noice noice, isogonal mittenpunkt gets remebered We use Vietnam TST 2003 Day 1 P2 to solve this. We know that $A, B, C$ are the $A_1, B_1, C_1 - $ excenters of the orthic triangle $\triangle A_1B_1C_1$ respectively. Also note that the incenter of the orthic triangle is the orthocenter of $\triangle ABC$, and that the $9-$point center of $\triangle ABC$ is the circumcenter of the orthic triangle. Thus, the Euler Line of $\triangle ABC$ is the line through the orthocenter and the incenter of the orthic triangle. Now set the orthic triangle as the reference triangle. $\textbf{RESTATED PROBLEM: }$Let $A_1B_1C_1$ be a triangle with circumcenter $N_9$, incenter $H$, $A, B, C$ as its $A_1, B_1, C_1 - $ excenters respectively. Let $A_2, B_2, C_2$ be the $A_1, B_1, C_1 - $ intouch points of $\triangle A_1B_1C_1$ respectively. Prove that $AA_2, BB_2, CC_2$ are concurrent at a point on the line $N_9H$. Firstly recall that by Midpoint of Altitudes Lemma, the set of points $\{A_0, D, I_A \}$, $\{B_0, E, I_B \}$ and $\{C_0, F, I_C \}$ are collinear. Thus it suffices to show that $A_{0}D$, $B_{0}E$, $C_{0}F$ and $OI$ are concurrent and we are done .

By cevian nest, they are obviously concurrent now note that $ABC$ and $A_2B_2C_2$ are homothetic and the centre of homothety is $AA_2 \cap BB_2 \cap CC_2=X$ so $X$, the circumcenter of $A_2B_2C_2 = H$ and the circumcenter of $ABC=O$ are collinear, done. $\blacksquare$

$B_2C_2\perp HA_1\Rightarrow BC \parallel B_2C_2$ Similarly other sides are parallel too. $ABC$ and $A_2B_2C_2$ are homothetic and homothety center $J$ sends $O$ to $H$ Q.E.D.

Note that $H$, the orthocenter of $\triangle ABC$, is also the incenter of $\triangle A_1B_1C_1$. Thus, $B_2C_2\perp HA_1$, so $B_2C_2\parallel BC$ etc, so $AA_2,BB_2,CC_2$ concur. Furthemore, the circumcenter of $A_2B_2C_2$ is $H$, so the center of homothety is collinear with the circumcenter and orthocenter of $\triangle ABC$, done.

Here is a related problem: https://artofproblemsolving.com/community/q1h2745277p23954950