Proof: By Euclid

For let some straight line $AB$ have been cut in extreme and mean ratio at point $C$.
And let $AC$ be the greater piece.
And let $AC$ have been cut in half at $D$.
I say that the (square) on $BD$ is five times the (square) on $DC$.

fig03e

For let the square $AE$ have been described on $AB$.
And let the figure have been drawn double.
Since $AC$ is double $DC$, the (square) on $AC$ (is) thus four times the (square) on $DC$ - that is to say, $RS$ (is four times) $FG$.
And since the (rectangle contained) by $ABC$ is equal to the (square) on $AC$ [Def. 6.3] , [Prop. 6.17], and $CE$ is the (rectangle contained) by $ABC$, $CE$ is thus equal to $RS$.
And $RS$ (is) four times $FG$.
Thus, $CE$ (is) also four times $FG$.
Again, since $AD$ is equal to $DC$, $HK$ is also equal to $KF$.
Hence, square $GF$ is also equal to square $HL$.
Thus, $GK$ (is) equal to $KL$ - that is to say, $MN$ to $NE$.
Hence, $MF$ is also equal to $FE$.
But, $MF$ is equal to $CG$.
Thus, $CG$ is also equal to $FE$.
Let $CN$ have been added to both.
Thus, gnomon $OPQ$ is equal to $CE$.
But, $CE$ was shown (to be) equal to four times $GF$.
Thus, gnomon $OPQ$ is also four times square $FG$.
Thus, gnomon $OPQ$ plus square $FG$ is five times $FG$.
But, gnomon $OPQ$ plus square $FG$ is (square) $DN$.
And $DN$ is the (square) on $DB$, and $GF$ the (square) on $DC$.
Thus, the (square) on $DB$ is five times the (square) on $DC$.
(Which is) the very thing it was required to show.
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