- #1
Lee33
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Homework Statement
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\}.##
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}##.