Indeed, it turns out to be easy Trivial that $(A_1,C_1,R,B_1)$ are concyclic. Let $I$ be the incenter of triangle $ABC$. Then $BI$ and $CI$ perpendicularly bisects $C_1A_1$ and $A_1B_1$ respectively. Hence, $I$ is the circumcenter of $\triangle A_1B_1C_1$. Let $\angle ACB=2\gamma$. $\angle A_1IB_1=2\angle B_1C_1A_1=2\angle A_1RB_1=2\angle AQB_1=2\angle B_1A_1C=180-2\gamma$ $\therefore (I,A_1,B_1,C)$ concyclic$\rightarrow IA_1\perp BC$, $IB_1\perp AC$ Similarly $IC_1\perp AB$, so circle $(A_1,C_1,R,B_1)$ is the incircle of $ABC$. Therefore $R$ lies on the incircle of $ABC$. $\blacksquare$