Q:

If $z$ is a point on the line $x+t(x-y)$ is it true that $\|x-y\|=\|x-z\|+\|z-y\|$?

In a normed space and given three points $x,y,z $ the norm satisfies the triangle inequality 
$$\|x-y \| \le \|x-z \| + \|z-y \|$$
But is also the intuitively obvious fact from $\mathbb R^2$ true that if $z$ is a point on the line $x+t(x-y)$, $0 < t < 1 $, between $x $ and $y $ then the distances betewwen $x $ and $z $ , and $z $ and $y $ add up to the distance between $x $ and $y $
$$\|x-y\|=\|x-z\|+\|z-y\|$$
? And how would we prove it?
Most grateful for any help!

A:

Since $z = x + t(x-y)$ then the expression
$$
\| x - y \| = \| x - z \| + \| z - y \| 
$$
becomes 
$$
\| x - y \| = \| x - x - t(x- y) \| + \| x + t(x-y) - y \| = |t| \| x - y \| + |1 - t| \| x - y \| \equiv \| x - y \|,
$$
where the last equality follows from the fact that $t \in (0, 1)$ and hence $|t| = t$; $|1 - t| = 1 - t$. 
QED.

A:

Yes this holds, here's the proof :
Let $t \in [0;1]$, $z:= x + t (y-x)$.
One has $z - x = t(y-x)$, and also $y - z = (1-t)(y-x)$. Hence,
$$\lVert x - z\rVert  + \lVert z - y\rVert = \lVert z - x\rVert  + \lVert y - z\rVert =\\
|t|\lVert y - x \rVert + |1-t| \lVert y - x \rVert = \lVert y - x \rVert$$

