Pythagoras's Theorem/Proof 3

Theorem

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

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.