The concurrency property of the diagonals (semidiagonals) of bicentric polygons. After answering a recent question where the collinearity of the points $X,O,I$ was proven in an elementary way, I rather soon came to the conclusion that the intersection point $X$ of the diagonals is completely fixed by the positions ($O,I$) and the radii ($R,r$) of the circumscribed and inscribed circles (i.e. the product $\sin\alpha\sin\beta$ in the cited answer is a constant for given $R,r$). After some algebra I obtained the following simple formula:
$$
XI=\frac rR\sqrt{\frac{R^2+d^2-2r^2}2},
$$
where $d$ is the distance between the centers of the circumscribed and inscribed circles $OI$ (given $R$ and $r$, it is fixed by the Poncelet porism).
To my surprise I did not find any mention of the above formula in the online sources. But during the search I realized that the formula follows from a much more general statement cited in the above reference:

For an even-sided [bicentric] polygon, the diagonals are concurrent at the limiting point of the two circles, whereas for an odd-sided polygon, the lines connecting the vertices to the opposite points of tangency are concurrent at the limiting point.

I am looking for a (possibly simple) proof of the above statement and/or reference to the original publication on this result.
 A: In 2020 I gave a partial, somewhat hand waving answer to this question . It's further on under the heading 'Original Answer'

Updated Answer
Even-sided polygons
For the case of even sided bicentric polygons there are some proofs in the literature.

*

*Lachlan, Modern Pure Geometry, 1893, pg 217, Ex. 4.  This is given as an exercise at the end of a chapter on coaxal circles that culminates in Poncelet's General Theorem.


*Halbeisen and Norbert, A Simple Proof of Poncelet’s Theorem (on the occasion of its bicentennial), 2014, Theorems 4.1,4.2. In my opinion it's not as simple as the title suggests.
Odd-sided polygons
The claim that the lines connecting the vertices to the opposite points of tangency are concurrent at the limiting point is incorrect.  For triangles the lines are concurrent at the Gergonne point, which is distinct (albeit close to) the limiting point of the incircle and circumcircle.  For pentagons, the lines are generally not concurrent.

Original Answer
The 4 vertex case isn't too difficult, and it suggests ways to attack the general case.

In the above diagram, $ABCD$ is a bicentric quadrilateral.  Points $K,L,M,N$ are the touch points with the incircle.  The diagonals of $ABCD$ and $KLMN$ meet at $X$. (see Yiu, Euclidean Geometry Notes, pg 157).
We want to show that $X$ is a limit point of the two circles.

To do this we construct the polars of $X$ for the two circles. (see Weisstein, Polar )  Build the complete quadrilateral $ABCDEF$.  Regarding the sides of $ABCD$ as tangents to the incircle, we get the construction of the polar $EF$ of $X$ with respect to the incircle.  Regarding $A,B,C,D$ as the points where two lines through $X$ cut the circumcircle, we get the polar $EF$ of $X$ with respect to the circumcircle.  Evidently the two polars are the same, which implies that both circles invert $X$ to the same point $X'$ (which lies on the polar). So $X$ is a limit point of the two circles.
Assume $n$ is even. For the case of the general $n$-sided bicentric polygon, if we assume that all principle diagonals are concentric at a point $X$, we can use a similar argument to show that $X$ is a limit point.  This of course is just a partial result, because it remains to prove that the diagonals are concurrent.
Some further empirical observations and speculations. The setup for the quadrilateral case suggests a construction for bicentric-adjacent polygons (they are tangential, but not necessarily cyclic) that may be a useful avenue to a proof.  Start with a circle $C$ (the incircle) and a line $p$(the polar).  For even $n$ place $\frac{n}{2}$ points $P_i$ on $p$ and draw the $n$ tangents from these points to $C$. Then the intersections of adjacent tangents form a tangential polygon $P$ with the property that the principal diagonals are concurrent.  But $P$ will generally not be cyclic.  For example, when $n=4$ the polygon $P$ will be cyclic only if $\angle{P_1IP_2}$, where $I$ is the center of $C$, is a right angle.  For general $n$ it remains to show that certain configurations of $P_i$ result in cyclic $P$, and that for a given combination of incircle and circumcircle if one bicentric polygon has concurrent diagonals they all do.
I've ignored the case $n$ odd.  With any luck it follows from $n$ even.
