Let $\mathcal{E}^2$ be an Euclidean affine space of dimension $2$ oriented by $R=(O,B=\{u_1,u_2\}$.
Let $$\mathcal{r}=\{P \in \mathcal{E}_2 \text{ such that } \overrightarrow{AP} \in \langle u \rangle \}$$ and $$\mathcal{s}=\{P \in \mathcal{E}_2 \text{ such that } \overrightarrow{BP} \in \langle v \rangle \}$$ be two lines of $\mathcal{E}_2$.
Let $\alpha$ be the angle between r and s and $\beta$ be the angle between $s$ and $r$.
Then I’ve been given the following facts (without any proof):
$\alpha$ depends on the orientation of $\mathcal{E}_2$ and on the reciprocal orientation of $r$ and $s$ and on the order in which you pick the lines. Furthermore, if you change the orientation of $r$, then, $\alpha’ = \alpha + \pi$
$\alpha + \beta = 2 \pi$
$ \cos \alpha = \cos \beta$ and depends only on the orientation of $\mathcal{E}_2$ and on the reciprocal orientation of the lines
$|\cos \alpha |= |\cos \beta|$ depends only on the orientation of $\mathcal{E}_2$.
$ \cos \alpha = \cos \beta = \frac{u \cdot v}{\lVert{ u }\rVert \lVert{ v }\rVert }$, where $u \cdot v$ is the scalar product of $u$ and $v$.
My question is:
The statements are clear, but surely the proofs don't seem obvious to me; how does one prove them?
We are in the vector plane $E$.
Part 1. Orientation. If $B,B'$ are two bases of $E$, then let $[B,B']$ be the matrix s.t. its $i^{th}$ column is the decomposition of the $i^{th}$ vector of $B$ in $B'$. We consider the equivalence relation: $B\sim B'$ iff $\det([B,B'])>0$; $\sim$ has two classes, that is, the possible $2$ orientations of $E$. According to the definition, if $\{u,v\}\sim B'$, then $\{-u,v\}\not \sim B'$ and $\{v,u\}\not \sim B'$.
Part 2. If we define a scalar product on $E$, then the orthogonal group is defined and we can define an orthonormal basis $B$ of $E$. Now we can define the angle $(u,v)$ between $2$ unitary vectors $u,v$: $(u,v)=\mathcal{O}$, where $\mathcal{O}\in O^+(2)$ satisfies $\mathcal{O}u=v$ (such $\mathcal{O}\in O^+(2)$ is unique and, moreover, its associated matrix $O$ only depend on the orientation of the chosen basis $B$). There is $\theta\in\mathbb{R}/2\pi\mathbb{Z}$ s.t. $O=\begin{pmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{pmatrix}$. We use the notation $(u,v)=O_{\theta}$; note that $O_{\alpha}\circ O_{\beta}=O_{\alpha+\beta}$.
If $(u,v)=O_{\theta}$, then $(v,u)={O_{\theta}}^{-1}=O_{-\theta}$, $(-u,v)=O_{\theta} \circ O_{\pi}=O_{\theta+\pi}$.
Assume that we choose another orientation for $E$; it is equivalent to say that in $B=[x,y]$, we change $x$ with $-x$. Then the matrix associated to $\mathcal{O}$ becomes $\begin{pmatrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{pmatrix}$, that is $O_{-\theta}$.
About the scalar product, $<u,v>=<u,O_{\theta}u>=<x,O_{\theta}x>=\cos(\theta)$. And if $u,v$ are not unitary, then $<u,v>=||u||||v||<\dfrac{u}{||u||},\dfrac{v}{||v||}>$.