Let $ABC$ be a triangle and $\Omega$ be its circumcircle, the internal bisectors of angles A, B, C intersect $\Omega$ at $A_1, B_1,C_1$. The internal bisectors of $A_1, B_1,C_1$ intersect Omega $A_2, B_2,C_2$. If the smallest angle of $\triangle ABC$ is $40$ degrees, find the smallest angle of $\triangle A_2B_2C_2$.
I took $A$ to be smallest angle and from my geogebra sketch I found out that the smallest angle of $\triangle A_2B_2C_2$ is $C_2$ which is equal to 55 degrees. Please help me with this problem. But don't give me the solution as I can find that elsewhere but rather please give me some hints such as which theorems I might need to use. Sequential hints that lead to the answer would be even more helpful.