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### Proof of Pythagorean Theorem: Pythagoras' Reasoning

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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/
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Associated Topics:
High School Geometry
High School Triangles and Other Polygons
Middle School Geometry
Middle School Triangles and Other Polygons

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