I want to prove, using the typical tools from a Hilbert plane, that the Wallis' axiom implies ($P_{\leq 1}$), where

Wallis' axiom: Given a triangle $\Delta ABC$ and given a line segment $DE$, there exists a similar triangle $\Delta A'B'C'$, having side $A'B' \geq DE$.

$P_{\leq 1}$: For each line $l$ and for each point $P\notin l$, there is at most one line containing $P$ that is parallel to $l$

I have already proved Proclo's axiom is equivalent to $P_{\leq 1}$, but I got no idea how to solve this problem...

Any help would be appreciate.

  • $\begingroup$ Hilbert space? Don't you mean Euclidean plane? $\endgroup$ – Han de Bruijn May 20 '18 at 19:53
  • $\begingroup$ That's a mistake. It's Hilbert plane, sorry... $\endgroup$ – user326159 May 20 '18 at 20:04

There's a proof on page 153 of Greenberg, which you may be able to access here.

If you can't access it or wish to fill in the details yourself, here's a proof sketch. Begin in the standard way: given l and P, we construct the parallel line m incident to P, and let Q be the foot of the perpendicular from P to l. Let n be any other line through P.

Now, we can pick a point R on n, and drop the parallel to PQ, letting S be the foot. Apply Wallis's postulate to the triangle PSR and the line segment PQ. This produces a point T which must lie on n and l.

  • $\begingroup$ Must point out that the citation only refers to the third edition of Greenberg's book. Statement and proof are on the page 216 of the fourth edition. $\endgroup$ – Akerbeltz May 9 at 15:34

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