Team selection tests for JBMO 2018

1) By definition of $E,F$ we can see that $EC$ is the angle bisector of $\angle MEF$ and $FB$ is the angle bisector of $\angle MFE$. This implies that $K$ is the incenter of triangle $MEF$. Hence, $MK$ is the angle bisector of $\angle EMF$ which passes through the midpoint $N$ of the $\mathrm{arc}EF$ of $(O)$. So by applying the well known property, we have $NE=NK=NI=NF$. This implies that $E,F,K,I$ are concyclic and two perpendicular bisectors of $EF,IK$ pass through the point $N\in (O)$.

2) We have $BK=BM=BE$ and $CK=CM=CE$. Hence, $K$ is the reflection of $M$ through $BC$. By similarly way, take $CD\cap AF=L$ then $L$ is the reflection of $M$ through $AC$ and $ML,EI$ intersect at midpoint $P$ of the arc $DF$ of $(O)$. Consider the hexagon $CDPNEM$ and apply Pascal's theorem, we can see that three intersections $$L=CD\cap MP,I=DN\cap PE,K=CE\cap MN\text{ are collinear. }$$ We also know that $KL$ is the Steiner's line of $M$ respect to $(O)$ then $KL$ passes through orthocenter $H$ of triangle $ABC$. This implies that $H,I,K$ are collinear. $\square$