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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..

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    $\begingroup$ -@Stranger Forever, you are a reputed user of the site and I hope that you know policy for asking questions. Please include your work or whatever you know regarding the context in the question so that we may suggest any method. :-) $\endgroup$
    – SarGe
    Jul 9 '20 at 14:28
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    $\begingroup$ I have always done that previously.. but I was really stuck in this question for long.. I am a bit weak in Euclidean geometry and I couldn't figure out much, and therefore there's no work.. but okay, I respect the terms of the site and if no answers come in some few hours, I will delete my question, don't worry.. @SarGe $\endgroup$ Jul 9 '20 at 14:30
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    $\begingroup$ @SarGe and if you see a bit, you will find out that I haven't asked for any answers.. I know I didn't show any work and therefore didn't have the right to ask for answers.. therefore, I have just asked for a hint, which can push me to start with the problem.. :) $\endgroup$ Jul 9 '20 at 14:34
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$\textbf{Hint:}$ 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)

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    $\begingroup$ The result that $B$, $C$, and $I$ lies on a circle centered at $M$ is a part of the Trillium Theorem. $\endgroup$ Jul 9 '20 at 16:00
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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 $\triangle IBM\cong\triangle IB''M$.

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  • $\begingroup$ Thank you so much for taking out some time for this.. I will surely work on it now.. If you finish, you may consider to post it here.. $\endgroup$ Jul 9 '20 at 14:52

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