Proving the converse of Pythagorus

Prove that if a triangle has sides of lengths a, b and c, such that a2 + b2 = c2 then the triangleis right angled.

The only thing i was thinking is cosing law. so

a2 + b2-2abcos(x)= c2 and so -2abcos(x)=0 therefore x=90degrees

Is there a more elegant way to prove this?

Re: Proving the converse of Pythagorus

Hey calculuskid1.

You could prove it with inner products and norms but the way you have used is pretty good and a lot easier.

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Re: Proving the converse of Pythagorus

Re: Proving the converse of Pythagorus

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Originally Posted by

**calculuskid1** Prove that if a triangle has sides of lengths a, b and c, such that a2 + b2 = c2 then the triangleis right angled.

The only thing i was thinking is cosing law. so

a2 + b2-2abcos(x)= c2 and so -2abcos(x)=0 therefore x=90degrees

Is there a more elegant way to prove this?

I think that your way is very elegant. Of course, it would mean that you have to prove the Cosine Rule...