In $\triangle A B C$, $\angle C=30^{\circ}, B M$ is a median, $A C=2 a$. If $\triangle A B C$ is folded along $B M$, the overlapping part of the two smaller triangles $A B M$ and $B C M$ is exactly one quarter of the area of $\triangle A B C$. Find the area of $\triangle A B C$.
iii) Solve by discussing two scenarios.
i) If in the original triangle, $\angle A B M >$ $\angle C B M$, after folding as shown in Figure (1), where $C^{\prime}$ is the position where point $C$ lands after folding. Then solve to get $S \triangle A_{B C}=\frac{\sqrt{3}}{2} a^{2}$.
ii) If

in the original triangle,
$\angle A B M$
$$
<\angle C B M \text {, }
$$

after folding as shown in Figure (2). Then solve to get $S \triangle A B C=\frac{a}{2}$.