Prove that if the incircle center of a triangle belongs to its height, then this triangle is isosceles.

The center of a circle inscribed in a triangle ABC is the point of intersection of the bisectors of the angles of this triangle. If the center of the inscribed circle O belongs to the height of the triangle BK, then this height of the triangle is also the bisector of the angle ABC. Consider two triangles ABK and BKС. The angles ABK and BKС are equal due to the fact that the segment BK is the bisector of the angle ABC. The angles AKB and СKB are equal to 90 degrees, since the segment BK is the height of the triangle. Side BK is common for triangles ABK and BKС. Therefore, by the sign of equality of triangles, these triangles are equal. Therefore | AB | = | BC | and triangle ABC is isosceles.