First, we will prove two following lemmas:

Lemma 1.

Let $ABC$ be the acute and non-isosceles triangle. The points $M,N$ are the feet of the perpendicular line from $A$ to the exterior angle bisector $B,C$ of triangle $ABC$. Let $P$ be the tangent point of line $BC$ and the excircle $(I)$ with respect to $BC$, $AQ$ is the altitude of triangle $ABC$. Then $MNPQ$ is the isosceles trapezoid and $P$ is the orthocenter of triangle $MNI$.

Prove.

Without loss of generality, we can assume that $AB < AC$. Let $D,E$ be the intersection of $AM,AN$ and the line $BC$. It is easy to see that triangle $BAD$ is isosceles with $BA = BD$ and $BM$ is the altitude of this triangle (with $M \in AD$) then $M$ is also the midpoint of segment $AD$. Similarly, $N$ is the midpoint of segment $AE$. Hence, the line $MN$ is parallel to the line $BC$.

We also have $DP = EP = \frac{AB + BC + CA}{2}$, then point $P$ is the midpoint of segment $DE$. Because $QN = \frac{1}{2}AE = MP$, then the quadrilateral $MNPQ$ is isosceles trapezoid. Furthermore, we have $PM \parallel AE, NI \perp AE$, so $PM \perp NI$; similarly, $PN \perp MI$ then $P$ is the orthocenter of triangle $MNI$. The lemma is proved.

Lemma 2.

Let triangle $ABC$ have the centroid $G$ and point $D$ is the projection of the Euler point of triangle $ABC$ on $AC$ and $E$ is the symmetric point of point $B$ about line $AC$. Prove that $\overrightarrow{GE} = 4\overrightarrow{GD}$.

Let $H,N,O$ be orthocenter, Euler point and circumcenter of triangle $ABC$.
Assume that $K,D,M$ in order are the projections of $H,N,O$ on the line $AC$.
Because $N$ is the midpoint of the segment $OH$ so $D$ is also the midpoint of segment $KM$. It is easy to see that
$$\frac{GN}{GH} = \frac{1}{4} \quad \text{and} \quad ND = \frac{1}{2}(OM + HK) = \frac{1}{4}BH + \frac{1}{4}(HE - BH) = \frac{1}{4}HE.$$
Hence, $\frac{GN}{GH} = \frac{ND}{HE}$ so $G,D,E$ are collinear and $GE = 4GD$. The lemma is proved.

Back to the original problem, let $E$ be the Euler point of triangle $ABC$. First, we will prove that the line $AA_1$ passes through the point $E$.

Indeed, we know that in a triangle, the circumcenter and the orthocenter are isogonal conjugate points. We have $\angle BAC' = \angle BAC = \angle B'AC$ then $AB', AC'$ are symmetric about the angle bisector of $\angle BAC$. Then, each pair of isogonal lines with respect to angle $A$ in triangle $ABC$ is also a pair of isogonal lines with respect to angle $A$ in triangle $AB'C'$; so we have to prove that the isogonal line of $AE$ with respect to angle $A$ in triangle $ABC$ passes through the orthocenter of triangle $AB'C'$. Let $B'', C''$ be the projections of point $E$ on the segment $AC, AB$. Based on Lemma 2, $B', C'$ are the image of $B'', C''$ under the dilation center $G$, ratio 4 with $G$ as the centroid of triangle $ABC$.

From this result, we get $B'C' \parallel B''C''$. And based on the properties of isogonal conjugates, the isogonal line of $AE$ is perpendicular to $B''C''$ and this exactly is the altitude from $A$ of triangle $AB'C'$. Then, the isogonal line of $AE$ in triangle $AB'C'$ passes through the orthocenter of triangle $AB'C'$, or the line $AA_1$ passes through point $E$.

Next, we will prove that $AA_2$ also passes through $E$. Let $O,H$ be the circumcenter and orthocenter of triangle $ABC$, respectively; $R,S$ be the projections of point $E$ on the sides $AB, AC$, respectively and $R,S$ are the midpoints of segment $AB, AC$, respectively. Let $M,N,P$ be the projections of $A_0$ on the lines $AB, BC, CA$, respectively.

It is easy to see that $OR \parallel A_0M, OS \parallel A_0P$ so if we set $k = \frac{AR}{AM} = \frac{AS}{AP}$ then the dilation $\Phi$ center $A$, ratio $k$ such that $\Phi: R \to M, \Phi: S \to P$ and $\Phi: O \to A_0$.

Then point $A$ is the center of the excircle opposite to $A_0$ of triangle $A_0BC$ and $M,P$ are the projections of $A_0$ on the exterior angle bisector of $B,C$ of triangle $A_0BC$.

Based on Lemma 1, $MP \parallel BC$ and if $K$ is the projection of $A$ on the line $BC$ then quadrilateral $MKNP$ is an isosceles trapezoid. Hence, $K$ belongs to the circle ($A_2$) and $K$ is the orthocenter of triangle $AMP$.

Let $T$ be the midpoint of segment $AH$ then $T \in (E)$. So $RT \parallel BH$ and $BH \parallel MK$ leads to $RT \parallel MK$ and $\Phi: T \to K$.

Therefore, $\Phi: \Delta RST \to \Delta MPK$ or $\Phi: E \to A_2$ (because $E$ is the center of the circle passing through three points $R,S,T$).

Then three points $A,E,A_2$ are collinear. Mixing with the above result, we get $A, A_1, A_2, E$ are all collinear or line $A_1A_2$ passes through point $E$.

It's quite similar to prove that $B_1B_2, C_1C_2$ also pass through point $E$.

So $A_1A_2, B_1B_2, C_1C_2$ are concurrent at the Euler point $E$ of triangle $ABC$. (Q.E.D)