(related to Proposition: 6.13: Construction of Mean Proportional)

- Let ($AB$ and $BC$) be laid down straight-on (with respect to one another), and let the semicircle $ADC$ have been drawn on $AC$ [Prop. 1.10].
- And let $BD$ have been drawn from (point) $B$, at right angles to $AC$ [Prop. 1.11].
- And let $AD$ and $DC$ have been joined.
- And since $ADC$ is an angle in a semicircle, it is a right angle [Prop. 3.31].
- And since, in the right-angled triangle $ADC$, the (straight line) $DB$ has been drawn from the right angle perpendicular to the base, $DB$ is thus in mean proportion3 to the pieces of the base, $AB$ and $BC$ [Prop. 6.8 corr.] 3.
- Thus, $DB$ has been found (which is) in to the two given straight lines, $AB$ and $BC$.
- (Which is) the very thing it was required to do.∎

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