proving a median in a triangle base on center of mass I was wondering: I know that the center of mass in a triangle is divided the medians in the ratio of $2:1$.
Is the opposite is true? 
I mean, if i have triangle $ABC$ and point $D$ is on $BC$, point $E$ is on $AB$ such that the point of intersection, divided $AD,CE$ in the ratio of $2:1$.
Can i say that $AD,CE$ are medians?
In Addition: what about if i have one median and a point on that median that dividing it by the ration of $2:1$, is it true to say that this point will definitely be the center of mass?
Thanks.  
 A: 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.

A: Consider the triangle ABC as shown with $AG:GD=2$ and $BG:GE=2$.

Now applying Menelaus' theorem in triangles $\Delta ADC$ and $\Delta BEC$ you will obtain the following relations:
$$\frac{BD\cdot EC}{BC\cdot AE}=\frac{1}{2}$$
$$\frac{AE\cdot DC}{AC\cdot DB}=\frac{1}{2}$$
Now take $\frac{EC}{EA}=p$ and $\frac{DC}{DB}=q$ and rewrie above relations as :
$$\frac{p}{1+q}=\frac{1}{2}$$
$$\frac{q}{1+p}=\frac{1}{2}$$
Solving for $p$ and $q$ will yield the required result.
ADDITION Draw $\Delta ABC$ and medians $AD$ and $BE$. Let $G$ be the point that divides $AD$ in the ratio $2:1$ and try proving that $BG$ and $BF$ are part of the same line. Doing so we can conclude that $G$ is the point of concurrency of the medians.
Hint: This can be done with many tools, including similarity, ceva's and menelaus' theorem.
A: this is true. 
let $AD$ and $BE$ cut at $G.$  let $GA = 2x, GD = x, GC= 2z, GE = z, GB = w, GF = y$ where extended $BG$ meets the side $AC$ at $F.$  we want to show that $w = 2y.$
we will use ceva's theorem that tells you $\frac{BD}{DC} \frac{CF}{FA} \frac{AE}{EB} = 1$  we will get the relations among the areas of six triangles:
$$\frac{AGE}{CGD} = 1, \frac{EGB}{CFG} = \frac{w}{2y}, \frac{BGD}{AFG} = \frac{w}{2y} = \frac{1}{\alpha} $$ 
$\begin{align}
1 &=\frac{BD}{DC} \frac{CF}{FA} \frac{AE}{EB} \\
&= \frac{BGD+BGE+AEG}{CDG+CGF+AGF}
\frac{CGF+CGD+BGD}{AGF+AGE+BGE} \frac{AGE+AGF+CGF}{BGE+BGD+CGD} \\
&=\frac{BGD+BGE+AEG}{AGE+\alpha EGB+\alpha BGD}
\frac{\alpha EGB+AGE+BGD}{\alpha BGD+AGE+BGE} \frac{AGE+\alpha BGD+\alpha EGB}{BGE+BGD+AGE} \\
&=\frac{\alpha EGB+AGE+BGD}{\alpha BGD+AGE+BGE} \\
\end{align}$
the last equality gives you $\alpha = 1$ and $w = 2y.$
