A Steiner Triple System is a set $\mathcal{S}$ of $v \geq 3$ elements together with a set $\mathcal{B}$ of $3$-subsets (triples) of $\mathcal{S}$ such that every $2$-subset of $\mathcal{S}$ occurs in exactly one triple of $\mathcal{B}$. As an example, the following forms a Steiner Triple System of order $7$ on the set $\{ 1,2,3,4,5,6,7 \}$.
$\mathcal{S}_7= \{ \{ 1,2,4\},\{2,3,5\},\{3,4,6\},\{4,5,7\},\{5,6,1\},\{6,7,2\},\{7,1,3\}\}$
This does not satisfy the parallel line postulate (where lines are represented here by the elements of $\mathcal{S}_7$), which I am defining here as: "Given a line $l$ and a point $P$ not on $l$, there is at least one line through $P$ and parallel (disjoint) to $l$".
Consider the point $1$ and the "line" $\{2,3,5\}$. Every "line" containing $1$ has a non-empty intersection with $\{2,3,5\}$, and thus the parallel line postulate proves false.
However, for a Steiner Triple System of order $9$, the parallel line postulate holds true. My question is, does this hold true for all Steiner Triple Systems of order $\geq 9$? My gut says yes, but I'm struggling with proving it. Thanks in advance for the help.
Consider a Steiner triple system on $13$ points. Consider a line $\ell=\{a,b,c\}$ and a point $p\notin\ell.$ There are exactly $6$ lines containing the point $p.$ Exactly one of those lines contains $a,$ one of them contains $b,$ one of them contains $c,$ and the other three are "parallel" to $\ell.$ This might or might not be a counterexample to the parallel line postulate, depending on just what statement you mean by the "parallel line postulate."
In general, if the Steiner triple system has $n$ points, then there are $\frac{n-1}2$ lines through $p;$ three of those lines intersect the line $\ell,$ and the remaining $\frac{n-7}2$ are "parallel." So there are no parallels if $n=7,$ exactly one parallel if $n=9,$ more than one parallel in all other cases.