Minkowski functional: proving $N(av)=|a|N(v)$

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This question is related to this one Norm in terms of an infimum obtained from a subset of a normed vector space.

It says:

Take a convex and bounded subset $A$ of a normed vector space $V$ over $\mathbb{C}$, so that the closed ball (defined with the norm of V) $B[0, r] \subset A$ for some $r>0$.

We have as well that if $v \in A$, then $-v \in A .$

We want to show that the map $N: V \rightarrow [0, 1)$, $ N(v) = \inf_k \left\{k>0: v/k \in A\right\}$ is another norm on $V$.

I have shown that (i) $N(v)=0$ iff $v=0$, and that (ii) $N(v+u) \le N(v)+N(u)$, but I'm having troubles showing that (iii) $N(av)=|a|N(v)$ with the given hypothesis.

I have already used convexness, boundedness and the closed ball to get (i) and (ii). But I don't see how to get (iii) with those assumptions and the symmetric property of $A$.