For three distinct points $X_1, X_2, X_3$, when the line $X_1X_2$ is rotated counterclockwise by an angle $\theta$ about $X_1$ to coincide with the line $X_1X_3$, this $\theta$ is denoted by $\angle X_2X_1X_3$. Note that differences of $180^\circ$ are disregarded. First, we show the following lemma concerning the Circle of Apollonius.

Lemma 1. Let $STU$ be a triangle satisfying $ST \neq SU$, and let $\omega$ be a circle passing through $S$. When an inversion with center $S$ is performed, mapping $T$ and $U$ to $T'$ and $U'$ respectively, the circle $\omega$ is the Apollonius circle for the segment $TU$ passing through $S$ if and only if $\omega$ is mapped by this inversion to the perpendicular bisector of the segment $T'U'$.

Proof of Lemma 1. Let $V$ and $W$ be the intersection points of the internal and external angle bisectors of $\angle TSU$ with the line $TU$ respectively, and let $V'$ and $W'$ be their respective images under the inversion. From $ST : SU = VT : VU = WT : WU$, the circumcircle of the triangle $SVW$ is the Apollonius circle for the segment $TU$ passing through $S$, so it suffices to show that the circumcircle of the triangle $SVW$ is mapped to the perpendicular bisector of the segment $T'U'$ by the inversion. Since $V'$ and $W'$ are the intersection points of the internal and external angle bisectors of $\angle T'SU'$ with the circumcircle of the triangle $ST'U'$ respectively, they are the midpoints of the arc $T'U'$ not containing $S$ and the arc $T'U'$ containing $S$ on the circumcircle of the triangle $ST'U'$ respectively. Therefore, the image of the circumcircle of triangle $SVW$ under the inversion, which is the line $V'W'$, is the perpendicular bisector of the segment $T'U'$. Thus, the lemma is proved. ■

Let $\Omega_A$ be the Apollonius circle for the side $BC$ passing through $A$, and similarly define $\Omega_B$ and $\Omega_C$. Note that $B_1$ and $B_2$ lie on $\Omega_B$, and $C_1$ and $C_2$ lie on $\Omega_C$.

Lemma 2. $A_1, A_2$ lie on $\Omega_A$.

Proof of Lemma 2. Since $\Omega_A$ is symmetric with respect to the line $BC$, $A_1$ is on $\Omega_A$ if and only if $A_2$ is on $\Omega_A$. We will show that $A_2$ lies on $\Omega_A$.
Let $P$ and $Q$ be the two intersection points of $\Omega_B$ and $\Omega_C$. Then,
$$BP : CP = \frac{AP \cdot BC}{AC} : \frac{AP \cdot BC}{AB} = AB : AC$$
Therefore, $P$ lies on $\Omega_A$, and similarly for $Q$.
Let $\sigma$ be the inversion with respect to a circle centered at $P$ with radius 1. Then, for any points $Y, Z$ different from $P$, the triangle $PYZ$ is similar to the triangle $P\sigma(Z)\sigma(Y)$, so we obtain
$$\sigma(A)\sigma(B) = \frac{AB}{AP \cdot BP} = \frac{BC}{BP \cdot CP} = \sigma(B)\sigma(C).$$
Similarly, $\sigma(A)\sigma(C) = \sigma(B)\sigma(C)$ also holds, so the triangle $\sigma(A)\sigma(B)\sigma(C)$ is equilateral. Also, by Lemma 1, $\sigma(\Omega_A)$ is the perpendicular bisector of the segment $\sigma(B)\sigma(C)$, and similarly for $\sigma(\Omega_B)$ and $\sigma(\Omega_C)$.
Therefore, $\sigma(Q)$ is the center of the equilateral triangle $\sigma(A)\sigma(B)\sigma(C)$, and the line $\sigma(B)\sigma(B_2)$ and the line $\sigma(C)\sigma(C_2)$ are symmetric with respect to the line $\sigma(A)\sigma(Q)$. Also, the four points $\sigma(B), \sigma(B_2), \sigma(C), \sigma(C_2)$ are concyclic, and since $\sigma(B)$ and $\sigma(C)$ are symmetric with respect to the line $\sigma(A)\sigma(Q)$, $\sigma(B_2)$ and $\sigma(C_2)$ are also symmetric with respect to the line $\sigma(A)\sigma(Q)$. Here, from the condition that $B_2, C_2$ do not lie on the circumcircle of the triangle $ABC$, note that the circumcircle of the triangle $\sigma(A)\sigma(B)\sigma(C_2)$ and the circumcircle of the triangle $\sigma(A)\sigma(B_2)\sigma(C)$ are distinct. Since $\sigma(A_2)$ is the intersection point of these two circles other than $\sigma(A)$, it lies on the line $\sigma(A)\sigma(Q)$, and this line is $\sigma(\Omega_A)$, so $A_2$ lies on $\Omega_A$.

Let $R$ be the intersection point, other than $A$, of the circumcircle of the triangle $ABC_1$ and the circumcircle of the triangle $AB_1C$. If the two circles are tangent, then let $R = A$, and the line $AR$ is taken to be the common tangent line to the two circles at $A$. Also, since $B_1, C_1$ are points in the interior of the triangle $ABC$, note that $R$ is different from $B$ and $C$. Then,
$$\begin{aligned} \angle BRC &= \angle BRA + \angle ARC \\ &= \angle BC_1A + \angle AB_1C \\ &= \angle AC_2B + \angle CB_2A \\ &= \angle AA_2B + \angle CA_2A \\ &= \angle CA_2B \\ &= \angle BA_1C. \end{aligned}$$
Therefore, the circumcircle of the triangle $A_1BC$ also passes through $R$. Here, since $A_1$ is a point in the interior of the triangle $ABC$, $R$ is also different from $A$.

