How do I prove reflections and rotations are bijective transformations in $\Bbb R^2$? I've got to prove that reflections and rotations in $\Bbb R^2$ are bijective but the thing is that I don't think I can use linear transformation notions or matrices since I haven't seen them yet, nor have I proved them in my lesson. 
The only thing I know is that isometries preserve the distance (and I still haven't proved that all isometries are bijective. I will do that in another exercise).
I've already proved that translations are bijective since it's quite easy but I don't know how I should handle the problem now.
 A: The simplest way is probably by exhibiting an inverse transformation (which will be the reflection itself or the rotation by the negated angle).
A: Here is one possible argument.  Let $\varphi:\mathbb{R}^2\to\mathbb{R}^2$ be distance-preserving; we wish to show $\varphi$ is a bijection.  Since $x=y$ iff $d(x,y)=0$, $\varphi$ is injective.  To show surjectivity, suppose $x\in \mathbb{R}^2$ is any point.  Fix two distinct points $a,b\in\mathbb{R}^2$ and suppose $\varphi(a)$, $\varphi(b)$, and $x$ are not collinear; if they are collinear, you just have to modify the argument slightly.   Consider the triangle formed by $\varphi(a)$, $\varphi(b)$ and $x$.  Note that there are exactly two points  $y\in\mathbb{R}^2$ such that $d(y,\varphi(a))=d(x,\varphi(a))$ and $d(y,\varphi(b))=d(x,\varphi(b))$: one of these points is $y=x$ and the other is the reflection $y=x'$ of $x$ across the line between $\varphi(a)$ and $\varphi(b)$ (this is a version of the "SSS theorem" in Euclidean geometry).  Similarly, there are exactly two points $c\in\mathbb{R}^2$ such that $d(c,a)=d(x,\varphi(a))$ and $d(c,b)=d(x,\varphi(b))$; call these points $c$ and $c'$.  Since $\varphi$ is an isometry, $\varphi(c)$ must be either $x$ or $x'$, and similarly for $\varphi(c')$.  Since $\varphi$ is injective so $\varphi(c)\neq\varphi(c')$, so one of them must be $x$ and the other must be $x'$.  In particular, this means $x$ is in the image of $\varphi$.
