Desargues' Theorem

Notes on Java and JavaSketchpad Malfunction:
The use of Java has become a browser and machine dependent issue.
Readers who might have difficulty running the Java applets are advised to use a book marklet that converts JavaSketchpad sketches on this page (or anywhere else on the internet) to work completely independently of Java.
Go to this site http://dn.kcptech.com/builds/804.12-r/cdn/bookmarklet.html to install the small tool in your web browser OR
click on the following link to "fix" the java. Convert JavaSketch

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=ABnnA'B'`; `Y=ACnnA'C'` ; and `Z=BCnnB'C'` all lie on the same line.

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 `Q=ABnnA'B'; P=ACnnA'C'` ; and `R=BCnnB'C'` all lie on the same line.

Sorry, this page requires a Java-compatible web browser.

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 `Q`*`=AB`*`nnA'B'`*; `P`*`=ACnnA'C'` ; and `R`*`=B`*`CnnB'`*`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`*`Q`*` nnABC = {Q}`, `P`*` = P` and `O`*`R`*` nnABC = {R}`. Thus the three points `P, Q` and `R` all lie on the line `l`. EOP.

Sorry, this page requires a Java-compatible web browser.

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.