# Two Pascals Merge into One

### What is this about?

### 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

- Pascal's Theorem
- Pascal in Ellipse
- Pascal's Theorem, Homogeneous Coordinates
- Projective Proof of Pascal's Theorem
- Pascal Lines: Steiner and Kirkman Theorems
- Brianchon's theorem
- Brianchon in Ellipse
- The Mirror Property of Altitudes via Pascal's Hexagram
- Pappus' Theorem
- Pencils of Cubics
- Three Tangents, Three Chords in Ellipse
- MacLaurin's Construction of Conics
- Pascal in a Cyclic Quadrilateral
- Parallel Chords
- Parallel Chords in Ellipse
- Construction of Conics from Pascal's Theorem
- Pascal: Necessary and Sufficient
- Diameters and Chords
- Chasing Angles in Pascal's Hexagon
- Two Triangles Inscribed in a Conic
- Two Triangles Inscribed in a Conic - with Solution
- Two Pascals Merge into One
- Surprise: Right Angle in Circle

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