$\newcommand\de\delta\newcommand\R{\mathbb R}$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, $$x\in C\iff \forall x^*\in\R^2\ x\cdot x^*\le\de^*_C(x^*).$$ 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. $\quad\Box$

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

$\newcommand\de\delta\newcommand\R{\mathbb R}$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 $$\de^*_C(x^*):=\sup\{x\cdot x^*\colon x\in C\}$$ for $x^*\in\R^2$ and $\cdot$ is the dot product.

By (say) Theorem 13.1, $$x\in C\iff \forall x^*\in\R^2\ x\cdot x^*\le\de^*_C(x^*).$$ 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. $\quad\Box$

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