Equality case in metric spaces

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I have a question. Let $(X,d)$ a metric space. Then, for all $x,y \in X$, we have the triangle inequality : $d(x,y)+d(y,z)\geq d(x,z)$.

But, in what case we have the equality in the triangle inequality ? What is a necessary and sufficient condition for : $d(x,y)+d(y,z) =d(x,z)$ ?

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In general, there is no geometric condition equivalent to that equality.

You can sometimes state one if you have more information about the space. You should be able to see one for $\mathbb{R}^n$.