Reflection through bisector of angle $\angle A$

Question

Let ABC be a triangle. Let B' and C' denote respectively the reflection of B and C in the internal angle bisector of $\angle A$, Show that the triangle ABC and AB'C' have the same in centre?

What should be the approach in this question? I don't even know how to initiate.

Answer 1

Hint: Since it is a reflection, the angle bisector at $B$ reflects to the angle bisector at $B',$ and similarly for the bisectors at $C$ and at $C'.$ In either triangle, the center of the incircle is at the intersection of the three bisectors. So each triangle's incenter lies on the bisector at $A$, and maybe the above about the other two bisectors in each triangle will lead to something to prove it.

Answer 2

In-center I is the point of concurrence. Since one bisector is fixed and other two bisectors are swapped by reflection about the first fixed line , position of I does not change.