This was a scenario I was trying to set up today. Suppose $V$ is an $n$-dimensional $\mathbb{R}$-vector space, and let $S$ be an $n-1$ dimensional subspace.
Then we can define a relation $\equiv$ on the set $V\setminus S$ by saying $u\equiv v$ if the 'segment' connecting them $ L(u,v)=\{\lambda u+(1-\lambda)v: \lambda\in[0,1]\} $ is such that $L(u,v)\cap S=\emptyset$.
It's not hard to see that $\equiv$ is reflexive and symmetric, but I can't show it is transitive. I believe that $V\setminus S=\{s+dv: s\in S, d\in \mathbb{R}^\times\}$, where $v$ is some fixed vector in $V\setminus S$, and $d$ is a nonzero scalar. I assumed that $u\equiv w$ and $w\equiv v$, so that $\lambda u+(1-\lambda)w\notin S$ and $\mu w+(1-\mu)z\notin S$ for any $\lambda,\mu\in[0,1]$, but I couldn't derive that $\rho u+(1-\rho)v\notin S$ for all $\rho\in[0,1]$.
I had the same difficulty showing that $\equiv$ partitions $V\setminus S$, into exactly two classes, corresponding to the two opposite 'sides' of $S$ in $V$.
Is there a way to show that $\equiv$ is transitive, and thus an equivalence relation that partitions $V\setminus S$ into two congruence classes? Thanks.