One. (20 points)
As shown in the figure, the square $E F G H$ is inscribed in $\triangle A B C$. Let $B C=\overline{a b}$ (this is a... digit number), $E F=c$, and the altitude $A D=d$. It is known that $a, b, c, d$ are exactly four consecutive positive integers in ascending order. Try to find the area of $\triangle A B C$.
From the figure, we know that
$S_{\triangle A B C}=S_{\triangle A E F}+S_{\triangle B E N}+S_{\triangle F C C}+S_{\text {square HEEFGH }}$, that is, $\frac{1}{2} \overline{a b} \times d=\frac{1}{2} c(d-c)+\frac{1}{2}(\overline{a b}-c) c+c^{2}$.
$$
\begin{array}{l}
\overline{a b} \times d=(\overline{a b}+d) c, \\
\therefore c=\frac{\overline{a b} \times d}{\overline{a b}+d} .
\end{array}
$$

Given $\overline{a b}=10 a+b, b=a+1, c=a+2, d=a+3$,
$$
\begin{array}{l}
\therefore a+2=\frac{(10 a+a+1)(a+3)}{10 a+a+1+a+3}, \\
(a+2)(12 a+4)=(11 a+1)(a+3) .
\end{array}
$$

Simplifying, we get $a^{2}-6 a+5=0$.
$\therefore a=1$ or 5.
When $a=1$, $S_{\triangle A B C}=\frac{1}{2} \overline{a b} \times d=\frac{1}{2} \times 12 \times 4=24$.
When $a=5$, $S_{\triangle A B C}=\frac{1}{2} \times 56 \times 8=224$.