Let $x,y,z$ three differents points in a vector space $E$, endowed with inner product. If $d(x,z)=d(x,y)+d(y,z)$, then $z-x=t(y-x)^{(*)}$, with $t\geq 1$.
My approach: Let $E$ a vector space with inner product, then $$\vert\vert y-z\vert\vert=\vert\vert (y-x)-(z-x)\vert\vert=\vert\vert (y-x) - t(y-x)\vert\vert^{(*)}=\vert\vert (y-x)(1-t)\vert\vert=$$ $$\vert (1-t)\vert \vert\vert y-x\vert\vert\implies t=1+\dfrac{\vert\vert y-z\vert\vert}{\vert\vert y-x\vert\vert}$$ $$\therefore \vert\vert z-x\vert\vert=\vert t\vert \vert\vert y-x\vert\vert=\left(1+\dfrac{\vert\vert y-z\vert\vert}{\vert\vert y-x\vert\vert}\right)\vert\vert y-x\vert\vert=\vert\vert x-y\vert\vert + \vert\vert y-z\vert\vert$$ $$\therefore d(x,z)=d(x,y)+d(y,z)$$
But in ${(*)}$, I occupy the hypothesis, therefore this is not rigurous. Any help pls
You want to show that if $\|x-z\|=\|x-y\|+\|y-z\|$ then $z-x$ and $y-x$ are colinear. Square the equation and deduce that $\langle z-x,y-x\rangle=\|z-x\|\|y-x\|$. You're done, because by CS-inequality we know that $|\langle a,b\rangle|=\|a\|\|b\|\iff a\parallel b$.