Assume $\angle QAB=\frac{\angle CAB}3$, let $D$ be the point where the perpendicular bisector of $BC$ cuts $AC$, and $E$ the projection of $A$ on this perpendicular bisector. Notice that $\frac{AE}{\frac{BC}2}=\frac{AD}{DC}=\frac{AB}{BC}$, so $AB=2AE\ (*)$. We also know that $\angle APE=30^\circ\ (\#)$ (a simple angle chase), so $AP=2AE$. Together with $(*)$, this shows that $AP=AB$. Now, since $Q$ lies on the bisector of $\angle BAP$, which is also the perpendicular bisector of $BP$, we have $QP=QB=QC$. This means that $PQ<AB\iff\angle AQB>\angle QAB$, and, since $\angle AQB+\angle QAB=30^\circ$ (a consequence of $(\#)$), we get $PQ<AB\iff\frac{\angle A}3<15^\circ$, and the conclusion follows.

,dr.trig avoids construction operation,does it with usual,but patiennt ease.