A Proof From Space

The lines AA', BB', CC' lie in a plane and meet in a point P. The segments AB, A'B' and BC, B'C' are pairwise parallel. How does it then follow that AC and A'C' are also parallel? The answer comes from space. Lifting P to some point Q outside the plane (say, vertically), one then constructs a triangle lying above A'B'C' by intersecting planes containing Q and appropriate sides of ABC. The proof is sketched on the diagram, and a more detailed argument is available here as a PDF-file.
Whereas beginnerers should primarily be motivated by the drive toward discovery and clarity, they might eventually be interested enough to try some mathematical "rock-climbing". One could, of course, go back to Euclid, but there a lot of time is initially spent on things which appear obvious; going back to Hilbert involves an even slower tempo and more tedious effort (say, on the "betweenness" relation). By contrast, the incidence geometry involved here is sufficiently challenging, and yet transparent once understood.