Problem 1459

1. Given: $\triangle ABC$ is an isosceles triangle with $AB = BC$. Let $\ell$ be a ray from $B$. Points $P$ and $Q$ on $\ell$ lie inside the triangle such that $\angle BAP = \angle QCA$.

2. Claim: Let $Q'$ be the reflection of $Q$ over the angle bisector of $\angle ABC$. Then $(P, Q')$ are isogonal conjugate pairs.

3. Proof of Claim:
- Since $\angle BAP = \angle QCA$ and $Q'$ is the reflection of $Q$ over the angle bisector of $\angle ABC$, we have $\angle BAP = \angle Q'AC$.
- This implies that lines $AP$ and $AQ'$ are isogonal with respect to $\angle BAC$.
- Similarly, lines $BP$ and $BQ'$ are isogonal with respect to $\angle ABC$.
- Therefore, $(P, Q')$ are isogonal conjugates. $\blacksquare$

4. Similarly, let $P'$ be the reflection of $P$ over the angle bisector of $\angle ABC$. Then $(Q, P')$ are also isogonal conjugates.

5. Angle Relationships:
- Since $\angle BAP = \angle Q'AC$, we have $\angle BAP + \angle PAP' = \angle QAQ' + \angle Q'AC$.
- This simplifies to $\angle PAP' = \angle QAQ'$.

6. Similarly, $\angle PCP' = \angle QCQ'$.

7. Symmetry:
- Due to the symmetry of the isosceles triangle $\triangle BAC$, we have $\angle QAQ' = \angle QCQ'$.

8. Conclusion:
- Therefore, $\angle PAP' = \angle QAQ' = \angle QCQ' = \angle PCP'$.

9. Final Step:
- Since $\angle Q'AC = \angle BCP'$ (both points are reflections over the angle bisector and $\triangle BAC$ is isosceles), we can conclude that:
- $\angle BAP + \angle PAQ + \angle QAQ' + \angle Q'AC = \angle BCP' + \angle P'CP + \angle PCQ + \angle QCA$.

10. Therefore, $\boxed{\angle PAQ = \angle PCQ}$.