A claim on the concurrency of area bisectors of planar convex regions

Question

We add a little bit to On 'fair bisectors' of planar convex regions and Bisectors and partitioning lines for convex regions defined with respect to the moment of inertia

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.

Answer 1

• 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.