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

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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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## marked as duplicate by William, M Turgeon, Fabian, Noah Snyder, tomaszOct 5 '12 at 19:48

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
@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
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
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. " –  mathguy Aug 20 '12 at 17:24

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 $$a^2+a^2=b^2.$$