Suppose we have triangle $ABC$ such that $\angle B$'s external angle bisector intersects $AC$'s extension at $P$ and $\angle A$'s external angle bisector intersects $BC$'s extension at $Q.$ Then, let the intersection of $BP$ and $AQ$ be $R.$ If the circumcircle of $\triangle PQC$ passes through $R$ and $PQ = 1,$ find the value of $PR + RQ.$
Due to the fact that $RPQC$ is cyclic, I got $$\angle RQP = \angle ACR = \angle RCB = \angle RPQ.$$ Therefore, I found that $PR = RQ.$ However, I was unsure how to proceed from here. Can somebody help please?
Here is a picture:
Since you have shown that $RQ = RP$, we just need to find $\angle QRP$ (the green angle).
All same-colored angles are equal since they relate to the external angle bisectors. Now consider:
\begin{align} \angle QRP &= \angle QCP\\ \text{In }\triangle ABC, \ \angle ACQ &= (180^\circ - 2\angle ABR) + (180^\circ - 2\angle BAR)\\ \text{In }\triangle ABR, \ \angle ARP &= \angle ABR + \angle BAR \end{align}
This gives $\angle QRP = \angle ABR + \angle BAR = 120^\circ$, and the rest of the question follows.