Show that the circumcentre of the $∆CB'I$ lies on the line $AI$ where $I$ is the incentre of $∆ABC$.

Question

QUESTION: In $∆ABC$ let $B'$ denote the reflection of $B$ in the internal angle bisector $l$ of $\angle A$. Show that the circumcentre of the $∆CB'I$ lies on the line $l$ where $I$ is the incentre of $∆ABC$.

MY APPROACH: I could not proceed much with the question.. Any hints? Thank you..

Answer 1

I'm still working on the solution, however, you can start with this :

Let $I'$ be the $A$-excenter (Note that $I'$ lies on internal angle bisector of $A$). Prove that $\square IBI'C$ is cyclic. Let us assume that circumcircle of quadrilateral $I B I' C$ intersect side $AC$ at point $B''$ and let $BB''$ intersect $II'$ at point $M$.

Finally prove that $B''$ is image of $B$ ($B'$) in internal angle bisector of $A$ by proving $\triangle IBM\cong\triangle IB''M$.

Answer 2

$\textbf{Hint:}$ The circumcenter of $\triangle CB'I$ is actually the midpoint of the arc $BC$ not containing $A$(say $M$) in the circumcircle of $\triangle ABC$

First prove that,$B,C,I$ three of them lies on a circle whose center is $M$ (which is simple angle chasing)