Suppose R is the line of reflection. Given a line L. Prove that the reflection of L through R is also a line. Moreover, prove that reflections over a line preserve angle measure.
How can I go about proving this rigorously? I want to avoid using the fact that a reflection in a line is an isometry.
On
How far will you go to avoid using isometries in general?
Personally I would apply a translation and rotation to reduce to the case where $R$ is in fact the $x$-axis, i.e. $\{y=0\}$ line (assuming 2 dimensions for brevity). This is if you like doing things algebraically.
In this case it is easy to see what happens to a line of the form, e.g. $\{y = mx + c\}$ or even $\{x = c\}$ (these are the lines perpendicular to $R$), since the reflection then takes the form:
$$(x,y) \mapsto (x,-y)$$
and you can be explicit about everything.
For example, take a point in the line $(c,t)\in \{(c,y)| y \in \mathbb R\}$, this gets reflected onto $(c,-t)$. Allowing $t$ to vary, this again gives you the line $\{(c,y)| y \in \mathbb R\}.$
It is even easy to say things about angles also, as this can be interpreted largely as a statement about gradients.
How satisfactory is this answer?
Choose three points $A<B<C$ on line $L$ and let $A'$, $B'$, $C'$ be their reflections. It is easy to prove that $\angle A'B'B=\angle ABB'$ and $\angle C'B'B=\angle CBB'$. Hence: $$ \angle A'B'B+\angle C'B'B=\angle ABB'+\angle CBB'=180° $$ and it follows that $A'B'C'$ are aligned.
Congruence of corresponding angles is a consequence of the congruence of corresponding triangles.