= Solution
Let $R=\{B_t:0\leq t\leq1\}$ 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, giving
$$
\lambda(A_1)\ \stackrel d=\ h\lambda(R)=\tfrac12\lambda(R).
$$
The increments $B_{h+t}-B_h$, $0\leq t\leq h$, form another standard <Brownian motion>. Its range has the same law as $A_1$, and translation by $B_h$ does not change its area. Hence
$$
\boxed{\lambda(A_1)\ \stackrel d=\ \lambda(A_2)\ \stackrel d=\ \tfrac12\lambda(R).}
$$
The equality concerns the laws of the areas. No <independence> between these two original areas is needed here.
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