Firstly see that a spiral similarity at $A$ takes $CO_1 \to DO_2.$ Claim: $A,C,E,D$ are concyclic. Proof: Since $\triangle ACO_1 \sim \triangle ADO_2 \implies \measuredangle ACO_1=\measuredangle ADO_2.$ Hence $$\measuredangle ACE+\measuredangle EDA=\measuredangle ACO_1+\measuredangle O_1CE+\measuredangle O_2DA+\measuredangle EDO_2=0$$So $E \in \omega=\odot(ACD).$ $\square$ Let $O_3$ be the center of $\omega$ and $T := \odot(AO_1O_2) \cap \omega \ne A.$ Then note that $\triangle AO_1C \sim \triangle AO_3E \sim \triangle AO_2D$ $\implies T,O_1,O_2,O_3$ are cyclic. Since $A$ is the center of spiral similarity taking $O_1C \to O_3E \to O_2D,$ hence $T=CO_1 \cap EO_3 \cap DO_2.$ In particular this implies $\angle TAF=\pi/2.$ So $FO_3 \perp TE$ and hence $FT=FE.$ But $FT$ is a diameter of $\odot(AO_1O_2)$ and so we are done. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -628.2844296030992, xmax = 832.7851954493207, ymin = -571.0310292852521, ymax = 372.8950177276852; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw(circle((-182.1759714551356,112.76345188182962), 1), linewidth(0.4)); draw(circle((-182.1759714551356,112.76345188182962), 144.0635706732429), linewidth(0.4)); draw(circle((148.2490833023254,107.88832812311297), 241.99720429758509), linewidth(0.4)); draw((-315.18405668710216,57.41761501896911)--(-80.26356088122945,-507.1474381088303), linewidth(0.4)); draw((-80.26356088122945,-507.1474381088303)--(328.0147750787007,-54.12075479512446), linewidth(0.4)); draw(circle((-20.04367671838516,-150.93064216623887), 361.2711449316043), linewidth(0.4) + linetype("2 2") + ccqqqq); draw((-72.75652709282883,206.47416413271645)--(-80.26356088122945,-507.1474381088303), linewidth(0.4)); draw((-315.18405668710216,57.41761501896911)--(328.0147750787007,-54.12075479512446), linewidth(0.4)); draw(circle((-18.11991673090931,31.942743419408018), 182.88350392543904), linewidth(0.4) + blue); draw((40.17620744445938,205.28615377635253)--(-80.26356088122945,-507.1474381088303), linewidth(0.4)); /* dots and labels */ dot((-182.1759714551356,112.76345188182962),dotstyle); label("$O_1$", (-229.30537880381806,108.76254507310773), NE * labelscalefactor); dot((-72.75652709282883,206.47416413271645),dotstyle); label("$A$", (-83.3374333894469,235.26809776556325), NE * labelscalefactor); dot((148.2490833023254,107.88832812311297),dotstyle); label("$O_2$", (165.5031592693382,104.59203234698283), NE * labelscalefactor); dot((-315.18405668710216,57.41761501896911),dotstyle); label("$C$", (-358.59127331368967,42.03434145510923), NE * labelscalefactor); dot((-75.56878663993477,15.865461495536955),linewidth(4pt) + dotstyle); label("$B$", (-51.363502489156076,22.571948733192997), NE * labelscalefactor); dot((-75.56878663993477,15.865461495536955),linewidth(4pt) + dotstyle); dot((328.0147750787007,-54.12075479512446),linewidth(4pt) + dotstyle); label("$D$", (342.0548646752919,-69.17933124155496), NE * labelscalefactor); dot((-80.26356088122945,-507.1474381088303),linewidth(4pt) + dotstyle); label("$E$", (-86.1177752068635,-547.398123837211), NE * labelscalefactor); dot((-72.75652709282883,206.47416413271645),linewidth(4pt) + dotstyle); dot((-76.41604090627781,-141.40066693753658),linewidth(4pt) + dotstyle); label("$F$", (-111.14085156361284,-167.88146575984442), NE * labelscalefactor); dot((40.17620744445938,205.28615377635253),linewidth(4pt) + dotstyle); label("$T$", (45.948461120424696,215.805705043647), NE * labelscalefactor); dot((-20.04367671838516,-150.93064216623887),linewidth(4pt) + dotstyle); label("$O_3$", (-15.219058862740363,-188.73402939046895), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Notice that A,C,D,E is cyclic, and triangle AO1O2 is similar to triangle ACD We use potolomy and the sine therom, So we can cauculate AE and the diameter of circlr AO1O2 easily.

