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.

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- Aug 21st 2010, 05:13 AMthe undertakerdifficult 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. - Aug 21st 2010, 07:37 AMmathaddict
- Aug 22nd 2010, 02:43 AMrazemsoft21
http://img104.herosh.com/2010/08/21/419164129.jpg

*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