Pythagoras's Theorem/Proof 3

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Theorem

Given any right triangle $\triangle ABC$ with $c$ as the hypotenuse, we have $a^2 + b^2 = c^2$.


Proof

Pythagoras3.png

The area of the big square is $c^2$.

It is also equal to $\displaystyle 4 \frac {ab} 2 + \left({a-b}\right)^2$.

So:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle c^2\) \(=\) \(\displaystyle \) \(\displaystyle 4 \frac {ab} 2 + \left({a-b}\right)^2\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle 2 a b + a^2 - 2 a b + b^2\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle a^2 + b^2\) \(\displaystyle \) \(\displaystyle \)                    

$\blacksquare$


Source of Name

This entry was named for Pythagoras of Samos.


Sources