Nice problem on mixtilinear stuff. (Also it should be $(CMI)$ lol) Let the $A$-mixtilinear incircle touch $(ABC), AC, AB$ at $T_A, N, N'$ respectively then $T_AN \cap (ABC)=M_B$ is the midpoint of minor arc $AC$ meaning $B,I,M_B$ are colinear, but since $AN=NC$ we have that $AT_A=T_AC$, now from homothety and midbase if we have that $Z$ is touchpoint of $A$-excircle with $BC$ then $AZ \parallel IM$ but also from $\sqrt{bc}$ we have that $T_A,Z$ swap so $AT, AZ_A$ are isogonal in $\angle BAC$ and $\triangle AT_AC \sim \triangle ABZ$ meaning that: \[180-\angle AIC=90-\frac{\angle B}{2}=\angle T_AAC=\angle BAZ=\angle AZB=\angle IMB \]Therefore $AI$ is tangent to $(CMI)$ as desired thus we are done .