Proof: By Euclid

Let the square $BC$ have been described on $AB$ [Prop. 1.46], and let the parallelogram $CD$, equal to $BC$, have been applied to $AC$, overshooting by the figure $AD$ (which is) similar to $BC$ [Prop. 6.29].
And $BC$ is a square.
Thus, $AD$ is also a square.
And since $BC$ is equal to $CD$, let (rectangle) $CE$ have been subtracted from both.
Thus, the remaining (rectangle) $BF$ is equal to the remaining (square) $AD$.
And it is also equiangular to it.
Thus, the sides of $BF$ and $AD$ about the equal angles are reciprocally proportional [Prop. 6.14].
Thus, as $FE$ is to $ED$, so $AE$ (is) to $EB$.
And $FE$ (is) equal to $AB$, and $ED$ to $AE$.
Thus, as $BA$ is to $AE$, so $AE$ (is) to $EB$.
And $AB$ (is) greater than $AE$.
Thus, $AE$ (is) also greater than $EB$ [Prop. 5.14].
Thus, the straight line $AB$ has been cut in extreme and mean ratio at $E$, and $AE$ is its greater piece.
(Which is) the very thing it was required to do.
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