In an isosceles triangle ABC with base AB on side CB, point D is chosen so that CD = AC-AB. Point M is the middle of AD. Prove that the BMC angle is obtuse.
Since ABC is isosceles, therefore, BC = AC. By condition AC = CD + AB. Since the side ВС = ВD + CD = AC, therefore ВD + CD = CD + AB. Then, BD = AB. From the last equality it follows that triangle ABD is isosceles. And since BM divides AD in half by condition, BM is the height of triangle ABD. So the BMD angle = 90º. Angle BMC = BMD + DMC = 90º + DMC> 90º. Therefore, the BMC angle is obtuse.
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