16. (16 points) As shown in Figure 7, in rectangle $ABCD$, points $E$ and $F$ are on $AB$ and $AD$ respectively. Lines $EH$ and $FG$ are drawn parallel to each other, intersecting $BC$ and $CD$ at points $H$ and $G$ respectively. Let the lengths of $AB$ and $AD$ be $a$ and $b$ ($a < b$).
(1) If $AE = AF = 1$ and $EH \parallel AC$, does there exist such integers $a$ and $b$ such that the lengths of $CH$ and $CG$ are also integers? If they exist, find the values of $a$ and $b$; if not, explain why.
(2) Compare the areas $S_{\triangle AHG}$ and $S_{\triangle CEF}$, and explain your reasoning.
16. (1) Does not exist.

Since $A B=a, A D=b, A E=A F=1$, we have
$$
B E=a-1, D F=b-1 \text {. }
$$

Also, since $E H / / A C$, then $\frac{B H}{B C}=\frac{B E}{A B}$, which means
$$
\frac{b-C H}{b}=\frac{a-1}{a} \Rightarrow C H=\frac{b}{a} \text {. }
$$

Similarly, we get $C G=\frac{a}{b}$.
Since $a<b$, then $C G$ is not an integer, hence there do not exist integers $a$ and $b$ such that the lengths of $C H$ and $C G$ are both integers.
(2) Let $B E=m, D F=n$, then
$$
\begin{array}{l}
S_{\triangle C E F}=a b-\left(S_{\triangle B C E}+S_{\triangle D C F}+S_{\triangle A E F}\right) \\
=a b-\frac{1}{2}[b m+a n+(a-m)(b-n)] \\
=\frac{1}{2}(a b-m n) .
\end{array}
$$

Let $B H=p, D G=q$, then
$$
\begin{array}{l}
S_{\triangle A H G}=a b-\left(S_{\triangle A B H}+S_{\triangle A D C}+S_{\triangle C H C}\right) \\
=a b-\frac{1}{2}[a p+b q+(b-p)(a-q)] \\
=\frac{1}{2}(a b-p q) .
\end{array}
$$

It is easy to prove that $\triangle B E H \backsim \triangle D G F$.
Thus, we have $\frac{m}{p}=\frac{q}{n}$, which means $m n=p q$.
Therefore, $S_{\triangle A H G}=S_{\triangle C E F}$.