# another circle geometry question~~~ really urgent

We have $\widehat{QPC}=\widehat{QAC}$, then the quadrilater $ACQP$ can be inscribed in a circle.
Yields $\widehat{PAQ}=\widehat{PCQ}=\widehat{QPC}$, so the triangle $PQC$ is isosceles, then $PQ=QC$