Pythagorean Theorem by Contradiction


Pythagorean theorem, proof 122,formulation


Let $h$ be the altitude from the right angle. The right triangle $(a,b,c)$ is split into two triangles, $(x,h,a)$ and $(h,y,b),$ both similar to the triangle $(a,b,c).$ In particular, $h^2=xy.$

Pythagorean theorem, proof 122

Assume to the contrary that $a^2+b^2\ne c^2.$ Let, e.g, $a^2+b^2\lt c^2.$ Then, by similarity, also

$x^2+h^2\lt a^2$ and $h^2+y^2\lt b^2.$

Summing up yields a contradiction:

$\begin{align}a^2+b^2\,&\gt (x^2+h^2)+(h^2+y^2)=x^2+y^2+2h^2\\ &=x^2+y^2+2xy=(x+y)^2=c^2, \end{align}$

so that $a^2+b^2\gt c^2,$ contrary to our assumption.


The proof by K. B. SUBRAMANIAM, Math. Gaz. 102 (2018) p. 128 was kindly pointed to me by Grégoire Nicollier.


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