Pythagorean Theorem Proof (Thabit ibn Qurra)

Date: 03/28/2002 at 20:39:21
From: Natalie Bramlett
Subject: Proving Pythag. th from cut-the-knot.com problem #24

Dr.Math,

I am working on a proof from cut-the-knot.com, #24. I don't understand 
exactly what it is saying that I have to prove. I tried making the 
base triangle and 3,4,5 triangle and then going from there, but I 
don't know how they are forming the other triangles.  If you could 
help me in any way I would greatly appreciate it.

Thanks,
Natalie

Date: 03/28/2002 at 23:26:24
From: Doctor Peterson
Subject: Re: Proving Pythag. th from cut-the-knot.com problem #24

Hi, Natalie.

I presume you are referring to

   http://www.cut-the-knot.org/pythagoras/   

   [Swetz] ascribes this proof to abu' l'Hasan Thabit ibn Qurra Marwan
   al'Harrani (826-901). It's the second of the proofs given by Thabit
   ibn Qurra. The first one is essentially the #2 above.

   The proof resembles part 3 from proof #12. ABC = FLC = FMC = BED
   = AGH = FGE. On the one hand, the area of the shape ABDFH equals
   AC^2 + BC^2 + area(ABC + FMC + FLC). On the other hand, area(ABDFH)
   = AB^2 + area(BED + FGE + AGH).

                  F+
                  /|    \     L
                 / |         +
                /  |        /     \     E
              G+---+-------/-----------+
              /|   |      /            |    \    D
             / |   |     /             |         +
            /  |   |    /              |        /
           /   |   |   /               |       /
          /    |   |  /                |      /
        M+     |   | /                 |     /
        /     \|   |/                  |    /
       /       |   + C                 |   /
      /        |  /     \              |  /
     +         | /           \         | /
   H      \    |/                 \    |/
               +-----------------------+
              A                         B

I'm not sure what you mean about a 3-4-5 triangle; do you just mean 
that you used a 3-4-5 right triangle to draw the figure as an example?

I believe the construction goes something like this:

Draw right triangle ABC. Extend sides AC and BC beyond C by lengths 
MC=AC and CL = CB, forming squares ACMH and BCLD. Lines HM and DL will 
meet at F. Erect perpendiculars AG and BE to AB, with G and E on HF 
and DF respectively; you can show that this is a square.

Now you can list various segments congruent to AB, BC, and AC (because 
they are part of of various parallelograms), and use those to prove 
the series of congruent triangles listed. (The vertices of these 
triangles are listed out of order; you will have to correct the 
naming.) The rest is just to dissect the whole figure two different 
ways.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/