Proof: By Euclid

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 $AB$.¹ 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.

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Here, use is made of the additional common notion that the squares of equal things are themselves equal. Later on, the inverse notion is used (translator's note). ↩