Answer:
AQ║PM and AP║QM, hence PAQM is a parallelogram
Explanation:
You want to show PAQM in the attached diagram is a parallelogram.
Circumcircle
Consider the circumcircle of ∆ABC. Equilateral triangle ABC divides the circle into three congruent arcs, each of which is 120°.
Angle CMQ is an interior angle of equilateral triangle MCQ, so is 60°. That makes angle QMP its supplement, or 120°. Vertical angle BMC is also 120°, the measure of arc BC. This means M is the center of the circumcircle.
Since MC is congruent to MQ in equilateral ∆MCQ, point Q lies on the circumcircle. Angle AQB is thus an inscribed angle that intercepts 120° arc AB, so its measure is 60°.
Angles AQM and QMP are supplementary, which requires AQ║PM. By symmetry, AP║QM, so PAQM is a parallelogram.