# Garfield's proof of Pythagorean theorem

## Primary tabs

Type of Math Object:
Proof
Major Section:
Reference
Groups audience:

## Mathematics Subject Classification

51-00 General reference works (handbooks, dictionaries, bibliographies, etc.)

It might be worth noting that the diagram used in this proof
is half of the diagram used in Bhaskara's proof of the same
result.

### Re: Half of Bhaskara's proof

I don't know that proof, and it doesn't seem to be posted. Perhaps you could either post it, or tell me how it goes and I will, then I'll modify my entry to refer to it. Thanks.

### Drawing right angles

In response to pahio's correction, I've managed to draw right angles on the left and right hand triangles, but I can't get xypic to work for the third triangle. (I'm trying to learn xypic as I go). Perhaps someone can help. I'm trying
$\begin{xy} ,(0,24) ;(42,18)**@{-} ;(42,0)**@{-} ;(42,2)**@{-} ;(40,2)**@{-} ;(40,0)**@{-} ;(42,0)**@{-} ;(0,0)**@{-} ;(0,2)**@{-} ;(2,2)**@{-} ;(2,0)**@{-} ;(0,0)**@{-} ;(0,24)**@{-} ;(18,0)**@{-} ;(19.4,.94)**@{-} ;(18.46,2.34)**@{-} ;(17.06,1.4)**@{-} ;(18,0)**@{-} ;(42,18)**@{-} ,(9,-2)*{a} ,(44,9)*{a} ,(-2,12)*{b} ,(30,-2)*{b} ,(9,16)*{c} ,(30,12)*{c} \end{xy}$

but nothing gets drawn for the elements using nonintegral coordinates. Is there a different way to do this?

Thanks,
Roger

### Re: Drawing right angles

You could try another package called pstricks. Google "pstricks tutorial" and you should be able to find a number of decent references. Let me know if you need more help.

Chi

### Re: Half of Bhaskara's proof

The diagram Bhaskara used was a square of length a+b which was
cut into four right traiangles of sides a,b,c and a square of
length c.

By the way, there is also a purely geometric way of presenting
Bhaskara's proof. Namely, one considers two differrent ways of
subdividing the square of length a+b. One is as above. The
other divides it into a square of side a, a square of side b and
four triangles of sides a,b,c. By comparing these subdivisions,
it is obvious that a^2 + b^2 = c^2.

Based on how you did your entry, I will add an account of this proof.

### Re: Half of Bhaskara's proof

Never mind, there already is an account of this proof:

http://planetmath.org/encyclopedia/ProofOfPythagoreasTheorem.html