14. Given that line $l$ intersects circle $\odot O$ at two distinct points $E$ and $F$, $CD$ is a diameter of $\odot O$, $CA \perp l$, $DB \perp l$, with the feet of the perpendiculars being $A$ and $B$ respectively. If $AB=7$, $BD-AC=1$, and $AE=1$, does there exist a point $P$ on segment $AB$ such that the triangle with vertices $P$, $A$, and $C$ is similar to the triangle with vertices $P$, $B$, and $D$? If such a point exists, find the length of $AP$; if not, explain why.
14. (1) If $l$ does not intersect with the diameter $C D$, as shown in Figure 9.
(i) Draw $O H \perp A B$ at $H$, it is easy to see that $A E=B F$. At this point, $\triangle A C E \backsim \triangle B E D, \triangle A F C \backsim \triangle B D F$. Therefore, $E$ and $F$ are the points that satisfy the conditions.
Thus, $A P=A E=1$ or
$$
A P=A F=A B-B F=6 \text{. }
$$
(ii) If, besides $E$ and $F$, there exists another point $P$ such that $\triangle A P C \backsim \triangle B P D$, let $A C=x, B D=y$, then $y-x=1$.

Since $\mathrm{Rt} \triangle A C E \backsim \mathrm{Rt} \triangle B E D$, we have $\frac{x}{A E}=\frac{E B}{y}$, leading to $x y=6$.
Thus, $x=2, y=3$ or $x=-3, y=-2$ (discard the latter).
Since $\triangle A P C \backsim \triangle B P D$, we have
$$
\frac{A P}{A C}=\frac{B P}{B D}, \frac{A P}{2}=\frac{7-A P}{3} \text{. }
$$

Solving this, we get $A P=\frac{14}{5}$.
Therefore, there exists a third point $P$ that satisfies the conditions, and $A P=\frac{14}{5}$.
Combining (i) and (ii), there are three points $P$ that satisfy the conditions, with $A P$ being $1, 6, \frac{14}{5}$.
(2) If $l$ intersects with the diameter $C D$ at point $Q$, as shown in Figure 10.
(i) Since $\angle A Q C = \angle D Q B$, we have $\mathrm{Rt} \triangle A C Q \sim \mathrm{Rt} \triangle B D Q$. Thus, point $Q$ satisfies the conditions.

Let $A C=x, B D=y$, then $y-x=1$.
Since $\angle D E B = \angle E C A$, we have $\mathrm{Rt} \triangle A C E \backsim \mathrm{Rt} \triangle B E D$. Thus, $\frac{A C}{B E}=\frac{A E}{B D}$, leading to $x y=8$.
Thus, $x=\frac{\sqrt{33}-1}{2}, y=\frac{\sqrt{33}+1}{2}$ or $x=\frac{-\sqrt{33}-1}{2}, y=\frac{-\sqrt{33}+1}{2}$ (discard the latter).
Since $\mathrm{Rt} \triangle A C Q \backsim \mathrm{Rt} \triangle B D Q$, we have $\frac{A Q}{Q B}=\frac{A C}{B D}$, solving this gives $A Q=\frac{231-7 \sqrt{33}}{66}$.
(ii) If, besides $Q$, there exists another point $P$ such that $\triangle A P C \backsim \triangle B D P$, then $\frac{A P}{B D}=\frac{A C}{P B}$.
Rearranging, we get $A P^{2}-7 A P+8=0$.
Solving this, we get $A P=\frac{7 \pm \sqrt{17}}{2}$.
Combining (i) and (ii), there are three points $P$ that satisfy the conditions, with $A P$ being $\frac{231-7 \sqrt{33}}{66}, \frac{7+\sqrt{17}}{2}, \frac{7-\sqrt{17}}{2}$.