Proving Triangle with 90° Angle: Length of Hypotenuse Segment

[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 length of the segment between the middle of the hypotenuse and the 90° angle is half the length of the hypotenuse itself?

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

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

john

 

[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?

 

[juef]

Well, I have to prove that if one of the two (the 90° angle or the length that is half the length of the hypotenuse) is true, then the other is too.

 

[metacristi]

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 length of the segment between the middle of the hypotenuse and the 90° angle is half the length 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 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]

 

[ahrkron]

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.

 

[robphy]

Try completing the rectangle.

 

[Gokul43201]

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.>

 

[metacristi]

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.

 

[juef]

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