We will apply this concept to Pascal's and Brianchon's theorems, using
in the first case *ABCDEF* for the points of the inscribed hexagon, and
abcdef for the lines of the circumscribed hexagon. For this we will
need to remember that lines lie on conics and can determine conics just
as points do.

In Pascal's theorem, what In Brianchon's theorem, what happens to a line AB as B happens to a point ab as b moves around the conic moves around the conic towards towards A? Gradually, the point a? Gradually, the point line approaches the line approaches the point which is which is tangent at A. the point of tangency on a. When B and A coincide, the When b and a coincide, the line through A and B has point on a and b has only one point in common only one line in common with with the conic; the line with the conic; the point is is thus the tangent at A. thus the point of tangency on In this special case, a. In this special case, Pascal's theorem says: Brianchon's theorem says: Say five points ACDEF lie Say five lines acdef lie on a conic. Then T(A).CE, on a conic. Then (t(a))(c.e), AC.EF, and CD.FA lie on a (a.c)(e.f), and (c.d)(f.a) lie line. on a point. We have used T(A) to denote We have used t(a) to denote the tangent at A. Note that the point of tangency on a. because B=A, we have used Note that because b=a, we have T(A) instead of AB, and have used t(a) instead of ab, and used AC instead of BC. have used ac instead of bc. This gives us a construction This gives us a construction with straightedge alone for with straightedge alone for the tangent to a conic at a the point of tangency to a point. Say we have a conic conic on a line. Say we have ACDEF, and want to find the a conic acdef, and want to tangent at A. We first draw find the point of tangency on the line through AC.EF and a. First we draw the point on CD.FA. Then we intersect (a.c)(e.f) and (c.d)(f.a). this line with DE. We draw We join this point to (d.e). a line through this We draw a point on this line intersection and point A, and and line a, and by the above by the above case of Pascal's case of Brianchon's theorem, theorem, this line will be this point will be the point the tangent to the conic at A. of tangency to the conic on a. If we take not only A=B but If we taken not only a=b but also D=E then the line AB also d=er then the line ab be- becomes the tangent at A, comes the point of tangency on and the line DE becomes the a, and the line de becomes the tangent at D. So the three point of tangency on d. So the points AB.DE, BC.EF, CD.FA lines (a.b)(d.e), (b.c)(e.f), become instead T(A).T(D), (c.d)(f.a) become instead t(a), AC.DF, CD.FA. We could .t(d), (a.c)(d.f), (c.d)(f.a). reformulate this case: say We could reformulate this case: we have a quadrilateral say we have a quadrilateral acdf ACDF inscribed in a conic; circumscribed to the conic; let let U=AC.DF and V=CD.FA. u=(a.c)(d.f) and v=(c.d)(f.a). Then the tangents to the Then the points of tangency to conic at A and C intersect the conic on a and c have a join on the line UV. which lies on the point u.v.We can use these special cases to establish by purely projective methods that any line conic is a point conic and vice versa.

Query: What happens in the case *A=B*, *C=D*? What about
the case *A=B*, *C=D*, *E=F*?

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Created: Nov 30 1995 ---
Last modified: Thu Nov 30 15:07:55 1995