we prove that this triangle spans $\Pi$.

\[triangle-span\] Let $\Pi$ be a proper plane and let $k_1, k_2, k_3$ be the sides of a proper triangle on $\Pi$. For every (proper or improper) point $a\in\Pi$ there is a proper line through $a$ that intersects two of the sides $k_1,k_2,k_3$ in two distinct proper points.

Let $a_1, a_2, a_3$ be the vertices of our triangle of sides $k_1, k_2, k_3$ with $a_i\notin k_i$ for $i = 1, 2, 3$ and let $a\in\Pi$. If $a\in k_i$ for some $i=1,2,3$, then $k_i$ is the required line. So, assume that $a\notin k_1,k_2,k_3$.

By \[fact:basic\] there are at least $\fixind$ proper points on $k_1$ distinct from $a_2, a_3$. Let $p_1,\dots, p_\lambda$ be such points where $\fixind\le\lambda$. Take $l_i := \LineOn(a, p_i)$ for $i=1,\dots, \lambda$. Each line $l_i$ is proper and intersect $k_2$ in some point $q_i$. Note that all these points $q_i$ are pairwise distinct. If $\fixind < \lambda$, then there is $j=1,\dots, \lambda$ such that $q_j$ is a proper point as by there are only up to $\fixind$ improper points on $k_2$. Thus, the line $l_j$ is the required one and we are through. If $\fixind = \lambda$, then either $q_j$ is a proper 