Difference between revisions of "Pascal's Theorem"
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''Proof.'' Since <math>ABNM, CDMN </math> are two sets of concyclic points and <math>A,M,C </math> and <math>B,N,D </math> are two sets of collinear points, | ''Proof.'' Since <math>ABNM, CDMN </math> are two sets of concyclic points and <math>A,M,C </math> and <math>B,N,D </math> are two sets of collinear points, | ||
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− | <math> \angle CAB \equiv \angle MAB \equiv 180^\circ -\angle MNB \equiv \angle MND \equiv \angle MCD \equiv \angle ACD </math>. | + | <math> \angle CAB \equiv \angle MAB \equiv 180^\circ -\angle MNB \equiv \angle MND \equiv 180^\circ - \angle MCD \equiv 180^\circ - \angle ACD </math>. |
</center> | </center> | ||
Because alternate interior angles <math>CAB</math> and <math>ACD</math> are congruent, <math>AB || CD.</math>{{Halmos}} | Because alternate interior angles <math>CAB</math> and <math>ACD</math> are congruent, <math>AB || CD.</math>{{Halmos}} |
Revision as of 21:57, 25 April 2020
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A diagram of the theorem |
Pascal's Theorem is a result in projective geometry. It states that if a hexagon is inscribed in a conic section, then the points of intersection of the pairs of its opposite sides are collinear:
Since it is a result in the projective plane, it has a dual, Brianchon's Theorem, which states that the diagonals of a hexagon circumscribed about a conic concur.
Proof
It is sufficient to prove the result for a hexagon inscribed in a circle, for affine transformations map this circle to any ellipse while preserving collinearity and concurrence in the projective plane, and projective transformations can map an ellipse to any conic while similarly preserving collinearity and concurrence in the projective sense. Thus we will prove the theorem for a cyclic hexagon, using directed angles mod .
Lemma. Let be two circles which intersect at
, let
be a chord of
, and let
be the second intersections of lines
with
. Then
and
are parallel.
Proof. Since are two sets of concyclic points and
and
are two sets of collinear points,
.
Because alternate interior angles and
are congruent,
∎
Theorem. Let be a cyclic hexagon, and let
,
,
. Then
are collinear.
Proof. Let be the circumcircle of
, and let
be the circumcircle of
. Let
be the second intersection of
with
, and let
be the second intersection of
with
. By lemma,
, and
, and
, see figure. It follows that triangles
and
are homothetic with a center
, because
. Therefore the line
passes through the center of homothety, i.e.
. ∎
Notes
In our proof, we never assumed anything about configuration. Thus the hexagon need not even be convex for the theorem to hold. In fact, many useful applications of the theorem occur with degenerate hexagons, i.e., hexagons in which not all of the points are distinct. In the case that two points are the same, we consider the line through them to be the tangent to the conic through that point. For instance, when we let a triangle be a "hexagon"
, Pascal's Theorem tells us that if
are the tangents to the circumcircle of
that pass through
, respectively, then
,
,
are collinear; the line they determine is called the Lemoine Axis. In fact, Pascal's Theorem tells us that
can be the tangent lines to any conic circumscribed about triangle
and the result still holds.