Inscribed Quadrilateral: Collinear Points

Question

$ABCD$ is an inscribed quadrilateral with $O$ being the circumcenter. The $\perp$ raised on $AB$ at $A$ cuts $CD$ at $F$ and the $\perp$ raised on $AD$ at $A$ cuts $BC$ at $E$. Show that $E, O, F$ are collinear.

Answer 1

Let the second point of intersection of $AE$ with the circumcircle be called $P$ and the second point of intersection of $AF$ with the circumcircle be called $Q$ (the first being $A$ for both lines). Then $O \in BQ$ because $\angle \, BAQ = 90^{\circ}$ implies that $BQ$ must be a diameter. Analogously, $O \in DP$ because $\angle \, DAP = 90^{\circ}$ implies that $DP$ must be a diameter. Consequently, $O = BQ \cap DP$.

Apply Pascal's theorem to the non-convex inscribed hexagon $BCDPAQ$. Then the intersection points of the opposite edges of the hexagon, $E = BC \cap AP, \,$ $F = CD \cap AQ$ and $O = BQ \cap DP$ must be collinear.