As shown in Figure 4, let quadrilateral $ABCD$ be a square, $E$ be any point on side $AD$ (not coinciding with $A$ or $D$), and $EC$ intersect $BD$ at point $F$. Let $x$, $y$, and $z$ be the areas of the shaded regions.
(1) Try to find the equation that $x$, $y$, and $z$ satisfy;
(2) If the values of $x$ and $y$ are given, try to find the area of square $ABCD$ (expressed in terms of $x$ and $y$).
(3) If a positive number $k$ is given, try to find the value of $\frac{DE}{EA}$ such that $y=kx$ (expressed in terms of $k$).
(1) There must be $x+y=z$. The reason is as follows.
From $z=S_{\triangle F B C}=S_{\triangle E B C}-S_{\triangle E B F}$
$$
=S_{\triangle A B D}-S_{\triangle E B F}=S_{\triangle D E F}+S_{\triangle A B E}=x+y,
$$

Therefore, the three numbers $x, y, z$ should satisfy the equation $x+y=z$.
(2) From $B C / / E D$, we get $S_{\triangle E B C}=S_{\triangle D B C}$.
Thus, $S_{\triangle E F B}=S_{\triangle C D F}=w$.
Additionally, $\frac{S_{\triangle D E F}}{S_{\triangle D D C F}}=\frac{E F}{F C}=\frac{S_{\triangle B E F}}{S_{\triangle B F C}}$, which means
$$
\frac{x}{w}=\frac{w}{z}, w^{2}=x z=x(x+y) \text {. }
$$

Therefore, $w=\sqrt{x(x+y)}$.
Thus, $S_{\text {quadrilateral } A B C D}=x+y+z+2 w$
$$
=2(x+y+w)=2[x+y+\sqrt{x(x+y)}] \text {. }
$$
(3) Let $A B=a$. From (2) and $y=k x$, we get
$$
\begin{array}{l}
a^{2}=2[x+y+\sqrt{x(x+y)}] \\
=2\left[\frac{y}{k}+y+\sqrt{\frac{y}{k}\left(\frac{y}{k}+y\right)}\right], \\
=2 y\left(\frac{1}{k}+1+\frac{\sqrt{k+1}}{k}\right) \\
=2 \times \frac{1}{2} a A E\left(\frac{1}{k}+1+\frac{\sqrt{k+1}}{k}\right) \\
=a A E\left(\frac{1}{k}+1+\frac{\sqrt{k+1}}{k}\right),
\end{array}
$$

Thus,
$$
\begin{array}{l}
\frac{A E}{a}=\frac{1}{\frac{1}{k}+1+\frac{\sqrt{k+1}}{k}}=\frac{k}{k+1+\sqrt{k+1}} . \\
\text { Therefore, } \frac{D E}{a}=1-\frac{A E}{a}=1-\frac{k}{k+1+\sqrt{k+1}} \\
=\frac{1+\sqrt{k+1}}{k+1+\sqrt{k+1}} .
\end{array}
$$

Thus, $\frac{D E}{A E}=\frac{1+\sqrt{k+1}}{k}$.