In triangle $ABC$, the perpendicular bisectors of sides $AB$ and $BC$ intersect side $AC$ at points $P$ and $Q$, respectively, with point $P$ lying on the segment $AQ$. Prove that the circumscribed circles of the triangles $PBC$ and $QBA$ intersect on the bisector of the angle $PBQ$.
Let $O$ be the circumcenter of $\triangle ABC.$
Claim: $O$ is the intersection of $(PBC)$ and $(QBA)$ other than $B$.
Proof: $\angle AOB = 2\angle ACB = \angle QCB + \angle QBC = \angle AQB$
So $O$ lies on $(QBA)$.
$\angle BOC = 2\angle BAC = \angle BAP + \angle ABP = \angle BPC$
So $O$ lies on $(PBC)$.
Thus, the claim is proved.
Now, $\angle PBO = \angle PCO = \angle ACO = \angle CAO = \angle QAO = \angle QBO$.
$O$ is on the bisector of $\angle PBQ$, and we are done. $\blacksquare$