(related to Proposition: Prop. 13.03: Area of Square on Lesser Segment of Straight Line cut in Extreme and Mean Ratio)

- 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 $DC$.

- 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.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"