(14) Let $F$ be the focus of the parabola $y^{2}=4 x$, and let $A$ and $B$ be two points on the parabola different from the origin $O$, satisfying $\overrightarrow{F A} \cdot \overrightarrow{F B}=0$. Extend $A F$ and $B F$ to intersect the parabola at points $C$ and $D$ (as shown in the figure). Find the minimum value of the area of quadrilateral $A B C D$.
(14) Let $A\left(x_{1}, y_{1}\right)$ and $C\left(x_{2}, y_{2}\right)$. From the problem, we know that the slope of line $AC$ exists and is denoted as $k$. Since line $AC$ passes through the focus $F(1,0)$, the equation of line $AC$ is $y=k(x-1)$. Solving the system of equations
$$
\left\{\begin{array}{l}
y=k(x-1), \\
y^{2}=4 x
\end{array}\right.
$$

eliminating $y$ gives
$$
k^{2} x^{2}-2\left(k^{2}+2\right) x+k^{2}=0 .
$$

By the relationship between roots and coefficients, we have:
$$
x_{1}+x_{2}=\frac{2 k^{2}+4}{k^{2}}, \quad x_{1} x_{2}=1 .
$$

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

Since $AC \perp BD$, the slope of line $BD$ is $-\frac{1}{k}$, so the equation of line $BD$ is $y=-\frac{1}{k}(x-1)$. Similarly, we can find that $|BD|=4\left(1+k^{2}\right)$. Therefore,
$$
\begin{aligned}
S_{A B C D} & =\frac{1}{2}|A C| \cdot|B D| \\
& =\frac{8\left(1+k^{2}\right)^{2}}{k^{2}} \\
& =8\left(k^{2}+\frac{1}{k^{2}}+2\right) \\
\geqslant & 8 \times(2+2) \\
& =32,
\end{aligned}
$$

with equality when $k= \pm 1$.
Thus, the minimum area of quadrilateral $ABCD$ is 32.