$\newcommand\de\delta\newcommand\R{\mathbb R}$As in the previous answer, assume that $C$ is a compact convex subset of $\R^2$. The condition that all the width bisectors of $C$ are concurrent means that the support function $\de^*_C$ of $C$ is even, where 
$$\de^*_C:=\sup\{x\cdot x^*\colon x\in C\}$$
and $\cdot$ is the dot product. 

By (say) [Theorem 13.1][1], 
$$x\in C\iff \forall x^*\in\R^2\ x\cdot x^*\le\de^*_C(x^*).$$
So, for any $x\in C$ and any $x^*\in\R^2$ we have 
$$(-x)\cdot x^*=x\cdot(-x^*)\le\de^*_C(-x^*)=\de^*_C(x^*),$$
so that $-x\in C$. Thus, $C$ is centrally symmetric. $\quad\Box$ 

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Again, almost the same reasoning proves the natural generalization of this "planar" result to the corresponding one for any dimension.



  [1]: https://press.princeton.edu/books/paperback/9780691015866/convex-analysis