First, we will show that $\angle C_1B_1A_1 = 90^\circ$. Let $K \in BC$ so that $B \in (KA_1)$, then $\angle ABK = 60^\circ$. Point $C_1$ is on the bisector of $\angle ACB$ and this implies that $d(C_1, BC) = d(C_1, AC)$ or $C_1F_1 = C_1F_3$, where $F_1$ is the projection of $C_1$ on $BC$ and $F_3$ is the projection of $C_1$ on $AC$. Segment $BA$ is the bisector of $\angle KBB_1$ implies that $d(C_1, KB) = d(C_1, BB_1)$ or $C_1F_1 = C_1F_2$, where $F_2$ is the projection of $C_1$ on $BB_1$. So, $C_1F_2 = C_1F_3$ and $C_1B_1$ is the bisector of $\angle BB_1A$. Let us denote $\angle BB_1C_1 = \alpha$. Likewise, we prove that $B_1A_1$ is the bisector of $\angle BB_1C$. Let $\angle BB_1A_1 = \angle CB_1A_1 = \beta$. Then, from $\angle AB_1C = 180^\circ$ we have $2\alpha + 2\beta = 180^\circ$ and $\alpha + \beta = 90^\circ$.

Let $r_1$ be the radius of inscribed circle to $\triangle A_1B_1C_1$, $r_2$ the radius of inscribed circle to $\triangle C_1B_1F$, and $r_3$ be the radius of inscribed circle to $\triangle A_1B_1F$, respectively. Considering the properties of right triangles, we have
$$\triangle C_1FB_1 \sim \triangle C_1B_1A_1 \Rightarrow \frac{r_2}{r_1} = \frac{B_1C_1}{C_1A_1} = \cos C_1$$
from which follows $r_2 = r_1 \cos C_1 = r_1 \sin A_1$. Likewise,
$$\frac{r_3}{r_1} = \frac{A_1 B_1}{C_1 A_1} = \cos A_1 \Rightarrow r_3 = r_1 \cos A_1 = r_1 \sin C_1$$
Now, we will see that $I_1R \parallel A_1B_1$ and $I_1S \parallel C_1B_1$, where $I_1$ is the projection of $I$ on $C_1A_1$. Let $I_1R_2 \perp C_1B_1$, $R_2 \in (C_1B_1)$ and $I_1R_2 \cap C_1I = \{R^*\}$. On account that $\triangle C_1II_1 \sim \triangle C_1R^*R_2$, then
$$\frac{R^*R_2}{II_1} = \frac{C_1R_2}{C_1I_1} = \cos C_1 \Rightarrow R^*R_2 = r_1 \cos C_1 = r_2$$
from which follows $R^* = R, I_1R \perp C_1B_1 \Rightarrow I_1R \parallel A_1B_1$. Likewise, we get $I_1S \parallel C_1B_1$.

In triangle $I_1RR_1$ we have $r_2 = I_1R \sin A_1 = r_1 \cos C_1 = r_1 \sin A_1$ and $I_1R = r_1$. In triangle $I_1SS_1$ we have $r_3 = I_1S \sin C_1 = r_1 \cos A_1 = r_1 \sin C_1$ from which follows $I_1S = r_1$. Finally, we get $I_1R = II_1 = I_1S = r_1$. Since $\triangle QSI_1 \sim \triangle QB_1C_1$, then
$$\frac{SI_1}{B_1C_1} = \frac{I_1Q}{C_1Q} \Rightarrow SI_1 = I_1Q$$
on account that $B_1C_1 = C_1Q$. Now, we can conclude that points $R, I, S$, and $Q$ lie on the same circle. $\square$