Prove that $z = tx + (1 − t)y$ if $d(x, y)= d(x, z) + d(z, y)$

Question

Let $x,y,z$ be elements of $\mathbb{R}^2$ Prove that $z = tx + (1 − t)y$ if $d(x, y)= d(x, z) + d(z, y)$ d is usual euclidean metric.

Answer 1

What can we use about the usual metric?

Heron's formula... If the sides of a triangle are $a,b,c$, then the area of the triangle is $$ A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} $$ Now if $a=b+c$, then that area is zero.