I'm struggling with the following elementary problem from Euclidean Geometry which is the last piece within a certain framework of a proof that congruent alternate interior angles implies parallel lines. I am trying to do this all without any notion of angle measure or the ability to sum angles, simply the following axiom: Suppose that point $D$ is interior to $\angle ABC$, then $\angle ABC \not \cong \angle DBC$. By "interior", I mean that $D$ is on the same side of line $BC$ as point $A$ (segment $\overline{AD}$ does not intersect line $BC$) and likewise, that $D$ is on the same side of line $AB$ as point $C$.
The missing piece of the proof of the above is the following very intuitive result: Suppose points $C$ and $D$ lie on the same side of line $AB$, then if segment $\overline{BD}$ meets ray $\overrightarrow{AC}$ and segment $\overline{BC}$ meets ray $\overrightarrow{AD}$, then $\overrightarrow{AC}=\overrightarrow{AD}$. In other words, if $D$ and $B$ are on opposite sides of line $AC$, then $C$ and $B$ are on the same side of line $AD$ unless $\overrightarrow{AC}=\overrightarrow{AD}$. This is equivalent to saying that of three rays all emanating from the same point, say $\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}$, one of the three points $B,C,D$ is interior to the angle formed by the other two rays.
This would be very easy with an angle measure version of the axiom quoted above since we could then use a simple triangle inequality argument, but I'd like to see if this is possible without appealing to angle measure. Any help would be appreciated.