Another implication 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 $o,a,b,c,x,y,z,p$ be eight distinct points in a Desarguesian affine plane $(X,\mathcal L)$ such that the lines $\overline{oc}$ and $\overline{oz}$ are distinct, $a,b\in \overline{oc}$ and $x,y\in\overline{oz}$, $\overline{xp}{\parallel} \overline{oc}$, $\overline{ap}{\parallel} \overline{oz}$, $\overline{xa}{\parallel} \overline{yb}$, $\overline{xb}{\parallel} \overline{pc}$, and $\overline{ay}{\parallel}\overline{pz}$. Is $\overline{zc}\parallel \overline{yb}$?

Remark 1. 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 introduce the operation of addition of scalars).

Remark 2. @AlexRavsky observed that in order to solve Problem 1 affirmatively, it suffices to prove that the lines $\overline{ay}$, $\overline{bx}$ and $\overline{op}$ have a common point $q$. Unfortunately, this is not always true (at least without the additional information provided by the points $c$ and $z$) as shown in the following picture.

In this case, however $c=z=o$, which contradicts the choice of the points $c$ and $z$. So, maybe in case $c\ne o\ne z$, the point $q\in\overline{op}\cap\overline{ay}\cap\overline{bx}$ indeed exists.

Answer 1

Just to close this question as answered, I will give a sketch of the proof of the parallelity of the lines $\overline{zc}$ and $\overline{yb}$.

As was noticed by @AlexRavsky, it suffices to prove that the lines $\overline{ay}$, $\overline{bx}$ and $\overline{op}$ have a common point $q$. Then we can apply the Affine Desargues Axiom to the triangles $pzc$ and $qyb$ and conclude that $\overline{zc}\parallel \overline{yb}$.

To prove that a point $q\in \overline{ay}\cap\overline{bx}\cap \overline{op}$ exists, we first prove that the lines $\overline{ay}$ and $\overline{bx}$ have a common point. In the opposite case the lines $\overline{ay}$ and $\overline{bx}$ are parallel and the quadruple $aybx$ is a parallelogram whose diagonals $\overline{ab}$ and $\overline{xy}$ intersect at the point $o$.

Since $oxpa$ is a parallelogram and $\overline{pc}\parallel \overline{xb}\parallel\overline{ay}$ the point $c$ coincides with the point $o$, which contradicts the choice of the (distinct) points $o,a,b,c,x,y,z$. This contradiction shows that the lines $\overline{ay}$ and $\overline{bx}$ have a common point $q$.

Using the Playfair Axiom, find a point $q'$ such $\overline{aq'}\parallel \overline{qa}$ and $\overline{xq'}\parallel \overline{qx}$. So that $q'aqx$ is a parallelogram. Using the Inverse Desargues Theorem (that follows from the Affine Desargues Axiom), it can be shown that the point $q'$ belongs to the line $\overline{oq}$.

It is known that the Affine Desargues Axiom implies its "parallel'' version (in which the lines $A,B,C$ are parallel). In affine planes satisfying this weaker "parallel'' version of the Affine Desargues Axiom, the notion of a vector (as an equivalence class of pairs of points) can be introduced. Moreover, we can define the operation of addition of vectors and show that it is associative and commutative.

Taking into account that $oapx$ and $q'aqx$ are parallelograms, we can see that $$\overrightarrow{pq}=\overrightarrow{qx}+\overrightarrow{xp}=\overrightarrow{aq'}+\overrightarrow{oa}=\overrightarrow{oa}+\overrightarrow{aq'}=\overrightarrow{oq'}$$ and hence the point $p$ belongs to the line $\overline{oq}$.

Therefore, the answer to the original question is affirmative but the proof is a bit complicated (in fact, we can fulfil the last step without vectors, just using the "parallel" Affine Desargues Theorem, but in this case more lines and triangles should be drawn).