proper points.

Let $a_m\in k_i\cap k_j$, where $\{m,i,j\} = \{1,2,3\}$, be the vertices of our triangle. Without loss of generality we can assume that the point $a_3$ is improper. By \[fact:basic\] we can take proper points $a\in k_1$ and $b_1,\dots, b_{\fixind+1}\in k_2$ such that $a\neq a_2$ and $b_i\neq a_1$ for all $i=1,\dots,\fixind+1$. Consider the lines $l_i := \LineOn(a, b_i)$ for $i=1,\dots,\fixind+1$. All these lines are proper and all they intersect the line $k_3$. As there are only up to $\fixind$ improper points on $k_3$, one of $l_i$ is the required line.

Now, we are able to give a definition of $\parallel$ purely in terms of $\fixout$.

\[prop:parallel\] Let $k_1, k_2\in\linesout$ with $k_1\neq k_2$. Then $$\begin{gathered}
\label{parallel:veblen}
    k_1\parallel k_2 :\iff (\exists k_3, l\in\linesout) (\exists c_1, c_2, c_3\in\pointsout)
      \bigl[\, \triangle(k_1, k_2, k_3) \Land \\
         c_1\in l, k_1 \Land c_2\in l, k_2 \Land c_3\in l, k_3 \,\bigr]
           \Land (\nexists\; x\in\pointsout)[\, x\in k_1, k_2 \,].
  \end{gathered}$$

Since $k_1, k_2$ are parallel we have $\emptyset\neq k_1\cap k_2\in\W$. So, $k_1, k_2$ share an improper point. Take two proper points $a_1\in k_1$ and $a_2\in k_2$. They must be distinct and the line $k_3 := \LineOn(a_1, a_2)$ is proper. We have $\triangle(k_1, k_2, k_3)$ with one improper vertex. 