Let $P$ be the midpoint of the minor arc $AB$ of $(ABC)$, we easily have $PA=PB=PI \Rightarrow \triangle PBI, \triangle PAB $ is isoceless triangle. Since $M,N$ is the midpoint of $AB,BI \Rightarrow \angle BMP=\angle BNP=90^o \Rightarrow P \in (BMN)$ Thus $BP$ is the diameter of $(BMN)$, since $\angle CPB=\angle CAB=90^0 \Rightarrow CI $ tangenst to $(BMN)$

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Let $M$ be the midpoint of $\overarc{BC}$ of $(ABC)$ so that $M$ lies on line $CI$. Then $M$ lies on $(BMN)$ by fact $5$ + homothety and $CM\perp BM$ since $BC$ is a diameter of $(ABC)$, as desired.