We add a little bit to [On 'fair bisectors' of planar convex regions](https://mathoverflow.net/questions/356279/on-fair-bisectors-of-planar-convex-regions) and [Bisectors and partitioning lines for convex regions defined with respect to the moment of inertia](https://mathoverflow.net/questions/437053/bisectors-and-partitioning-lines-for-convex-regions-defined-with-respect-to-the)


**Definitions:** Given a planar convex region C (could be smooth or polygonal), an area bisector of C
is any line that partitions C into 2 pieces of equal area. A perimeter bisector is a line that partitions C into 2 pieces of equal perimeter. Obviously, thru every point on the boundary of C we can draw an area bisector and a perimeter bisector.


**Question:** Are the following claims easy to prove/counter?


* 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.


Note: Higher dimensional analogs of these claims are easy to state.



A claim on the concurrency of area bisectors of planar convex regions
* A planar convex region is centrally symmetric if and only if its area bisectors are all concurrent.


Yes. Let all area bisectors of our region $K$ pass through a point $O$. Note that any line $\ell$ through $O$ is an area bisector of $K$: otherwise there exists a parallel area bisector (by continuity), and it does not pass through $O$. Assume that $K$ is not $O$-symmetric. It yields that some chord $AB$ through $O$ is not bisected by $O$. Move it slightly to get a chord $A\_1B\_1$ ($A\_1$ close to $A$). The curvilinear triangles $AOA\_1$ and $BOB\_1$ have unequal area, thus at least one of chords $AB$, $A\_1B\_1$ is not an area bisector.


