Halfspace separation for the local-geodesic mesh

Fix $y_i\in D(x_i,x_1)$ for $i=0,2$, and write $E_i=d(x_i,y_i)$, $\eta =d(y_0,y_2)$. Let $u$ be the midpoint of a chosen segment from $y_0$ to $y_2$. Triangle inequalities and the halfspace assumptions give $d(u,x_i)\le E_i+\eta \mathrm{/}2$ and $d(u,x_1)\ge E_i-\eta \mathrm{/}2$ for each $i=0,2$.

For every $h>0$, slimness of triangle $(x_0,u,x_2)$ at $x_1$ supplies $v$ on $[x_0,u]$ or $[x_2,u]$ with $d(v,x_1)<\delta +h$. Suppose it is on $[x_i,u]$. Then $d(u,v)>E_i-\eta \mathrm{/}2-\delta -h$, while $d(x_i,u)\le E_i+\eta \mathrm{/}2$, so $d(x_i,v)<\eta +\delta +h$. Therefore $3\delta =d(x_i,x_1)<\eta +2\delta +2h$. Since this holds for every $h>0$, $\eta \ge \delta$.

Taking the infimum over $y_0,y_2$ gives the first bound; both sets are nonempty since they contain $x_0,x_2$, respectively. They are disjoint because $\delta >0$. Thus any $z\in D(x_2,x_1)$ does not belong to $D(x_0,x_1)$, which is precisely the asserted strict inequality.

If $y\in D(x_1,x_0)$, then $3\delta \le d(x_0,y)+d(y,x_1)\le 2d(x_0,y)$, proving the endpoint bound by taking an infimum. Interchanging $x_0,x_2$ gives the other bound. No closest point to a halfspace or bisector has been selected, and all segment selections are finite.