[Math] Prove that if a ray has its endpoint on a line, where the ray is not on the line, then the points of the ray are on the same side of the line.

euclidean-geometrygeometry

I have been having trouble with the following proof:

Suppose we have a line $L$ in the plane $P$ (where
$L \subset P$), a ray $r$ with vertex
$V\in$ $L$, and $r\subset P$
but $r\not\subset L$. Then the points of
$r$, are all on the same side of $L$ except for
the vertex $V$.

To give a visual, this is how it would look (credit to GeoGebra):

Key definitions I'm working with (the labels of these points don't correspond to my picture):

$\bf{Def. 1}$: A point $B$ is said to be $\bf{between}$ the points $A$ and $C$ iff:

$(1)$ $A$,$B$, and $C$ are distinct collinear points

$(2)$ $AB+BC=AC$.

$\bf{Def. 2}$: If $A$ and $B$ are two points, the set consisting of all points of the segment $\overline {AB}$ and all points, $X$, such that $B$ is between $A$ and $X$ is called a $\bf{ray}$

Sketch of my proof:

Suppose $A$ and $B$ are two points of the ray, $r$. Assume $A$ and $B$ are on the opposite sides of the line, $L$, which means they lie in different half-planes. Now, since $A$ and $B$ are on opposite sides, then the segment $\overline {AB}$ intersects the line $L$ and 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. But this implies that $V$ is between $A$ and $B$ 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 $L$.

Any thoughts?

Best Answer

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.

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