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Geometry Difficulty 8.4 Shortlist Prove it Turkey

Let $ABC$ be a triangle, $\omega$ be its circumcircle and $I$ be its incentre. Let the line $BI$ meet $AC$ at $E$ and $\omega$ at $M$ for the second time. The line $CI$ meet $AB$ at $F$ and $\omega$ at $N$ for the second time. Let the circumcircles of $BFI$ and $CEI$ meet at $K$ for the second time. Prove that the lines $BN$ , $CM$ , $AK$ are concurrent.

Solution

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 $\angle BK_A = \angle BEA = B/2 + C$ 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.

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