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Geometry Difficulty 6.3 National olympiad Prove it Saudi Arabia

Let $ABC$ be an acute triangle with $O, H$ are its circumcenter, orthocenter. Take a point $M$ belongs to the minor $\operatorname{arc} BC$ of $(O)$ (not coincide to $B, C$ ) and denote $D, E, F$ as reflection of $M$ through $OA, OB, OC$ . Suppose that $BF$ meets $CE$ at $K$ and let $I$ be the incenter of triangle $DEF$ .

1. Prove that perpendicular bisectors of $EF$ and $IK$ meet on circle $(O)$ .

2. Prove that three points $H, K, I$ are collinear.

Solution

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 $\operatorname{arc} EF$ 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
$L = CD \cap MP, \quad I = DN \cap PE, \quad K = CE \cap MN \text{ are collinear. }$
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. $\square$

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