Let $A_1A_2$ intersect $\omega$ at $A_3$. Consider the common tangent of circles $\omega$ and $\omega_A$ passing through $A_2$ and let $U$ be a point on this common tangent which is at the same side of the line $A_1A_2$ with respect to $B$. Then by tangency, we have $\angle A_1A_2U = \angle BA_1A_2$ and $\angle BA_2U = \angle BA_3A_2$. Therefore
$$\angle CA_2A_3 = \angle A_3BC = \angle BA_1A_2 - \angle BA_3A_2 = \angle A_1A_2U - \angle BA_2U = \angle BA_2A_1 = \angle A_3CB$$
which means that $A_3B = A_3C$. By the sine law, we obtain that
$$\frac{A_1B}{A_1C} = \frac{A_3B}{A_3C} \cdot \frac{\sin \angle BA_3A_2}{\sin \angle CA_3A_2} = \frac{\sin \angle BA_3A_2}{\sin \angle CA_3A_2}$$

and since $\angle BAA_2 = \angle BA_3A_2$ and $\angle CAA_2 = \angle CA_3A_2$, we obtain that
$$\frac{A_1B}{A_1C} = \frac{\sin \angle BAA_2}{\sin \angle CAA_2}$$
Similarly, we get
$$\frac{B_1C}{B_1A} = \frac{\sin \angle CBB_2}{\sin \angle ABB_2} \quad \text{and} \quad \frac{C_1A}{C_1B} = \frac{\sin \angle ACC_2}{\sin \angle BCC_2}$$
Therefore, we have
$$\frac{A_1B}{A_1C} \cdot \frac{B_1C}{B_1A} \cdot \frac{C_1A}{C_1B} = \frac{\sin \angle BAA_2}{\sin \angle CAA_2} \cdot \frac{\sin \angle CBB_2}{\sin \angle ABB_2} \cdot \frac{\sin \angle ACC_2}{\sin \angle BCC_2}$$

Since $AA_1, BB_1, CC_1$ are concurrent, by Ceva Theorem, we have
$$\frac{A_1B}{A_1C} \cdot \frac{B_1C}{B_1A} \cdot \frac{C_1A}{C_1B} = 1$$
and hence
$$\frac{\sin \angle BAA_2}{\sin \angle CAA_2} \cdot \frac{\sin \angle CBB_2}{\sin \angle ABB_2} \cdot \frac{\sin \angle ACC_2}{\sin \angle BCC_2} = 1$$
which means that the lines $AA_2, BB_2, CC_2$ are also concurrent by the Converse Trigonometric Ceva Theorem.