22nd Chinese Girls' Mathematical Olympiad

Proof: Let $EX\parallel AF$, intersecting $AP$ at $X$, and $DP$ at $Y$. Extend $XQ$ and $YQ$ to intersect $BC$ at $S$ and $T$, respectively. From $\frac{PQ}{PE}=\frac{PE}{PF}=\frac{PX}{PA}$, we get $XQ\parallel AE$, similarly $YQ\parallel DE$. Since $\angle EXS=\angle AEX=\angle DAC=\angle EBS$, we have $X,E,S,B$ are concyclic. From $\angle PXE=\angle PAD=\angle PBE$, we find $X,E,P,B$ are concyclic, hence $X,E,S,P,B$ are concyclic. Similarly, $Y,E,T,P,C$ are concyclic. Since $\angle PQT=\angle PEB=\angle PST$, we have $P,S,Q,T$ are concyclic. Given $SQ\parallel CE,TQ\parallel BE$, and $CE\perp BE$, we get $SQ\perp TQ$. Since $\angle RQS=90^{\circ }-\angle QXY=90^{\circ }-\angle QTS=\angle RSQ$, it is known that $R$ is the center of circle $\odot PSQT$, thus $RP=RQ$. $\square$