# another circle geometry question~~~ really urgent

• Jul 2nd 2007, 10:40 PM
shosho
another circle geometry question~~~ really urgent
in a traingle ABC, <BAC=80 and Q is the point on BC such that AQ bisects the angle BAC. There is also a point P on AB such that <QPC=40 and P is distinct from A. Prove that PQ=QC

any help would be awesome!!
• Jul 2nd 2007, 11:04 PM
red_dog
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 $PQ=QC$