# Pythagorean Theorem, Proof 115

The proof below is by Nileon M. Dimalaluan, Jr. and was published in the online journal Chiaroscuro of a High IQ Society (March 2004.)

The proof is based on the following diagram:

Since triangles $ABC\;$ and $AED\;$ are similar $\displaystyle \frac{r}{a+c}=\frac{a}{b}.$

Triangles $EBD,EBC$ are equal, each with the area of $\displaystyle\frac{1}{2}ra.$

If $[F]\;$ denotes the area of shape $F,\;$ then $[\Delta AED]=[\Delta ABC]+2[\Delta EBC]\;$ so that

$\displaystyle\frac{1}{2}r(a+c)=2\cdot \frac{1}{2}ra+\frac{1}{2}ab,$

or, $\displaystyle r(c-a)=ab.\;$ Substituting $\displaystyle r=\frac{a(c+a)}{b}\;$ results in $\displaystyle \frac{c^2-a^2}{b}=b\;$ which is the Pythagorean theorem, $a^2+b^2=c^2.$

Copyright © 1996-2018 Alexander Bogomolny

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