(199 As shown in the figure, in the right-angled coordinate plane $x O y$, $\triangle A_{i} B_{i} A_{i+1}(i=1,2, \cdots, n, \cdots)$ are equilateral triangles, and satisfy $\overrightarrow{O A}_{1}=\left(-\frac{1}{4}, 0\right), \overrightarrow{A_{i} A_{i+1}}=$ $(2 i-1,0)$.
(1) Prove that points $B_{1}, B_{2}, \cdots, B_{n}, \cdots$ lie on the same parabola $\Gamma$, and find the equation of this parabola;
(2) Through the focus $F$ of the parabola $\Gamma$ found in (1), draw two mutually perpendicular chords $A C$ and $B D$. Find the minimum area of the quadrilateral $A B C D$.
(19) (1) Let the coordinates of point $B_{n}$ be $(x, y)$, then
$$
\left\{\begin{array}{l}
x=-\frac{1}{4}+1+3+\cdots+(2 n-3)+\frac{2 n-1}{2}=\left(n-\frac{1}{2}\right)^{2} \\
|y|=\frac{\sqrt{3}}{2}(2 n-1)
\end{array}\right.
$$

Eliminating $n$, we get $y^{2}=3 x$. Therefore, points $B_{1}, B_{2}, \cdots, B_{n}, \cdots$ lie on the same parabola $\Gamma$ : $y^{2}=3 x$.
(2) It is easy to see that the focus of the parabola $\Gamma$ is $F\left(\frac{3}{4}, 0\right)$, and neither $A C$ nor $B D$ is perpendicular to the coordinate axes. Let the equation of line $A C$ be $x=m y+\frac{3}{4}$, substituting into $y^{2}=3 x$, we get
$$
y^{2}-3 m y-\frac{9}{4}=0 .
$$

By Vieta's formulas, we have
$$
y_{A}+y_{C}=3 m, y_{A} \cdot y_{C}=-\frac{9}{4} .
$$

Thus,
$$
\begin{aligned}
|A C| & =\sqrt{1+m^{2}}\left|y_{A}-y_{C}\right| \\
& =\sqrt{1+m^{2}} \cdot \sqrt{\left(y_{A}+y_{C}\right)^{2}-4 y_{A} y_{C}} \\
& =3\left(1+m^{2}\right) .
\end{aligned}
$$

Replacing $m$ with $-\frac{1}{m}$ in the above equation, we get
$$
|B D|=3\left(1+\frac{1}{m^{2}}\right)
$$

Therefore,
$$
\begin{aligned}
S_{A B C D} & =\frac{1}{2}|A C| \cdot|B D| \\
& =\frac{9}{2}\left(1+m^{2}\right)\left(1+\frac{1}{m^{2}}\right) \\
& \geqslant \frac{9}{2} \cdot 2 m \cdot \frac{2}{m} \\
& =18
\end{aligned}
$$

The equality holds if and only if $m= \pm 1$. Hence, the minimum area of quadrilateral $A B C D$ is 18.