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| The median is shorter (Posted on 2011-04-04) |
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Each median in a triangle is shorter than the average of the two adjacent sides.
Prove it.
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Submitted by Ady TZIDON
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Solution:
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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 |
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Date |
 | solution | Jer | 2011-04-04 15:10:47 |
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