In $\square A B C D$, $C E \perp A B, C F \perp A D$, with feet of the perpendiculars at $E$ and $F$. Let $E F$ intersect the diagonal $B D$ at point $P$. If $A B: A D=2: 3$, find $P F: P E$.
Solution: As shown in the figure,
2, connect $A C$ intersecting
$B D$ at $O$, draw $E H / / B D$
intersecting $A C$ and
$A F$ at $G$ and $H$ respectively,
draw $O M \perp E F$ at $M$, connect $G M, M C, C P$.
Since $C E \perp A B, C F \perp A D$, therefore, quadrilateral $A E C F$ is a cyclic quadrilateral with $A C$ as its diameter.

Given that $O$ is the midpoint of $A C$, $O M \perp E F$, it is known that $M$ is the midpoint of $E F$.

Also, since $E H / / B D$ and $O$ is the midpoint of $B D$, $G$ is the midpoint of $E H$.
Thus, $G M / / A D$.
Therefore, $\angle G M E = \angle A F E$.
Since $A, E, C, F$ are concyclic,
$\angle A C E = \angle A F E$.
Hence, $\angle G M E = \angle A C E$.
Thus, $E, G, M, C$ are concyclic.
So, $\angle G E M = \angle G C M$.
Since $E H / / B P$, then $\angle G E M = \angle M P O$.
Thus, $\angle G C M = \angle M P O$.
Therefore, $M, O, C, P$ are concyclic.
Noting that $\angle O M P = \angle P C M = 90^{\circ}$, i.e., $P C \perp A C$, hence $P C$ is the tangent to the circumcircle of quadrilateral $A E C F$.
So, $\angle P C F = \angle F E C$.
Since $\angle F P C = \angle C P E$, then
$\triangle F P C \sim \triangle C P E$.
Thus, $\frac{P C}{P E} = \frac{P F}{P C} = \frac{F C}{E C}$.
Therefore, $\left(\frac{F C}{E C}\right)^{2} = \frac{P C}{P E} \cdot \frac{P F}{P C} = \frac{P F}{P E}$.
Given $S_{\square A B C D} = A D \cdot C F = A B \cdot C E$, we get
$$
\frac{F C}{E C} = \frac{A B}{A D} = \frac{2}{3} \text{. }
$$

Hence, $P F: P E = 4: 9$.
(Kuozhengping, Anhui Normal University High School, 241001)