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 The median is shorter (Posted on 2011-04-04)
Each median in a triangle is shorter than the average of the two adjacent sides.
Prove it.

 Submitted by Ady TZIDON No Rating Solution: (Hide) Jer's solution is impecable: Call the triangle ABC and M the midpoint of BC so that AM is the median from A to BC. We are to prove AM<(AB+AC)/2. Construct point D such that ABDC is a parallelogram. So we have BD=AC, CD=AB, and AM=MD. The diagonals of a parallelogram bisect each other so A, M, and D are on the same line and AD=2*AM Looking at triangle ABD we see AB+BD>AD by the triangle inequality. Substituting we get AB+AC>2AM. or AM<(AB+AC)/2. q.e.d.

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 Subject Author Date solution Jer 2011-04-04 15:10:47
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