Join BB' and CC'. Let the internal angle bisector l of ∠A meet BB' in E and CC' in F. Since B' is the re ection of B in l, we observe that BB'⊥ l and BE = EB'. Hence B' lies on AC. Similarly, C' 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 intesection point D of its diagonal lies on the perpendicular bisector of its parallel sides. Thus ` passes through D. We also observe that CD = C'D.
Let I be the incentre of ΔABC. This means that CI bisects ∠C. Hence AI/ID = AC/CD. But AC' = AC' and CD = C'D. Hence we also get that AI/ID = AC'/C'D. This implies that C'I bisects ∠AC'B'. Therefore the two angle bisectors of ΔAC'B' meet at I. This shows that I is also the incentre of ΔAC'B'.
