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!

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 $$|t| = t$$; $$|1 - t| = 1 - t$$.
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, $$\lVert x - z\rVert + \lVert z - y\rVert = \lVert z - x\rVert + \lVert y - z\rVert =\\ |t|\lVert y - x \rVert + |1-t| \lVert y - x \rVert = \lVert y - x \rVert$$