:300px Let $AB$ be the side of the sum of two medial areas divided at $C$. Let $AC > CB$. Let $DE$ be a rational straight line. Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth. It is to be demonstrated that $DG$ is a sixth binomial straight line. From {{EuclidPropLink|book = II|prop = 4|title = Square of Sum}}: :$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$ Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$. Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$. Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$. Let $MG$ be bisected at $N$. Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing). Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$. We have that $AB$ is the side of the sum of two medial areas divided at $C$. Therefore, by definition, $AC$ and $CB$ are straight lines which are incommensurable in square such that: :$AC^2 + CB^2$ is medial :$AC \cdot CB$ is a medial rectangle :$AC^2 + CB^2$ is incommensurable with $AC \cdot CB$. Since $AC^2 + CB^2$ is medial, $DL$ is medial. Therefore from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$DM$ is rational and incommensurable in length with $DE$. We have that: :$2 \cdot AC \cdot CB$, which equals $MF$, is medial. We also have that $MF$ is applied to the rational straight line $ML$. Therefore from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$MG$ is rational and incommensurable in length with $DE$. Therefore $DM$ and $MG$ are rational straight lines which are incommensurable in length with $DE$. We have that $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$. Thus $DL$ is incommensurable with $MF$. So 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}} it follows that: :$DM$ is incommensurable in length with $MG$. That is, $DM$ and $MG$ are rational straight lines which are  commensurable in square only. Therefore by definition $DG$ is binomial. It remains to be proved that $DG$ is a sixth binomial straight line. From {{EuclidPropLink|book = X|prop = 60|qual=Lemma|title = Square on Binomial Straight Line applied to Rational Straight Line}}: :$AC^2 + CB^2 > 2 \cdot AC \cdot CB$ Therefore $DL > MF$. From {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$DM > MG$ Also $DK \cdot KM = MN^2$. Because: :$AC^2$ is incommensurable with $CB^2$ it follows that: :$DH$ is incommensurable with $KL$. So 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}} it follows that: :$DK$ is incommensurable in length with $KM$. So from {{EuclidPropLink|book = X|prop = 18|title = Condition for Incommensurability of Roots of Quadratic Equation}}: :$DM^2$ is greater than $MG^2$ by the square on a straight line incommensurable in length with $DM$. Also, neither $DM$ nor $MG$ is commensurable in length with the rational straight line $DE$. Therefore $DG$ is a sixth binomial straight line. {{qed}} {{Euclid Note|65|X|{{EuclidNoteConverse|prop = 59|title = Root of Area contained by Rational Straight Line and Sixth Binomial}}}}