Collinearity in bicentric polygons Can you provide a proofs for the following two claims?

Claim 1. The circumcenter, the incenter, and the intersection of the principal diagonals in a bicentric even-sided polygon are collinear.


Claim 2. The circumcenter, the incenter, and the incenter of the polygon formed by long diagonals in a bicentric odd-sided polygon  are collinear.

Picture for the case when the polygon is pentagon:

Picture for the case when the polygon is hexagon:

GeoGebra applets for: quadrilateral, pentagon,hexagon,heptagon and octagon.
The case where the polygon is a quadrilateral is already known in the literature and the case where the polygon is a pentagon has been discussed here.
 A: The short answer is that the circles are part of a coaxal system (also known as Apollonian circles) and therefore their centers are collinear.
For the odd sided case, the circle (J) is coaxal with (O) and (I).  For the even sided case the intersection of the diagonals is a zero-radius circle (J) that is coaxal with (O) and (I).  It is also known as a limiting point (or limit point) of the coaxal system.
This leads to two questions:

*

*why do the long diagonals in the odd-sided case describe a coaxal circle?

*why are the principal diagonals in the even-sided case concurrent at a limiting point of the coaxal system generated by (O) and (I)?

The answers are related to Poncelet's General Theorem.  Rather than prove those assertions here I will give some references and try to contextualize them.
Poncelet's Porism asserts that if there is a bicentric polygon with a given incircle and circumcircle, there are an infinite number of such polygons.  But the General Theorem says the same thing about a set of coaxal circles - that if there is a polygon whose vertices are on one of these circles such that the sides are all  tangent to members of the coaxal family, then there are an infinite number.  This can also be generalized to pencils of conics, and even dualized.
There are plenty of survey articles on this, but for the purposes of your question I recommend Lachlan, Modern Pure Geometry (published 1893), Chapter XIII.  This is a chapter on coaxal circles that works its way up to $\S 341$ Poncelet's General Theorem (PGC).

PGC: If $A_1, A_2,\dots A_n$ be any number of points taken in order on a
circle of a given coaxal system, so that
$A_1A_2, A_2A_3,\dots
A_{n-1}A_n$ touch respectively $(n — 1)$ fixed circles
$X_1,X_2,\dots X_{n-1}$ of the system, then $A_nA_1$ must touch a fixed
circle, $X_n$, of the system [...]

For odd-sided bicentric polygons:
Consider the $(n-1)$-gon $A_1A_2A_3\dots A_{n-1}$.  The first $n-2$ edges touch the incircle, so by PGC the last edge $A_{n-1}A_1$ must touch a fixed circle of the coaxal system generated by the circumcircle and incircle, thus showing Claim 2 because the centers of circles in a coaxal system are collinear.
For even-sided bicentric polygons:
$\S 335$(pg 210-211) demonstrates the following theorem:

Theorem 1: If any four points be taken on a circle of a given coaxal system, so
that one pair of opposite connectors of the tetrastigm formed by them
intersect in a limiting point of the system, the other pairs of
opposite connectors will each touch two circles of the system.

(Connector is a general term that describes both edges and diagonals. Tetrastigm is Lachlan's term for a complete quadrangle.)
The converse to Theorem 1 is given in Exercise 4 on pg 212:

Theorem 2: If $ABCD$ be a tetrastigm inscribed in a circle of a coaxal system,
and if $AB, CD$ touch one circle of the system at the points $Q, Q'$,
and $AD, BC$ another circle of the system at $R, R'$, [...] the
connectors AC, BD will intersect in a limiting point of the system,
provided that $Q, Q', R,R'$ are not collinear.

Given an even-sided bicentric $2n$-gon, choose two consecutive diagonals and the enclosing quadrangle, say $A_1A_2A_{n+1}A_{n+2}$.  One pair of opposite edges are polygon edges and touch $(I)$, and by an argument similar to the one in the odd-sided case, the other two edges are polygon connectors touching some circle coaxal with $(I)$ and $(O)$.  By Theorem 2, the diagonals intersect at a limiting point of the coaxal system generated by the $(I)$ and $(O)$, thus showing Claim 1.
(Full disclosure: I have not tried to prove Theorem 2.  I assume it should follow from Theorem 1 and some of the other exercises.)
