Let \(ABC\) be a triangle. Let \(B'\) denote the reflection of \(b\) in the internal angle bisector \(l\) of \(\angle A\).Show that the circumcentre of the triangle \(CB'I\) lies on the line \(l\) where \(I\) is the incentre of \(ABC\).
We have \[\angle DB'A = \angle ABD = \angle DBC\]and $B,C,B',I$ are concyclic. We know that the circumcenter of $(BIC)$ is the midpoint of arc $BC$ not containing $A$ of $(ABC)$.
TheDarkPrince wrote:
We have \[\angle DB'A = \angle ABD = \angle DBC\]and $B,C,B',I$ are concyclic. We know that the circumcenter of $(BIC)$ is the midpoint of arc $BC$ not containing $A$ of $(ABC)$.
What is $D$ here?