1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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! :)

    john
     

    Attached Files:

  2. jcsd
  3. Sep 13, 2004 #2

    Alkatran

    User Avatar
    Science Advisor
    Homework Helper

    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:

    [PA3]*[A1A2]2+[PA2]*[A1A3]2=[PA1]2*[A2A3]+[PA2]*[PA3]*[A2A3]
     
    Last edited: Sep 13, 2004
  6. Sep 13, 2004 #5

    ahrkron

    User Avatar
    Staff Emeritus
    Gold Member

    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

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A little geometry
  1. Atom of argon charge (Replies: 5)

  2. A little entropy (Replies: 13)

  3. A little doubt (Replies: 2)

  4. A little help (Replies: 2)

Loading...