18.89 As shown in the figure, $ABCD$ is a square with an area of 1, and $\triangle BPC$ is an equilateral triangle. Then the area of $\triangle BPD$ is 
(A) $\frac{\sqrt{3}-1}{2}$.
(B) $\frac{2 \sqrt{3}-1}{8}$.
(C) $\frac{\sqrt{3}}{4}$.
(D) $\frac{\sqrt{3}-1}{4}$.
(E) $\frac{1}{4}$.
(China Beijing Junior High School Mathematics Competition, 1984)
[Solution] Connect $AC$ to intersect $BD$ at $O$, connect $PO$, then $PO \parallel DC$. Thus, $S_{\triangle DPO} = S_{\triangle CPO}$.
$$
\begin{aligned}
\therefore S_{\triangle BPD} & = S_{\triangle BPC} - S_{\triangle BCC} = \frac{\sqrt{3}}{4} - \frac{1}{4} \\
& = \frac{\sqrt{3} - 1}{4}.
\end{aligned}
$$

Therefore, the answer is $(D)$.