The square on a second bimedial (straight line) applied to a rational (straight line) produces as breadth a third binomial (straight line).^{1} * Let $AB$ be a second bimedial (straight line) having been divided into its (component) medial (straight lines) at $C$, such that $AC$ is the greater segment. * And let $DE$ be some rational (straight line). * And let the parallelogram $DF$, equal to the (square) on $AB$, have been applied to $DE$, producing $DG$ as breadth. * I say that $DG$ is a third binomial (straight line).
(not yet contributed)
Proofs: 1
Proofs: 1
In other words, the square of a second bimedial is a third binomial. See [Prop. 10.56] (translator's note). ↩