# Pythagoras's Theorem/Proof 2

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## 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.