= Solution
Assume $u(0)=0$. Applying midpoint preservation to $2v$ and $0$ gives
$$
u(v)=\frac{u(2v)+u(0)}2=\frac12u(2v),
$$
so $u(2v)=2u(v)$. Consequently,
$$
u(v+w)
=u\left(2\frac{v+w}{2}\right)
=2u\left(\frac{v+w}{2}\right)
=u(v)+u(w).
$$
Thus $u$ is additive. It follows successively that
$$
u(nv)=nu(v)
\quad(n\in\mathbb Z),
\qquad
u(qv)=qu(v)
\quad(q\in\mathbb Q).
$$
An isometry is continuous. For any $\lambda\in\mathbb R$, choose rationals $q_j\to\lambda$. Then
$$
u(\lambda v)
=\lim_{j\to\infty}u(q_jv)
=\lim_{j\to\infty}q_ju(v)
=\lambda u(v).
$$
Together with additivity, this proves that $u$ is real-linear. Hence the origin-fixing case of the <Mazur-Ulam theorem> gives
$$
\boxed{u(av+bw)=au(v)+bu(w)}
$$
for all $a,b\in\mathbb R$ and $v,w\in V$.
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