Since, by condition, AM = BC / 2, then AM = BM = CM, therefore, the vertices of the triangle, points A, B, C, are equidistant from point M, and then they lie on a circle with a radius BM.
Then BC = 2 * BM is the diameter of the circumscribed circle around the triangle ABC.
Since the side of the BM is the diameter of the circle, and the inscribed angle BAC rests on it, then the angle BAC = 180/2 = 90, and then the triangle ABC is rectangular, which was required to be proved.
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