17.168. Let $M$ be the midpoint of side $AB$ in $\triangle ABC$, and let $P$ be a point on $AB$ between $A$ and $M$. Draw $MD$ parallel to $PC$ and intersecting $BC$ at $D$. If $r$ represents the ratio of the area of $\triangle BPD$ to the area of $\triangle ABC$, then
(A) $\frac{1}{2}<r<1$, depending on the position of $P$.
(B) $r=\frac{1}{2}$, regardless of the position of $P$.
(C) $\frac{1}{2} \leqslant r<1$, depending on the position of $P$.
(D) $\frac{1}{3}<r<\frac{2}{3}$, depending on the position of $P$.
(E) $r=\frac{1}{3}$, regardless of the position of $P$.
(17th American High School Mathematics Examination, 1966)
[Solution] Connect $C M, D P$. Given $P C / / M D$, then we have $S_{\triangle M D C}=S_{\triangle M D P}$.
Therefore, $S_{\triangle B P D}=S_{\triangle B M D}+S_{\triangle M D P}=S_{\triangle B M D}+S_{\triangle M D C}$
$$
=S_{\triangle B M C}=\frac{1}{2} S_{\triangle A B C} \text {. }
$$

Notice that $C M$ is the median on $A B$.
Thus,
$$
r=\frac{S_{\triangle B P D}}{S_{\triangle A B C}}=\frac{1}{2} .
$$

Therefore, the answer is $(B)$.