Problem: Let $\displaystyle{S}$ be a subset of $\displaystyle{\mathbb{E}^{n}}$. Suppose that $\displaystyle{S}$ contains at least three points that are not collinear, and suppose that for every three noncollinear points $\displaystyle{x , y , z}$ in $S$ there exists a unique point $p$ in $S$ such that $\displaystyle{\overline{xp} \text{ , } \overline{yp}}$, and $\displaystyle{\overline{zp}}$ all lie in $S$. Prove that the kernel of $S$ consists of a single point.
Solution: The kernel of the set $S$ is $\displaystyle{K = \ker(S) = \big \{ x \in S : \overline{x y} \in S , \forall y \in S \big \}}$. \ We know that: $K$ is convex set. \ Since $\displaystyle{S}$ contains at least three points $\displaystyle{x , y , z \in S}$ that are not collinear, then there exists unique $p \in S$ such that $\displaystyle{\overline{p x} \text{ , } \overline{p y} \text{ , } \overline{p z} \in S }$. \ Suppose by contradiction that: $|K| \geqslant 2$. Let $\displaystyle{a , b \in K}$. \ But then $\displaystyle{ \begin{cases} \overline{a x} \text{ , } \overline{a y} \text{ , } \overline{a z} \in S \\ \\ \overline{b x} \text{ , } \overline{b y} \text{ , } \overline{b z} \in S \end{cases}}$, which is a contradiction, because there is an unique $\displaystyle{p}$ with this property. Therefore $\displaystyle{| K | \leqslant 1 \Longrightarrow K = \varnothing}$ or $\displaystyle{| K | = 1}$.
Assume by contradiction that: $\displaystyle{K = \varnothing \Longrightarrow \forall u \in S : \exists v \in S , \overline{u v} \notin S}$.
My question is: How can we prove that $|K|=1$?
I was thinking that $p \in K$, but is it correct?