Proving the Pythagorean Theorem using Congruent Squares

Date: 12/5/95 at 15:14:58
From: Anonymous
Subject: Proof of Pythagorean Theorem

Dear Dr. Math,

A friend of mine is irked because of constant use of the Pythagorean
Theorem, which he has not seen proven.  I have seen it proven before,
but could not quite recount the proof.  I have heard the theorem
called the "most proved theorem," so this ought to be an easy and
a very prolific search, but could you find a proof of the theorem.

Thanks,
Johnny Vogler

Date: 3/6/96 at 15:15:45
From: Doctor Dusty
Subject: Re: Proof of Pythagorean Theorem

This is a very neat little proof of the Pythagorean Theorem, but it
might look a little funky because of the graphics.  Here it goes:


        b    a             b    a
       _____.__           _____.__
  a   |        | a     a |        |
      .        .         .        |  b
      |        |         |        | 
  b   |        | b     b |        .
      |_____.__|         |__._____|  a
        b    a            a    b

First of all, construct two congruent squares, both with sides a + b.
You know that the area of each square should be (a + b)^2

Now connect the points in the first square to create two squares and 
two rectangles.  When you add the areas of those four up, you get
a^2 + ab + ab + b^2 which is also equal to the area of the entire 
square, (a + b)^2.

Connect the points of the second square to form another square and call 
the length of each side of the inner square c.  You now have four right
triangles and one square.  Add the areas of the 5 figures together and 
you have .5ab + .5ab + .5ab + .5ab + c^2

The areas of both squares are equal; we agreed to that in the beginning.
Set the area equation of the first equal to that of the second.

a^2 + 2ab + b^2 = 2ab + c^2

Subtract 2ab from both sides and you have

a^2 + b^2 = c^2

where a is one leg of the right triangle, b is the other leg, and c
is the hypotenuse.  (I got that from looking at the second square.)  

I just think that this proof is really neat.  I hope it calms your 
friend down.  

-Doctor Dusty,  The Math Forum