Olimpiada Matemática del Cono Sur

By considering the inscribed angle $\angle BAC$ in the circumcircle of the triangle $ABC$, we have that $\angle BOC = 2 \cdot \angle BAC = 120^\circ$; then, $\angle BO'C = 60^\circ$. Since $O$ is a point of the perpendicular bisector of $BC$, then $O'$ is also on this line. Therefore, the triangle $BO'C$ is equilateral.

On the other hand, we have that $\angle BIC = 90^\circ + \frac{1}{2} \angle BAC = 120^\circ$. It follows that $I$ is on the circumcircle of $BOC$. By applying Ptolemy's theorem to the quadrilateral $BICO'$, we obtain:
$$IO' \cdot BC = BI \cdot O'C + CI \cdot O'B$$
and, recalling that $BC = O'C = O'B$, we conclude that
$$IO' = BI + CI.$$