Elementary proof for a generalized version of Pascal's hexagon theorem? I saw a generalization version of Pascal's theorem on a book, which is actually proved by deep algebraic geometry theorems (e.g., Cayley-Bacharach-Chasles theorem, or Bézout's theorem). I just wonder whether we can prove it by elementary/synthetic methods (e.g., perspective points, or invariance of cross-ratios and properties of second order point-rows)?
Given six points $A,B,C,D,E,F$ on a conic, we take another point $G$ and draw two conics, one (denoted as $a$) constructed from $A,B,G,E,F$ and another (denoted as $b$) from $B,C,D,G,E$. Two conics $a$ and $b$ intersect on four points $B,G,E,H$. Suppose $AF\cap CD=O$.
Prove: $G$, $O$ and $H$ are co-linear.


I don't know about synthetic proof but this theorem is the radical axis theorem, seen from a different projective point of view (I.e. Gaged differently). Extend this configuration from the real projective plane to the complex one. Then by applying a complex projective transformation, send the two points $B$ and $E$ to the two complex points at infinity $[1:i:0]$ and $[1:-i:0]$ respectively. Then the three conics on the picture get transformed into three conics from the pencil defined by $[1:i:0]$ and $[1:-i:0]$ which means they become three circles when we restrict ourselves back to the real projective plane. The three lines determined by pairwise intersecting conics are mapped to the radical axes of the three corresponding pairs of circles. By the radical axis "theorem" the three radical axes are concurrent. 
