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<BAC + <MIN = 180^o, incenter, AC = NC, AB = MB

Source: 2022 Moldova JBMO TST p8

November 4, 2022
geometryincenterangles

Problem Statement

Let ABCABC be the triangle and II the center of the circle inscribed in this triangle. The point MM, located on the tangent taken to the point BB to the circumscribed circle of the triangle ABCABC, satisfies the relation AB=MBAB = MB. Point NN, located on the tangent taken to point CC to the same circle, satisfies the relation AC=NCAC = NC. Points M,AM, A and NN lie on the same side of the line BCBC. Prove that ∠BAC+∠MIN=180o.\angle BAC + \angle MIN = 180^o.
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