Claim 1: $PQ=PA$ Proof: By the spiral of similarity on $Q$ that takes $PA$ to $OB. \square$ Let $E:=PQ \cap \omega$ Claim 2: $PE = PJ$ Proof: By the Claim 1 and cyclic $AJEQ$, we have that $\angle JEQ = \angle PAQ = \angle PQA = \angle AJE. \square$ Let $R' := PQ \cap BC $ Claim 3: $BROQ$ is cyclic Proof: By proof of Claim 1: $\angle QPA = \angle QOB$. Since $AP \parallel BR$: $\angle RPA = \angle PRB \implies \angle QOB = \angle QR'B \square$ Claim 4: $Q-M-C$ Proof: $C' := QM \cap \omega$. By the spiral of similarity on $J$ that takes $PE$ to $MC' \implies MJ =MC' \iff C' \equiv C. \square$ Claim 5: $F$ is midpoint of $BC$ Proof: By PoP and Claim 2: $QE=AJ=CJ \implies AE=QJ, QC = JE$. By SSS criterion, $\triangle QJE \cong \triangle AEJ$. Now, we can rotate the figure in $90^{\circ}$ counterclockwise such that $A \mapsto B, B \mapsto C, C \mapsto J, J \mapsto A, Q \mapsto E \implies QC \cap AB \mapsto EJ \cap BC \implies M \mapsto F$, since $M$ is the midpoint $AB \implies F$ is the midpoint of $BC. \square$ By Reim's Theorem on $\odot (BROQ)$ and $\omega$, we have that $OR \parallel JE \implies \frac{1}{2} = \frac{BO}{BJ} = \frac{BR'}{BF} \implies BR' = \frac{BF}{2} = \frac{BC}{4} \implies R' \equiv R. \blacksquare$