Let $ABCDEF$ be a cyclic hexagon with $AB=CD=EF$. Let $AC\cap BD=P, CE\cap DF=Q, EA\cap FB=R$. Prove that $\triangle PQR\sim\triangle BDF$.
This problem seems simple, but I'm having trouble figuring out how to use the equal side length condition. All I figured out was that $\triangle APD$ is isosceles with $AP=PD$, since $AB=CD\implies \angle PAD=\angle PDA $. It follows that $\triangle FCQ, \triangle ERB$ are isosceles with $FQ=QC,ER=RB$ respectively. Does anyone have any ideas about how to prove the similarity? Any observations are appreciated!
The equal length is used to perform angle equalities.
First since arc $CD=EF$ we have arc $CE=DF$ and hence $\angle CAE=\angle DBF$. Or in other words $\angle PBR=\angle PAR$. From this we can conclude $ABPR$ are cocylic. Similarly $EFRQ$ are cocyclic as well.
Now $\angle PRQ=180^{\circ}-\angle ARP-\angle QRE=\angle ABP-\angle QRE=\angle ABP-\angle QFE= arc(AED)-arc(ED)=arc(AE)=arc(BF)=\angle BDF$.
Similarly for the other two angles and the similarity between triangles $BDF$ and $QRP$ are proven by angle equality.