$P A \perp B C, P A=B C=l, P A, B C$ have a common perpendicular line $E D=h$. Prove: the volume of the pyramid $P-A B C_{i}$


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Proof 1: Draw $B L^{\prime} \therefore A P, C C^{\prime} \mathbb{\|} A P$, then the pyramid $P-A B C$ is supplemented to form a triangular prism $A B C-P B^{\prime} C^{\prime}$ (as shown in Figure 5). Given
$\left.\begin{array}{l}P A \perp D E \\ P A \perp B C\end{array}\right\} \Rightarrow P A \perp$ section $E B C$.
$\therefore V$ of the triangular prism $\mathrm{ABC}-\mathrm{PB}^{\prime} \mathrm{C}^{\prime}=S$ of the section $\mathrm{ERC} \cdot A P$
$$
\begin{array}{l}
=\left(\frac{1}{2} B C \cdot D E\right) \cdot A P \\
=\frac{1}{2} l \cdot h \cdot l=\frac{1}{2} l^{2} h .
\end{array}
$$
$V$ of the pyramid $\mathrm{P}-\mathrm{ABC}=\frac{1}{3} V$ of the triangular prism $\mathrm{APC} \mathrm{PB}^{\prime} \mathrm{C}^{\prime}$
$$
=\frac{1}{3}\left(\frac{1}{2} l^{2} h\right)=\frac{1}{6} l^{2} h \text {. }
$$

Proof 2: Draw $A^{\prime} D P^{\prime} A E P$, such that $A^{\prime} D=$ $A E$, draw: $B^{\prime} E C^{\prime} \Perp B D C$, such that $B^{\prime} E=B D$, then the pyramid $P-A B C$ is supplemented to form a right quadrilateral prism with the quadrilateral $A^{\prime} B P^{\prime} C$ as the base, and $B C \perp$
$A^{\prime} P^{\prime}, B C=A^{\prime} P^{\prime}$
(as shown in Figure 6). Given
$$
\begin{array}{l}
E D \perp B C \\
E D \perp A P \\
A^{\prime} P^{\prime} \| A P
\end{array}
$$
$A^{\prime} P^{\prime} \| A P$ )
$\Rightarrow E D \perp$ base $A^{\prime} B P^{\prime} C$.
Also, $B C \perp A^{\prime} P^{\prime}, B C=A^{\prime} P^{\prime}=l$,
$\therefore S$ of the quadrilateral $\mathrm{A}^{\prime} \mathrm{BP}^{\prime} \mathrm{C}$
$$
=\frac{1}{2} A^{\prime} P^{\prime} \cdot B C=\frac{1}{2} l^{2} \text {. }
$$
$=S$ of the quadrilateral $\mathrm{A}^{\prime} \mathrm{BP}^{\prime} \mathrm{C} \cdot E D=\frac{1}{2} l^{2} \cdot h$.
But $V$ of the pyramid $\mathrm{P}-\mathrm{ABC}$
$$
\begin{array}{l}
=\frac{1}{3} \cdot \frac{1}{2} l^{2} h \\
=\frac{1}{6} l^{2} h_{\text {. }} \\
\end{array}
$$