, 2019

Leaving the trivial case $AB = AC$ aside, we show that $I$, $M$, $O$ all lie on the Euler line of the triangle formed by the three excenters $I_A$, $I_B$, $I_C$. Recall that the Euler line of a triangle is the line through the orthocenter, the center of the nine-point circle and the circumcenter of that triangle. Since $I$ is the orthocenter of the triangle $I_A I_B I_C$, and $O$ is the center of its nine-point circle, the line $IO$ is indeed the Euler line of this triangle.

We now show that, if the midpoint of the segment $KL$ lies on the circle $ABC$, then $M$ is the circumcenter of the triangle $I_A I_B I_C$. The conclusion then follows by the preceding.

To prove that $M$ is the circumcenter of the triangle $I_A I_B I_C$, we show that it lies on the perpendicular bisectrix of the segment $I_A I_B$; similarly, it lies on the perpendicular bisectrix of the segment $I_A I_C$, so it is indeed the circumcenter of the triangle $I_A I_B I_C$.

Let $P$ be the midpoint of the segment $KL$. Since $AK = AL$, the point $P$ lies on the bisectrix $AI$ of the angle $BAC$, so it is the midpoint of the circular arc $BPC$, and therefore lies on the perpendicular bisectrix of the segment $BC$; and since $B$ and $C$ both lie on the circle on diameter $II_A$ (the angles $IBI_A$ and $ICI_A$ are both right), it follows that $P$ is the midpoint of the segment $II_A$.

Clearly, the line $I_A CI_B$ is the perpendicular bisectrix of the segment $LM$, so it crosses the latter at its midpoint $Q$. Since $PQ$ is a midline in the triangle $KLM$, it is parallel to $KM$, and since $KM$ and $BII_B$ are both perpendicular to $I_A BI_C$, it follows that $PQ$ and $BII_B$ are parallel. Recall that $P$ is the midpoint of the segment $II_A$, to infer that $PQ$ is a midline in the triangle $II_A I_B$, so $Q$ is the midpoint of the segment $I_A I_B$. Consequently, $M$ lies on the perpendicular bisectrix of the segment $I_A I_B$, as desired. This ends the proof.