Proof: By Euclid

• For let the squares $AE$ and $DF$ have been described on $AB$ and $DC$ (respectively).
• And let the figure in $DF$ have been drawn.
• And let $FC$ have been drawn across to $G$.
• 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 $FH$ the (square) on $AC$.
• Thus, $CE$ (is) equal to $FH$.
• And since $BA$ is double $AD$, and $BA$ (is) equal to $KA$, and $AD$ to $AH$, $KA$ (is) thus also double $AH$.
• And as $KA$ (is) to $AH$, so $CK$ (is) to $CH$ [Prop. 6.1].
• Thus, $CK$ (is) double $CH$.
• And $LH$ plus $HC$ is also double $CH$ [Prop. 1.43].
• Thus, $KC$ (is) equal to $LH$ plus $HC$.
• And $CE$ was also shown (to be) equal to $HF$.
• Thus, the whole square $AE$ is equal to the gnomon $MNO$.
• And since $BA$ is double $AD$, the (square) on $BA$ is four times the (square) on $AD$ - that is to say, $AE$ (is four times) $DH$.
• And $AE$ (is) equal to gnomon $MNO$.
• And, thus, gnomon $MNO$ is also four times $AP$.
• Thus, the whole of $DF$ is five times $AP$.
• And $DF$ is the (square) on $DC$, and $AP$ the (square) on $DA$.
• Thus, the (square) on $CD$ is five times the (square) on $DA$.
• Thus, if a straight line is cut in extreme and mean ratio then the square on the greater piece, added to half of the whole, is five times the square on the half.
• (Which is) the very thing it was required to show.

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