Post Number:#4 by ahmedittihad » Fri Mar 31, 2017 10:07 pm
As both of you didn't show the proof of $A$,$I$,$O$ being colinear, I'm just showing that. Let $AI$ intersect the circumcircle of $\triangle ABC$ at $X$. We know that $X$ is the midpoint of arc $BC$. So, $BX=CX$.
Now we will show that $BX=XI$.
We have, $\angle XBI = \angle XBC + \angle CBI = \angle (A/2) + \angle (B/2) = \angle BIX$. We get that $BX=XI=XC$. So, $X$ is the circumcenter of $\triangle BIC$.
Frankly, my dear, I don't give a damn.