Let $ABC$ be a triangle and let $I,I_a, I_b,I_c$ denote the incenter, the A-excenter, the B-excenter and the C-excenter of ABC respectively. Let $M$ and $M'$ be the midpoints of $II_a$ and $I_bI_c.$
- Prove that $M$ and $M'$ are antipodal on the circumcircle of ABC and $MM'$ is the perpendicular bisector of BC.
- Prove that the circumradius of the excentral circle is $2R.$
For reference, below is a picture to help visualize the problem.
I know that $ABC$ is the orthic triangle of $I_a I_bI_c$ and its circumcircle is the nine-point circle $C$ of $I_a I_bI_c$. Also $C$ passes through $M$ and $M'$ as well as the midpoint of $I_bI_c$, of $I_a I_c$, of $I_c I,$ and of $I_bI.$ $IBI_aC$ is also cyclic with $M$ at its center. I need to show that $M,O,M'$ are collinear and $MM'$ is perpendicular to $BC$ and passes through its midpoint, say $M_1.$ This follows from the fact that $OB = OC = OM,$ where $O$ is the circumcenter of $ABC$ and the fact that $BM=MC$ since $I B I_aC$ is inscribed in a circle centred at M. But why are $M$ and $M'$ antipodal? It suffices to show that $M'$ is on the perpendicular bisector of $BC,$ since then $M,O,M'$ would all be collinear.
For the proof of 2), I found a proof here, but I don't fully understand solution 1. In solution 1, how would one prove the nine-point circle's center get sent by the homothety to the circumcenter of $I_aI_a I_c$?
As I mentioned in the comments, we need to show that $M'$ is the midpoint of major arc $BC$. Its sufficient to prove that in a general triangle $XYZ$, the midpoint of $YZ$ is the midpoint of the major arc formed by the altitude feet of $Y$ and $Z$ on the nine point circle. If the feet are $E,F$ and the midpoint is $G$ its sufficient to show $GE=GF$ (because $G$ is on the circle as mentioned in the comments). This is true since $G$ is the center of cyclic $YZEF$. The second problem is then trivial because $IOM$ and $II'I_a$ are homothetic by one of the other problems you posted.