Consider an acute triangle $XYZ$ with incenter $I$. Let the incircle touch $YZ, ZX, XY$ at $A, B, C$ respectively. Label the incenters of the circles inscribed in $XCIB$ and $YCIA$ as $P$ and $Q$. (This is possible since $CI$+$XB$ = $CX$+$BI$). Let $PQ$ meet $XY$ at $T$. Let $TI$ meet $CZ$ at $R$. Prove that $ZBRIA$ is cyclic.
Diagram:
Lemma 1. The three points $T, \, A$ and $B$ are collinear.
Proof: Observe that point $P$ lies on $XI$ and point $Q$ lies on $YI$. By Menelaus' Theorem applied to triangle $XYI$ and line $PQ$ $$\frac{YT}{TX} \cdot \frac{XP}{PI} \cdot\frac{IQ}{QY} = 1$$ or in other words $$\frac{YT}{TX} = \frac{PI}{XP} \cdot\frac{QY}{IQ}$$ However, $CP$ is an angle bisector at vertex $C$ in triangle $CIX$ so $$\frac{PI}{XP} = \frac{CI}{XC}$$ as well as $CQ$ is an angle bisector at vertex $C$ in triangle $CIY$ so $$\frac{QY}{IQ} = \frac{CY}{CI}$$ Therefore, after combining all of these facts $$\frac{YT}{TX} = \frac{CY}{XC}$$
Next, we apply the converse direction of Menelaus' Theorem to the triangle $XYZ$ and the three points $T,A, B$. First form the ratio $$\frac{YT}{TX} \cdot \frac{XB}{BZ} \cdot\frac{ZA}{AY}$$ in which $BZ = ZA$, also $AY=CY$ and $XB=XC$ so $$\frac{YT}{TX} \cdot \frac{XB}{BZ} \cdot\frac{ZA}{AY} = \frac{YT}{TX} \cdot \frac{XC}{ZA} \cdot\frac{ZA}{CY} = \frac{YT}{TX} \cdot \frac{XC}{CY} = \frac{CY}{XC} \cdot \frac{XC}{CY} = 1$$ which is possible exactly when the three points $T, A$ and $B$ are collinear.
Let $CZ$ intersect the incircle $k_I$ at point $D$.
Lemma 2. Line $TD$ is tangent to the incircle $k_I$ at the point $D$.
Proof: Let $k_T$ be the circle centered at point $T$ and of radius $TC$. Let $k_Z$ be the circle centered at point $Z$ and of radius $ZA = ZB$. Then by construction, because $TC$ is tangent to $k_I$ at point $C$, circle $k_I$ is orthogonal to $k_T$ and therefore is mapped to itself under inversion with respect to $k_T$. Since the points $T, A$ and $B$ are collinear by Lemma 1, point $A$ is mapped to point $B$ under inversion with respect to $k_T$. Consequently the circle $k_Z$, which by construction passes through both $A$ and $B$, is orthogonal to circle $k_T$. Moreover, since $\angle \, IAZ = 90^{\circ}$ circle $k_Z$ is also orthogonal to the incircle $k_I$. Since the circle $k_Z$ is orthogonal to both $k_I$ and $k_T$ its center $Z$ lies on the radical axis of $k_I$ and $k_T$, which means that $CZ$ is the radical axis in question. However, by construction, point $D$ is the intersection point of the radical axis $CZ$ and circle $k_I$. Therefore point $D$ is the second intersection point of $k_T$ and $k_I$, apart from $C$. Consequently, as circles $k_T$ and $k_I$ are orthogonal, $\angle \, TDI = 90^{\circ}$ and thus $TD$ is tangent to $k_I$ at point $D$.
Remark: Lemma 2 can also be proved by using some results from projective geometry, something like a version of Brianchon's theorem or something like poles and polars.
Completing the proof: As already proved in Lemma 2, $TC$ and $TD$ are the two tangents from point $T$ to circle $k_I$, where $D$ lies on line $CZ$. Thus, one concludes that $TI$ is orthogonal to $CD$ and thus to $CZ$. Recall that $R$ is the intersection point of $CZ$ and $TI$. Therefore, angle $\angle \, IRZ = \angle \, IBZ = \angle \, IAZ = 90^{\circ}$ so the points $I, R, B, Z$ and $A$ lie on a common circle.
Remark: This proof has variations and one can modify some arguments in alternative way, but this is the essence.