Let $D$ be the contact of $(I)$ with $BC$, $D'=DI\cap (I)$ and $M'$ a point on $AC$ such that $M'D'\parallel BC$. We see that $IM'$ is a bisector of $\angle CM'D'=180-\angle M'CB$, then $\angle CM'I=90-(1/2)\angle M'CB$ and $\angle M'CI=(1/2)\angle M'CB$ therefore $\angle M'IC=90$. Let $h$ and $r$ be the height through $A$ and the inradius of $ABC$, respectively, $\frac{BC(h)}{2}=[ABC]=\frac{36}{2}r$, therefore $\frac{h}{r}= 2\frac{36}{2}\frac{1}{BC}=4$. Since $DD'=2r\rightarrow DD'=\frac{h}{2}$, then $M'=M$ and $\angle MIC=90$.