Proof: By Euclid

For let the straight line $AB$ have been cut in extreme and mean ratio at point $C$.
And let $AC$ be the greater piece.
And let $AD$ be [made] equal to $AC$.
I say that the straight line $DB$ has been cut in extreme and mean ratio at $A$, and that the original straight line $AB$ is the greater piece.

fig05e

For let the square $AE$ have been described on $AB$, and let the (remainder of the) figure have been drawn.
And since $AB$ has been cut in extreme and mean ratio at $C$, the (rectangle contained) by $ABC$ is thus equal to the (square) on $AC$ [Def. 6.3] , [Prop. 6.17].
And $CE$ is the (rectangle contained) by $ABC$, and $CH$ the (square) on $AC$.
But, $HE$ is equal to $CE$ [Prop. 1.43], and $DH$ equal to $HC$.
Thus, $DH$ is also equal to $HE$.
Let $HB$ have been added to both. Thus, the whole of $DK$ is equal to the whole of $AE$.
And $DK$ is the (rectangle contained) by $BD$ and $DA$.
For $AD$ (is) equal to $DL$.
And $AE$ (is) the (square) on $AB$.
Thus, the (rectangle contained) by $BDA$ is equal to the (square) on $AB$.
Thus, as $DB$ (is) to $BA$, so $BA$ (is) to $AD$ [Prop. 6.17].
And $DB$ (is) greater than $BA$.
Thus, $BA$ (is) also greater than $AD$ [Prop. 5.14].
Thus, $DB$ has been cut in extreme and mean ratio at $A$, and the greater piece is $AB$.
(Which is) the very thing it was required to show.
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