High school problem from IMO selection round 
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.

 A: Here is a hint.  For any triangle $XYZ$ inscribed in $\Omega$, let $x$, $y$, and $z$ denote the measures of the angles at $X$, $Y$, and $Z$, resp, of $\triangle XYZ$.  Let $X'$ be the point on $\Omega$ s.t. $XX'$ internally bisects the angle $X$.  The points $Y'$ and $Z'$ are defined similarly.  If $x'$, $y'$, and $z'$ are the angles at $X'$, $Y'$, and $Z'$ of $\triangle X'Y'Z'$, then
$$x'=\frac{y+z}{2}\wedge y'=\frac{z+x}{2}\wedge z'=\frac{x+y}{2}.$$
Apply the above result with $\triangle ABC$, and then $\triangle A_1B_1C_1$.  (Note that $55=\frac{180+40}{4}$.)
A: Observe that the full circumference is the sum of the four arcs $CB, BC_1,\> C_1B_1,\>B_1C$, which respectively correspond to the angles $A, \>\frac C2,\> A_1, \> \frac B2$, i.e.
$$180= A_1+A+\frac{B+C}2\implies A_1=90-\frac A2=70$$
assuming $A=40$, $B,\> C \in(40,100)$ without loss of generality. Then,
$$B_1=90- \frac B2\in (40,70) \>\>\>\>\> C_1= 90- \frac C2 \in (40,70) 
$$
and 
$$A_2=90-\frac {A_1}2=55,\>\>\>\>\>
B_2=90-\frac {B_1}2\in(55,70),\>\>\>\>\> C_2=90-\frac {C_1}2\in(55,70)$$
Thus, the smallest angle of △$A_2B_2C_2$ is 55 degrees.
