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### Pythagorean Theorem - Euclid's Proof

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Date: 12/27/98 at 21:14:19
From: Gayle R.
Subject: Pythagorean Theorem - Euclid's Proof

Dear Dr. Math,

The books I have read don't really explain Euclid's proof to the degree
I need. I understand why they want to prove that the areas of the
rectangles are equal to the squares of the legs of the right triangle,
but I can't understand how they got to that conclusion.

Thank you.
```

```
Date: 12/28/98 at 11:52:16
From: Doctor Peterson
Subject: Re: Pythagorean Theorem - Euclid's Proof

Hi, Gayle.

Euclid's proof on the Pythagorean Theorem is an interesting one. His
proposition was:

In right-angled triangles the square on the side opposite the right
angle equals the sum of the squares on the sides containing the
right angle.

See David Joyce's Web presentation of Euclid's Elements:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI47.html

It sounds as if the part you are having trouble with is:

Now the parallelogram BL is double the triangle ABD, for they have
the same base BD and are in the same parallels BD and AL. And the
square GB is double the triangle FBC, for they again have the same
base FB and are in the same parallels FB and GC.

Therefore the parallelogram BL also equals the square GB.

The idea here is that we have a rectangle "BL" (BDLX) and a triangle
ABD, and another rectangle "GB" (GFBA) and a triangle FBC. In each
pair, we want to show that the rectangle has twice the area of the
square; since we've proved that the two triangles are congruent (and
therefore have the same area), the rectangles will therefore have the
same area. It may be easier to follow by pulling those figures out
from the whole figure:

D         B
+---------+
|         |  \
|         |     \
+---------+-------+
L         X       A

F      B
+------+
|      |\
|      | \
|      |  \
+------+---+
G      A   C

The proof is the same for each pair. I'll work with the first. The area
of the rectangle BDLX is twice the area of triangle ADB by Euclid's
Proposition I.41 ("If a parallelogram has the same base with a triangle
and is in the same parallels, then the parallelogram is double the
triangle"); in modern terms, rectangle BDLX has twice the area of
triangle BDL, which has the same area as ABD because they share the
same base DB and the same altitude DL.

If this still isn't enough detail, please write back and point out the
part you don't understand, either from this copy of the proof or from
the version you have. I'll be glad to help!

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

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