This is my short proof from 1963:
In the triangle ABC draw medians BE, and CF, meeting at point G.
Construct a line from A through G, such that it intersects BC at point D.
We are required to prove that D bisects BC, therefore AD is a median, hence medians are concurrent at G (the centroid).
Proof:
Produce AD to a point P below triangle ABC, such that AG = GP.
Construct lines BP and PC.
Since AF = FB, and AG = GP, FG is parallel to BP. (Euclid)
Similarly, since AE = EC, and AG = GP, GE is parallel to PC
Thus BPCG is a parallelogram.
Since diagonals of a parallelogram bisect one another (Euclid),
therefore BD = DC.
Thus AD is a median. QED
Corollary: GD = AD/3.
Proof:
Since AG = GP and GD = GP/2, AG = 2GD.
AD = (AG + GD) = (2GD + GD) = 3GD.
Hence GD = AD/3. QED
$\color{red}{\text{Method 1}}$: All three medians of a triangle pass through the centroid of the triangle hence the centroid of $\triangle OAB$ having vertices $O(0, 0)$, $A(0, 6)$ & $B(6, 0)$ is given as $$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\equiv\left(\frac{0+0+6}{3}, \frac{0+6+0}{3}\right)\equiv(2, 2)$$
Thus the given point $(2, 2)$ (coincident with the centroid) lies on all three medians of $\triangle OAB$.
$\color{red}{\text{Method 2}}$: Calculating the mid-points of all the sides of $\triangle OAB$ & finding the equations of all three medians of $\triangle OAB$ as follows
$\color{blue}{\text{Median 1}}$: drawn from the vertex $A(0, 6)$ to the mid-point of side $OB$ i.e. $\left(\frac{0+6}{2}, \frac{0+0}{2}\right) $ or $(3, 0) $ given as $$y-6=\frac{0-6}{3-0}(x-0)$$ $$\implies \color{blue}{2x+y-6=0}$$
Substituting the coordinates of the given point $(2, 2)$ in the above equation of the median, as follows $$2(2)+2-6=0$$ $$\implies 0=0$$ Hence, point $(2, 2)$ lies on the median through the vertex $A$. Similarly,
$\color{blue}{\text{Median 2}}$: drawn from the vertex $B(6, 0)$ to the mid-point of side $OA$ i.e. $\left(\frac{0+0}{2}, \frac{0+6}{2}\right) $ or $(0, 3) $ given as $$y-0=\frac{3-0}{0-6}(x-6)$$ $$\implies \color{blue}{x+2y-6=0}$$
Substituting the coordinates of the given point $(2, 2)$ in the above equation of the meridian, as follows $$2+2(2)-6=0$$ $$\implies 0=0$$ Hence, point $(2, 2)$ lies on the median through the vertex $B$.
$\color{blue}{\text{Median 3}}$: drawn from the vertex $O(0, 0)$ to the mid-point of side $AB$ i.e. $\left(\frac{0+6}{2}, \frac{6+0}{2}\right) $ or $(3, 3) $ given as $$y-0=\frac{3-0}{3-0}(x-0)$$ $$\implies \color{blue}{y=x}$$
Substituting the coordinates of the given point $(2, 2)$ in the above equation of the median, as follows $$2=2$$ Hence, point $(2, 2)$ lies on the median through the vertex $O$.
Thus the given point $(2, 2)$ lies on all three medians of $\triangle OAB$
Best Answer
The easiest proof I know is using analytic (coordinate) geometry.
Given that CE and AD are divided as given, then $$ \frac23E+\frac13C=F=\frac23D+\frac13A\tag{1} $$ Add $\frac13B$ to both sides and multiply by $\frac32$ and subtract $D+E$: $$ \underbrace{\frac12(C+B)-D}_{\text{parallel to $CB$}}=\underbrace{\frac12(A+B)-E}_{\text{parallel to $AB$}}\tag{2} $$ Since $CB\not\parallel AB$, both sides of $(2)$ must be $0$. That is $$ D=\frac{C+B}2\quad\text{and}\quad E=\frac{A+B}2\tag{3} $$
Additional Problem
At the beginning of your question, it is given that the center of mass divides each median into two segments whose lengths have a ratio of $2:1$. There are only two points that divide a given line segment into lengths with a ratio of $2:1$. By what was given at the beginning of the question, one of those points must be the center of mass, and the other must be halfway between the center of mass and one of the vertices.