The inscribed circle of triangle ABC is tangent to side BC at point D. Prove that if AD is the median of the triangle, then AB = AC.
From the point O, the center of the circle, construct the segments OK and OM to the points of tangency of the circle and the lateral sides of the triangle.
By the property of tangents drawn from one point, the segment BK = BD, CM = CD, similarly to AK = AM.
Since, by condition, point D is the middle of BC, then BD = CD, which means BK = CM.
Then (ВK + AK) = (CM + AM), which means AB = AC, and the triangle ABC is isosceles, which was required to prove.
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