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Math Help - Midpoint Theorem

  1. #1
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    Midpoint Theorem

    A triangle has vertices A (a,b) , B (c,d) and C (e,f). Prove two things - algebra only - that the line segment connecting the midpoint of two sides of the triangle is parallel to the third side, and ​the line segment is also equal to one-half the length of the third (parallel) side.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Midpoint Theorem

    Do you know how to find the coordinates of the mid-point of a side of the triangle?
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  3. #3
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    Re: Midpoint Theorem

    It would be just using the midpoint formula

    So let's say the midpoint of A and B is ( (a+c)/2 , (b+d)/2 ) and the midpoint of A and C is ( (a+e)/2 , (b+f)/2 ).

    I believe.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Midpoint Theorem

    Yes, that's correct. Now, how can we show that two line segments are parallel?
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  5. #5
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    Re: Midpoint Theorem

    Well let me call AB's midpoint D and AC's midpoint E.

    Having a line go through BC and DE and producing similar angles would prove this to be true... and convenient since AB and AC do happen to this. But i'm supposed to prove it algebraically, which is where I am stuck.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Midpoint Theorem

    Hint: Two lines are parallel if their slopes are equal...
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  7. #7
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    Re: Midpoint Theorem

    Well the Slope of the line DE is change in y / change in x so: ( (b+d)/2 - (b+f)/2 ) / ( (a+c)/2 - (a+e)/2 ) )

    Simplify once i get ( (b+d-b-f/2) / (a+c-a-e/2) )

    Simplify again i get ( (d-f/2) / (c-e/2) )

    and again (d-f/c-e)

    Slope of the line BC is ( (d-f) / (c-e) )

    Well isn't math just magic

    So all that leaves is to prove the line segment DE is half the size of BC, and I am sure I can accomplish that with the distance formula?
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Midpoint Theorem

    Yes, the distance formula will work wonders for you!
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  9. #9
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    Re: Midpoint Theorem

    So the distance formula for DE is sqt { [(a+c)/2 - (a+e)/2]2 + [(b+d)/2 - (b+f)/2]2 }

    So sqt {
    (c-e/2)2 + (d-f/2)2 }

    And distance formula for BC is sqt { (c-e)2 + (d-f)2 }

    But that doesn't look like definite proof to me ( I could be wrong ), is it possible to simplify this further?
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: Midpoint Theorem

    Yes, you may write:

    \sqrt{\left(\frac{c-e}{2} \right)^2+\left(\frac{d-f}{2} \right)^2}=\sqrt{\left(\frac{1}{2} \right)^2((c-e)^2+(d-f)^2)}=

    \frac{1}{2}\sqrt{(c-e)^2+(d-f)^2)}
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  11. #11
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    Re: Midpoint Theorem

    Oh wow. Thanks again. So just one quick question, how do you display your math like that, with the fancy text? I was thinking that hopefully I would be able to help people at lower levels of math rather than just leach off the forum for answers. Hopefully I can contribute more and knowing how to do that would help plenty.
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: Midpoint Theorem

    It is accomplished using \LaTeX. Teaching its usage is really beyond the scope of a single post, but if you do a search here and online for using latex, you will find lots of good information.

    One technique I used when I was learning to use it is to quote the posts of others and look at the code they use without submitting the quoted post.
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