I am reading Hilbert's foundations of geometry, translated by Townsend. The theorem 17 and 18 don't have proofs, probably because they are too obvious, I don't know. And I haven't been able to see why they are correct.

Hilbert puts them down as "most general theorems relating to congruences in a plane".

Here is theorem 17:

If $( A, B, C, \dots )$ and $(A_0 , B_0, C_0, \ldots )$ are congruent plane figures and $P$ is a point in the plane of the first, then it is always possible to find a point $P_0$ in the plane of the second figure so that $(A, B, C, \ldots, P) $ and $( A_0 , B_ 0, C_ 0, \ldots, P_0 )$ shall likewise be congruent figures. If the two figures have at least three points not lying in a straight line, then the selection of $P_0$ can be made in only one way.

Could someone explain why three points?

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    $\begingroup$ I recommend some books that do Hilbert's axioms in a more relaxed manner, such as Hartshorne, Geometry:Euclid and Beyond, also Marvin Jay Greenberg, Euclidean and Non-Euclidean Geometry (I'm in the fourth edition). When I was writing my article, I relied on Martin, The Foundations of Geometry and the Non-Euclidean Plane. These books are also not "easy" but are intended for use in the classroom, the audience being students $\endgroup$
    – Will Jagy
    Oct 19, 2017 at 22:12
  • $\begingroup$ That's great! Thanks for the references. My next goal is going to be learning some more about the basis for quantum mechanics equations, that's why I wanted to scan Hilbert's foundations first. Will, for example Hartshorne give the generality I need to continue with next steps? $\endgroup$ Oct 19, 2017 at 22:37
  • $\begingroup$ A guess: if the two figures don't have at least 3 non-colinear points, we are in the degenerate case of lines... $\endgroup$
    – user403337
    Oct 20, 2017 at 0:09
  • $\begingroup$ Hilbert spaces are something quite different. en.wikipedia.org/wiki/Hilbert_space $\endgroup$
    – Will Jagy
    Oct 20, 2017 at 0:10
  • $\begingroup$ Chris, I couldn't understand what you meant by degenerate case. If it's the same as what I posted, could you post it instead and I'll accept it as answer? $\endgroup$ Oct 20, 2017 at 0:24

2 Answers 2


You can find the proof with all details in "Foundations of geometry" by Karol Borsuk and Wanda Szmielew in paragraph 34 of chapter II (starting from page 131).


A guess: if the two figures don't have at least 3 non-colinear points, we are in the degenerate case of lines...


And in degenerate case, you can have a $P_0$ on both sides of the figure and will have the same figure (for example, you can rotate the plane).


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