If $A, B,$ and $C$ are non-collinear points, then $A$ and $B$ are distinct, and $C$ does not lie on the line $\overleftrightarrow{AB}$.
The axioms are the following:
- Incidence Axiom 1: There exist at least three distinct noncollinear
points. - Incidence Axiom 2: Given any two distinct points, there is at least
one line that contains both of them. - Incidence Axiom 3: Given any two distinct points, there is at most
one line that contains both of them. - Incidence Axiom 4: Given any line, there are at least two distinct
points that lie on it.
Proof:
I attempted to prove this by the contrapositive by assuming that $A$ and $B$ are equal points and arriving at a contradiction.
Suppose $A, B,$ and $C$ are non-collinear points and that $A=B$ then there is a line, $l$, that contains them. From Incidence Axiom 4, we know given any line $l$ there are at least two distinct points that lie on it, let this distinct point be $D$ therefore $\overleftrightarrow{AD}$ (or similarly $\overleftrightarrow{BD}$ since $A=B$) is a line with two distinct points. Now since $C$ is collinear from $A$, we have that $C \notin \overleftrightarrow{AD}$ therefore $C$ is distinct from $A$ and $B$, however, since $A=B$ and $C$ are two distinct points we have from Incidence Axiom 3 that there is at most one line that contains both of them, however, this a contradiction since $C$ is collinear from $A$ therefore the assumption that $A=B$ is false.
Is my proof correct?
How can I expand on this theroem to prove that, $A$ does not lie on $\overleftrightarrow{BC}$, $B$ does not lie on $\overleftrightarrow{AC}$ and $C$ does not lie on $\overleftrightarrow{AB}$?
Best Answer
I think you can conclude more quickly: if $A=B\ne C$ there is a line $l$ passing through $A$ and $C$ (Axiom 3), whence contradiction.
Once you know $A\ne B$, if $C\in \overleftrightarrow{AB}$ we get a contradiction, hence $C\not\in \overleftrightarrow{AB}$.