Solution: Given that $d$ is a metric on a normed linear space $V$. We need to show that given $x,y,z,\in V$ such that \[ d(x+z,y+z) = d(x,y). \] Since the metric $d$ is induced by norm on $V$, so \[ d(x,y) = \lVert x-y \rVert. \] Now note that, \begin{align*} d(x+z,y+z) & = \lVert (x+z)-(y+z) \rVert \\ & = \lVert x-y \rVert \\ & = d(x,y). \end{align*} Thus, \[ \textcolor{blue}{\boxed{ d(x+z,y+z) = d(x,y) }} \]