Let $\odot(AOB) \cap BC\equiv D$ and $\odot(AOC) \cap BC\equiv E$.We see that $D,E$ are the inverse of $A_1,A_2$ w.r.t $\odot(ABC)$.Let $X \equiv \odot(ABC) \cap \omega_A $ Now, we will work in complex plane, See that $AD=DC => d = ac\bar d$ and $D \in BC => bc \bar d + d =b+c => \bar d = \frac{b+c}{c(a+b)}=> a_1 = \frac{c(a+b)}{b+c}$.Similarly $a_2 = \frac{b(c+a)}{b+c}$.See that $X$ is the center of spiral similarity $A_1A_2 \mapsto BC => x = \frac{\frac{(c^2a+c^2b) -(b^2a+b^2c)}{b+c} }{(c-b)+\frac{a(c-b)}{b+c}} = \frac{ab+bc+ca}{a+b+c}$ which is symmetric in $a,b,c$ done.

After making an inversion with center at $A$ and radius $\sqrt{\dfrac{1}{2}AB \cdot AC}$, then reflect everything over the bisector of angle $\angle{BAC}$: 1. $\odot(BOC)$ goes to the nine point circle. 2. If we let $M,N$ be the midpoints of sides $AB,AC$, respectively, and $E,F$ be the foots of the perpendiculars from $B,C$, then the image of second point of intersection of $\odot(AA_1A_2)$ and $\odot(ABC)$ will be just $MN\cap EF=\{X\}$. 3. Let $D$ be the foot of the perpendicular from $A$ to $BC$. The image of $\odot(AOC)$ is the line $DM$, because the image of $O$ is $D$. 4. Let $\odot(AMN) \cap DM$ be $Y$ and $DM \cap AC$ be $Z$. The image of $\odot(BB_1B_2)$ is $\odot(NYZ)$. 5. We are left to prove that $X$ lies on $\odot(NYZ)$. But this is simple. We will prove that $Y$ lies on the circle with diameter $AC$. Take $\odot(AMN) \cap \odot(AC)=\{Y’\}$. Then $\angle{MY’N}+\angle{NY’D}=180^{\circ}$. Now observe that $\angle{ZDC}=\angle{B}-\angle{C}=\angle{FXN}$, so it’s enough to prove that $F, Y$ are symmetric with respect $MN$. But that’s obvious because $NY=NF$. $ \textbf{DONE.}$

Wait, I have created this problem in a private forum 5 months ago without proposing to somewhere else ?! Let $H$ be the orthocenter of $\Delta ABC$ and let $P=\odot(AA_1A_2)\cap\odot(ABC)$. The concurrency point is isogonal conjugate of ${\infty}_{\perp OH}$ which lies on $\odot(ABC)$ since $\odot(ABC)$ is the isogonal conjugate of the line at infinity. To prove the claim, we perform the inversion at $A$ with radius $\Delta ABC$ and denote the inverse with prime. The inversion swaps $\odot(BOC)$ and $\odot(BHC)$ since it maps $O$ to the reflection of $A$ across $BC$ which is known to be lie on $\odot(BHC)$. Therefore $A_1' = \odot(BHC)\cap AC$ and $A_2'=\odot(BHC)\cap AB$. And $P'=A_1'A_2'\cap BC$. We must prove that $AP'\cap OH$. Note that the homothety centered at $A$ with ratio $0.5$ takes $\odot(BHC)$ to the nine-point circle of $\Delta ABC$. Therefore the homothety sends $A_1', A_2'$ to the projection of $B, C$ to $AC, AB$ respectively which we call $E, F$. Now note that the circumcenter of $\Delta AEF$ is midpoint of $AH$ therefore by homothety, the circumcenter of $\Delta AA_1'A_2'$ is precisely $H$. Now note that $P'$ the the radical center of $\odot(AA_1'A_2'), \odot(ABC), \odot(BHC)$. Therefore $AP'$ is the radical axis of $\odot(AA_1'A_2'), \odot(ABC)$ but the centers of two circles are $H, O$ respectively, yielding the conclusion.

Lemma (known). Circumcircles of $AA_1A_2$, $OBC$ has equal radiuses. Now, let the lines through $A, B, C$ and parallel to $BC, AC, AB$ form a triangle $A'B'C'$ with circumcenter $O'$. Then the nine-point circle $\omega_a$ of $O'B'C'$ is tangent to $B'C'$ at $A$ and has radius equal to the radius of $OBC$. From Lemma we get that radiuses of the circumcircles of $AA_1A_2$, $\omega_a$ are equal. Also we know that these two circles are tangent at $A$, so they coincide, i.e. $\omega_A = \omega_a$. Similarly we get that $\omega_B$, $\omega_C$ coincides with the nine-point circles of $O'A'C'$, $O'A'B'$ and so $\omega_A, \omega_B, \omega_C$ are concurrent at the Poncelet point of $A'B'C'O'$ which lies on the circumcircle of $ABC$.

There also is another (maybe even simpler proof) : Lemma. Point $O$ is orthocenter of $AA_1A_2$. Let $\omega_a$ meet the circumcircle of $ABC$ at $U$. So from Lemma we get that $AU$ is perpendicular to the Euler line of $AA_1A_2$. But $AA_1A_2$ is similar to $ABC$, so if we reflect $AU$ respect to angle bisector of $\angle A$, then resulting line will be perpendicular to the Euler line of $ABC$. Rest of the proof is obvious.