Example 4 Given that the base of the quadrilateral pyramid $P-ABCD$ is a parallelogram, with height $h$, a section is made through the base side $BC$ intersecting the side face $PAD$ at $EF$. If the section divides the pyramid into two parts of equal volume, find the distance from $EF$ to the base.

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Solution 1: Let the distance between $EF$ and the base be $x$, then $x$ is the height of the tetrahedrons $E-ABD$ and $F-BCD$. Let $S$ be the area of the base, then
$$
V_{E-ABD}=V_{F-BCD}=\frac{1}{3} \times \frac{1}{2} S x=\frac{x S}{6}.
$$

Since the tetrahedrons $B-ADE$ and $B-DEF$ have the same height, and $\triangle ADE$ and $\triangle DEF$ have the same height, then
$$
\frac{V_{B-ADE}}{V_{B-DEF}}=\frac{S_{\triangle ADE}}{S_{\triangle DEF}}=\frac{AD}{EF'}=\frac{h}{h-x},
$$

which means $V_{B-DEF}=\frac{h-x}{h} \cdot \frac{x S}{6}=\frac{(h-x) x S}{6 h}$. 
Let the volume of the polyhedron $ABCD EF$ between the section and the base be $V$, then
$$
\begin{array}{l}
V=V_{E-ABD}+V_{F-BCD}+V_{B-DEF} \\
=\frac{x S}{6}+\frac{x S}{6}+\frac{(h-x) x S}{6 h}=\frac{(3 h-x) x S}{6 h}.
\end{array}
$$

Given $V=\frac{1}{2} V_{P-ABCD}=\frac{1}{2} \times \frac{1}{3} h S=\frac{h S}{6}$, we have
$$
\frac{(3 h-x) x S}{6 h}=\frac{h S}{6},
$$

which simplifies to $x^{2}-3 h x+h^{2}=0$.
Solving for $x$ gives $x=\frac{3 \pm \sqrt{5}}{2} h$.
Since $\frac{3+\sqrt{5}}{2}>1$, it should be discarded.
Therefore, the required distance is $\frac{3-\sqrt{5}}{2} h$.
Solution 2: As shown in Figure 5, extend $FE$ to $G$ such that
$FG=AD$. Clearly,
the volume of the triangular prism $ABG-DCF$
$V_{1}=\frac{1}{2} S x$
(can be expanded into
a parallelepiped, with a base area of $S$ and height $x$). Since the tetrahedron $E-ABG$ and the triangular prism $ABG-DCF$ have the same base, its volume is
$$
V_{E-ABC}=V_{2}=\frac{1}{3} V_{1} \cdot \frac{EG}{AD}=\frac{1}{3} V_{1} \cdot \frac{AE}{PA}=\frac{x}{3 h} V_{1}.
$$

Thus, the volume of the polyhedron $ABCD EF$ is
$$
V=V_{1}-V_{2}=\left(1-\frac{x}{3 h}\right) V_{1}=\frac{(3 h-x) x S}{6 h}.
$$

The rest is the same as Solution 1.