Locus of centroid given base fixed and vertex moves along a line. Consider triangle $ABC$ with base $AB$ fixed in length and position. As the vertex $C$ moves on a straight line (which hasn't to be parallel to line $AB$),what is the locus of the centroid of the triangle ?
My effort:
After a little bit of playing i see that the locus of the centroid is a line parallel to the base $AB$,but how would i prove it  with means of synthetic geometry .
 A: $$ G = \frac{A+B+C}{3} = \frac{A+B}{3}+\frac{C}{3}, $$
hence if $A,B$ stay fixed and $C$ moves on some line $l$, $G$ moves on a line parallel to $l$.
A: There is another way to think about this question.
Let $E$ be the midpoint of side $BC$.
Using Ceva's theorem on $\triangle CDB$ and line $AXE$,
$\frac{CX}{XD}\frac{DA}{AB}\frac{BE}{EC}=1$
$\frac{CX}{XD}=\frac{2}{1}$ (which is a well-known result concerning medians)
Another way this can be stated is: $\frac{CD}{XD}=\frac{3}{1}$
It follows that a scaling of factor $\frac{1}{3}$ with centre $D$ will bring the locus of $C$ to the locus of $X$.
Hence the locus of $X$ is a straight line parallel to the locus of $C$ where the distance from the locus of $X$ to $D$ is $\frac{1}{3}$ the distance from the locus of $C$ to $D$.
A: Geometrical Approach: 
Let the fixed vertices of $\triangle ABC$ be $A(0, 0)$ (at the origin) & $B(a, 0)$ be lying on the x-axis & the vertex $C(x, b)$ is movable parallel to the base $AB$ say on the line $y=b$ 
where, $x$ is variable & $a, b$ are arbitrary constants. 
then the centroid $G(h, k)$  of $\triangle ABC$ is given as $$(h, k)\equiv\left(\frac{0+a+x}{3}, \frac{0+0+b}{3}\right)\equiv \left(\frac{x+a}{3},\ \frac{b}{3}\right)$$ 
Comparing the corresponding coordinates, we get $h=\frac{x+a}{3}$  & $k=\frac{b}{3}$ which shows that the locus of the centroid $G(h, k)$ is a straight line $\color{red}{y=\frac{b}{3}}$ which is parallel to the fixed base $AB$ i.e. the centriod $G$ moves on a straight line parallel to the fixed base $AB$ 
