48th Austrian Mathematical Olympiad National Competition (Final Round, part 1)

Let $S$ denote the common point of $RP$ and $AE$, see Figure 1. Since we are given a regular pentagon, the angles in triangle $ABE$ are well known as $\angle BAE=108^{\circ }$ and $\angle ABE=\angle AEB=36^{\circ }$. Since $BE$ and $CD$ are parallel, $RP$ is perpendicular to $BE$, and we therefore have $\angle ASP=126^{\circ }$ and $\angle QSP=54^{\circ }=\angle ASR$. From this, $$\angle SPA=54^{\circ }-\angle SAP=\angle PAB-54^{\circ }=\angle PBA-54^{\circ }=126^{\circ }-\angle AQP=126^{\circ }-\angle SQP=\angle SPQ$$ follows, since $ABPQ$ is inscribed. We therefore see that $SP$ (or $RP$) bisects the angle $\angle APQ$, which implies that $AR$ and $QR$ must be of equal length, as claimed.