If the square on one of the sides of a triangle is equal to the (sum of the) squares on the two remaining sides of the triangle then the angle contained by the two remaining sides of the triangle is a right angle. * For let the square on one of the sides, $BC$, of triangle $ABC$ be equal to the (sum of the) squares on the sides $BA$ and $AC$. * I say that angle $BAC$ is a right angle.
If the square on one side (\(\overline{BC}\)) of a triangle (\(\triangle{ABC}\)) equals the sum of the squares on the remaining sides (\(\overline{BA}\), \(\overline{AC}\)), then the angle (\(\angle{CAB}\)) opposite to that side is a right angle.
Proofs: 1