Problem setting
Let $(M,d)$ be a metric space.
We will say that $l\subset M$ is a line if there is a surjective function $f:\mathbb{R}\to l$ such that $d(f(x),f(y))=|x-y|$.
If $l$ is a line, being $f$ its asociated function, we will say that the set $f[x,y]\subset l$ is a segment with endpoints $f(x)$ and $f(y)$.
Then, suppose that the following properties hold:
- Given two poitns $P$ and $Q$, there is an unique line $l$ such that $P$, $Q\in l$.
- Given a line $l$, $M\setminus l=U\cup V$, where $U$ and $V$ are nonempty disjoint convex sets. Also, if a segment has one endpoint in $U$ and the other in $V$, the segment intersects $l$. $U$ and $V$ will be called opposite sides of $l$.
- For each line $l$ with opposite sides $U$ and $V$, there exists an isometry $\tau_l$ such that $\tau_l|_{l}=\text{id}$ and $\tau_{l}(U)=V$, $\tau_l(V)=U$. We will say that $\tau_l$ is a reflection around $l$.
This is just the setting for euclidean geometry given in this abstract.
The problem itself
I am trying to prove that if $A$, $B$, $C$ are three distinct points such that $d(A,C)=d(A,B)+d(B,C)$, then they are collinear.
My approach
Let $A$, $B$ and $C$ be three non collinear points such that $d(A,C)=d(A,B)+d(B,C)$. Consider the point $D$ in the segment $\overline{AC}$ such that $d(A,D)=d(A,B)$. Also, let $B'$ be the reflection of $B$ over the line $l:=\overleftrightarrow{AC}$, as long as $B$ and $B'$ are in opposite sides of $l$, the segment $\overline{BB'}$ intersects $l$ in $B^*$.
Let's assume that $d(A,B^*)>d(A,D)$, then \begin{equation} \begin{aligned} d(B,B')&\leq d(B,A)+d(A,B')=2d(A,B)=2d(A,D)\\&<2d(A,B^*)=d(B,B')\text{, } \end{aligned} \end{equation} so, obviously $d(A,B^*)\leq d(A,D)$, and, it is easy to conclude that if $d(A,B^*)= d(A,D)$ then \begin{equation} d(A,B^*)=d(B,A)=d(B,C)=d(B,B^*)=d(B^*,C)\text{, } \end{equation} from here, I really don't know how to proceed, it seems that a contradiction will arise, but I don't see how.