"Only if" part (for a symmetric region, bisectors are concurrent) is rather clear.
Now "if" part. Assume that the common point of all bisectors exists, denote it by $O$. Clearly every chord $AOB$ through $O$ is a bisector, otherwise we may find a bisector not passing through $O$. Denote by $X$ our region and by $Y$ the region obtained from $X$ by a symmetry with respect to $O$. One of $X, Y$ can not be a proper subset of another, and we should prove $X=Y$. Assume the contrary. Consider the set $\alpha$ of points on the boundary $\gamma$ of $X$ which belong to the boundary of $Y$. They exist and they form a closed symmetric (with respect to $O$) set. Our assumption $\alpha \ne \gamma$ allows to find an inclusion-maximal interval $PQ$ in the complement $\gamma\setminus \alpha$. Consider two sectors, each formed by $PO, QO$ and arcs of the boundaries of $X, Y$ respectively joining $P$ and $Q$. They must have equal area or perimeter, but one is contained in another. A contradiction.