:300px Let $AB$ be a second bimedial straight line divided into its medials 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 third 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 second bimedial which has been divided into its medials at $C$. Therefore, by definition, $AC$ and $CB$ are medial straight lines which are commensurable in square only such that $AC \cdot CB$ is a medial rectangle. Thus, by definition, $AC^2$ and $CB^2$ are also medial. From: :{{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}} and: :{{EuclidPropLink|book = X|prop = 23|qual = Porism|title = Straight Line Commensurable with Medial Straight Line is Medial}} it follows that: :$DL$ is medial. We have that $DL$ has been applied to the rational straight line $DE$. Therefore from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$MD$ is rational and incommensurable in length with $DE$. For the same reason: :$MG$ is also rational straight line and incommensurable in length with $ML$, which equals $DE$. Therefore each of $DM$ and $MG$ is rational straight line and incommensurable in length with $DE$. We have that $AC$ is incommensurable in length with $CB$. Also: :$AC : CB = AC^2 : AC \cdot CB$ Therefore by {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AC^2$ is incommensurable with $AC \cdot CB$. Hence by: :{{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}} and: :{{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}} it follows that: :$AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$. That is: :$DL$ 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 $MG$. But $DM$ and $MG$ are rational straight lines which are incommensurable in length with each other. Therefore by definition $DG$ is binomial. It remains to be proved that $DG$ is a third 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}}: :$DM > MG$ Since: :$AC^2$ is commensurable with $CB^2$ it follows that: :$DH$ 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 $KM$. Also $DK \cdot KM = MN^2$. 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 $DM$. Also: :Neither $DM$ nor $MG$ is commensurable in length with $DE$. Therefore $DG$ is a third binomial straight line. {{qed}} {{Euclid Note|62|X|{{EuclidNoteConverse|prop = 56|title = Root of Area contained by Rational Straight Line and Third Binomial}}}}