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the conic in at most two points; the points A, B, C, D, E, F are assumed to
be d istin ct a nd therefore exhaust all possible intersections. If X = Y ,then
AB and BC meet in both B and Y , and therefore the lines are equal. But
this means that A = C , contradiction. Similarly, X
= Z and Y
= Z .
PROOF
Define the following lines:
1 = EF, 2 = AB, 3 = CD, m 1 = BC, m 2 = DE, m 3 = FA.
We have the following table of intersections:
1
2
3
m 1
YBC
m 2
EXD
m 3
FAZ
Let q ( x, y ) = 0 be the ane equation of the conic. In order to apply The-
orem 2.6, we change q ( x, y ) to its homogeneous form Q ( x, y, z ). Let ( x, y, z )
be a linear form giving the line through X and Y .Then
C ( x, y, z )= Q ( x, y, z ) ( x, y, z )
is a homogeneous cubic polynomial. The curve C = 0 contains all of the
points in the table, with the possible exception of Z . Itiseasilycheckedthat
the only singular points of C are the points of intersection of Q =0and
= 0, and the intersection of the two lines comprising Q = 0 in the case
of a degenerate conic. Since none of these points occur among the points
we are considering, the hypotheses of Theorem 2.6 are satisfied. Therefore,
C ( Z ) = 0. Since Q ( Z )
=0,wemusthave ( Z )=0,so Z lies on the line
through X and Y . Therefore, X , Y , Z are collinear. This completes the proof
of Pascal's theorem.
COROLLARY 2.15 (Pappus's Theorem)
Let and m be two distinctlines inthe plane. Let A, B, C be distinctpoints
of and let A ,B ,C be distinct pointsof m . A ssu m e that none of t hese
poin tsisthe intersection of an d m . Let X b e the intersection of AB and
A B ,le t Y be th e intersection of B C and BC ,and let Z be the intersection
of CA and C A .Then X, Y, Z are collin ear (see F igure 2.5).
PROOF
This is the case of a degenerate conic in Theorem 2.13.
The
“hexagon” is AB CA BC .
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