I am trying to understand relationship between Median and barycentric coordinates.
here is a figure (fig_1) comes from wiki
that post gives a good explanation and I am trying to connect the notation in that post to fig_1.
Vertex of such small triangle at the first median has B.C.(baricentric coordinates) like
(g,f,f), whereg+f+f=1, at the second median(f,g,f), and(f,f,g) at the third one. We can see that forg=1/3triangle size is zero (coefficientcf= 0), forg=1triangle is equal to large triangle(cf = 0).
in this context, is it correct to consider point E as (g,f,f)?
Yes, a way to understand the situation is as follows. Let $x,y,z$ be the normed barycentric coordinates of $O$, $x+y+z=1$. By definition: $$ O = xA+yB+zC\ . $$ (Understand this either using affix points $A,B,C$, or using a "testing point" $P$ and rewrite with vectors the relation as $PO =xPA+yPB+zPC$.)
These $x,y,z$ are barycentric coordinates, thus coordinates "in the plane", and the argument reduces them, their symmetry to one "on the line". Let us see how. Just rewrite: $$ \begin{aligned} O &= xA+(y+z)E\ ,\text{ where}\\ E &= \frac y{y+z}B+\frac z{y+z}C\ . \end{aligned} $$ (From the first relation, $E$ is on $OA$, from the second relation it is on $BC$, so it is indeed the point $E$ in the picture.)
Since $E$ is the middle of $BC$, we must have the corresponding coordinates "in the line" equal, so $y/(y+z)=z/(y+z)$, so $y=z$.
By symmetry to $y=z$ we also get $x=y$ (and $x=z$), so $x=y=z=1/3$.
Note that already knowing the porportion $OE:AE$ lets us know the $x$.