$22 \cdot 11$ Point $P$ is outside the plane of $\triangle A B C$, $P A \perp$ plane $A B C$, and $S_{1}, S_{2}$ represent the areas of $\triangle A B C$ and $\triangle P B C$ respectively, then the value of $S_{1}: S_{2}$ 
(A) is greater than 0, and no more than 1.
(B) is greater than 0 and less than 1.
(C) is greater than 1.
(D) is no less than 1.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] Draw $A D \perp B C$, connect $P D$. By the theorem of three perpendiculars, we know $P D \perp B C$. Then
$$
\begin{aligned}
\frac{S_{1}}{S_{2}} & =\frac{\frac{1}{2} B C \cdot A D}{\frac{1}{2} B C \cdot P D} \\
& =\frac{A D}{P D}=\cos \angle P D A \in(0,1),
\end{aligned}
$$

Therefore, the correct choice is $(B)$.