# A little geometry

1. Sep 13, 2004

### juef

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?

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

john

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2. Sep 13, 2004

### Alkatran

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. Sep 13, 2004

### juef

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. Sep 13, 2004

### metacristi

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 $$b^2+c^2=a^2$$ 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
5. Sep 13, 2004

### ahrkron

Staff Emeritus
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. Sep 13, 2004

### robphy

Try completing the rectangle.

7. Sep 13, 2004

### Gokul43201

Staff Emeritus
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. Sep 14, 2004

### metacristi

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
9. Sep 15, 2004