Stupid geometry question I'm doing a proof in geometry. 
There is a line AC, with a point B also on the line, between A and C. In other words, the three points on the line are ABC, in that order. 
Another line, let's call it DA, intersects with ABC at A. This creates an angle CAD. 
CAD is the same angle obviously as BAD. 
But what is the justification in a geometric proof (something about the nature of lines?) that says that these two are the same.
(NB BAD and CAD are separately important as part of different triangles).
Thanks!
 A: This question can be answered from Hilbert's axioms standpoint.
Among primitive notions we have a ternary betweenness relation which describes exactly the situation "point $B$ lies between points $A$ and $C$ (on the same line)". Let's denote $A*B*C$ for such situation.
To answer this question in detail, I will need the precise definition of a halfline in terms of betweenness relation. First look at the theorem:
Theorem. For any line $l$ and any point $O\in l$ the binary relation $\sim$ on the set $l\setminus\{O\}$ defined by the formula:
$$A\sim B :\iff \left(O*A*B \vee A=B \vee O*B*A\right)$$
is an equivalence relation with exactly two equivalence classes.
We will call the set $r$ a halfline whenever $r$ is an equivalence class of relation $\sim$ for some line $l$ and some point $O\in l$. It can be proved that both the line $l$ and the point $O\in l$ are uniquely determined by a halfline $r$. $O$ is called the origin of $r$.
Also let's denote $\overrightarrow{OA}:=\{B: O*A*B \vee A=B \vee O*B*A\}$ for any distinct points $O$ and $A$. It is quite obvious that $\overrightarrow{OA}$ is a halfline with origin $O$ contained in the line $\overleftrightarrow{OA}$.
A very simple consequence of all above statements is the following property:
Property 1. If $O*A*B$ then $\overrightarrow{OA}=\overrightarrow{OB}$.
Basically if follows from the fact that $O*A*B$ implies that $A\sim B$ where $\sim$ is the relation introduced before for line $\overleftrightarrow{OA}=\overleftrightarrow{OB}$ and point $O$. The well known fact about equivalence relations is that if $A\sim B$ then the equivalence classes of $A$ and $B$ are equal.
Another definition needed here is the angle definition:
If $r,s$ are halflines with the same origin, then an unordered pair $\{r,s\}$ is called an angle.
Also whenever $A\neq O\neq B$ we denote $\angle AOB:=\{\overrightarrow{OA},\overrightarrow{OB}\}$

Answer to the question
From assumptions we have $A*B*C$.
By the property 1, $\overrightarrow{AB}=\overrightarrow{AC}$.
Next $\angle BAD=\{\overrightarrow{AB},\overrightarrow{AD}\}=\{\overrightarrow{AC},\overrightarrow{AD}\}=\angle CAD$.
A: An angle is between two lines. One line is $D\vee A$. The other line is (i) $A\vee C$ and (ii) $A\vee B$. We therefore have to prove that $A\vee C=A\vee B$. You have said that "$B$ is in the middle of $A$ and $C$". This implies that $B\in A\vee C$, but $B\ne A$. Now there is exactly one line containing $B$ and $A$, namely $A\vee B$. Since $B$ is also on the line $A\vee C$ we can conclude that in fact $A\vee B=A\vee C$.
