Affine Planes – Does Affine Pappus Axiom Imply Affine Desargues Axiom?

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I am interested in the affine version of the well-known Hessenberg's Theorem (saying that Pappian projective planes are Desarguesian).

First I introduce all necessary definitions.

Definition L. A linear space is a pair $(X,\mathcal L)$ consisting of a set $X$ and a family $\mathcal L$ of subsets of $X$ satisfying three axioms:

(L1) for any distinct points $x,y\in X$ there exists a unique line $L\in\mathcal L$ containing $x$ and $y$;

(L2) every set $L\in\mathcal L$ contains at least three points;

(L3) $X\notin\mathcal L$.

Elements of the family $\mathcal L$ are called lines.

For any distinct points $x,y\in X$ of a linear space $(X,\mathcal L)$, the unique line $L\in\mathcal L$ containing the points $x,y$ will be denoted by $\overline{xy}$.

Lines $L_1,\dots,L_n$ in a linear space are called concurrent if $\bigcap_{i=1}^n L_i$ is a singleton.

Definition. A linear space $(X,\mathcal L)$ is called

$\bullet$ a projective plane if any distinct lines in $X$ are concurrent;

$\bullet$ an affine plane if for every line $L\in\mathcal L$ and point $x\in X\setminus L$ there exists a unique line $L'\in\mathcal L$ such that $x\in L'$ and $L'\cap L=\emptyset$.

Definition P. A projective plane $(X,\mathcal L)$ is called

$\bullet$ Desarguesian if for every concurrent lines $A,B,C\in\mathcal L$ and points $a,a'\in A\setminus(B\cup C)$, $b,b'\in B\setminus(A\cup C)$, $c,c'\in C\setminus(A\cup B)$, the set $(\overline{ab}\cap\overline{a'b'})\cup(\overline{ac}\cap\overline{a'c'})\cup(\overline{bc}
\cap \overline{b'c'})$
is contained in some line;

$\bullet$ Pappian if for every lines $L,L'\in\mathcal L$ and distinct points $a,b,c\in L\setminus L'$ and $a',b',c'\in L'\setminus L$, the set $(\overline{ab'}\cap\overline{a'b})\cup(\overline{ac'}\cap\overline{a'c})\cap(\overline{bc'}\cap\overline{b'c})$ is contained in some line.

By the famous Hessenberg's Theorem, every Pappian projective plane is Desarguesian.

Now let us define affine counterparts of Desarguesian and Pappian projective planes.
Given two lines $L,L'\in\mathcal L$ we write $L\parallel L'$ if $L=L'$ or $L\cap L'=\emptyset$.

Definition A. An affine plane $(X,\mathcal L)$ is called

$\bullet$ Desarguesian if for every concurrent lines $A,B,C\in\mathcal L$ and points $a,a'\in A\setminus(B\cup C)$, $b,b'\in B\setminus(A\cup C)$, $c,c'\in C\setminus(A\cup B)$,
if $\overline{ab}\parallel\overline{a'b'}$ and $\overline{bc}\parallel\overline{b'c'}$, then $\overline{ac}\parallel \overline{a'c'}$;

$\bullet$ Pappian if for every lines $L,L'\in\mathcal L$ and distinct points $a,b,c\in L\setminus L'$ and $a',b',c'\in L'\setminus L$, if $\overline{ab'}\parallel \overline{a'b}$ and $\overline{bc'}\parallel\overline{b'c}$, then $\overline{ac'}\parallel\overline{a'c}$.

Problem. Is every Pappian affine plane Desarguesian?

Remark. In the lecture notes of Adrien Deloro "Affine and Projective Geometry"
I found Theorem 6.2.1 which contains the proof of Hessenberg's Theorem and also mentions (unfortunately without a convincing proof) that the affine counterpart of the Hessenberg's Theorem also is true. So, what I would like to see is a correct and complete proof of the fact that Pappian affine planes are indeed Desarguesian. Is it published anywhere? If yes, then where exactly?

Best Answer

There is German book Plane geometry (2007) which contains equivalent adapted Hessenberg Theorem for affine planes.

It's too long to translate and adapt proof to your terminology, so here are screenshots of formulations and proof.

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