Pythagoras's Theorem/Proof 3

Theorem

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

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 $

$\blacksquare$

This entry was named for Pythagoras of Samos.

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