Construction of a Conic from the 5 points A,B',C,A',B via (the converse of)
Pascal's Theorem.
Let x be an arbitrary line through B, then we're looking for the locus of the
points C', where C' is a point on the conic, and is the 6th point of the hexagon
inscribed in the conic. The line x is thus the line BC'.
Find A'' as the intersection of BC' (x) and B'C. FInd C'' as the intersection of
AB' and B'A. The pascal line then goes through A'' and C''. B'' is the
intersection of A'C and the pascal line. Finally, C' is the intersection of
BC' (x) and AB''.
In this Cinderella construction, the points A,B',C,A',B are free. The red
conic is the locus of point C' as the orange line x rotates.
This construction is described in "Introduction to Projective Geometry" by
C. R. Wylie, Jr. (Dover, 2008).
-- David Bakin
