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Proof of Pythagorean Theorem: Pythagoras' Reasoning


Date: 02/23/98 at 11:49:20
From: Pat Hagedorn
Subject: Geometry

I would like to know how Pythagoras reasoned his theorem. What he drew 
on and how he came to figure it out. Why does it work the way that it 
does? For example: say the hypotenuse is 20", the other known is 16".

How is it that 20" is squared and so is 16"?

Why is 400 relevant to 20?
Why is 256 relevant to 16?
How are both related to 12?

How does he know that this process produces the unknown exactly? 
It is not an obvious thing for me... it is a reasoning thing. How did 
he figure this out in the first place? He had to have based his 
thoughts on something that led him to this conclusion. Do you know?
Thank you in advance for any explanation you can provide as this is 
getting to my ability to go on and just rely on a formula.

Sincerely,

Pat


Date: 02/23/98 at 16:14:11
From: Doctor Rob
Subject: Re: Geometry

400 is 20^2, and is the area of a square of side 20.
256 is 16^2, and is the area of a square of side 16.
Sqrt[400-256] = Sqrt[144] = 12.

The process produces the number exactly because the following proof 
is, in fact, a valid proof.  Every step can be logically justified 
using the postulates and axioms of geometry and rules of logical 
inference.

Start with the triangle ABC, where C is the right angle.  Let the 
length of side AB be c, of side BC be a, and of sice AC be b.

The traditional proof goes like this.  

    On each side of the triangle construct a square whose side length  
    is the length of the side of the triangle:  ABDE, BCFG, and CAHI.   
    Drop a perpendicular from C to AB, meeting it at K, and extended 
    to meet DE at J.  Draw lines CE and BH.  
    
    Now prove triangles ACE and ABH are congruent by S.A.S. 
 
    Now show that the area of square ACIH is twice the area of   
    triangle ABH, and the area of rectangle AEJK is twice the area of  
    triangle ACE.  
 
    Conclude that the area of square ACIH equals the area of rectangle  
    AEJK.  
 
    Similarly, draw lines CD and AG. Now prove triangles BCD and ABG  
    are congruent by S.A.S.  Now show that the area of square BCFG is  
    twice the area of triangle ABG, and the area of rectangle BDJK is  
    twice the area of triangle BCD. Conclude that the area of square  
    BCFG equals the area of rectangle BDJK.  

    Now add your two conclusions:  the area of square BCFG (a^2) plus   
    the area of square ACIH (b^2) equals the area of the two  
    rectangles AEJK plus BDJK (which togethermake up square ABDE, of  
    area c^2).


This proof gives us some insight into Pythagoras's train of thought.  
The fact that the square on each leg has the same area as the adjacent
rectangle which is part of the square on the hypotenuse is the key.  
We can guess that, via drawing many examples, he divined that this 
might be the case, and then was able to prove it as above.

There are other proofs which use the equal-area-square-and-rectangle 
idea. There are many other proofs which use entirely different ideas.

Aside from that, there is no description surviving of Pythagoras' 
reasons for his discovery or reasoning leading to it.

-Doctor Rob,  The Math Forum
 http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
Middle School Geometry
Middle School Triangles and Other Polygons

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