The radar round trip of an inertial observer halves at the midpoint of the reception interval

Set $d:=q-e_{-}$ and $r:=e_{+}-q$, so that $d+r=e_{+}-e_{-}=({\lambda }_{+}-{\lambda }_{-})n$. By [F1] write $d=-\eta (n,d)n+h(d)$ with $h(d)\in n^{\perp }$; by [F2] and [F3] a nonzero vector $v$ is future null exactly when $\eta (v,v)=0$ and $\eta (n,v)<0$, and for such a $v$ the decomposition satisfies $\eta (n,v{)}^2=\mathrm{\mid }h(v){\mathrm{\mid }}^2$ by [F3] applied to $\eta (v,v)=0$. Also $r=({\lambda }_{+}-{\lambda }_{-})n-d=({\lambda }_{+}-{\lambda }_{-}+\eta (n,d))n-h(d)$ by [F1], because $-d=\eta (n,d)n-h(d)$.

(ii) implies (i): assume $q=o+\frac{{\lambda }_{-}+{\lambda }_{+}}{2}n+u$ with $u\in n^{\perp }$ and $\mathrm{\mid }u\mathrm{\mid }=({\lambda }_{+}-{\lambda }_{-})\mathrm{/}2=:\rho$. Then $q-e_{-}=\rho n+u$ by the affine action, and by [F3] $\eta (q-e_{-},q-e_{-})=-{\rho }^2+\mathrm{\mid }u{\mathrm{\mid }}^2=0$ while $\eta (n,q-e_{-})=-\rho <0$, so $q-e_{-}$ is future null by [F2]; likewise $e_{+}-q=\rho n-u$ satisfies $\eta (e_{+}-q,e_{+}-q)=-{\rho }^2+\mathrm{\mid }u{\mathrm{\mid }}^2=0$ and $\eta (n,e_{+}-q)=-\rho <0$, so $e_{+}-q$ is future null.

(i) implies (ii): assume $d$ and $r$ are future null. By step 1.1 applied to $d$ one has $\eta (n,d)<0$; put $\mu :=-\eta (n,d)>0$, so $d=\mu n+h(d)$ and $\mathrm{\mid }h(d){\mathrm{\mid }}^2=\eta (n,d{)}^2={\mu }^2$, hence $\mathrm{\mid }h(d)\mathrm{\mid }=\mu$ by [F1]. Applying step 1.1 to $r$, whose expression is $({\lambda }_{+}-{\lambda }_{-}-\mu )n-h(d)$: its future nullness gives $\eta (n,r)=-({\lambda }_{+}-{\lambda }_{-}-\mu )<0$, and $\eta (r,r)=-({\lambda }_{+}-{\lambda }_{-}-\mu {)}^2+\mathrm{\mid }h(d){\mathrm{\mid }}^2=0$ by [F3]. Hence $({\lambda }_{+}-{\lambda }_{-}-\mu {)}^2={\mu }^2$, and since $\eta (n,r)=-({\lambda }_{+}-{\lambda }_{-}-\mu )<0$ we have ${\lambda }_{+}-{\lambda }_{-}-\mu >0$, so the positive root applies: ${\lambda }_{+}-{\lambda }_{-}-\mu =\mu$, that is $\mu =({\lambda }_{+}-{\lambda }_{-})\mathrm{/}2$. Therefore $q=e_{-}+d=o+({\lambda }_{-}+\mu )n+h(d)=o+\frac{{\lambda }_{-}+{\lambda }_{+}}{2}n+u$ with $u:=h(d)\in n^{\perp }$, $\mathrm{\mid }u\mathrm{\mid }=\mu =({\lambda }_{+}-{\lambda }_{-})\mathrm{/}2$; this is (ii).

Midpoint and distance: assume (i), hence (ii) with $q=o+\frac{{\lambda }_{-}+{\lambda }_{+}}{2}n+u$ as produced in step 2.1. Put $m:=o+\frac{{\lambda }_{-}+{\lambda }_{+}}{2}n$; since $e_{\pm }=o+{\lambda }_{\pm }n$, the displacement identity $m-e_{-}=\frac{{\lambda }_{+}-{\lambda }_{-}}{2}n=e_{+}-m$ shows that $m$ is the affine midpoint $\frac{e_{-}+e_{+}}{2}$. Then $q-m=u\in n^{\perp }$, so $\eta (n,q-m)=0$, that is $q\in H(m,n)$; and $d_n(m,q)=\sqrt{\eta (u,u)}=\mathrm{\mid }u\mathrm{\mid }=({\lambda }_{+}-{\lambda }_{-})\mathrm{/}2$ by [F4] and step 2.1.

Conclusion: step 2.1 proves (i)$\Rightarrow$(ii) and step 1.2 proves (ii)$\Rightarrow$(i), so the two conditions are equivalent; step 3.1 shows that whenever they hold the reflection event satisfies $q-e_{-}=\frac{{\lambda }_{+}-{\lambda }_{-}}{2}n+u$ and $e_{+}-q=\frac{{\lambda }_{+}-{\lambda }_{-}}{2}n-u$ with the same horizontal vector $u$, lies in the slice $H(m,n)$ of the fixed midpoint $m=\frac{e_{-}+e_{+}}{2}$, and has the fixed rest distance $({\lambda }_{+}-{\lambda }_{-})\mathrm{/}2$. Since the data $e_{-}$, $e_{+}$ determine $m$ and the norm $\rho =({\lambda }_{+}-{\lambda }_{-})\mathrm{/}2$, two admissible reflection events differ only by the choice of $u\in n^{\perp }$ with $\mathrm{\mid }u\mathrm{\mid }=\rho$; such $u$ exist because $n^{\perp }$ is three-dimensional and $\rho >0$. This is the full assertion.