, 2022

We adopt the standard notation for $\triangle ABC$. Denote by $P$ and $Q$ the midpoints of $AC$ and $BC$, respectively. Let $J$ be the center of the excircle, tangent to the segment $AB$. It is well known that $I$ is the midpoint of $CJ$, therefore $PI$ is a midsegment for $\triangle AJC$ and $PI \parallel AJ$. Hence $\angle AIP = 90^\circ$. Analogously, $\angle BIQ = 90^\circ$. We have $\triangle AIP \cong \triangle AMI$, respectively $\triangle BIQ \cong \triangle BIN$, i.e., $AP = AM$ and $BQ = NB$. Moreover, $r_c = 2r$. Finally
$$\begin{aligned} S_{ABC} &= p \cdot r = (p-c) \cdot r_c \quad \Rightarrow \quad p = 2(p-c) \quad \Rightarrow \quad 2c = \frac{a+b+c}{2} \\ &\Rightarrow \quad AC + BC = 3AB \quad \Rightarrow \quad AM + AN = AP + BQ = \frac{AC+BC}{2} = \frac{3AB}{2}. \end{aligned}$$
But $AM + AN = AB + MN$, thus $MN = AB/2$.

Remark. The opposite claim holds also true, i.e., $CI = 2 \cdot IL$ whenever $AB = 2 \cdot MN$.