Solution:

Because $O$ is the midpoint of $\overline{B C}$, it suffices to show that $R O$ is parallel to $B I$, as the result follows by simply looking at a homothety centered at $C$.

So first of all by looking at the problem we instantly recognise the point $P$ as the circumcenter of $B I C$, as this is a well known configuration. So from this we $|P B|=|P I|$ and $\angle B O P=90^{\circ}$. Next up we find $A Q$ to be parallel to $I R$, as they both are perpendicular to $P Q$. From this we get $\angle B I R=\angle I B A=\angle I B C$. We now have two same angles at $B$ and $I$ and also $|P B|=|P I|$. This gives us motivation to look at perpendicular bisector of $\overline{B I}$, as it goes through $P$ and $X:=R I \cap B C$. If we can show that $R O X$ is also isosceles, then it'll instantly follow that $\angle B O R=\angle O B I$, which then will give us $R O$ parallel to $B I$ as wanted. So now for the fun stuff; more angle chasing.

We easily calculate $\angle A P Q=\angle O P B=45^{\circ}$. From this we get that $\angle B P Q=\angle O P I$ and thereby that the angle bisector of $\angle I P B$ and $\angle O P R$ coincide. And as the angle bisector of $\angle I P B$ is equal to the perpendicular bisector of $B I$, we have $\angle X P R=\angle O P R$. Finally using the cyclic quadrilateral $O P R X$ we have $\angle O R X=\angle O P X=\angle X P R=\angle X O R$, which finishes the proof as explained before.

Let $S$ be the intersection of $I C$ and $O R$