Pythagoras's Theorem/Proof 3

Theorem

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

Then:

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

Proof

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

It is also equal to $4 \dfrac {a b} 2 + \paren {a - b}^2$.

So:

\(\ds c^2\)

\(=\)

\(\ds 4 \frac {a b} 2 + \paren {a - b}^2\)

\(\ds \)

\(=\)

\(\ds 2 a b + a^2 - 2 a b + b^2\)

\(\ds \)

\(=\)

\(\ds a^2 + b^2\)

$\blacksquare$

Source of Name

This entry was named for Pythagoras of Samos.

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.1$: The Pythagorean Theorem