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### 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/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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