Pythagoras's Theorem/Proof 6

Theorem

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

Then:

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

Proof

We have that $CH = BS = AB = AJ$.

Hence the result follows directly from Pythagoras's Theorem for Parallelograms.

$\blacksquare$

Source of Name

This entry was named for Pythagoras of Samos.

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.8$: Pappus (fourth century A.D.)
• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.1$: The Pythagorean Theorem