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Homework Help: A little geometry

  1. Sep 13, 2004 #1
    Hi all,

    I'm having a hard time proving something. Let's say you have a triangle with a 90° angle. How can I prove that the lenght of the segment between the middle of the hypotenuse and the 90° angle is half the lenght of the hypotenuse itself?

    Here's a little pic to help you...

    Thanks a lot, and sorry for my pityful english! :)


    Attached Files:

  2. jcsd
  3. Sep 13, 2004 #2


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    Uh... isn't that what the picture says anyways? Why do you need to prove it if the picture is telling you it... or do you need to proce it for all cases?
  4. Sep 13, 2004 #3
    Well, I have to prove that if one of the two (the 90° angle or the lenght that is half the lenght of the hypotenuse) is true, then the other is too.
  5. Sep 13, 2004 #4

    I do not know your level of knowledge in geometry,the simplest solution I see at first sight is to use Stewart's relation (knowing also that [tex]b^2+c^2=a^2[/tex] and that the hypothenuse is split into two equal segments a/2).

    See http://mathworld.wolfram.com/StewartsTheorem.html

    With the notations used there we have:

    Last edited: Sep 13, 2004
  6. Sep 13, 2004 #5


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    I think Stewart's theorem is an overkill in this case.

    At the middle point of the hypothenuse, draw a segment parallel to one of the other two sides, and look for congruent triangles.
  7. Sep 13, 2004 #6


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    Try completing the rectangle.
  8. Sep 13, 2004 #7


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    How about using the (converse of the) fact that the angle in a semicircle is always a right angle ?

    <Not sure if you've done circles yet. If not, I defer to robphy's suggestion.>
  9. Sep 14, 2004 #8
    Well depends on the knowledge level.After all Stewart's relation can be easily deduced by applying Pitagoras' generalized theorem two times,rearranging a bit the equations and taking also into account that cosX+cos[π-X]=0.
    Last edited: Sep 14, 2004
  10. Sep 15, 2004 #9
    Thank you everybody for your help, every advice was very helpful. :D
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