Question

Let ABC be a triangle. Let B’ denote the reflection of B in the internal angle bisector l of ∠A. Show that the circumcentre of the triangle CB’ I lies on the line l , where I is the incentre of the triangle ABC.

Answer 1

In given ΔABC, Let I be the excentre and I₁ be the except corresponding to vertex A.

As we know that CI ⊥ CI₁ (external and internal angle bisectors are perpendicular)

Let us assume that circumcircle of quadrilateral I B I₁ C intersect side AC at point B” and let BB” intersect II ₁ at point M

[ Angle made by arc in same segment are equal and IC is the internal angle bisector of C]

=> B” is the mirror image B’ of B in line II₁ 

Hence B” is the given reflection B’ of point B in II₁ 

=> I, B’ and C are concyclic with II₁ as diameter of their circum circle

i.e centre of circum circle of ΔIB'C lies on II₁ i.e bisector l of ∠A