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What is the most elegant proof of the Pythagorean theorem?

How do we prove that the Pythagorean theorem holds for a right angled isoceles triangle with sides, $a,b,a$.

For a right angled triangle with sides $a,b,c$, where $\angle C = 90^{\circ}$, we have $a^2+b^2=c^2$

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HOLYBIBLETHE
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  • I do not understand if you want to prove Pythagora's theorem or something else. – Siminore Aug 20 '12 at 17:07
  • What is it about the many standard methods of proof that you think don't work in the isoceles case? – Robert Mastragostino Aug 20 '12 at 17:08
  • It follows immediately from the general Pythagorean theorem. Or are you asking whether there’s a simpler proof for the special case of an isosceles right triangle? – Brian M. Scott Aug 20 '12 at 17:08
  • You can find some proofs at http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/pythagorean.html – Siminore Aug 20 '12 at 17:08
  • In a $\triangle ABC$ with $\angle C=90^{\circ}$, Pythagoras stated $a^2+b^2=c^2$, assuming $a \ne b \ne c$. – HOLYBIBLETHE Aug 20 '12 at 17:10
  • Well the square on each of the short sides can be made from two copies of the original triangle. The square on the long side takes four copies. – Mark Bennet Aug 20 '12 at 17:11
  • I don't know what the basis of your claim about Pythagoras and $a\ne b\ne c$ is, since all the writings of the Pythagorean school are lost. Euclid's proposition 47 says simply "In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle" and the proof goes through just fine for isosceles right triangles. – MJD Aug 20 '12 at 17:15
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    @RajeshKSingh You do see that $a\neq b\neq c$ does not prohibit $a=c$, right? It kind of sounds like you're treating $\neq$ as a transitive relationship. – rschwieb Aug 20 '12 at 17:15
  • $a \ne b, a \ne c, b \ne c$ – HOLYBIBLETHE Aug 20 '12 at 17:21
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    Downvoters and closers: I'm half-sure that you're all missing this person's mistake. I'm pretty sure he has just misread $a\neq b \neq c$ as meaning "a, b and c are all unequal." He just thinks that the isosceles case is still unproven. – rschwieb Aug 20 '12 at 17:24
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    you say that "Pythagoras stated $a^2+b^2=c^2$, assuming a≠b≠c. ", however, I think a more correct statement is "Pythagoras stated $a^2+b^2=c^2$, including cases where a≠b≠c. " – Sidd Singal Aug 20 '12 at 17:24

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Make a paper square with sides $b$. Divide it into $4$ triangles by drawing the two diagonals. Cut along the diagonals.

We get $4$ congruent isosceles right-angled triangles. Let the right-angled triangles we get have legs $a$. They each have hypotenuse $b$.

We can put these triangles together in pairs along their hypotenuses to form two $a\times a$ squares. So we have cut a $b\times b$ square into four pieces and reassembled the pieces to make two $a\times a$ squares. Since area is conserved, it follows that $$a^2+a^2=b^2.$$

Remark: This is a simple dissection proof of a very special case of the Pythagorean Theorem. There are several general dissection proof of the Theorem.

André Nicolas
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