A little geometry

  • Thread starter juef
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  • #1
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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
 

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Answers and Replies

  • #2
Alkatran
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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?
 
  • #3
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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.
 
  • #4
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juef said:
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


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:
  • #5
ahrkron
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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.
 
  • #6
robphy
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Try completing the rectangle.
 
  • #7
Gokul43201
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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.>
 
  • #8
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ahrkron said:
I think Stewart's theorem is an overkill in this case.

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.
 
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  • #9
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Thank you everybody for your help, every advice was very helpful. :D
 

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