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Midpoint of the internal bisector

Source: ITAMO 2019 #5

May 3, 2019
geometryITAMO 2019

Problem Statement

Let ABCABC be an acute angled triangle.. Let DD be the foot of the internal angle bisector of ∠BAC\angle BAC and let MM be the midpoint of AD.AD. Let XX be a point on segment BMBM such that ∠MXA=∠DAC.\angle MXA=\angle DAC. Prove that AXAX is perpendicular to XC.XC.
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