An isosceles triangle with $AB=AC$ has an inscribed circle $O$, which touches its sides $BC,CA,AB$ at $K,L,M$ respectively. The lines $OL$ and $KM$ intersect at $N$; the lines $BN$ and $CA$ intersect at $Q$. Let $P$ be the foot of the perpendicular from $A$ on $BQ$. Suppose that $BP=AP+2\cdot PQ$. Then, what values can the ratio $\frac{AB}{BC}$ assume?
1. **Given Information and Setup:**
   - We have an isosceles triangle \( \triangle ABC \) with \( AB = AC \).
   - The incircle \( O \) touches \( BC, CA, AB \) at \( K, L, M \) respectively.
   - The lines \( OL \) and \( KM \) intersect at \( N \).
   - The lines \( BN \) and \( CA \) intersect at \( Q \).
   - \( P \) is the foot of the perpendicular from \( A \) to \( BQ \).
   - It is given that \( BP = AP + 2 \cdot PQ \).

2. **Lemma Application:**
   - Using the lemma, we know that the incircle \( C(I, r) \) of \( \triangle ABC \) touches \( BC, CA, AB \) at \( D, E, F \) respectively.
   - Denote \( M \) as the midpoint of \( AC \), and \( P \in EF \cap BI \), \( R \in EF \cap CI \), \( S \in BR \cap CP \).
   - The lines \( EF, ID, AM \) are concurrent.
   - Points \( P \) and \( R \) lie on the circle with diameter \( BC \), and \( S \in ID \).

3. **Simplification of the Problem:**
   - Given the symmetry and properties of the isosceles triangle, we can simplify the problem to:
     - In \( \triangle ABC \) with \( AB = AC \), denote \( M \) as the midpoint of \( AC \).
     - \( P \in BM \) and \( AP \perp BM \).
     - Prove that \( BP = AP + 2 \cdot PM \).

4. **Mathematical Formulation:**
   - Let \( AB = AC = b \), \( BC = a \), and \( \angle BAC = 2x \).
   - Denote \( \angle ABM = \phi \), thus \( \angle CBM = 90^\circ - x - \phi \).
   - Using the sine rule in \( \triangle ABC \), we have \( 2b \sin x = a \), hence \( \frac{AB}{BC} = \frac{b}{a} = \frac{1}{2 \sin x} \).

5. **Trigonometric Relationships:**
   - Using the sine and cosine rules, we get:
     \[
     \frac{MA}{MC} = \frac{BA}{BC} \cdot \frac{\sin \angle MBA}{\sin \angle MBC} = \frac{b}{a} \cdot \frac{\sin \phi}{\cos (x + \phi)} = 1
     \]
     \[
     \Rightarrow 2 \sin x \cos (x + \phi) = \sin \phi
     \]
     \[
     \Rightarrow \sin (2x + \phi) = 2 \sin \phi
     \]
     \[
     \Rightarrow \tan \phi = \frac{\sin 2x}{2 - \cos 2x}
     \]

6. **Condition \( BP = AP + 2 \cdot PM \):**
   - Using the given condition:
     \[
     BP = AP + 2 \cdot PM \Rightarrow b \cos \phi = \frac{b}{2} \sin (2x + \phi) + b \cos (2x + \phi)
     \]
     \[
     \Rightarrow \cos \phi = \sin \phi + \cos (2x + \phi)
     \]
     \[
     \Rightarrow (\cos \phi - \sin \phi)^2 = 1 - \sin^2 (2x + \phi)
     \]
     \[
     \Rightarrow \sin 2\phi = 4 \sin^2 \phi
     \]
     \[
     \Rightarrow \tan \phi = \frac{1}{2}
     \]
     \[
     \Rightarrow \frac{\sin 2x}{2 - \cos 2x} = \frac{1}{2}
     \]
     \[
     \Rightarrow 2 \sin 2x + \cos 2x = 2
     \]
     \[
     \Rightarrow x \in \{45^\circ, \arctan \frac{1}{3}\}
     \]
     \[
     \Rightarrow A \in \{90^\circ, \arctan \frac{3}{4}\}
     \]

7. **Conclusion:**
   - For the initial problem, the ratio \( \frac{AB}{BC} \) can assume the values:
     \[
     \frac{AB}{BC} \in \left\{ \frac{\sqrt{2}}{2}, \frac{\sqrt{10}}{2} \right\}
     \]

The final answer is \( \boxed{ \frac{AB}{BC} \in \left\{ \frac{\sqrt{2}}{2}, \frac{\sqrt{10}}{2} \right\} } \)