Lemma 1

$$AI\cdot AI_A=AB\cdot AC$$proof

Lemma 2 Lines $AA'$ and $AO$ are symmetric wrt. line $AI$.

proof

Lemma 3 $$[ABC]=\frac{AB\cdot BC\cdot CA}{4R}$$Let $D$ be the reflection of point $A'$ wrt. to line $AI$. By lemma 2 we have $D\in AO^\rightarrow$. $$AD\cdot AO=2h_A\cdot R=\frac{4[ABC]}{BC}\cdot\frac{AB\cdot BC\cdot CA}{4[ABC]}=AB\cdot CA=AI\cdot AI_A$$Thus points $I,I_A,D,O$ are concyclic. Point $D$ lies on the same side of line $AI$ as the point $O$, so $$\angle IOI_A=IDI_A=IA'I_A$$

Dear, nice proof It can be shorter... Sincerely Jean-Louis

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/30.%201.%20Angles%20egaux.pdf p. 3… Sincerely Jean-Louis

Initial remark. The problem's conclusion basically asks us to prove that $O$ and $A'$ swap under $\sqrt{bc}$ inversion. This is very natural to figure out- we already know that $I$ and $I_a$ swap (the proof of this fact was also explicitly showcased above), so the angles in question themselves "stay preserved". I think this is more of a motivational step than a part of the proof itself, so I won't formalize it. However, since the next part is pretty cute, I'll write it out (although it is pretty much incorporated in the proof above as well). Actual proof. For brevity, we first introduce some notations. Let $R$ be the circumradius of $\Delta ABC$, let $h_a$ be the length of the altitude of $\Delta ABC$ onto side $BC$ and finally, let $[ABC]$ denote the area of $\Delta ABC$. We want to prove that $A'$ and $O$ swap under an inversion centred at $A$ with radius $\sqrt{bc}$, followed by a reflection over the bisector of $\angle BAC$. Let $A^{*}$ be the image of $A'$ under this inversion. We first note that $AA'$ and $AO$ are isogonal wrt $\angle BAC$ (i. e. $AO$ is the reflection of $AA'$ over the bisector of $\angle BAC$). Recalling the way that we defined the inversion, it becomes sufficient to prove that $AA^{*}=AO=R$. Since, by our inversion's definition, we have $AA^{*}\cdot AA'=AA^{*}\cdot 2h_a=bc$, it further becomes sufficient to prove that $$R\cdot 2h_a=bc$$but this last equality immediately follows from the fact that $$\frac{h_a\cdot a}{2}=[ABC]=\frac{abc}{4R}\Leftrightarrow R\cdot 2h_a=bc$$which ends our proof.