Concurrency problem involving incircles of a triangle and sub-triangles The incircle of triangle $A_1 A_2 A_3$ touches the sides $A_2 A_3$, $A_3 A_1$, and $A_1 A_2$ at $S_1$, $S_2$, and $S_3$, respectively. Let $O_1$, $O_2$, and $O_3$ be the incenters of triangles $A_1 S_2 S_3$, $A_2 S_3 S_1$, and $A_3 S_1 S_2$, respectively. Prove that $O_1 S_1$, $O_2 S_2$, $O_3 S_3$ are concurrent.
I have tried drawing a big diagram, as accurate as possible, as a start, and saw that $O_1$, $O_2$, and $O_3$ seem to touch the incircle of triangle $A_1A_2A_3$. However, I am not sure of how to prove this, or how this helps with proving the concurrency. Can I have a little hint as to how to prove this and how it helps answer the question? Much appreciated.


You're correct about $O_1$, etc.
Here's a hint to prove it.
Take two points $S_2$ and $S_3$ on a circle. Draw the tangent line at $S_2$, and take a bisector between the tangent line and $S_2S_3$. Prove that the bisector meets the circle midway along the arc from $S_2$ to $S_3$. (You'll want to use the version of the inscribed angle theorem that works for tangent lines.) 
Once you have that, the problem is simplified somewhat because you can forget about $A_1$, $A_2$ and $A_3$ and just focus on the points $S_i$ and $O_i$ on the circle, with the points $O_i$ placed on the circle at the midpoints. In this case, say $S_2 O_2$ and $S_3 O_3$ meet at $P_1$, and define $P_2$ analogously. Prove that $P_1O_3 = S_2 O_3 = S_1 O_3 = P_2 O_3$. 
