When doing another geometry proof, I have this problem: Let $A$ is the point outside the circle $(O;R)$ such that $OA=2R$. The line $d$ goes through $A$ and cut the circle at $2$ distinct points $E$ and $F$. What is the locus of the centroid point of the triangle $OEF$?
I used Geogebra to draw the trace of this centroid, it's an arc of a circle but i can't find out which point is the center.
Assume $O$ is origin. Let $M$ be the midpoint of all chords $EF$. $\angle OMA = 90$. $M$ lies on arc of circle of diameter $OA$.
The centroid $G$, of $\triangle OEF$ lies two-thirds away from $O$ along $OM$.
Find $B$ on $OA$ so that $GB\parallel MA$. Clearly $\triangle OGB \sim \triangle OMA$. So $B$ lies two-thirds away from $O$ along $OA$. $$B=\frac{2A}{3}$$
Hence $G$ lies on the circle whose diameter is $OB$. Its center $C$ is midpoint of $OB$.
$$C=\frac{O+B}{2}=\frac{A}{3}$$
Thus $C$ is $2R/3$ away from $O$ along $OA$.