Proof of the associative property of segments (axiomatic geometry)

Question

Show that the sum of the segments is associative, i.e (AB + CD) + EF = AB + (CD + EF). P.S.: Here we don't speak in congruente segments, but in equal segments. P.S.²: I don't even know how to start to solve this exercise =/. (sorry about my bad english)

Answer 1

To sum two or more segments they need to be disposed on the same line and each pair of summand segments must have exactly one endpoint in common. Once you have that, the sum is the same as the union of sets, which is associative.

Answer 2

What do you mean by the sum of segments? There may be two answers:

1. Sum of measures.

In this case answer is very simple, because it is implied by associative property of real numbers.

2. Sum of free segments (i'm not sure sure about the translation). I define free segment as an equivalence class of a congruence relation. We say that a free segment $\mathfrak{c}=[ab]$ is a sum of free segments $\mathfrak{a}$ and $\mathfrak{b}$ (we write $\mathfrak{c}=\mathfrak{a}+\mathfrak{b}$ ) if there exists point $p$ such that $B(apb)$ and $\mathfrak{a}=[ap], \mathfrak{b}=[pb]$. It can be proved that this definition doesn't depend on the choice of segment $ab$ and $+$ is a function.

Lemma: $\left(B(abc) \wedge B(acd)\right)\implies \left(B(abd)\wedge B(bcd)\right)$

Proof of assosiative property:

Let $\mathfrak{a}+\mathfrak{b}=[p_{1}p_3]$

So for some point $p_2$ such that $B(p_1p_2p_3)$ we have $\mathfrak{a}=[p_1p_2], \mathfrak{b}=[p_2p_3]$.

Now we find point $p_4$ such that $B(p_1p_3p_4)$ and $\mathfrak{c}=[p_3p_4]$. (We can do this by one of the axioms)

We have $(\mathfrak{a}+\mathfrak{b})+\mathfrak{c}=[p_1p_4]$

By lemma we have $B(p_2p_3p_4)$ which gives $\mathfrak{b}+\mathfrak{c}=[p_2p_4]$.

Again by lemma we have $B(p_1p_2p_4)$ which gives $\mathfrak{a}+(\mathfrak{b}+\mathfrak{c})=[p_1p_4]$

I denote $B(abc)$ when $b$ lies between $a$ and $c$ on the same line.