Desargues' Theorem in 3-space
and the plane.
Notes by M. Flashman
Based on proof found in Hilbert&Cohn-Vossen's
Geometry
and The Imagination.
We define a perspective relation: Two points P and P' are perspectively
related by the center O if O is on the line PP" . Two triangles ABC and
A'B'C' are perspectively related by the center O if O is on the lines
AA',
BB', and CC'.
Desargues' Theorem in Space:
- Desargues' Theorem: (in projective 3 space).
If two non co-planar
triangles ABC and A'B'C' are perspectively related by the center O,
then
the points of intersection X=AB*A'B'; Y=AC*A'C' ; and Z=BC*B'C' all lie
on the same line.[* is used here to indicate the intersection of the
lines.]
- Proof of this result in space: The intersection of
the
planes determined by the two triangles is a line. This line has the
points
X, Y, and Z on it, because these points must lie on the intersection of
the two planes.
In a "projective spatial geometry" any pair of distinct planes would
intersect in a line.
- Notice that if we use a central projection of the
spatial configuration
of lines and vertices from the spatial Desargues' Theorem we have a
planar
configuration of lines and vertices, satisfying the same hypotheses and
illustrating the same conclusion. This basic idea of projecting a
result
from 3-space onto the plane is used in the proof of the planar version
of Desargues' Theorem as provided in the book Geometry and The
Imagination by Hilbert and Cohn-Vossen.
- Desargues' Theorem: (in the (projective) plane). If
two
coplanar
triangles ABC and A'B'C' are perspectively related by the center O,
then
the points of intersection X=AB*A'B'; Y=AC*A'C' ; and Z=BC*B'C' all lie
on the same line.
Proof: [Based on H&C-V ].
Choose O* not in the plane ABC. Choose B* a point on BO*.
Let B'* denote the intersection of O*B' with OB* in the plane OBO*.
Now triangles AB*C and A'B'* C' are perspectively related
in space by the center O.
So by Desargues' Theorem in space, the points X*=AB*#A'B'*;
Y*=AC#A'C' ; and Z*=B*C#B'*C' all lie on the same line l* in
space
which
does not pass through the point O*.
Now the plane determined by l* and O* meets the plane ABC
on a line l.
Consider that O*X* #ABC = {X}, Y* = Y and O*Z* #ABC =
{Z}.
Thus the three points X, Y and Z all lie on the line l.
EOP.
-
- Comments: This theorem is a result of
projective geometry in its
use of the fact that any pairs of line will have a point of
intersection.
This can be transferred to ordinary Euclidean Geometry by using the
connecting
geometry of the affine plane where parallel lines meet on the ideal
line.
The result allows that one pair of the lines determining P,Q, or R may
be parallel, but if two of the pairs are parallel, then the line
deteremined
by those two pairs is the ideal line, so the third point of
interesection
is also an ideal point.
- The consequence in Euclidean geometry are these
special cases of Desargues' Theorem:
- Desargues' Theorem:
(in the Euclidean plane).
If two coplanar triangles ABC and A'B'C' are perspectively related by
the
center O, then either
(i) there are three points of intersection
P=AB*A'B'; Q=AC*A'C' ; and R=BC*B'C' which all lie on the same line;
(ii) one pair of sides, say AB and A'B'
are
parallel to the line determined by the points of intersection of the
other
pairs of sides Q=AC*A'C' ; and R=BC*B'C';
or (iii) if two pairs of sides , say AB
and
A'B' are parallel and AC and A'C' are parallel, then the third
pair
of sides BC and B'C' are also parallel.
[You can use the first java sketch to illustrate
this from the projective, affine and euclidean views in the plane.]
For a related result about circles see Tom
Banchoff's web page about Monge's Theorem.
