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Math Help - College geometry - Proof

  1. #1
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    College geometry - Proof

    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, (not sure exactly where to put this question, this isn't differential geometry, I would say it's closer to Abstract Algebra, but there was a terrible warning sticky there.... so...)
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  2. #2
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    Here is some bad news for you. If this is indeed college geometry (axiomatic) then there will probability be little we do to help. That is not because we donít want to help, but because we do not have access to your set of axioms and sequences of theorems. Although I have not taught the course in some years, I may have a copy of the text; what is it?

    Unless you take the time to type out all relevant definitions, axioms, theorems there is almost no way to help you.
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  3. #3
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    . .
    `__

    the book we are using is "Euclidean Geometry and Transformations" by Clayton W Dodge

    We've only covered three chapters, and I'm not sure which of the theorms I need to use to solve this. It's not obvious to me, which is what led me here.

    Most of the theorms in these chapters deal with collinear points, concurrent lines, and ratios between them.

    The only things that deal with acctual Triangles are calculating the areas.

    with triangle ABC
    Area = (1/2) AB* BC * sin (angle B)

    a ratio example: with Triangle ABC, let the be a point L on line BC. then

    d(BC)/d(LC) = AB * sin( d( angle BAL))/CA * sin ( d(angle LAC))

    the d(__) indicates that its directional, so that d(AB) = - d(BA)
    it depends on the positive direction.

    * this one I thought might help me, but I couldn't figure out how to use it, I have pages and pages of rearranging, substitueing, etc. and still no answer.

    then there is the area ratio: with triangle ABC, and point L on line BC,
    and h is "height" - length of the altitude from vertex A.

    K = K1 + K2
    K1 = (1/2)BL * h = (1/2) AB * AL * sin (angle BAL)
    K2 = (1/2)LC * h = (1/2) AL * CA * sin (angle LAC)

    and provided C doesn't equal L,

    K1/K2 = BL/-LC = (AB * sin (BAL))/ (CA * sing (LAC))
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  4. #4
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    Sorry, I have never even seen that textbook.
    It looks like a mixture of classical and vector geometry.
    Those textbooks are all unique.
    I hope you find someone with that very text to help you.
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