# Two Pascals Merge into One

### Problem

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.

### Hint

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.

### Solution

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 $D.$

### Acknowledgment

This problem has been posted and proved by Dao Thanh Oai at the CutTheKnotMath facebook page. Dao Thanh Oai formulated the problem for a circle, but his proofs applies to any conic without change. ### Pascal and Brianchon Theorems 