If $AB=AC$, we are done. WLOG $AB<AC$. As the midpoint of $KM$ lies on the circumcircle, $KIMI_a$ is a rhombus where $I_a$ is the $A-$ excentre. We need to show $O$ is the midpoint of $IN$. Thus we need to show $r=2\cdot \frac{a}{2}\cot A=a\cot A$. Thus we need \[r=(s-a)\tan \frac{A}{2} = a\cot A \Leftrightarrow s-a = s\cos A.\]Now $KIMI_a$ is a rhombus, thus $\angle AKI = 90^{\circ}-A$. Thus \[\frac{s}{\cos \frac{A}{2}}= \frac{AI}{\cos A}=\frac{s-a}{\cos A\cos \frac{A}{2}}\implies s = \frac{s-a}{\cos A}.\] Proved.

Dear Mathlinkers, also at https://artofproblemsolving.com/community/c6t48f6h1668321_nice_geo Sincerely Jean-Louis