(related to Proposition: Prop. 13.05: Straight Line cut in Extreme and Mean Ratio plus its Greater Segment)

- 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.

- 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.∎

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