# Pythagoras's Theorem/Proof 3

From ProofWiki

## Contents

## Theorem

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

## Proof

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

- George F. Simmons:
*Calculus Gems*(1992): Chapter $\text {B}.1$