# 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.)

## Comments

### Half of Bhaskara's proof

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