Let $ABC$ be a triangle with centroid $G$. Let the circumcircle of triangle $AGB$ intersect the line $BC$ in $X$ different from $B$; and the circumcircle of triangle $AGC$ intersect the line $BC$ in $Y$ different from $C$. Prove that $G$ is the centroid of triangle $AXY$. I successfully solves this problem by assuming that $X$ lies between points $D$ and $B$ and $Y$ lies between points $D$ and $C$, Where $D$ is the foot of median from $A$ to $BC$. Is it necessarily true that $X$ lies Between $B$ and $D$?
My Thoughts: X cant't coincide with D because a circle can't pass through 3 colinear points (which can be proved easily). I think x can't lie beyond D (I can't prove this right now but I think convexity may be involved).
Pretty simple to brute-force:
$$ DG\cdot DA = DX\cdot DB = DX\cdot \frac{a}{2} = \frac{m_a^2}{3} = \frac{2b^2+2c^2-a^2}{12} \tag{1}$$ $$ DX = \frac{2b^2+2c^2-a^2}{6a} = DY \tag{2}$$ and by trilinear/baricentric coordinates $X,G$ and the midpoint of $AY$ are collinear. Similarly, $Y,G$ and the midpoint of $AX$ are collinear, proving the claim. In a simpler way: $AD$ is a median of $AXY$ and $\frac{AG}{GD}=2$, hence $G$ is the centroid of $AXY$.