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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.

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.