Simple Geometry Proof (Betweenness Axiom)

[dancergirlie]

Homework Statement

Given a line l, a point A on l, and a point B not on l. Then every point of the ray AB (except A) lies on the same side of l as B.

Homework Equations

The Attempt at a Solution

I understand why this is true, however I'm having some trouble wording my proof. Any help would be great!

Alright, this is what I have so far:

Suppose there exists a point c on the ray AB such that c lies on the opposite side of l than B.
However, By definition of a ray, A*B*C

(this is where I don't know how to continue wording my proof, I know that since B is between a and c, and the line goes through A, that means that C must be on the same side of l as B, however I don't know how to word that mathematically)

 

[mighty2000]

First, we can define the ray AB as a line segment that starts at A and extends infinitely in the direction of B. Therefore, any point on this ray can be represented as A*B*C, where C is any point on the ray except for A.

Now, let's consider the line l and the point B. Since B is not on l, it must lie on one side of the line, which we will call side X.

Next, we can draw a line segment from A to B, which intersects line l at point D. By the definition of a line, we know that D must also lie on side X of l.

Now, let's consider a point C on the ray AB that lies on the opposite side of l from B, which we will call side Y. This means that C*A*D, since A and C are on the same side of l and D is on the opposite side.

However, since A*B*C, we know that C must also lie on side X of l, since B is on side X and A is between B and C. This contradicts our previous statement that C lies on side Y.

Therefore, our initial assumption that there exists a point C on the ray AB that lies on the opposite side of l from B is false. This means that every point on the ray AB (except for A) must lie on the same side of l as B, which proves the statement.