point for some $j=1,\dots, \lambda$ and $l_j$ is the required line, or all the points $q_i$, and only those, are improper points of $k_2$. In the latter case take $l := \LineOn(a, a_2)$. It is a proper line as $a_2$ is proper and it intersect $k_2$ in a point distinct from $q_i,\dots, q_\lambda$ thus, in a proper point as required.

Immediately from \[triangle-span\] we get

\[cor:planes\] Proper planes are definable in terms of $\fixout$.

Next lemmas may look technical and superfluous until Section \[sec:cliques\].

\[lem:bundle\] There are at least three pairwise distinct non-coplanar proper lines through every (improper) point of $\fixproj$.

Let $a$ be a point of $\fixproj$. If $a$ is proper, then the claim follows directly by \[fact:basic\]. So, assume that $a$ is improper. By \[fact:basic\] there is some proper point $a_1$, thus $a\neq a_1$. Every line through $a_1$ is proper so, it is $k_1 := \LineOn(a, a_1)$. Take another line $l$ through $a_1$ distinct from $k_1$. By \[fact:basic\] there is a proper point $a_2\in l$ with $a_1\neq a_2$. The line $k_2 := \LineOn(a, a_2)$ is proper, $k_1\neq k_2$, and $a\in k_1, k_2$. So, we have a proper plane $\Pi = \gen{k_1, k_2}$. Now, as $\fixproj$ is at least 