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!!

thanx in advnace

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- Jul 2nd 2007, 10:40 PMshoshoanother 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!!

thanx in advnace - Jul 2nd 2007, 11:04 PMred_dog
We have $\displaystyle \widehat{QPC}=\widehat{QAC}$, then the quadrilater $\displaystyle ACQP$ can be inscribed in a circle.

Yields $\displaystyle \widehat{PAQ}=\widehat{PCQ}=\widehat{QPC}$, so the triangle $\displaystyle PQC$ is isosceles, then $\displaystyle PQ=QC$