Four concyclic points inside bicentric quadrilateral Can you provide a proof for the following proposition:

Proposition. Let quadrilateral $ABCD$ be inscribed into a circle with center $O$ and circumscribed around a circle with center $I$. Let $X$ be a common point of inscribed circle and side $AB$ , $Y$ a common point of inscribed circle and side $BC$ , $Z$ a common point of inscribed circle and side $CD$ , $W$ a common point of inscribed circle and side $DA$ and let $S$ be the intersection point of the line segments $XZ$ and $YW$ . I claim that the incenters of the $\triangle SAB$, $\triangle SBC$, $\triangle SCD$ and $\triangle SDA$ are concyclic.


GeoGebra applet that demonstrates this proposition can be found here.
My attempt:
Let $I_1,I_2,I_3,I_4$ be the incenters of  $\triangle SAB,\triangle SBC,\triangle SCD$ and $\triangle SDA$ , respectively. My idea is to apply Pythagorean theorem on  $\triangle SI_1I_2$, $\triangle SI_2I_3$,$\triangle SI_3I_4$ and $\triangle SI_4I_1$ in order to express lengths of the line segments $I_1I_2,I_2I_3,I_3I_4,I_4I_1$ using lengths of the line segments $SI_1,SI_2,SI_3,SI_4$ and then to apply Ptolomy's theorem on quadrilateral $I_1I_2I_3I_4$ , but I have difficulty proving that points $I_1,I_2,I_3$ and $I_4$ lie on the line segments $XZ$ and $YW$.
 A: We can prove $I_1 \in \overline{SX}$ using a property of bicentric quadrilaterals, $\frac{AX}{XB}=\frac{DZ}{ZC}$. Since $\triangle ASD \sim \triangle BSC$,
$$\frac{AX}{XB}=\frac{AX+DZ}{XB+ZC}=\frac{AD}{BC}=\frac{AS}{BS}$$
so $\overline{SX}$ bisects $\angle ASB$ by the angle bisector theorem. Now by the angle bisector theorem and equal tangents,
$$\frac{SI_1}{I_1X}=\frac{SB}{BX}=\frac{SB}{BY}=\frac{SI_2}{I_2Y}.$$
Analogous reasoning shows
$\frac{SI_1}{I_1X}=\frac{SI_2}{I_2Y}=\frac{SI_3}{I_3Z}=\frac{SI_4}{I_4W}$
so the concyclicity follows from a dilation of the incircle at $S$.
A: This is a well known theorem.  $ABCD$ doesn't have to be bicentric, it only needs to be tangential.
The paper Josefsson, More Characterizations of Tangential Quadrilaterals has a good write up, specifically on the OP problem in Section 4, along with references.
For example, https://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/415.shtml.
And https://cms.math.ca/publications/crux/issue/?volume=25&issue=4 (pg 243-245)
The original conjecture had $ABCD$ cyclic, and it is possible that the proof is easier if that is a condition.
A quick summary of the subtriangles: https://en.wikipedia.org/wiki/Tangential_quadrilateral#Characterizations_in_the_four_subtriangles
Update: See also Theorem 20 of Grinberg, Circumscribed quadrilaterals revisited
