Converse part of triangular inequality hold equality when third point lies in the both point

Question

I was reading following proof.

In that I understand first part But In converse part I do not understand Why if equality hold then d(x,z) is scalar times d(z,y)? As shown in yellow highlight.
I do not find any reasoning for that.
Any Help will be appreciated

Answer 1

Let $u=x-z,v=z-y$ so $\|u+v\|=\|u\|+\|v\|$. Squaring both sides and expanding $\|u+v\|^{2}$ as $\langle u+v,u+v\rangle =\langle u,u\rangle+\langle v,v\rangle+2\langle u,v\rangle$ we see that $\langle u,v\rangle =\|u\|\|v\|$. Now consider $\|u-tv\|^{2}=\|u\|^{2}+\|v\|^{2}-2t\langle u,v\rangle=\|u\|^{2}+\|v\|^{2}-2t\|u\|\|v\|$ where $t$ is a real number. We can always choose $t\geq 0$ so that the right side is $0$ and then we get $u=tv$. So $x-z=t(z-y)$ which gives $z=\frac 1 {1+t} x +\frac t {1+t} y$. This means that $z$ lies in the line segment joining $x$ and $y$.