$18 \cdot 90$ As shown in the figure, given a square $ABCD$ with side length $a$, $E$ is the midpoint of $AD$, $P$ is the midpoint of $CE$, and $F$ is the midpoint of $BP$. Then the area of $\triangle BFD$ is
(A) $\frac{1}{64} a^{2}$.
(B) $\frac{1}{32} a^{2}$.
(C) $\frac{1}{16} a^{2}$.
(D) $\frac{1}{8} a^{2}$.
[Solution] Connect $P D, B E$. Given that $E, P$ are the midpoints of $A D, E C$ respectively, we can prove $B E / / P D$ using the properties of the diagonals of a rectangle (see figure (2)).
$\therefore S_{\triangle B F D}=\frac{1}{2} S_{\triangle B D P}$,
Also, $S_{\triangle E P D}=S_{\triangle B D P}$, so $S_{\triangle B F D}=\frac{1}{2} S_{\triangle E P D}=\frac{1}{4} S_{\triangle C D E}$.
Since $S_{\triangle C D E}=\frac{1}{4} S_{A B C D}$, then $S_{\triangle B F D}=\frac{1}{16} a^{2}$. Therefore, the answer is $(C)$.