Pythagoras's Theorem/Proof 2

Theorem

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

Then:

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

Proof

We have:

$\dfrac b c = \dfrac d b$

and:

$\dfrac a c = \dfrac e a$

using the fact that all the triangles involved are similar.

That is:

$b^2 = c d$

$a^2 = c e$

Adding, we now get:

$a^2 + b^2 = c d + c e = c \paren {d + e} = c^2$

$\blacksquare$

Source of Name

This entry was named for Pythagoras of Samos.

Historical Note

This proof was demonstrated by Bhaskara II Acharya in the $12$th century.

It was rediscovered in the $17$th century by John Wallis.

Sources

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

• For a video presentation of the contents of this page, visit the Khan Academy.