# Rays, points and line segments

1. Feb 17, 2014

### Lee33

1. The problem statement, all variables and given/known data

Let A, B and D be points. If $D\notin \vec{AB}$ in a metric geometry, prove that $\vec{AD}\cap\vec{AB} = \{A\}.$

2. Relevant definitions

Ray: If A and B are distinct points then the ray from A toward B is the set $\vec{AB} = \overline{AB} \cup \{ C \in P \ | \ A-B-C\},$ where $A-B-C$ means B is between A and C.

Line segment: If A and B are distinct points in a metric geometry then the line segment from A to B is the set $\overline{AB} = \{C\in P\ | \ A-C-B \ \text{or} \ C=A \ \text{or} \ C = B\}.$

3. The attempt at a solution

I am not sure how to go about proving this.

If I let $Z$ be a point where $Z\in\vec{AD}$ then $Z\notin\vec{AB}$ since $Z$ was arbitrary then there are no points in $\vec{AD}$ that lie in $\vec{AB}$. So $A$ is the only point in $\vec{AD}\cap\vec{AB}$.

2. Feb 18, 2014

### tiny-tim

Hi Lee33!
Haven't you just assumed exactly what you have to prove?

And don't say "Z is arbitrary", it's almost meaningless … say "if Z ≠ A then A – Z – D"

(However, I don't understand the question … why can't D be along the extended line AB, the other side of B ??)

3. Feb 18, 2014

### Lee33

Well the question is asking if the point D is not in the ray $\vec{AB}$ then I must prove that $\vec{AB}\cap\vec{AD}$ equals the set {A}.

Last edited: Feb 18, 2014
4. Feb 18, 2014

### tiny-tim

but there's no C in the question

5. Feb 18, 2014

### Lee33

Opps. I meant A instead of C. I have edited the post. Sorry about that.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted