20. As shown in Figure 7, the side length of square $ABCD$ is 1, point $F$ moves on segment $CD$, and $AE$ bisects $\angle BAF$ intersecting side $BC$ at point $E$.
(1) Prove:
$$
AF = DF + BE \text{.}
$$
(2) Let $DF = x (0 \leqslant x \leqslant 1)$, and let $S$ be the sum of the areas of $\triangle ADF$ and $\triangle ABE$. Does $S$ have a maximum value? If it exists, find the value of $x$ and $S$; if not, explain why.
20. (1) As shown in Figure 11, extend $C B$ to point $G$ such that $B G = D F$, and connect $A G$.

Since quadrilateral $A B C D$ is a square, in the right triangles $\triangle A D F$ and $\triangle A B G$,
$$
\begin{array}{l}
A D = A B, D F = B G, \\
\angle A D F = \angle A B G = 90^{\circ}.
\end{array}
$$

Therefore, $\triangle A D F \cong \triangle A B G$.
Thus, $A F = A G$, and $\angle D A F = \angle B A G$.
Since $A E$ is the angle bisector of $\angle B A F$, then
$\angle E A F = \angle B A E$.
Hence, $\angle D A F + \angle E A F = \angle B A G + \angle B A E$, which means $\angle E A D = \angle G A E$.
By $A D \parallel B C \Rightarrow \angle G E A = \angle E A D$
$\Rightarrow \angle G E A = \angle G A E \Rightarrow A G = G E$
$\Rightarrow A G = B G + B E$.
Therefore, $A F = D F + B E$.
(2) $S = S_{\triangle M F} + S_{\text{ABE}}$
$$
= \frac{1}{2} D F \cdot A D + \frac{1}{2} B E \cdot A B.
$$

Since $A D = A B = 1$, we have
$S = \frac{1}{2}(D F + B E)$.
From (1), we know $A F = D F + B E$, so $S = \frac{1}{2} A F$.
In the right triangle $\triangle A D F$, $A D = 1$, $D F = x$, then
$A F = \sqrt{x^{2} + 1}$, $S = \frac{1}{2} \sqrt{x^{2} + 1}$.
From the above equation, we know that $S$ is maximized when $x^{2}$ is maximized.
Since $0 \leqslant x \leqslant 1$, when $x = 1$, the maximum value of $S$ is $\frac{1}{2} \sqrt{x^{2} + 1} = \frac{\sqrt{2}}{2}$.
(Zou Shouwen provided)