Hilbert's foundations of geometry theorem 17

Question

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?

Answer 1

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

addition:

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).

Answer 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).