Problem

Source:

Tags: inequalities, geometry, circumcircle, trigonometry, angle bisector, geometry proposed, Geometric Inequalities



Let $ ABC $ be a triangle and the point $ D $ is on the segment $ BC $ such that $ AD $ is the interior bisector of $ \angle A $. We stretch $ AD $ such that it meets the circumcircle of $ \Delta ABC $ at $ M $. We draw a line from $ D $ such that it meets the lines $ MB,MC $ at $ P,Q $, respectively ($ M $ is not between $ B,P $ and also is not between $ C,Q $). Prove that $ \angle PAQ\geq\angle BAC $.