Let $AB$ meet $CD$ at $G$. Then because $ABCD$ is cyclic, we have a famous property that the Miquel point of $ABCD.EG$ is the projector of $O$ on $FG$ and $E$ is the orthocenter of $FGO$. Then let $GE$ meet $FO$ at $H$. Now we have $\angle O_{1}EB+\angle EDC = 90^{o}-\angle EAB+\angle EDC = 90^{o}$ so $O_1E \perp DC$ then $O_1E \parallel OO_2$ and $OO_1 \parallel O_2E$, hence $O_1EO_2O$ is a parallelogram, therefore $H$ lie on $(OO_1O_2)$. Now it is trivial that $\overline{FH} \cdot \overline {FO} = \overline{FI} \cdot \overline{FG} = \overline{FB} \cdot \overline{FC}$. Because $PQ$ is the radical axis of $(OO_1O_2)$ and $(O)$, we hence the conclusion.

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