The four point condition implies slim triangles

Write $\mathrm{\ell }=d(a,b)$, $t=d(a,p)$ and $\alpha =(b\mathrm{\mid }c{)}_a$. Suppose first that $t\le \alpha$. Since $\alpha \le d(a,c)$ there is $q\in [a,c]$ with $d(a,q)=t$. Products along a radial geodesic give $(p\mathrm{\mid }b{)}_a=t$ and $(c\mathrm{\mid }q{)}_a=t$. Applying the product inequality first through $b$, then through $c$, gives $(p\mathrm{\mid }c{)}_a\ge t-\kappa$ and $(p\mathrm{\mid }q{)}_a\ge t-2\kappa$. Hence $d(p,q)=2t-2(p\mathrm{\mid }q{)}_a\le 4\kappa$.

If $t\ge \alpha$, use $b$ as basepoint. Indeed $(a\mathrm{\mid }c{)}_b=\mathrm{\ell }-\alpha$ by expansion, and $d(b,p)=\mathrm{\ell }-t\le \mathrm{\ell }-\alpha$. The argument of step 1.1 with $a$ and $b$ interchanged gives a point on $[b,c]$ at distance at most $4\kappa$ from $p$. At $t=\alpha$ either construction works.

Thus every point of $[a,b]$ is within $4\kappa$ of the other two sides; relabeling vertices proves this for all sides and all specified triangles. Zero side lengths require only $t=0$, and when $\kappa =0$ the produced point equals $p$. There were only finitely many segment choices.