Suppose we have a triangle $abc$ and point $x$ on $bc$. I think the barycentre $B = (a+b+c)/3$ should be about $\|x-a\|/3$ distance away from $x$.
This is not true because measuring the blue segment says its about 3.5 times the length of the red segment.
Is there any constant $C>0$ such that $|xB| > C |xa|$ for all triangles?
This might be completely obvious but I'm just not seeing it right now.
Let $|ac|=|bc|=1$, $\angle acb=\pi-\epsilon$ and $|cx|=\epsilon$. As $\epsilon\rightarrow 0$, you can easily show $C$ should go to $0$.
(you can prove it using $|xB|>C|xa|\Rightarrow 2\epsilon\ge|xc|+|Bc|\ge|xB|>C|xa|\ge C(|ca|-|cx|)=C(1-\epsilon)$)