2.38); draw(D--E); draw(D--F); draw(circumcircle(A,B,C) , magenta); draw(circumcircle(A,E,F) , orange); pair G = (2.7 , 4.91); pair H = (3.66 , 3.46); pair L = (-2.59 , 1.96); pair K = (3.61 , 2.31); draw(L--G); pair M = (-0.94 , 3.41); draw(C--G); pair O = circumcenter(A,B,C); draw(circumcircle(D,G,L) , purple); pair P = circumcenter(D,G,L); draw(circumcircle(D,G,M) , cyan); pair Q = circumcenter(D,G,M); // Dots and Labels // dot("$A$" , A , N); dot("$B$" , B , W); dot("$C$" , C , SE); dot("$D$" , D , S); dot("$E$" , E , NE); dot("$F$" , F , NE); dot("$G$" , G , NE); dot("$H$" , H , NE); dot("$L$" , L , W); dot("$K$" , K , NE); dot("$M$" , M , W); dot("$O$" , O , NE); dot("$P$" , P , NE); dot("$Q$" , Q , SE); [/asy][/asy] Claim : $P , Q \in EF$ Proof : We have $AE = BF$ and therefore $OE = OF$. As a result $GE=FA=ED$ and also $GF=AE=FD$. $\blacksquare$ Also $\angle FKD = \angle FAE = \angle FDE$ , therefore $FK = FB = FD$ and $BO$ intersects $\omega_2$ at $K_1$. Then $FK_1 = FB$ which implies $K = K_1$. Since we have $OE \parallel GM$ we have $EM = OG = OC$ and also $OM = EG = EC$. Thus $CM$ bisects $OE$. Since $HQ \parallel CG , HO = HE$ and hence $QO = QE$. Thus we have \begin{align*} & \angle OPE = \angle KGA = \angle KOA=\angle A =2 \angle OEG = \angle OQF \\ \implies &OQ = OF \\ \implies & AEPQ \; \text{is an isosceles trapezoid} \\ \end{align*}