In triangle $ABC$, where $B'$ and $C'$ are the feet of the interior angle bisectors of angles $B$ and $C$, $I_a$ is the intersection point of the exterior angle bisectors, and $O$ is the circumcenter of the circumscribed circle, we are to prove that the line segment $I_aO$ is perpendicular to $B'C'$.
First I made a drawing in Geogebra
Since $B'$ and $C'$ are the feet of the interior angle bisectors then $B'A=B'C=\frac{AC}{2}$ and $C'A=C'B=\frac{AB}{2}$
$C'B' \parallel BC$ because it is the midline of the triangle.
From the midline theorem we get: $\frac{C'B'}{BC}=\frac{AC'}{AB}=\frac{AB'}{AC}=\frac{1}{2}$
From here I'm stuck, what would be the next step to solve this problem?