In $\triangle ABC$, the circumcenter is $O$, the interior center is $I$, and center of gravity is $G$. If $IG \parallel BC$, prove $OI \perp AI$.

Question

The diagram is

What is the quickest way to prove this? Can we use the fact that the centroid G divides the median from A to BC in a 2:1 ratio.

Let M be the midpoint of BC and let AM intersect OI at point P. Since O is the circumcenter of △ABC, we know that OI is the perpendicular bisector of BC. Therefore, OM is perpendicular to BC, and OIMP is a rectangle.

Since IG is parallel to BC and G divides AM in a 2:1 ratio, we know that AG = 2GM. But we also know that GM = MC, so AG = 2MC. Therefore, AM = 3MC.

Since P lies on both OI and AM, we know that AP = 3PM. But PM is half of BC, so AP is perpendicular to BC. Therefore, OI is perpendicular to AI, and we have proved that OI is perpendicular to the line passing through A and the centroid G, which is equivalent to saying that OI is perpendicular to AI.