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> 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 $\triangle A\_2B\_2C\_2$.
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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.

![Geogebra Sketch](https://i.stack.imgur.com/k8iNE.png)