:400px Let the area $AB$ be contained by the rational straight line $AC$ and the sixth apotome $AD$. It is to be proved that the "side" of $AB$ is a straight line which produces with a medial area a medial whole. Let $DG$ be the annex of the sixth apotome $AD$. Then, by definition: :$AG$ and $GD$ are rational straight lines which are commensurable in square only :neither the whole $AG$ nor the annex $GD$ is commensurable with the rational straight line $AC$ :the square on the whole $AG$ is greater than the square on the annex $GD$ by the square on a straight line which is incommensurable in length with $AG$. Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $GD$ and deficient by a square figure. From {{EuclidPropLink|book = X|prop = 18|title = Condition for Incommensurability of Roots of Quadratic Equation}}: :that parallelogram divides $AG$ into incommensurable parts. Let $DG$ be bisected at $E$. Let the rectangle contained by $AF$ and $FG$ be applied to $AG$ which is equal to the square on $EG$ and deficient by a square figure. Therefore $AF$ is incommensurable with $FG$. We have that $AF$ is incommensurable in length with $FG$. But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$AF : FG = AI : FK$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AI$ is incommensurable with $FK$. We have that $AG$ commensurable in length with $AC$, and both are rational. Therefore from {{EuclidPropLink|book = X|prop = 21|title = Medial is Irrational}}: :the rectangle $AK$ is medial. We have that $DG$ is incommensurable in length with $AC$, while both are  rational. Therefore from {{EuclidPropLink|book = X|prop = 21|title = Medial is Irrational}}:  :the rectangles $DK$ is medial. But $AG$ and $GD$ are rational straight lines which are commensurable in square only. Therefore $AG$ is incommensurable in length with $GD$. But: :$AG : GD = AK : KD$ Therefore from: :{{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}} and: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AK$ is incommensurable with $KD$. Let the square $LM$ be constructed equal to $AI$. Let the square $NO$ be subtracted from $LM$ having the common angle $\angle LPM$ equal to $FK$. Therefore from {{EuclidPropLink|book = VI|prop = 26|title = Parallelogram Similar and in Same Angle has Same Diameter}}: :the squares $LM$ and $NO$ are about the same diameter. Let $PR$ be the diameter of $LM$ and $NO$. It is to be shown that $LN$ is the "side" of the area $AB$. We have that $AK$ is medial and equals $LP^2 + PN^2$. Therefore $LP^2 + PN^2$ is medial. We have that the rectangle $DK$ is medial. But $DK$ equals twice the rectangle contained by $LP$ and $PN$. Therefore twice the rectangle contained by $LP$ and $PN$ is medial. We have that $AK$ is incommensurable with $DK$. Therefore $LP^2 + PN^2$ is incommensurable with $2 \cdot LP \cdot PN$. We have that $AI$ is incommensurable with $FK$. Therefore $LP^2$ is incommensurable with $PN^2$. Therefore $LP$ and $PN$ are straight lines which are incommensurable in square such that $LP^2 + PN^2$ is medial and such that $2 \cdot LP \cdot PN$ is also medial. Further, $LP^2 + PN^2$ is incommensurable with $2 \cdot LP \cdot PN$. Therefore by definition $LN$ is a straight line which produces with a medial area a medial whole. But $LN$ is the "side" of the area $AB$. Hence the result. {{qed}} {{Euclid Note|96|X|{{EuclidNoteConverse|prop=102|title=Square on Straight Line which produces Medial Whole with Medial Area applied to Rational Straight Line}}}}