solution: let $I_{a}$ is $A$-excenter. $O$ is the circumcenter of triangle $ABC$.let $AI,BI,CI$ meets circumcircle at $D,E,F$. easy to see $I_{a}O$ is perpendicular to $B_{2}C_{2}$.(http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=387658) $OI_{a}\perp B_{2}C_{2},A'A_{1}\perp B_{2}C_{2}\Rightarrow OI_{a}\parallel A_{1}A'$ $OD\perp BC,IA_{1}\perp BC\Rightarrow OD\parallel A_{1}I\Rightarrow ODI_{A}\sim A_{1}IA'\Rightarrow IA'=\frac{r}{R}.DI_{a}=\frac{r}{R}.BD$ now it is easy to see $A'B'BA$ is cycle.(by power of point theorem) now it is easy to see $A'B'=\frac{IA'.AB}{IB}=A_{1}A_{2}$ so we are done

Let $\triangle I_AI_BI_C$ be the excentral triangle and consider $\mathcal{H} : \triangle I_AI_BI_C\mapsto \triangle A_1B_1C_1$.Now it's well known that $I_AO\perp B_2C_2$ $\implies$ $A_1I'$ passes thru the nine-point circl of $\triangle A_1B_1C_1$ $\implies$ $A'$ is the reflection of $I$ over $C_1B_1$ and thus the conclusion follows immediately.$\blacksquare$

lemma: let $\triangle ABC$ a triangle with circumcenter $O$ and ninepointcenter $N$ and $\triangle XYZ$ be its tangency triangle let $OX \cap AN=D$ define $E,F$ similary prove that $\triangle ABC$ IS similar to $\triangle DEF$ proof: we will assume $O$ is the origin and let $D'$ the reflection if $A$ around $N$ $\vec AN= \vec ND' $ and since $\vec N=\vec A+ \vec C+\vec B \implies \vec D'=\vec B +\vec C$ thus $OBCD'$ is a rhombus then $OD'\perp BC \implies D=D'$ thus there exist a hmothety sending $ABC $to $DEF$ with factor $1$ back to our problem It's well-known that $I_aO \perp B_2C_2$ and also $O$ is the nine-point center of$I_aI_bI_c$ now consider $\mathcal{H} : \triangle I_AI_BI_C\mapsto \triangle A_1B_1C_1$$\implies$ $A_1I'$ passes through the nine-point circle of $\triangle A_1B_1C_1$ now use the lemma above