Any idea?

Suppose that $I_{1a}, I_{2a}$ be center of $\omega_1, \omega_2$. Note that $I, A_1$ lie on the same side WRT $AB$ so $I_{1a}$ lies on the same side with $I, A_1$ WRT $AB$. But $\omega = \omega_1$ and these 2 circles tangent to $AB$ then $II_{1a} \parallel AB$. Similarly, $II_{2a} \parallel CA$. Since $A_1I = A_1I_{1a} = A_1I_{2a} = r,$ we have $A_1$ is center of $(II_{1a}I_{2a})$. We also have $A_1I \parallel AO$ so it's easy to prove that $\triangle AOB \sim \triangle IA_1I_{1a}, \triangle COA \sim \triangle I_{2a}A_1I,$ hence $\dfrac{II_{1a}}{II_{2a}} = \dfrac{II_{1a}}{A_1I} \cdot \dfrac{A_1I}{II_{2a}} = \dfrac{AB}{OA} \cdot \dfrac{OA}{AC} = \dfrac{AB}{AC}$. From this, we have $\triangle ABC \sim \triangle II_{1a}I_{2a},$ therefore $I_{1a}I_{2a} \parallel BC$. We define $I_{1b}, I_{1c}$ similarly with $I_{1a};$ $I_{2b}, I_{2c}$ similarly with $I_{2a}$. So $\triangle ABC \sim \triangle I_{2b}II_{1b} \sim \triangle I_{1c}I_{2c}I$. But $A_1I = B_1I = C_1I = r$ then $(A_1, A_1I) = (B_1, B_1I) = (C_1, C_1I),$ hence $\triangle II_{1a}I_{2a} = \triangle I_{2b}II_{1b} = \triangle I_{1c}I_{2c}I,$ then $II_{2b} = I_{2c}I_{1c}$. Combine with $II_{2b} \parallel AB \parallel I_{2c}I_{1c},$ we have $II_{2b}I_{1c}I_{2c}$ is parallelogram, so $I_{2b}I_{1c} \parallel II_{2c} \parallel BC$. On the other side, $B_1I \parallel BO, C_1I \parallel CO$ then $\angle{B_1IC_1} = \angle{BOC},$ so $\triangle BOC \sim \triangle B_1IC_1,$ hence $B_1C_1 \parallel BC$. Let $U, V$ be midpoint of $I_{1b}I_{2b}, I_{1c}I_{2c}$. Since $I_{2b}I_{1c} \parallel II_{2c} \parallel BC$ then $UV \parallel BC$. But $B_2, C_2$ are reflections of $B_1, C_1$ through $I_{1b}I_{2b}, I_{1c}I_{2c}$ so $B_2, C_2$ are also reflections of $B_1, C_1$ through $U, V$ hence $B_2C_2 \parallel BC$. Similarly, $C_2A_2 \parallel CA,$ $A_2B_2 \parallel AB$. Therefore $\triangle ABC$ and $\triangle A_2B_2C_2$ have corresponding parallel side, from this we have the result that mentioned above