# Points being collinear

If $ABC$ and $ABD$ are true and $C \neq D$, then either $ABCD$ or $ABDC$ is true. $ABC$ means $C$ is an element of $AB$ and $ABD$ means $D$ is an element of $AB$. Since $C \neq D$, then $A,$ $B,$ $C,$ and $D$ are collinear. If $A,$ $C,$ and $D$ are collinear and different points, then $CDA$, $DAC,$ or $ACD$ are true. If $CDA$ is true, then $ADC$ is true and $CAD$ is false. To prove $CDA$ is true, assume $CAD$ is true and deduce a contradiction. If $CAD$ is true, then $A$ is between point $C$ and $D,$ $B$ is between $A$ and $D.$ $B$ is also between $C$ and $A,$ here is the contradiction. Therefore $CDA$ is true and $DAC$ is false.

Am I on the right track here?

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could you draw a picture – binn Sep 3 '12 at 16:34
I'm not sure how to draw a picture on here... do you? I think the basic picture is that there are points $A$,$B$,$C$, and $D$ and they are collinear and we know that $A$$B$$C$ and $A$$B$$D$ are true. Then $D$ is between either $B$ and $C$ or $C$ is between $B$ and $D$. Does that make sense? – user23793 Sep 3 '12 at 16:37
@user23793 You can upload a picture that you have drawn using any standard drawing program. For something like this, I would suggest inkscape. – process91 Sep 3 '12 at 17:14

I refer to Hilbert's axioms and assume that $ABC$ is the inbetween relation "$B$ is between $A$ and $C$" and that $ABCD$ is defined as $ABC\land BCD$. We are given $ABC$ and $ABD$ and $C\ne D$. Then expecially all four points are distinct. Let $a$ be the unique line detemined by $A$ and $B$ according to axiom I.1. Then $C$ is on $a$ because $ABC$ and $D$ is on $a$ because $ABD$ (using axiom II.1). Your conlusion that $CDA$ or $DAC$ or $ACD$ is justified by II.3 and II.1 and your indirect proof of $CDA$ looks ok. However, I'd rather expect a proof of $BCD$ (as we want to show $ABC\land BCD$ - however your local definition of $ABCD$ may differ; is ist $ABC\land ACD$?)

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