I'm trying to understand Euclidean geometry the hard way. I don't want to start with analytical geometry, building on coordinates and vector spaces, nor on the Euclidean approach of synthetic geometry which, implicitly, builds on rigid motions which need length and rotation to be predefined.
I tried this:
Let $(\mathbb P, \mathbb L)$ be a (lightwigth, skipping the $\mathbb I$) incidence structure with $\mathbb L \subset \frak{P}(\mathbb P)$ with the following properties:
$\forall_{a,b\in \mathbb L}\;\;a\cap b=\emptyset \;\lor\; \exists_{x\in\mathbb P}\,a\cap b=\{x\} \;\lor\; a\cap b=a$
and
$\forall_{a,b\in \mathbb L}\;\;a\subset b\implies a=b$.
Thus $\mathbb L$ mimics lines in the Euclidean plane, but with very few assumptions: lines intersect in exactly one point, not at all or in one of the lines, and, lines are maximal in the sense that no line is proper part of another line.
Then mimic the reflection of points by a line:
Define a family of functions $\mu_a: \mathbb P\to \mathbb P\;\;\forall a\in \mathbb L$ with the following properties:
$\forall_{x\in \mathbb P}\;\;\mu_a(x)=x\iff x\in a$ (The points on the line are the only fixpoints)
$\forall_{b \in\mathbb L}\;\;\mu_a(b)\in\mathbb L$ (The image of a line is again a line)
$\forall_{b \in\mathbb L}\;\;\mu_a(\mu_a(b))=b$ (Applying a reflection twice maps lines onto themselves)
In the Euclidean plane, two different lines are perpendicular if a reflection by $a$ maps $b$ onto itself.
Then for $a\neq b \in\mathbb L$ define $a \perp b \iff \mu_a(b)=b$.
I fail either to show that $a\perp b\implies b\perp a$ or to find a counterexample.
