characterization of uniquely geodesic normed vector spaces I am trying to understand the proof of Proposition 1.1.6 in Bridson-Haefliger. They deal with the notion of geodesics in metric spaces, as per the definition here:
http://en.wikipedia.org/wiki/Geodesic#Metric_geometry
They note that given a normed vector space $V$, for any two vectors $u,v\in V$ the path $t\mapsto (1-t)u+tv$ is a geodesic path with respect to the metric $d(x,y)=\|x-y\|$, which they denote $[u,v]$. Then they formulate a property:
For any $v,v',v''$ in $V$, the condition $d(v,v')+d(v',v'')=d(v,v'')$ implies that $v'$ belongs to $[v,v'']$.
Then they claim that the normed vector space $V$ is uniquely geodesic if and only if the above property holds. 
I cannot see why $V$ being uniquely geodesic implies the above property. Any hints?
 A: The implication in question is proven in Athanase Papadopoulos's book "Metric Spaces, Convexity and Nonpositive Curvature", Proposition 7.2.1, (i)=>(ii)=>(iii)=>(iv). In a shortened form, the reasoning goes as follows.
Let $[v,v']$ be a geodesic parameterized by interval $[0,a]$, and $[v',v'']$ be a geodesic parameterized by $[0,b]$. Consider their concatenation $c=[v,v'].[v',v'']$ parameterized by the interval $[0,a+b]$. Let's show that $c$ is also a geodesic. For that, we need to prove that for any $t,t'\in [0,a+b]$, one has $d(c(t),c(t'))=|t-t'|$. If both $t$, $t'$ belong to $[0,a]$ or to $[a,a+b]$ then this is true due to the fact that both $[v,v']$ and $[v',v'']$ are geodesics. Now let $t\in[0,a]$ and $t'\in [a,a+b]$. Then $d(c(t),c(t'))\leq d(c(t),v')+d(v',c(t'))=(a-t)+(t'-a)=t'-t$. If the last inequality were strict, then $$d(v,v'')\leq d(v,c(t))+d(c(t),c(t'))+d(c(t'),v'') = t + d(c(t),c(t')) + (a+b-t') <$$
$$t + (t'-t) + (a+b - t') = a+b.$$ But $d(v,v'')=d(v,v')+d(v',v'')=a+b$, by assumption. The contradiction shows that $d(c(t),c(t'))=|t-t'|$ for all $t,t'$, which means that $c$ is a geodesic. Now, $V$ being a unique geodesic space, $c$ should coincide with $[v,v'']$, and therefore $v'$ belongs to $[v,v'']$.
