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.