:300px Let $AB$ be a binomial straight line divided into its terms 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 first 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 a binomial which has been divided into its terms at $C$. Therefore, by definition, $AC$ and $CB$ are rational straight lines which are commensurable in square only. Therefore $AC^2$ and $CB^2$ are commensurable and rational. Thus by {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$AC^2 + CB^2$ is a rational area. But $AC^2 + CB^2 = DL$. Therefore $DL$ is a rational area. We have that $DL$ is applied to the rational straight line $DE$. Therefore from {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :$DM$ is rational and commensurable in length with $DE$. We have that $AC$ and $CB$ are rational straight lines which are commensurable in square only. So from {{EuclidPropLink|book = X|prop = 21|title = Medial is Irrational}}: :$2 \cdot AC \cdot CB = MF$ is medial. We 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 also rational and incommensurable in length with $DE$. But $MD$ is also rational and commensurable in length with $DE$. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$DM$ is incommensurable in length with $MG$. But $DM$ and $MG$ are both rational straight lines which are commensurable in square only. Therefore, by definition, $DG$ is binomial. It remains to be proved that $DG$ is first binomial. From {{EuclidPropLink|book = X|prop = 54|qual = Lemma|title = Root of Area contained by Rational Straight Line and First Binomial}}: :$AC \cdot CB$ is a mean proportional between $AC^2$ and $CB^2$. Therefore $MO$ is a mean proportional between $DH$ and $KL$. Therefore by {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$DK : MN = MN : MK$ Therefore from {{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}}: :$DK \cdot KM = MN^2$ We have that $AC^2$ is commensurable with $CB^2$. Therefore $DH$ is commensurable 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 commensurable in length with $KM$. 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: :$DL > MF$ and so from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$DM > MG$ And: :$DK \cdot KM = MN^2 = \dfrac {MG^2} 4$ and: :$DK$ is commensurable in length with $KM$. Thus from {{EuclidPropLink|book = X|prop = 17|qual = Lemma|title = Condition for Commensurability of Roots of Quadratic Equation}}: :$DM^2$ is greater than $MG^2$ by the square on a straight line commensurable in length with $DM$. We also have that: :$DM$ and $MG$ are rational and: :the greater term $DM$ is commensurable in length with the rational straight line $DE$. Therefore by definition $DG$ is a first binomial straight line. {{qed}} {{Euclid Note|60|X|{{EuclidNoteConverse|prop = 54|title = Root of Area contained by Rational Straight Line and First Binomial}}}}