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Owner confidence rating: High Entry average rating: No information on entry rating
Desargues' theorem (Theorem)

If $ABC$ and $XYZ$ are two triangles in perspective (that is, $AX,BY$ and $CZ$ are concurrent or parallel) then the points of intersection of the three pairs of lines $(BC,YZ), (CA,ZX), (AB,XY)$ are collinear.

Also, if $ABC$ and $XYZ$ are triangles with distinct vertices and the intersection of $BC$ with $YZ$, the intersection of $CA$ with $ZX$ and the intersection of $AB$ with $XY$ are three collinear points, then the triangles are in perspective.

\includegraphics[scale=0.75]{desargues}
(XEukleides source code for the drawing)



"Desargues' theorem" is owned by drini. [ full author list (2) | owner history (1) ]
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proof of Desargues' theorem (Proof) by drini
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Cross-references: vertices, collinear, lines, intersection, points, parallel, concurrent, triangles
There are 4 references to this entry.

This is version 7 of Desargues' theorem, born on 2001-10-20, modified 2006-07-29.
Object id is 424, canonical name is DesarguesTheorem.
Accessed 4749 times total.

Classification:
AMS MSC: 51A30 (Geometry :: Linear incidence geometry :: Desarguesian and Pappian geometries)
Pending Errata and Addenda
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spelling by pbruin on 2002-11-28 02:46:57
This theorem is (as far as I know) called Desargues' Theorem (with the apostrophe after the s). It is named after a French architect.
[ reply | up ]
pictures by drini on 2001-10-20 22:38:29
as soon as I get xfig installed I'll update my geometry items with figures
 f
G -----> H G
p \ /_ ----- ~ f(G) 
 \ / f ker f 
 G/ker f 
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