Thread: difficult hsc geometry

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.

Originally Posted by the undertaker

In triangle ABC, angle BAC = 80 degrees and Q is the point on BC such that AQ bisects angle BAC.

Point B also lies on AB such that angle QPC = 40 degrees.
P is distinct from A.

Prove that PQ = QC.

angle QPC=angle QAC=40

This shows that the quadrilateral APQC is bounded by the circle and with the use of one of the properties of circle, you will get your prove.

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