A corollary of the affine Desargues axiom

Question

Definition 1. An affine plane is a pair $(X,\mathcal L)$ consisting of a set $X$ and a family $\mathcal L$ of subsets of $X$ called lines which satisfy the following axioms:

1. Any distinct points $x,y\in X$ are contained in a unique line $L\in\mathcal L$;

2. Any line $L\in \mathcal L$ contains at least three points;

3. $X\notin \mathcal L$;

4. For every line $L\in\mathcal L$ and point $x\in X\setminus L$ there exists a unique line $\Lambda\in\mathcal L$ such that $x\in \Lambda$ and $\Lambda\cap L=\emptyset$.

For two distinct points $x,y\in X$ of an affine space $(X,\mathcal L)$ by $\overline{xy}$ we denote the unique line containing the points $x,y$.

Two lines $A,B\in\mathcal L$ are called parallel (denoted by $A\parallel B$ ) if either $A=B$ or $A\cap B=\emptyset$.

Definition 2. An affine plane $(X,\mathcal L)$ is called Desarguesian if it satisfies the affine Desargues axiom: for every lines $A,B,C\in\mathcal L$ with $A\cap B\cap C\ne\emptyset$ and any 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 $\overline{ac}\parallel \overline{a'c'}$.

Problem. Let $a,b,c,d$ be four distinct points in a Desarguesian affine plane $(X,\mathcal L)$ and $p,q,r,s$ be distinct points in $X$ such that $p\in\overline{ab}$, $q\in\overline{bc}$, $r\in\overline{cd}$, $s\in\overline{da}$, $\overline{pq}\parallel \overline{ac}$, $\overline{qr}\parallel \overline{bd}$, $\overline{rs}\parallel \overline{ac}$. Is $\overline{ps}\parallel \overline{bd}$?

Remark. I hope that the answer is affirmative. In this case, it would be desirable to have a direct proof from the axioms, which does not use the algebraization of Desarguesian affine planes (because I need this fact to prove that a Desarguesian affine plane is isomorphic to the affine plane over a skew-field).

Answer 1

Under one additional condition, the answer to this problem is affirmative. The proof involves the following implication of the Affine Desargues Axiom:

The Affine Moufang Axiom: for every parallel lines $A,B,C$ in an affine plane $(X,\mathcal L)$ and any 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 $\overline{ac}\parallel \overline{a'c'}$.

The implication of the Affine Moufang Axiom (called the little Desargues Theorem) from the Affine Desargues Axiom (called the Affine Desargues Theorem) is discussed in this MO-post.

Also we shall use the transitivity of the parallelity relation in the affine planes.

Proposition. Let $A,B,C$ be three lines in an affine plane $(X, \mathcal L)$. If $A\parallel B$ and $B\parallel C$, then $A\parallel C$.

Proof. If $A\nparallel C$, then the lines $A,C$ have a common point $b$. If $b\in B$, then $A\parallel B\parallel C$ implies $A=B=C$ and hence $A\parallel C$. So, $b\notin B$ and $A=C$ by the axiom 4 of an affine plane. $\quad\square$

Now we can prove the partial answer to the MO-problem.

Theorem. Let $a,b,c,d$ be distinct points of a Desarguesian affine space $(X,\mathcal L)$ and $p,q,r,s\in X$ be points such that $p\in\overline{ab}$, $q\in\overline{bc}$, $r\in\overline{cd}$ and $s\in\overline{ad}$. If the lines $\overline{ca}$, $\overline{cb}$, $\overline{cd}$ are distinct and $\overline{pq}\parallel \overline{ac}$, $\overline{qr}\parallel \overline{bd}$, $\overline{rs}\parallel \overline{ac}$, then $\overline{ps}\parallel \overline{bd}$.

Proof. It follows from $\overline{qr}\parallel \overline{bd}$ and $q\ne r$ that $q\ne c\ne r$ and hence $q\notin\overline{ac}$. By the axiom 4 of an affine space, there exists a unique point $t\in \overline{ac}$ such that $\overline{ab}\parallel \overline{qt}$. Since $\overline{tq}\parallel\overline{ab}$ and $\overline{qr}\parallel\overline{bd}$, the Affine Desargues Axiom guarantees that $\overline{tr}\parallel \overline{ad}$. Since $\overline{pq}$, $\overline{at}$, $\overline{sq}$ are three parallel lines, the Affine Moufang Axiom guarantees that $\overline{qr}\parallel \overline{ps}$ and hence $\overline{ps}\parallel \overline{bd}$, by the transitivity of the parallelity relation. $\square$

Remark: If $\overline{ca}=\overline{cb}\ne\overline{cd}$, then it is easy to find points $p,q\in \overline{ca}=\overline{cb}$, $r\in\overline{cd}$, and $s\in\overline{ad}$ such that $\overline{qr}\parallel \overline{bd}$ and $\overline{rs}\parallel\overline{ca}$, but $\overline{ps}\nparallel \overline{bd}$.