# Parallel lines in 2-simplex

I do have a problem in understanding a statement in the following argumentation.

Consider a 2-simplex $\Delta := \{ (x_1, x_2) : x_1, x_2 \geq 0, x_1+x_2\leq 1 \}$.

Assume that for every $P,Q,R \in \Delta$, and for every $\alpha \in (0,1)$, $P \succsim Q$ iff $\alpha P + (1- \alpha)R \succsim \alpha Q + (1 - \alpha ) R$.

Take two points $P,Q \in \Delta$ such that $P \sim Q$, and construct another point $R$ such that $S= R + (P-Q)$ is also in $\Delta$.

Then, $R \sim S$.
Moreover, by $P-Q = R-S$, the indifference curve going through $R,S$ is parallel to the one going through $P,Q$.

The problematic statement is the one in bold. Indeed, why should we see that the indifference curves are parallels?
Moreover, isn't problematic the fact that $P-Q$ can actually be outside the 2-simplex?
I think the answer lies in some argument related to affine sets or so, but I do not really know.

As always, thanks in advance for any help or feedback!

PS: For completeness and reference sake, this argumentation comes from the Expected Utility Theorem by von Neumann and Morgenstern).

EDIT: Here there is a reasoning.

Simply, $P - Q = R - S$ means that the translation of $P$ by $Q$ (namely $P-Q$) and the translation of $R$ by $S$ (that is, $R-S$) are equal. We have to remember that $P$ and $Q$ are on the same indifference curve (line), and the same applies to $R$ and $S$. Said so, take the two flats $P$ and $R$: they are parallel, because one is the translate of the other. That is, $P= R - S + Q$. Thus, for instance $R$ is parallel also to $Q$ by $P \sim Q$. The same reasoning applies to $S$.

However I feel that it still doesn't not fill all the gaps. Indeed, why is $P-Q = R-S$ the point that make us infer that they are parallel and not simply the manipulation $P = R-S + Q$ of the initial statement?

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