Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Pythagorean Theorem - Euclid's Proof


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

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/