 Solution

Let $R=\{B_t:0\le t\le 1\}$ and $h=1/2$. For planar Brownian motion, the process $h^{-1/2}{B}_{ht}$ is again standard Brownian motion. Thus its time-$h$ range has law $\sqrt{h}R$. Two-dimensional Lebesgue measure scales by the square of the spatial factor, givingThe increments $B_{h+t}-B_h$, $0\le t\le h$, form another standard Brownian motion. Its range has the same law as $A_1$, and translation by $B_h$ does not change its area. HenceThe equality concerns the laws of the areas. No independence between these two original areas is needed here.