Claim. $EFZY$ is a cyclic quadrilateral. Proof. $$\angle ZFE + \angle ZYE = \angle BFD + \angle C + 90^{\circ}-\angle AYZ= 90^{\circ}-\angle FDB+ \angle C + 90^{\circ}-\angle ADZ=180^{\circ}\square$$ Now by the radical axis theorem on $\odot(ABC)$, $\odot(BFEC)$, and $\odot(EFZY)$ we have that $YZ$, $EF$, and, $BC$ concur at a point.$\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -22.772584895993297, xmax = 16.656389049878932, ymin = -4.260015997360821, ymax = 12.887480484307561; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((-8,10)--(-12,2)--(1,2)--cycle, linewidth(0.8) + zzttqq); /* draw figures */ draw(circle((-5.5,3.75), 6.731456008918131), linewidth(0.8) + ffxfqq); draw((-8,10)--(-12,2), linewidth(0.8) + zzttqq); draw((-12,2)--(1,2), linewidth(0.8) + zzttqq); draw((1,2)--(-8,10), linewidth(0.8) + zzttqq); draw((-3,-2.5)--(-6.826143756350537,10.349533524234307), linewidth(0.8) + qqwuqq); draw((-3,-2.5)--(-10.262495372084413,8.507219548315437), linewidth(0.8) + qqwuqq); draw((-12,2)--(-3,-2.5), linewidth(0.8)); draw((-3,-2.5)--(1,2), linewidth(0.8)); draw((-6.826143756350537,10.349533524234307)--(-10.262495372084413,8.507219548315437), linewidth(0.8) + qqwuqq); draw((-9.4,7.2)--(-6.2620689655172415,8.455172413793104), linewidth(0.8) + qqwuqq); draw(circle((-8.256933115823818,8.892332789559545), 2.0422027745350433), linewidth(0.8) + red); /* dots and labels */ dot((1,2),linewidth(4pt) + dotstyle); label("$C$", (1.1260965777092078,1.4643727981661987), NE * labelscalefactor); dot((-12,2),linewidth(4pt) + dotstyle); label("$B$", (-12.53029283583508,1.515712607991553), NE * labelscalefactor); dot((-8,10),linewidth(4pt) + dotstyle); label("$A$", (-8.62846728910814,10.089460848825745), NE * labelscalefactor); dot((-3,-2.5),linewidth(4pt) + dotstyle); label("$D$", (-2.6473794444543453,-2.950850846814283), NE * labelscalefactor); dot((-6.2620689655172415,8.455172413793104),linewidth(4pt) + dotstyle); label("$E$", (-6.164156417491126,8.65194617371582), NE * labelscalefactor); dot((-9.4,7.2),linewidth(4pt) + dotstyle); label("$F$", (-10.065981964218066,7.009072259304478), NE * labelscalefactor); dot((-6.826143756350537,10.349533524234307),linewidth(4pt) + dotstyle); label("$Y$", (-6.7288943255700255,10.551519137253933), NE * labelscalefactor); dot((-10.262495372084413,8.507219548315437),linewidth(4pt) + dotstyle); label("$Z$", (-10.810409206685705,8.549266554065111), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Claim: $YZFE$ is cyclic. Proof. Since $AH, AA'$ are isogonal and $A'C \parallel BE$, we know $$\angle YZF = \angle YZA' = \angle YZC + \angle CZA' = \angle YA'C + \angle CAA'$$$$= \angle EA'C + \angle DAB = \angle A'EH + \angle HAF = \angle A'EH + \angle HEF = \angle A'EF$$implying the result. $\square$ Now, the Radical Axis Theorem on $(ABC)$, $(BCEF)$, and $(YZFE)$ finishes. $\blacksquare$ GGB Here: https://www.geogebra.org/geometry/audfx87z. Remark: The only claim in this problem can be reverse engineered from the truth of the problem statement.

By radical axis, it is sufficient to show that $YZEF$ is cyclic. Let $R=AD\cap EF$. Note that $AO\perp EF$ due to isogonality. Hence, $ARZE$ and $ARFY$ are cyclic. Thus, by radical axis we conclude that $YZEF$ is cyclic. [asy][asy] import olympiad; size(6cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,D,E,F,Y,Z,R; O=(0,0);A=dir(110);B=dir(200);C=dir(340);path w=circumcircle(A,B,C);D=-A; E=foot(B,A,C);F=foot(C,A,B);Y=intersectionpoints(w,D--100F-99D)[0];Z=intersectionpoints(w,D--100E-99D)[0]; R=extension(A,O,E,F); draw(A--B--C--cycle,deep);draw(w,deep); draw(E--F,med);draw(A--D,med);draw(Y--D--Z,deep);draw(circumcircle(Y,Z,E),deep); draw(foot(A,B,C)--A,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$O$",O,dir(90)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$R$",R,dir(R)); [/asy][/asy]

By angle chasing, we have $AO\perp EF$. But it's trivial that $DD\perp AO$. Hence $\measuredangle YZF=\measuredangle(YD,DD)=\measuredangle(YD,EF)\quad\blacksquare$

$EFYZ$ cyclic. $$DE^2+AE*EC=DF^2+AF*FB$$