12th Czech-Polish-Slovak Mathematics Competition

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 $$MI=MA=MB\text{and}NJ=NA=ND=NK(1)$$ (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 $$\angle BFM=\angle EFN=\angle KFN.(2)$$

But clearly we have $$\angle BMF=\angle BNF=\angle KNF.(3)$$ 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 $$\angle IFM=\angle JFN\text{and}\angle IFJ=\angle MFN.$$ Hence $$\angle IFJ=\angle MFN=\angle MCN=\angle ICJ,$$ and the points $C$, $F$, $I$, $J$ are indeed concyclic.