[ Mutual arrangement of altitudes, medians, bisectors, and others. ]
Prove that in any non-isosceles triangle, the bisector lies between the median and the altitude drawn from the same vertex.
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Solution
Use the circumcircle of the triangle.
## Solution
Let H, D, and M be the feet of the altitude, angle bisector, and median, respectively, drawn from vertex B in triangle ABC. Describe a circumcircle around triangle ABC. Let P be the point of intersection of line BD with this circle. Then P is the midpoint of arc AC. Therefore, the line drawn through point P parallel to BH is perpendicular to chord AC and passes through its midpoint, i.e., point M. Since points B and P lie on opposite sides of line AC, point D lies between the projections of the endpoints of segment BP, i.e., between points H and M.
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