8.3.11 ** Given $\triangle A B C, \angle B=90^{\circ}$, the incircle touches $B C$, $C A$, $A B$ at points $D$, $E$, $F$, respectively, and $A D$ intersects the incircle at another point $P$, with $P F \perp P C$, find the ratio of the three sides of $\triangle A B C$.
As shown in the figure, connect $F D$, $P E$, and $E D$. It is easy to see that $\triangle F B D$ is an isosceles right triangle. By the inscribed angle theorem, $\angle F P D = \angle F D B = 45^{\circ}$, thus $\angle D P C = 45^{\circ}$. Also, $\angle P D C = \angle P F D$, so $\triangle P F D \sim \triangle P D C$, which means $\frac{P F}{F D} = \frac{P D}{C D}$. Since $\triangle A P F \sim \triangle A F D$ and $\triangle A P E \sim \triangle A E D$, we have $\frac{P E}{D E} - \frac{A P}{A E} = \frac{A P}{A F} = \frac{P F}{F D}$. Therefore, $\frac{P E}{D E} = \frac{P D}{C D}$. Also, $\angle E P D = \angle E D C$, so $\triangle E P D \sim \triangle E D C$. Thus, $\triangle E P D$ is also an isosceles triangle, so $\angle P E D = \angle E P D = \angle E D C$. Therefore, $P E \parallel B C$, and $\frac{A E}{A C} = \frac{P E}{C D} = \frac{P E}{E D} \cdot \frac{E D}{C D} = \left(\frac{E D}{C D}\right)^2 = 4 \sin^2 \frac{C}{2} = 2(1 - \cos C) = 2\left(1 - \frac{B C}{A C}\right) = 2 \frac{A C - B C}{A C}$.

Also, $\frac{A E}{A C} = \frac{\frac{1}{2}(A B + A C - B C)}{A C}$, so $A B + A C - B C - 4(A C - B C) = A B = 3(A C - B C)$. Squaring both sides, we get $A B^2 = 9(A C - B C)^2 = A C^2 - B C^2$, which means $9(A C - B C) = A C + B C$, so $\frac{B C}{A C} = \frac{4}{5}$. Therefore, $A B : B C : A C = 3 : 4 : 5$.