Two Pascals Merge into One
What is this about?
Points $A, B, C$ lie on a conic, $D$ elsewhere. The conic meets $AD,$ $BD,$ $CD$ the second time in $A',$ $B',$ $C'.$ $N$ is also on the conic. $A'N,$ $B'N,$ $C'N$ cross $BC,$ $AC,$ $AB$ in $A_0,$ $B_0,$ $C_0.$
Points $A_0,$ $B_0,$ $C_0$ and $D$ are collinear.
There is a conic with a good deal of points on it. Some intersection are to be proved collinear. Pascal's Theorem is likely to be involved here.
Apply Pascal's Theorem once to the hexagon $ACC'NB'B$ and then also to $CBB'NAA'.$
The first shows that the points $B_0,$, $C_0,$ and $D$ are collinear; the second that so are the points $A_0,$, $B_0,$ and $D.$
Note that Pascal's theorems applied to the hexagon $BAA'NC'C$ (redundantly) shows that points $A_0,$, $C_0,$ and $D.$