Pythagoras's Theorem/Proof 3

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Theorem

Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.

Then:

$a^2 + b^2 = c^2$


Proof

Pythagoras3.png

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

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

So:

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

$\blacksquare$


Source of Name

This entry was named for Pythagoras of Samos.


Sources