In triangle ABC, angle BAC = 80 degrees and Q is the point on BC such that AQ bisects angle BAC.
Point P also lies on AB such that angle QPC = 40 degrees.
P is distinct from A.
Prove that PQ = QC.
CPQ and CAQ are 2 angles drawn on the same base and on the same site
and both are equal then they lie on the same circle and the base is a chord
on the circle, this means that APQC is a quadrilateral pie
In quadrilateral pie each opposite angles have the sum of 180 degrees.
angle PAC + angle PQC = 180 degrees
80 + angles PQC = 180
angle PQC = 180 - 80 = 100 degrees
angle QCP = 180 - (100 + 40) = 40 deg.
then PQ = QC
Faraj Razem