4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
+50
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.