Let $R=\overline{B_1C_1}\cap\overline{B_2C_2}$. By DDIT on $B_1C_1C_2B_2A_2R$, we get that $\overline{RA}$ is tangent to $(ABC)$. Let $R'$ be intersection of the tangent from $A$ to $(ABC)$ and the line through $A_2$ parallel to $\overline{BC}$. Observe that by angle chase, $R'A=R'A_2$, thus there exists a circle tangent to $\overline{R'A},\overline{R'A_2}$, let that circle be $\omega$. Also, as $\measuredangle C_2A_2R'=\measuredangle C_2CB=\measuredangle C_2B_2A_2$, we get that $\overline{R'A_2}$ is tangent to $(A_2B_2C_2)$. By radical axis theorem on $(ABC),(A_2B_2C_2),\omega$, we get that $R'$ lies on $\overline{B_2C_2}$, thus $R\equiv R'$. As $(R,\overline{A_2M}\cap\overline{B_2C_2};C_2,B_2)=-1$, projecting this through $A_2$ onto $\overline{BC}$ yields that $\overline{A_2M}$ passes through the midpoint of $\overline{BC}$.