Hi, so I am doing a proof but I need some help proving one part of it. I'm having trouble proving that angle $ D'Bi = $ angle $ D'iB $. point $ i $ is the incenter of triangle ABC and D' is the point at which ray Ai intersects the circumcircle. I'm trying to prove that $ D' $ is the circumcenter of triangle $ BiC $, however like I said before I'm having trouble proving angle $ D'Bi = $angle $ D'iB $.
Note $\alpha, \beta, \gamma$ the angles of the triangle $ABC$ on $A, B, C$.
Use the property of inscribed angles two times : $BD'A = \gamma$, and $D'BC = D'AC = \alpha/2$. So $D'BI = D'BC + D'CI = \alpha/2 + \beta/2$. Then the sum of the angles of the triangle $D'IB$ is $180$° $= \alpha + \beta + \gamma$ ; you obtain that the angle $D'IB$ is also $\alpha/2 + \beta/2$. Whence, $D'BI = D'IB$ and your problem is solved.
(Thank you for this beautiful property, I did not know it)