Join BB' and CC’. Let the internal angle bisector l of ∠A meet BB’ in E and CC’ in F. Since B’ is the reflection of B in we observe that BB’ ⊥ l and BE = EB’. Hence B’ lies on AC. similarly , CC’ lies on the line AB.

Let D be the point of intersection of BC and B’C’. Observe that BB’ C’C . moreover the triangles ABC is congruent to AB’C’; this follows from the observation that AB = AB’ and AC = AC’ and the included angle ∠A is common. Hence BC’ = B’C so that C’CB’B is an isosceles trapezium. This means that the intersection D of its diagonal lies on the perpendicular bisector of its parallel side. Thus passes through D. We also observe that CD=C’D

Let I be the in centre of triangle ABC. This means that C I bisects ∠C. Hence A / D = AC/CD. But AC = AC’ and CD = C’D. Hence we also get that A/D = AC’/C’D this implies that C’ bisects ∠AC'B' therefore the two angles bisectors of ∠AC'B' meet at I. This shows that I is also the incentre of triangle AC’B’.