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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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