An angle is not a measurement of openness; it measures how much you "spin" going from one ray to the other. A full spin - going back to where you started - is 360 degrees. So a quarter spin, more commonly called a right angle, is 90 degrees no matter which protractor you use. Protractor markings are spaced differently based on their size (which you can see by putting one protractor on top of another) so no matter what, the same angle has the same measurement.
360 is a completely arbitrary number - we just chose it because it's easy to divide by a lot of things. You could also think about angles in fractions of a full turn (actually, there are several other systems that don't use 360 degrees as a full turn!) The important thing is what angles measure: they measure how much you turn.
Mathematically, you're hitting all the key notes. There are two facts that I'm not quite sure you should be taking for granted or proving them - it depends among other things on whether you've proved them before. They are:
- That $V$ lies on line $AB$ by "definition of ray and betweenness". This is pretty easy to spell out.
- That if $V$ is the endpoint of $r$ (I take this to mean that $r$ is was defined using $V$ and another point $W$ as the segment $VW$ plus all points $X$ such that $W$ is between $V$ and $X$?) it cannot be between $A$ and $B$. This is some arithmetic with segment lengths. In particular, at some point it needs to be shown that at most one of "$A$ is between $B$ and $C$", "$B$ is between $A$ and $C$", "$C$ is between $A$ and $B$" can hold.
Apart from these two points, your proof is solid, and it's not hard to fill in both of these gaps if you decide you need to.
Stylistically, I want to complain a bit about your overuse of proofs by contradiction when a direct proof would do nicely. You do this twice. First, when you write
...that intersection is a point. Now if the segment is on the ray, then the intersection is the vertex $V$. If the segment is not on the ray then the intersection is a new point, call it $V'$. It can't be the latter since it would contradict that $V$, $A$, and $B$ are collinear, by definition of ray and betweenness. Hence, it's the former.
There's no point in introducing the idea of $V'$ if you're going to dismiss its existence immediately. I would instead write that the point $V$ lies on line $L$ (by assumption) as well as on line $AB$ (by "definition of ray and betweenness"), so it's the intersection point of $AB$ with $L$.
This applies to the outer argument as well: when you write
Assume $A$ and $B$ are on the opposite sides of the line, $L$, which means they lie in different half-planes. ...which contradicts that $V$ is the endpoint of $r$. Therefore the points $A$, $B$, and all points of $r$ are on the same side as $L$.
that's also unnecessarily indirect. We have everything we need to just let $A$ and $B$ be two arbitrary points of $r$ and prove that they're on the same side of $L$.
I mean, we probably do - you haven't defined "on the same side of $L$" but I assume it's something along the lines of "no point of $L$ is between $A$ and $B$".
So if that's what we're going for, we can just argue that the only point of $L$ that could possibly be between $A$ and $B$ is $V$ (since it's the only point that lies both on $L$ and on line $AB$), and even $V$ is not between $A$ and $B$ because it's the endpoint of ray $r$.
Anyway, using contradiction is fine, but if there's no reason to use contradiction - if we're not getting anything out of it that we couldn't say just as easily in a direct proof - then the direct proof is shorter and easier to read, and I encourage you to think along those lines instead.
Best Answer
Zero degrees actually does fall valid under that definition because any single ray or line can be drawn as two overlapping rays or lines (just copy the other one!). In that essence, the two lines have an angle of zero because there is no space between them, they are the same.
I'm not sure if you are familiar with the topic of vectors, but you can also define angles as the cosine of this formula
$$ \theta = \arccos\left(\dfrac{\vec u\cdot \vec v}{||\vec u|| *||\vec v||}\right) $$
And if $\vec u = \vec v$ then you get
$$ \theta = \arccos \left(1\right) = 0 $$