On centrally symmetric convex figures on the hyperbolic plane

Question

A planar region C such that there is an interior point that bisects all chords of C that passes through it may be termed centrally symmetric. It appears that such figures exist in non-Euclidean geometry as well.

Question: Are these claims (shown to be valid in Euclidean plane at A claim on the concurrency of area bisectors of planar convex regions) meaningful and valid in the hyperbolic plane (assumption: area bisector and perimeter bisector are definitions that go over to hyperbolic geometry)?

• A planar convex region is centrally symmetric if and only if its area bisectors are all concurrent.

• A planar convex region is centrally symmetric if and only if its perimeter bisectors are all concurrent.

Answer 1

"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.