8.3.18 In isosceles $\wedge A B C$, $A B=A C$, $\odot O$ is the incircle of $\triangle A B C$, touching the sides $B C$, $C A$, and $A B$ at points $K$, $L$, and $M$ respectively. Let $N$ be the intersection of line $O L$ and $K M$, $Q$ be the intersection of line $B N$ and $C A$, and $P$ be the foot of the perpendicular from $A$ to line $B Q$. If $B P=A P+2 P Q$, find all possible values of $\frac{A B}{B C}$.
(1) First prove that $Q$ is the midpoint of segment $A C$.
As shown in Figure $\mathrm{a}$, let the line through point $N$ parallel to $A C$ intersect $A B$ and $B C$ at points $M_{1}$ and $K_{1}$, respectively. Since $\angle A L O=\angle M_{1} N O=\angle O M M_{1}=90^{\circ}$, quadrilateral $M M_{1} N O$ is a cyclic quadrilateral; therefore, $\angle M N M_{1}=\angle M O M_{1}$. Since $\angle O K K_{1}-\angle O N K_{1}-90^{\circ}$, quadrilateral $O N K K_{1}$ is also a cyclic quadrilateral. Therefore, $\angle K O K_{1}=\angle K N K_{1}$. Since $\angle M N M_{1}=\angle K N K_{1}$, it follows that $\angle M O M_{1}=\angle K O K_{1}$. Thus, $\triangle O M M_{1} \cong \triangle O K K_{1}$, so $O M_{1}=O K_{1}$. Hence, $\triangle O M_{1} K_{1}$ is an isosceles triangle. Since $\angle M_{1} N O=90^{\circ}$, $N$ is the midpoint of $M_{1} K_{1}$, and $M_{1} K_{1} \parallel A C$, it follows that $Q$ is the midpoint of $A C$.
(2) As shown in Figure a, consider the case where point $P$ is inside $\triangle A B C$.
Let $R$ be the foot of the perpendicular from $C$ to $B Q$, then $\triangle A P Q$ and $\triangle C R Q$ are both right triangles and congruent. Therefore, $P Q=Q R, A P=C R$. Assume $B P=P R+C R$. Take a point $S$ on line $B Q$ such that $C R=R S$, then $B P-P S$, since $A P \perp B Q$, so $A B=A S-A C$, hence $A$ is the circumcenter of $\triangle B S C$. By the relationship between the central angle and the inscribed angle, we know $\angle B A C=2 \angle B S C=90^{\circ}$, so $\triangle A B C$ is an isosceles right triangle. Thus, $\frac{A B}{B C}=\frac{\sqrt{2}}{2}$.
Figure a
Figure b
(3) As shown in Figure b, consider the case where point $P$ is outside $\triangle A B C$. Take a point $T$ on line $B Q$ such that $Q P=P T$. Since $\triangle A P Q \cong \triangle C R Q \cong \triangle A P T$, then $C R=A P$.
Since $B P=B R+2 P Q$, and $B P=A P+2 P Q$, it follows that $A P=B R$.
(2) From equations (1) and (2), we know that $\triangle B C R$ is an isosceles right triangle, so $\frac{B C}{A P}=\frac{B C}{B R}=\sqrt{2}$, (3) and $\angle A B Q=\angle Q C R$.

Since $\triangle A P Q \cong \triangle C R Q \cong \triangle A P T$, it follows that $\angle A B Q=\angle Q C R=\angle P A Q=\angle P A T$. Therefore, $\angle B A T=90^{\circ}$. Thus, $\triangle C Q R \sim \triangle B I A$. Hence, $\frac{A P}{P Q}=\frac{C R}{Q R}=\frac{A B}{A T}=\frac{A B}{A Q}=\frac{2}{1}$.
Therefore, $A P=2 P Q$.
In $\triangle A P Q$, we have $\frac{1}{2} A B=A Q=\sqrt{A P^{2}+P Q^{2}}=\sqrt{5} P Q$.
(5) From equations (3) and (4), we have $B C=\sqrt{2} A P=2 \sqrt{2} P Q$. (6) From equations (5) and (6), we get $\frac{A B}{B C}=\sqrt{\frac{5}{2}}=\frac{\sqrt{10}}{2}$.