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

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..

• -@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. :-) Jul 9 '20 at 14:28
• 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 Jul 9 '20 at 14:30
• @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.. :) Jul 9 '20 at 14:34

$$\textbf{Hint:}$$ 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)

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

• 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.. Jul 9 '20 at 14:52