In right-angled triangles,  the square on the side subtending the right-angle is equal to the (sum of the) squares on the sides containing the right-angle. Let $ABC$ be a right-angled triangle having the angle $BAC$a right-angle. I say that the square on $BC$ is equal to the (sum of the) squares on $BA$ and $AC$.  <prf>  Thus, the square on the side $BC$ is equal to the (sum of the) squares on the sides $BA$ and $AC$. Thus, in right-angled triangles,  the square on the side subtending the right-angle is equal to the (sum of the) squares on the sides surrounding the right-[angle].