I am not good at presenting proofs properly but here is my logic.
Let X be the centr of the circle.
Equal sectors cut of equal segments so the perpendicular heights from the centre of each segment chord to the circumference is going to be equal.
AX=BX=CX=DX=EX=FX all equal radii
So
\(\triangle ABX \cong \triangle CDX \cong\triangle EFX\)
and all these triangles are isoscles with AB, CD, and EF respectively being the base sides.
The perpendicular bisector of the base sides will pass throught the centre of the circle. They are all altitudes off the base sides.
Since the altitudes are all the same, the circle that passes through these bisector points is concentric with the original circle.
Hence the centre of the incircle is also the centre of the original circle.
This definitely needs to be tidied up but do you understand what I am saying?