In right-angled triangles, the figure (drawn) on the side subtending the right angle is equal to the (sum of the) similar, and similarly described, figures on the sides surrounding the right angle. * Let $ABC$ be a right-angled triangle having the angle $BAC$ a right angle. * I say that the figure (drawn) on $BC$ is equal to the (sum of the) similar, and similarly described, figures on $BA$ and $AC$.
This is a generalization of the Pythagorean theorem, in which the squares are replaced by similar rectangles.
Proofs: 1