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.  For let $AD$ have been drawn from point $A$ at right-angles to the straight-line $AC$ [Prop.~1.11], and let $AD$ have been made equal to $BA$ [Prop.~1.3], and let $DC$ have been joined. Since $DA$ is equal to $AB$, the square on $DA$ is thus also equal to the square on $AB$. Let the square on $AC$ have been added to both. Thus, the (sum of the) squares on $DA$ and $AC$ is equal to the (sum of the) squares on $BA$ and $AC$. But, the (square) on $DC$  is equal to the (sum of the squares) on $DA$ and $AC$. For angle $DAC$ is a right-angle [Prop.~1.47]. But, the  (square) on $BC$ is equal to (sum of the squares) on $BA$ and $AC$. For (that) was assumed. Thus, the square on $DC$ is equal to the square on $BC$. So  side $DC$ is also equal to (side) $BC$. And since $DA$ is equal to $AB$, and $AC$ (is) common, the two (straight-lines) $DA$, $AC$ are equal to the two (straight-lines) $BA$, $AC$. And the base $DC$ is equal to the base $BC$. Thus, angle $DAC$ [is] equal to angle $BAC$ [Prop.~1.8].  But $DAC$ is a right-angle. Thus, $BAC$ is also a right-angle.  Thus, if the square on one of the sides of a triangle is equal to the (sum of the) squares on the remaining two sides of the triangle then the angle contained by the remaining two sides of the triangle is a right-angle. (Which is) the very thing it was required to show.