Given $ABCD$ is a unit square, $O$ is its center, $P$ is a point on $CD$, line $AP$ intersects the extension of $BC$ at point $Q$, intersects $DO$ at point $E$, and $OQ$ intersects $PC$ at point $F$. If $EF \| AC$, find the length of $AP$.
Solution: As shown in Figure 1, let $DP$
$$
=x(0<x<1), CQ=y,
$$

then $AP=\sqrt{1+x^{2}}, PC=$
1 - $x$. From the similarity of Rt $\triangle PCQ \sim \text{Rt}$
$\triangle PDA$, we get $\frac{CQ}{AD}=\frac{PC}{PD}$.
Therefore, $CQ=y=$
$$
\frac{AD \cdot PC}{PD}=\frac{1-x}{x}.
$$

Since $EF \parallel AC$, we have $\frac{PE}{FC}=\frac{PE}{EA}$.
Because $DE$ bisects $\angle ADP$, we have $\frac{PE}{EA}=\frac{DP}{AD}=x$. Hence,
$$
\frac{PF}{FC}=x.
$$

In $\triangle DBC$ and its transversal $OFQ$, by Menelaus' Theorem, we get $\frac{CF}{FD} \cdot \frac{DQ}{OB} \cdot \frac{BQ}{QC}=1$.
Therefore, $\frac{CF}{FD}=\frac{OC}{BQ}=\frac{y}{1+y}=\frac{\frac{1}{x}-1}{\frac{1}{x}}=1-x$.
Thus, $\frac{FD}{FC}=\frac{1}{1-x}$, which means $\frac{PF+PD}{FC}=\frac{1}{1-x}$. So,
$$
\frac{PF}{FC}=\frac{1}{1-x}-\frac{x}{FC}.
$$

Draw $OH \perp BC$, with $H$ being the midpoint of $BC$, then $\frac{CF}{OH}=\frac{CQ}{HQ}$. Hence $CF=\frac{OH \cdot CQ}{HQ}=\frac{\frac{1}{2} y}{\frac{1}{2}+y}=\frac{y}{1+2y}=\frac{1-x}{2-x}$.
From (1) ② we get
$x=\frac{1}{1-x}-x \cdot \frac{2-x}{1-x}$.
Rearranging and simplifying, we get $2x^{2}-3x+1=0$, which is
$(2x-1)(x-1)=0$.
Since $0<x<1$, we have $x=\frac{1}{2}$. Therefore, $AP=\sqrt{1+x^{2}}=$ $\frac{\sqrt{5}}{2}$, at this time $P$ is the midpoint of $CD$.
(Wu Weizhao, Department of Mathematics, College of Science, Guangzhou University, 510405)