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I have got very soft question. It's true that all plane geometry problems had analytical (easy) solutions. And everyone can take Cartesian coordinate plane and count all problems for $<\infty$ years. Is the next "anti-Cartesian" method true?

We define the next geometry:

1) Take all Euclidian axioms about angles and lines, parallel lines and points (without distances!).

2) Take some lemmas from elementary geometry and make their formulation as an axiom. (for example: common chords of three given circles intersects at one point, points A, B, C, D are on same circle <=> angle ABC = angle ADC, e.c.)

Now in this "geometry" you can solve some (not all) problems using ONLY elementary method. Hope that my question is clear.

solver6
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    The question is not clear to me. Which Euclidean axioms do you want to keep, and which do you want to omit? – Mårten W Mar 11 '13 at 19:15
  • To be more clear i'll try to make example. We have points and lines, say some of points can lie on line. We have function f:(A,B,C)->[0,1) and f(A,B,C)=-f(C,B,A),if points A,B,C are lie on line, then f(D,A,B)=f(D,A,C) for every other point D. Say that points A,B,C,D are on same circle iff f(A,B,C)=f(A,D,C). If given points A,B,C,D then there exists point E: C,E,D are on same line and A,B,C,E are on same circle. If given points A,B,C,D, then if f(A,B,C)=f(D,C,B), then f(D,A,B)=f(A,D,C). – solver6 Mar 12 '13 at 18:30
  • From this "axioms" we can get for example next problem: A,B,C,D,E,F are points A,B,C are lie on line D,E,F are lie on line f(C,A,E)=f(C,B,F), f(A,B,D)=f(A,C,E), then prove that f(C,A,D)=f(B,C,F) – solver6 Mar 12 '13 at 18:31

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The Euklid (or preferably: Hilbert) axioms of planar geometry are good enough to allow us to introduce a notion of distance (proportions) and a two-dimensional orthogonal coordinate system. However, coordinates are not necessarily from $\mathbb R$, but from a field obtained from $\mathbb Q$ by repeated addition of square roots. All coordinate calculations (as long as they use rationanal functions only) can be described in terms of this (essentially, "constructed"). But we loose limits such as arc length of the circle or the existence of a length ratio $\sqrt[3]2$, I think.

Hagen von Eitzen
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  • Yes, but if we haven't got "old" interpretation of circle (we can't construct circle with given center and radius), then we can't construct 1/2 and other lengths too. – solver6 Mar 11 '13 at 19:05
  • For example from axioms in my previous post we can get that heights of triangle intersects at one point. – solver6 Mar 11 '13 at 19:12