I was studying this paper on hyperbolic geometry by Jouni Parkkonen.
This is the part that confuses me (proposition 0.1): euclidean space is uniquely geodesic (page 2). The proof goes like this: let $z$ be an element in the image of a geodesic segment with extremes $x$ and $y$. Then $||x-z|| + ||z-y|| = ||x-y||$. But,
using the Cauchy inequality from linear algebra, it is easy to see that the Euclidean triangle inequality becomes an equality if and only if $z$ is in the image of the linear segment $j|_{[0,||x−y||]}$.
where $j(t)$ is $x+ t* \frac{y-x}{||y-x||}$.
I know that if $||x-z|| + ||z-y|| = ||x-y||$ holds, then the vectors $x-z$ and $z-y$ are multiples of each other. But that's not quite what I want, I want to show that $x-z$ is a multiple (with an appropiate constant) of $x-y$, because then $z = x +\lambda (y-x)$. So all in all, I don't understand that "if and only if".
Finally, even if we prove that, then what we would have shown is that for every $0 \leq t \leq ||x-y||$, there is a $0 \leq t' \leq ||x-y||$ such that $g(t) = x+ t'* \frac{y-x}{||y-x||}$. That still doesn't quite prove that $g(t) = x+ t* \frac{y-x}{||y-x||}$ for all appropiate $t$.. This follows from the equation $t = ||g(t)-x||$, that is common to all geodesic segments.
For some positive $\lambda, \mu$, $\mu(x-z) = \lambda(z-y)$ and so $(\mu+\lambda)z = \mu x+\lambda y$ and so (since $\lambda +\mu >0$) we have $z = \frac{\mu}{\mu+\lambda}x + \frac{\lambda}{\mu+\lambda}y$ : $z$ is a barycenter of $x$ and $y$, which amounts to saying that it is on the segment $[x,y]$ (if you want an explicit $t$ such that $z=x+t(y-x)$, simply put $t=\frac{\lambda}{\mu+\lambda}$).
Now you get $g(t) = x +t'(x-y)$ for some $t'$ and apply $||g(t)-x|| = t$ to get $t'=\frac{t}{||y-x||}$