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Math Help - proof: triangle ABC, M midpoint of BC...

  1. #1
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    proof: triangle ABC, M midpoint of BC...

    IF you have a triangle, ABC, with a point M as the midpoint between B and C, then prove that AM < (1/2)(AB + AC)


    I've been working on this problem for days, and I'm sure the solution is going to be so simple that I'll tear out my hair when I figure it out, but as of now, I need help.

    I've already shown that IF ABC are co-linear points, then AM = (1/2)(AB + AC).

    I now need to show, that when ABC aren't colinear and they form a triangle, that AM< (1/2)(AB + AC).


    I'm trying to do it in a way that keeps two of the values constant, (like AB and BC ooor AC and BC oooor, AB and AC... etc), and moving only one of the points around. that way most of the values in the equation AM < (1/2) (AB +AC) will stay the same....

    for instance, if I keep AB and AC the same, then the distance between BC will be change. technically it can be many values, (assuming AC > AB)
    then BC falls in the range:

    (AB + AC) > BC > (AC- AB)

    .......

    can anyone tell me if I'm headed in the right direction, if so, what's the next step(s) if not, what is the right direction?

    Thanks,
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  2. #2
    MHF Contributor
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    hi fruitbodywash,

    you can certainly reason geometrically,
    by showing that the longest AM occurs when A, B, M, C are co-linear.

    That's the only time AM stretches to midway between the circles.
    Attached Thumbnails Attached Thumbnails proof: triangle ABC, M midpoint of BC...-abcm.jpg  
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