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So the axioms for a projective plane are given by:

  1. Any two “points” are contained in a unique “line.”
  2. Any two “lines” contain a unique “point.”
  3. There exist four “points”, no three of which are in a “line.”

Consider now this question: Suppose that $A,B,C,D$ are four “points” in a projective plane, no three of which are in a “line.” Consider the “lines” $AB,BC,CD,DA$. Show that if $AB$ and $BC$ have a common point $E$, then $E = B$.

I feel like I'm not approaching this correctly. To me it just seems that if $E\neq B$, then $AB$ and $BC$ intersect at $B$, and also intersect at $E$, which contradicts axiom 2. But this does not use the hypothesis at all. I think I'm missing something very, very obvious.

Vasting
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  • Statement 2. is sloppy and wrong (perhaps an English problem). You mean, of course, that two distinct lines have a unique point in common. I agree with your confusion. Of course $\overleftrightarrow{AB}$ and $\overleftrightarrow{BC}$ have the point $B$ in common. Something is amiss. – Ted Shifrin Jun 11 '23 at 21:47
  • I agree with @TedShifrin about axiom 2 being inaccurate. Similarly, axiom 1 should require the two points to be distinct. Fortunately the question and your solution still make sense after these corrections. First note that every two of the four points $A,B,C,D$ are distinct, for otherwise those two and any third one would lie on a line. So lines like AB make sense. Furthermore, $AB$ and $BC$ are distinct lines, for if they were a single line then that line would contain the three points $A,B,C$. And now your proof works even with the corrected axiom 2. – Andreas Blass Jun 12 '23 at 00:59

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