= Solution
The defining formulas imply the matching-distance identities
$$
d(z,p)=d(z,r),
\qquad
d(x,p)=d(x,q),
\qquad
d(y,q)=d(y,r).
$$
For example,
$$
d(z,r)=d(y,z)-d(y,r)
=\frac{d(x,z)+d(y,z)-d(x,y)}2
=d(z,p),
$$
and the other two follow cyclically. Thus $p,q,r$ are the <tripod points of a geodesic triangle>.
By the <thin geodesic triangle> condition, $p$ lies within $\delta$ of some point $p'$ on $[x,y]$ or $[y,z]$. If $p'\in[x,y]$, then the <triangle inequality> gives
$$
|d(x,p')-d(x,p)|\leq\delta.
$$
Since $q$ is the point of $[x,y]$ at distance $d(x,p)$ from $x$, the geodesic parametrization gives $d(p',q)\leq\delta$, and hence $d(p,q)\leq2\delta$. If instead $p'\in[y,z]$, comparison of distances from $z$ gives $d(p,r)\leq2\delta$.
Applying the same argument cyclically, each of $p,q,r$ lies within $2\delta$ of at least one of the other two. The graph on these three points whose edges join pairs at distance at most $2\delta$ therefore has no isolated vertex, so it is connected. Any two vertices are joined by at most two edges, and the <triangle inequality> yields
$$
d(p,q),\ d(q,r),\ d(r,p)\leq4\delta.
$$
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