Triangle inscribed in a circle,2 points fixed and 1 moving. The track of centroid makes a circle but how do I prove it without cartesian coordinate? Triangle ABC and circle O. A and B are fixed, but C is moving on the circle.
So I have triangle ABC and circle O. A and B are fixed on the circle, but C is moving around the circle. Let G is the centroid of ABC, G' is the centroid of OAB, and $r$ is the radius of O. Then the track of G makes a circle, and its center is G' and radius is $\frac{r}{3}$.
It is easy to prove with Cartesian coordinate. Let O($0,0$), A($a_x,a_y$), B($b_x,b_y$), C($c_x,c_y$), G($g_x,g_y$). Then $$a_x^2+a_y^2=r^2$$ $$b_x^2+b_y^2=r^2$$ $$c_x^2+c_y^2=r^2$$ 
Since G is the centroid of ABC, $$g_x=\frac{a_x+b_x+c_x}{3}\quad  \therefore c_x=3g_x-a_x-b_x$$ $$g_y=\frac{a_y+b_y+c_y}{3}\quad \therefore c_y=3g_y-a_y-b_y$$Then $$c_x^2+c_y^2=(3g_x-a_x-b_x)^2+(3g_y-a_y-b_y)^2=r^2$$ $$(g_x-\frac{a_x+b_x}{3})^2+(g_y-\frac{a_y+b_y}{3})^2=(\frac{r}{3})^2$$ so G$(g_x,g_y)$ makes a circle, center of which is $(\frac{a_x+b_x}{3},\frac{a_y+b_y}{3})$ and radius $\frac{r}{3}$. Also, $(\frac{a_x+b_x}{3},\frac{a_y+b_y}{3})$ is the centroid of triangle OAB.
But there must be a way that proves this without cartesian coordinate but with pure geometry. Problem is, I know little of geometry and can't find the way. Could you enlighten me and show me the way?

This is not too hard to see using ordinary geometry.
Let $A$, $B$ be fixed points on circle with center $O$, and $C$ any other point on the circle.
In triangle $ABC$, bisecting $CB$, $AB$ at $E$, $F$, and joining $AE$, $CF$, then $G$ is the centroid of $\triangle ABC$. 
In fixed triangle $AOB$, bisect $AO$ at $D$, and join $BD$, $OF$, giving $K$ the centroid of $\triangle AOB$. Finally, join $GK$.

Assuming as well known, that the centroid divides the median lines of a triangle in a $\frac{2}{1}$ ratio, then$$\frac{CG}{GF}=\frac{OK}{KF}=\frac{2}{1}$$Hence$$\frac{CF}{GF}=\frac{OF}{KF}=\frac{3}{1}$$Therefore$$GK\parallel CO$$whence$$\triangle CFO\sim\triangle GFK$$and$$\frac{CO}{GK}=\frac{3}{1}$$And since $CO$ has a fixed length for all positions of $C$, so does $GK$. 
Therefore, since $K$ is fixed in position, $G$ lies always on the circumference of a circle centered on $K$.   
