Let $M$ and $N$ be the midpoints of the arcs $AB$ and $AD$ (not containing any other vertices of $ABCD$), respectively. Then the incenter $I$ lies on $CM$, the incenter $J$ lies on $CN$, and the incenter $K$ lies on $BN$. Moreover, we have
$$MI = MA = MB \quad \text{and} \quad NJ = NA = ND = NK \quad (1)$$
(these well-known relations follow from an easy angle chasing).

Observe also that the incenter $K$ necessarily lies in the interior of the isosceles triangle $BDE$, hence the line $EK$ intersects the segment $BD$ and the point $F$ lies on the same side of $BD$ as $C$ (Fig. 2).

Fig. 2

Consider now the arc $BD$ of $\omega$ containing $A$. The point $E$ is the midpoint of this arc, and the points $M$ and $N$ are the midpoints of the subarcs $BA$ and $AD$, respectively. It follows that the subarcs $BM$ and $EN$ are of equal length (and similarly the subarcs $ME$ and $ND$ are of equal length). Thus
$$\angle BFM = \angle EFN = \angle KFN. \quad (2)$$

But clearly we have
$$\angle BMF = \angle BNF = \angle KNF. \qquad (3)$$
The equalities (2) and (3) imply that the triangles $MBF$ and $NKF$ are similar, and with the same orientation. Therefore
$$\frac{MB}{MF} = \frac{NK}{NF},$$
which by an application of (1) can be rewritten as
$$\frac{MI}{MF} = \frac{NJ}{NF}.$$
This, together with the relation
$$\angle IMF = \angle CMF = \angle CNF = \angle JNF,$$
proves that the triangles $MIF$ and $NJF$ are similar with the same orientation. This orientation-preserving similarity implies that
$$\angle IFM = \angle JFN \quad \text{and} \quad \angle IFJ = \angle MFN.$$
Hence
$$\angle IFJ = \angle MFN = \angle MCN = \angle ICJ,$$
and the points $C$, $F$, $I$, $J$ are indeed concyclic.