Mehhhhhhh nice problem Lemma 1:

Lemma 2:

Lemma 3:

Lemma 4:

Solution: Because of the parallelism, $\angle PO_2F=\angle PDE=90$. Hence $FP$ is a diameter ! So we just need to prove $FE=FP$. But this is true since $\angle FEP=\angle AEP=\angle ADP=\angle O_2AD=\angle GAE=\angle GAF=\angle GDF$, where $G=\Gamma \cap PE$ and since there's a spiral similarity centered at $A$ that takes $O_1O_2$ to $CD$. Note: The spiral similarity is because $AO_1C\sim AO_2D$ (since their angles are equal because $ADEC$ is cyclic.

Nice problem. Let \(X=\overline{CO_1}\cap\overline{DO_2}\). Since \(\triangle AO_1C\sim\triangle AO_2D\), we know \(A\) is the Miquel point of \(CDO_2O_1\), implying \(AO_1FO_2X\) and \(ACEDX\) are cyclic. Moreover \(\measuredangle XAE=\measuredangle XCE=90^\circ\), so \(\overline{XF}\) is a diameter of \(\omega\). [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair O1,O2,A,B,C,D,X,EE,F; O1=(-3,0); O2=(9,0); A=(0,4); B=reflect(O1,O2)*A; C=O1+abs(O1-A)*dir(200); D=2*foot(O2,B,C)-B; X=extension(C,O1,D,O2); EE=2*circumcenter(X,C,D)-X; F=2*foot(circumcenter(A,O1,O2),A,EE)-A; draw(A--X,linewidth(.8)); draw(circumcircle(A,C,D),gray+dashed); draw(A--EE,gray+linewidth(.4)); draw(C--X--D,linewidth(.4)); draw(C--EE--D,linewidth(.8)); draw(circumcircle(A,O1,O2),dashed); draw(circle(O1,abs(A-O1))); draw(circle(O2,abs(A-O2))); dot("\(A\)",A,dir(110)); dot("\(B\)",B,dir(-10)); dot("\(C\)",C,W); dot("\(D\)",D,dir(-30)); dot("\(E\)",EE,dir(250)); dot("\(F\)",F,dir(210)); dot("\(X\)",X,dir(75)); dot("\(O_1\)",O1,dir(140)); dot("\(O_2\)",O2,dir(200)); [/asy][/asy] From \(\measuredangle XFE=\measuredangle XFA=\measuredangle XO_1A=\measuredangle CO_1A\) and \(\measuredangle FEX=\measuredangle AEX=\measuredangle ACX=\measuredangle ACO_1\), we have \(\triangle XFE\sim\triangle AO_1C\), and since \(O_1A=O_1C\), we have \(FX=FE\). This is sufficient.

Cute. Let $R=\overline{CO_1}\cap\overline{DO_2}$. Claim: $R$ lies on $(AO_1O_2)$ and on $(CDE)$. Proof. Indeed, $\measuredangle O_1RO_2=\measuredangle O_1CB+\measuredangle BDO_2=\measuredangle O_2BO_1=\measuredangle O_1AO_2$, which yields $R\in (AO_1O_2)$. On top of that $R$ lies on $(ACD)$, as $\measuredangle O_1CA=90^\circ+\measuredangle CBA=\measuredangle O_2DA$. On the other hand, $A$ lies on $(CDE)$, as $\measuredangle ECA=\measuredangle ABC=\measuredangle ABD=\measuredangle ADE$, we conclude that $R\in (CDE)$. $\square$ By the claim, we get that $R$ is the antipode of $F$ wrt $(AO_1O_2)$. Finally, $\measuredangle RFA=\measuredangle RO_1A=2\measuredangle RCA=2\measuredangle REF$, and hence $EF=RF$, we are done.

ptolomey sinus lemma

Solved with Jeffrey Chen and Rich Wang Denote $T = CO_1\cap DO_2$. First, observe that \[\angle CED = \angle O_1CB + \angle O_2DB = \angle O_1BC + \angle O_2BD = 180 - \angle O_1BO_2 = 180- \angle O_1AO_2 = \angle O_1FO_2\]We also have \[\angle AO_1C = 90 - \angle ABC = \angle ABD - 90 = \angle AO_2D\]so $\angle CAD = \angle O_1AO_2 = 180 - \angle CED$, which means $(ECDA)$ is cyclic. Next, observe \[\angle CTD = 180 - \angle TCD - \angle TDC = 180 - \angle CED\]\[\angle O_1TO_2 = 180 - \angle CTD = \angle CED = 180 - \angle CAD = 180 - \angle O_1AO_2\]so $T$ lies on $(ACED)$ and $(AO_1O_2F)$. This means \[\angle FO_2T = \angle FAT = \angle ECT = 90\]so $FT$ is a diameter of $\omega$ and it suffices to show that $FT = EF$, or $2\angle FET = \angle AFT$. However, this is true because \[2\angle FET = 2\angle ACT = \angle AO_1T = \angle AFT\]so we are done.