Proof of Lemma 3. Let $\Gamma_A$ be the arc $BC$ of the circumcircle of the triangle $A_1BC$ that contains $A_1$. Let $\Gamma_B$ be the arc $CA$ of the circumcircle of the triangle $AB_1C$ that contains $B_1$. Let $\Gamma_C$ be the arc $AB$ of the circumcircle of the triangle $ABC_1$ that contains $C_1$. Then, since $A_1$ is in the interior of the triangle $ABC$, $\Gamma_A$ is inside the circumcircle of the triangle $ABC$, and the part of the circumcircle of the triangle $A_1BC$ not containing $\Gamma_A$ is outside the circumcircle of the triangle $ABC$. Since the same holds for $\Gamma_B$ and $\Gamma_C$, either all of $\Gamma_A, \Gamma_B, \Gamma_C$ contain $R$, or none of them contain $R$. Also, if none of them contain $R$, then $R$ must be on the opposite side of the line $BC$ from $A_1$, on the opposite side of the line $CA$ from $B_1$, and on the opposite side of the line $AB$ from $C_1$. But since $A_1, B_1, C_1$ are all in the interior of the triangle $ABC$, this is a contradiction. Thus $\Gamma_A, \Gamma_B, \Gamma_C$ all pass through $R$.
Here, if we assume $P = R$ (where $P$ is from Lemma 2), then $P = A_1 = B_1 = C_1$, which contradicts the assumption that $A_1, B_1, C_1$ are distinct. So note that $P \neq R$. Let $\tau$ be the operation of applying $\sigma$ defined in Lemma 2 followed by an inversion centered at $\sigma(R)$. By Lemma 1, $\tau(\Omega_A)$ is the Apollonius circle for the segment $\tau(B)\tau(C)$ passing through $\tau(A)$.
Furthermore, since the four points $A_1, B, C, R$ are concyclic, $\tau(A_1)$ lies on the line $\tau(B)\tau(C)$. Therefore, $\tau(A_1)$ is the intersection of the internal or external bisector of $\angle\tau(B)\tau(A)\tau(C)$ with the line $\tau(B)\tau(C)$, and similarly for $\tau(B_1)$ and $\tau(C_1)$. Here, since $\Gamma_A, \Gamma_B, \Gamma_C$ all contain $R$, $\tau(\Gamma_A), \tau(\Gamma_B), \tau(\Gamma_C)$ all contain the point at infinity $\tau(R)$. Therefore, $\tau(A_1), \tau(B_1), \tau(C_1)$ are all intersection points of external angle bisectors with the opposite sides. Also,
$$\frac{\tau(B)\tau(A_1)}{\tau(A_1)\tau(C)} \cdot \frac{\tau(C)\tau(B_1)}{\tau(B_1)\tau(A)} \cdot \frac{\tau(A)\tau(C_1)}{\tau(C_1)\tau(B)} = \frac{\tau(A)\tau(B)}{\tau(C)\tau(A)} \cdot \frac{\tau(B)\tau(C)}{\tau(A)\tau(B)} \cdot \frac{\tau(C)\tau(A)}{\tau(B)\tau(C)} = 1.$$
Therefore, by the converse of Menelaus's Theorem, the three points $\tau(A_1), \tau(B_1), \tau(C_1)$ are collinear. Thus the lemma is proved. ■

Let $D$ be the intersection of the internal angle bisector of $\angle BAC$ with the line $BC$. This point $D$ lies on $\Omega_A$. Also, let $O_A$ be the center of $\Omega_A$. Then $O_A$ lies on the line $BC$. Therefore,
$$\angle O_A A B = \angle O_A A D + \angle D A B = \angle A D O_A + \angle C A D = \angle A C D$$
from which it follows that the line $AO_A$ is tangent to the circumcircle of triangle $ABC$ at $A$, so $O_A A^2 = O_A B \cdot O_A C$ holds. From this and $O_A A = O_A A_2$, $O_A A_2^2 = O_A B \cdot O_A C$ holds. Therefore, the line $A_2 O_A$ is tangent to the circumcircle of the triangle $A_2 BC$ at $A_2$. Also, under the inversion with respect to $\Omega_A$, $B$ and $C$ are interchanged, and since $P, Q$ defined in Lemma 2 are invariant, $\Omega_B$ and $\Omega_C$ are interchanged by this inversion.
Therefore, if $B'_2$ is the image of $C_2$ under this inversion, then $B'_2$ lies on $\Omega_B$. Here, the four points $B, B'_2, C, C_2$ are concyclic, and this circle is invariant under the inversion with respect to $\Omega_A$, so $B'_2 = B_2$. Therefore, $O_A$ lies on the line $B_2 C_2$, and since $O_A A_2^2 = O_A B_2 \cdot O_A C_2$, the circumcircle of the triangle $A_2B_2C_2$ is tangent to the line $A_2O_A$. Therefore,
$$\begin{aligned} \angle A_1B_1C_1 &= \angle A_1RC_1 \\ &= \angle A_1RB + \angle BRC_1 \\ &= \angle A_1CB + \angle BAC_1 \\ &= \angle BCA_2 + \angle C_2AB \\ &= \angle BCA_2 + \angle C_2A_2B \\ &= \angle BA_2O_A + \angle C_2A_2O_A + \angle O_A A_2B \\ &= \angle C_2A_2O_A \\ &= \angle C_2B_2A_2 \end{aligned}$$

and, similarly $\angle A_1C_1B_1 = \angle B_2C_2A_2$, so the result is shown.