(related to Proposition: Prop. 10.078: That which produces Medial Whole with Medial Area is Irrational)

- For let the straight line $BC$, which is incommensurable in square $AB$, and fulfils the (other) prescribed (conditions), have been subtracted from the (straight line) $AB$ [Prop. 10.35].
- I say that the remainder $AC$ is the irrational (straight line) called that which makes with a medial (area) a medial whole.

- For let the rational (straight line) $DI$ be laid down.
- And let $DE$, equal to the (sum of the squares) on $AB$ and $BC$, have been applied to $DI$, producing $DG$ as breadth.
- And let $DH$, equal to twice the (rectangle contained) by $AB$ and $BC$, have been subtracted (from $DE$) [producing $DF$ as breadth].
- Thus, the remainder $FE$ is equal to the (square) on $AC$ [Prop. 2.7].
- Hence, $AC$ is the square root of $FE$.
- And since the sum of the squares on $AB$ and $BC$ is medial, and is equal to $DE$, $DE$ [is] thus medial.
- And it is applied to the rational (straight line) $DI$, producing $DG$ as breadth.
- Thus, $DG$ is rational, and incommensurable in length with $DI$ [Prop. 10.22].
- Again, since twice the (rectangle contained) by $AB$ and $BC$ is medial, and is equal to $DH$, $DH$ is thus medial.
- And it is applied to the rational (straight line) $DI$, producing $DF$ as breadth.
- Thus, $DF$ is also rational, and incommensurable in length with $DI$ [Prop. 10.22].
- And since the (sum of the squares) on $AB$ and $BC$ is incommensurable with twice the (rectangle contained) by $AB$ and $BC$, $DE$ (is) also incommensurable with $DH$.
- And as $DE$ (is) to $DH$, so $DG$ also is to $DF$ [Prop. 6.1].
- Thus, $DG$ (is) incommensurable (in length) with $DF$ [Prop. 10.11].
- And they are both rational.
- Thus, $GD$ and $DF$ are rational (straight lines which are) commensurable in square only.
- Thus, $FG$ is an apotome [Prop. 10.73].
- And $FH$ (is) rational.
- And the [rectangle] contained by a rational (straight line) and an apotome is irrational [Prop. 10.20], and its square root is irrational.
- And $AC$ is the square root of $FE$.
- Thus, $AC$ is irrational.
- Let it be called that which makes with a medial (area) a medial whole.
- (Which is) the very thing it was required to show.∎

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