Proof: Converse of Pythagoras' theorem using vectors

I am trying to prove the converse of the Pythagorean theorem using vector geometry and vector algebra. Here's a picture:

http://draw.to/static/d/19raZB.png?v=531d0

Take notice that . The given is that

The goal is to show that . I dotted with , thus yielding

I then attemped to find an expression for from the equation by multiplying both sides by and dot every vector with it.

is the angle formed between and . I then I figured that I could compare the equation with and then identify the terms. This however did not work out as it lead to showing that is the hypotenuse whereas is not, thus contradicting the very premise of the proof.

Anyone got an idea?

PS: I've already seen the proof using congruence but that particular proof is rather irrelevant, so don't bother referencing it.

Re: Proof: Converse of Pythagoras' theorem using vectors

Quote:

Originally Posted by

**MathCrusader** I am trying to prove the converse of the Pythagorean theorem using vector geometry and vector algebra. Here's a picture:

http://draw.to/static/d/19raZB.png?v=531d0
Take notice that

. The given is that

The goal is to show that

. I dotted

with

, thus yielding.

From the given

and .

.

But that mean

Re: Proof: Converse of Pythagoras' theorem using vectors

Thank you. It works just as well with as with

Re: Proof: Converse of Pythagoras' theorem using vectors

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