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    right. consider a triangle ABC. call D the midpoint of BC. i want to prove the area of triangle ABD equals area of triangle ACD.

    right, so consider the line AD. it bisects the angle BAC. now if the areas are equal, 0.5absinC will be the same for both of them... or in this case:

    0.5(AB)(AD)sin(BAC/2) = 0.5(AD)(AC)sin(BAC/2)
    AB = AC

    but this isn't neccessarily true:confused:
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    (Original post by chewwy)
    right. consider a triangle ABC. call D the midpoint of BC. i want to prove the area of triangle ABD equals area of triangle ACD.

    right, so consider the line AD. it bisects the angle BAC. now if the areas are equal, 0.5absinC will be the same for both of the...
    this isnt necessarily true
    see attachment
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    (Original post by Undry1)
    this isnt necessarily true
    see attachment
    ok yeah:rolleyes:. i see you do it by considering the common side and the fact sin(180-x) = sin(x).
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    Just because the line AD bisects <BAC it does not mean the that the triangle itself has been split into two equal triangles, thus areas wont be the same.
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    (Original post by Malik)
    Just because the line AD bisects <BAC it does not mean the that the triangle itself has been split into two equal triangles, thus areas wont be the same.
    the areas will be the same.
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    Here is a solution.
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