Team Selection Test for EGMO 2024

Let $AK\cap MN=L$. Since $K$ is the Miquel point of the quadrilateral $AFIE$ we get that $ABKE$ and $AFKC$ are conicyclic. Therefore, we have $\angle B{K}_A=\angle BEA=B/2+C$ and since $\angle BNM=B/2+A$ we get that quadrilateral $BNKL$ is cyclic. Similarly the quadrilateral $MCKL$ is also cyclic. By using of the radical axis theorem on the circles $(BNKL)$, $(MCKL)$ and $(ABC)$ we get the desired concurrency.