Given two non parallel lines $l$ and $l'$ in $\mathbb{E}^2$, I have to proof that there are exact four transformations $F: \mathbb{E}^2 \rightarrow \mathbb{E}^2$ with $F(l) = l'$ and $F(x)= x$ with $ x = l \cap l'$.
Does anyone know how I can proof this?
Since the lines $l$ and $l'$ define $4$ regions in the Euclidean plane, we can consider the bisectors of these regions, say $b_1$ and $b_2$. Notice that $b_1\cap b_2=\{x\}$. For $i=1,2$, define $R_i$ the reflection with respect to the bisector $b_i$. Both reflections satisfy the property $R_i(x)=x$ and $R_i(l)=l'$. Moreover, each combinations of them satisfy that conditions. Now it remains to see that the group generated by $R_1$ and $R_2$ has exactly $4$ elements.