Proof: By Euclid

For since the straight line $CB$ has been cut, at random, at (point) $D$, the (sum of the) squares on $CB$ and $BD$ is thus equal to twice the rectangle contained by $CB$ and $BD$, and the square on $DC$ [Prop. 2.7].
Let the square on $DA$ have been added to both.
Thus, the (sum of the) squares on $CB$, $BD$, and $DA$ is equal to twice the rectangle contained by $CB$ and $BD$, and the (sum of the) squares on $AD$ and $DC$.
But, the (square) on $AB$ (is) equal to the (sum of the squares) on $BD$ and $DA$.
For the angle at (point) $D$ is a right angle [Prop. 1.47].
And the (square) on $AC$ (is) equal to the (sum of the squares) on $AD$ and $DC$ [Prop. 1.47].
Thus, the (sum of the squares) on $CB$ and $BA$ is equal to the (square) on $AC$, and twice the (rectangle contained) by $CB$ and $BD$.
So the (square) on $AC$ alone is less than the (sum of the) squares on $CB$ and $BA$ by twice the rectangle contained by $CB$ and $BD$.
Thus, in acute-angled triangles, the square on the side subtending the acute angle is less than the (sum of the) squares on the sides containing the acute angle by twice the (rectangle) contained by one of the sides around the acute angle, to which a perpendicular (straight line) falls, and the (straight line) cut off inside (the triangle) by the perpendicular (straight line) towards the acute angle.
(Which is) the very thing it was required to show.
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