I think it's not possible...you must have AB=BC. Edit: I was interpreting the circle inscribing ABC as the incircle instead of the circumcircle.

[asy][asy] import graph;import geometry;import olympiad;size(7cm); pair A=expi(pi/3),B=(0,0),C=(3,0),D,E,F,G; D=A+C-B; path c=circumcircle(A,B,C);pair O=circumcenter(A,B,C); E=tangent(D,O,abs(O-A),1);F=tangent(D,O,abs(O-A),2); G=C+(expi(pi/3)*(D-C)); draw(A--B--C--D--cycle,heavymagenta);draw(A--E--D--F--C--G--E--F,heavycyan);draw(c,green); dot(A);dot(B);dot(C);dot(E);dot(F);dot(D);dot(G);draw(anglemark(E,D,A),orange);draw(anglemark(C,D,F),orange); label("$A$",A,NNW);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,N);label("$F$",F,ESE);label("$G$",G,S); [/asy][/asy] Let the second inetersection of $DA$ with $\odot (ABC)$ be $G$. The reflection about the perpendicular bisector of $EF$ fixes $\odot (ABC)$ and $D$, but swaps the lines $DA,DC$. So it swaps $G$ and $C$. Therefore $DG=DC$. On the other hand, $DC=AB=GC$, since $ABCG$ is an isosceles trapezium. So $DG=CD=GC\implies \triangle DGC$ is equilateral $\implies \angle CDG=60^{\circ}$, so $\boxed{\angle ABC=60^{\circ}}$. $\blacksquare$

Let $\odot (ABC)$ meet rays $DA,DC$ again at $X,Y.$ It's obvious that $ABCX,AYCX,ABYC$ are isosceles trapezoids, therefore $\angle ABC=\frac{\widehat{AC}}{2}=\frac{\widehat{AC}+\widehat{AB}+\widehat{BY}+\widehat{CY}}{6}=60^\circ.$

You’ve got to be kidding me. [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) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -28.812152524048628, xmax = 36.378721606980605, ymin = -12.308710368581648, ymax = 23.54627040348473; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((9.277124501573722,-3.7329560648768045)--(-9.247398913784334,-3.805997021719843)--(-1.4057274673923552,9.900703524872293)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((0,0), 10), linewidth(0.5) + qqwuqq); draw((9.277124501573722,-3.7329560648768045)--(-9.247398913784334,-3.805997021719843), linewidth(0.5) + ccqqqq); draw((-9.247398913784334,-3.805997021719843)--(-1.4057274673923552,9.900703524872293), linewidth(0.5) + ccqqqq); draw((-1.4057274673923552,9.900703524872293)--(9.277124501573722,-3.7329560648768045), linewidth(0.5) + ccqqqq); draw((9.277124501573722,-3.7329560648768045)--(17.1187959479657,9.973744481715332), linewidth(0.5) + ffvvqq); draw((17.1187959479657,9.973744481715332)--(-1.4057274673923552,9.900703524872293), linewidth(0.5) + ffvvqq); draw((0,0)--(17.1187959479657,9.973744481715332), linewidth(0.5) + ffvvqq); draw((9.277124501573722,-3.7329560648768045)--(0,0), linewidth(0.5) + ffvvqq); draw((0,0)--(-1.4057274673923552,9.900703524872293), linewidth(0.5) + ffvvqq); draw(circle((7.871397034181364,6.1677474599954865), 10), linewidth(0.5) + qqwuqq); /* dots and labels */ dot((9.277124501573722,-3.7329560648768045),dotstyle); label("$A$", (9.487486027931045,-3.243104434735317), NE * labelscalefactor); dot((-9.247398913784334,-3.805997021719843),dotstyle); label("$B$", (-9.051168803080394,-3.2940348051501838), NE * labelscalefactor); dot((-1.4057274673923552,9.900703524872293),dotstyle); label("$C$", (-1.2078917591909386,10.406234836449043), NE * labelscalefactor); dot((0,0),dotstyle); label("$O$", (0.2181586124253258,0.525742975964842), NE * labelscalefactor); dot((17.1187959479657,9.973744481715332),dotstyle); label("$D$", (17.3307630718205,10.508095577278777), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Notice that the condition implies $DE$ and $DF$ are isogonal in $\angle ADC$, however noticing that if $O$ is the circumcentre of $\triangle ABC$, then $\triangle DFO\cong\triangle DEO$, so $DO$ bisects $\angle EDF$. In effect this means that $DO$ also bisects $\angle ADC$. Note however that $AO=OC$ and $AD\neq DC$, so it’s well-known $ADOC$ cyclic. From there $\angle ABC=\angle ADC=180^\circ-\angle AOC=180^\circ-2\angle ABC$, and therefore $\angle ABC=\boxed{60^\circ}$ as desired.