XXVII Olimpiada Matemática Rioplatense

Let $M$ be the midpoint of the segment $BC$.



Let $\alpha =\angle AEF$. Since $AE=AF$, we have that $\angle AFE=\angle AEF=\alpha$. Considering the cyclic quadrilaterals $APEF$ and $APDH$, we have that $$\alpha =\angle AEF=\angle APF=\angle APH=\angle ADH.$$ Now, since $\angle AFI=\angle ADK$, then the quadrilateral $IFKD$ is cyclic. Hence, $\angle FKD+\angle FID=180^{\circ }$. Also, $\angle EAI=\angle FAI$, because $D$ is the midpoint of the arc $BC$; then, $\angle AIF=90^{\circ }$. Therefore, $\angle FKD=90^{\circ }$ and, then, $DK$ is perpendicular to $AC$.

On the other hand, the quadrilaterals $APEF$ and $APGD$ are cyclic and $$\angle ADG=180^{\circ }-\angle APG=180^{\circ }-\angle APE=\angle AFE=\alpha .$$ Since $\angle AEI=\angle ADJ=\alpha$, we have that the quadrilateral $DJEI$ is cyclic and $\angle AJD=180^{\circ }-\angle EID=90^{\circ }$. Then, $DJ$ is perpendicular to $AB$.

Finally, since $D$ is the midpoint of the arc $BC$ and $M$ is the midpoint of the corresponding chord, $DM$ is perpendicular to $BC$.

Using the Simson line of the point $D$, we conclude that $J$, $M$ and $K$ are collinear.