Proof: By Euclid
(related to Proposition: Prop. 13.02: Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio)
 For let the squares $AF$ and $CG$ have been described on each of $AB$ and $CD$ (respectively).
 And let the figure in $AF$ have been drawn.
 And let $BE$ have been drawn across.
 And since the (square) on $BA$ is five times the (square) on $AC$, $AF$ is five times $AH$.
 Thus, gnomon $MNO$ (is) four times $AH$.
 And since $DC$ is double $CA$, the (square) on $DC$ is thus four times the (square) on $CA$  that is to say, $CG$ (is four times) $AH$.
 And the gnomon $MNO$ was also shown (to be) four times $AH$.
 Thus, gnomon $MNO$ (is) equal to $CG$.
 And since $DC$ is double $CA$, and $DC$ (is) equal to $CK$, and $AC$ to $CH$, [$KC$ (is) thus also double $CH$], (and) $KB$ (is) also double $BH$ [Prop. 6.1].
 And $LH$ plus $HB$ is also double $HB$ [Prop. 1.43].
 Thus, $KB$ (is) equal to $LH$ plus $HB$.
 And the whole gnomon $MNO$ was also shown (to be) equal to the whole of $CG$.
 Thus, the remainder $HF$ is also equal to (the remainder) $BG$.
 And $BG$ is the (rectangle contained) by $CDB$.
 For $CD$ (is) equal to $DG$.
 And $HF$ (is) the square on $CB$.
 Thus, the (rectangle contained) by $CDB$ is equal to the (square) on $CB$.
 Thus, as $DC$ is to $CB$, so $CB$ (is) to $BD$ [Prop. 6.17].
 And $DC$ (is) greater than $CB$ (see lemma).
 Thus, $CB$ (is) also greater than $BD$ [Prop. 5.14].
 Thus, if the straight line $CD$ is cut in extreme and mean ratio then the greater piece is $CB$.
 Thus, if the square on a straight line is five times the (square) on a piece of itself, and double the aforementioned piece is cut in extreme and mean ratio, then the greater piece is the remaining part of the original straight line.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"