Problem 2

Since $M$ is the midpoint of both $AO$ and $DJ$, and also of $EF$, the quadrilateral $DFJE$ is a parallelogram. Thus $\angle DEF=\angle EFJ$. The equal chords $AE=AF$ show that $A$ is the midpoint of the arc $EF$ not containing $C$, so $AC$ is the internal angle bisector of $\angle ECF$; consequently the incenter of triangle $CEF$ lies on $AC$, just as $J$ does. It remains to prove that $FJ$ is the internal bisector at $F$. Because $D,E,F$ lie on $\omega$, the inscribed-angle theorem and the equilateral triangle $AFO$ give $\angle DEF=\frac12\angle DOF=\frac12(\angle DOA+60^\circ)$. From the previous step $\angle DOA=\angle OAC$, while $\angle EAO=60^\circ$, so $\angle DEF=\frac12(\angle OAC+\angle EAO)=\frac12\angle EAC$. The cyclic points $E,A,F,C$ give $\angle EAC=\angle EFC$. Therefore $\angle EFJ=\frac12\angle EFC$, so $FJ$ bisects the angle at $F$. The two internal angle bisectors meet at $J$, proving that $J$ is the incenter.