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Proving that Central Median of Triangle is... watch

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    Well, here I had a little homework - to prove that the central median line MN=AB/2,

    which I did using vectors:
    MN=MC+CN
    MN=MA+AB+BN
    When I sum the equations:
    2MN=MC+MA+CN+BN+AB
    As MC & MA and CN & BN have contrary directions, their sum is 0 and therefore:
    2MN=AB
    MN=AB/2
    As I have no place to check if I did this right, please reply whether you think it's right or wrong.
    Thanks in advance.
    NOTE: All of the above is in vectors, consider there's an arrow above MN, MA etc. I'm not quite familiar with latex
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    That's fine, although your wording here:
    As MC & MA and CN & BN have contrary directions, their sum i 0 and therefore ...
    is a bit dodgy. I mean it's correct, but the point is that M and N are defined so that AM=MC and BN=NC; in particular, AM=-CM and BN=-CN, hence the result. The only reason I say this is because you haven't said that MC and MA have the same length (which is obvious, but still worth mentioning to justify cancelling them out).
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    (Original post by nuodai)
    That's fine, although your wording here:

    is a bit dodgy. I mean it's correct, but the point is that M and N are defined so that AM=MC and BN=NC; in particular, AM=-CM and BN=-CN, hence the result. The only reason I say this is because you haven't said that MC and MA have the same length (which is obvious, but still worth mentioning to justify cancelling them out).
    Thanks, I forgot to mention that here Though I wrote it in my homework.
 
 
 
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Updated: April 6, 2011
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