:450px Let $A$ be a rational straight line. Let $BC$ be a binomial. Let $DC$ be the greater term of $BC$. Let $BC \cdot EF$ be the rectangle on $BC$ equal to $A^2$. It is to be demonstrated that: :$EF$ is an apotome whose terms are commensurable with $CD$ and $DB$ and in the same ratio and: :the order of $EF$ is the same as the order of $BC$. Let $BD \cdot G$ be the rectangle on $BD$ equal to $A^2$. Then: :$BC \cdot EF = BD \cdot G$ So from {{EuclidPropLink|book = VI|prop = 16|title = Rectangles Contained by Proportional Straight Lines}}: :$CB : BD = G : EF$ But: :$CB > BD$ and so from: :{{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}} and: :{{EuclidPropLink|book = V|prop = 14|title = Relative Sizes of Components of Ratios}} it follows that: :$G > EF$ Let $EH = G$. Then: :$CB : BD = HE : EF$ So from {{EuclidPropLink|book = V|prop = 17|title = Magnitudes Proportional Compounded are Proportional Separated}}: :$CD : BD = HF : FE$ Let it be contrived that $HF : FE = FK : KE$. Therefore from {{EuclidPropLink|book = V|prop = 12|title = Sum of Components of Equal Ratios}}: :$HK : KF = FK : KE$ Therefore from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$FK : KE = CD : DB$ and: :$HK : KF = CD : DB$ But from the definition of binomial, $CD^2$ is commensurable with $DB^2$. Therefore from: :{{EuclidPropLink|book = VI|prop = 22|title = Similar Figures on Proportional Straight Lines}} and: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}} it follows that: :$HK^2$ is commensurable with $KF^2$. We have that $HK$, $KF$ and $KE$ are proportional. So: :$HK^2 : KF^2 = HK : KE$ Thus $HK$ is commensurable in length with $KE$. Therefore from {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$HE$ is commensurable in length with $EK$. We have that $A^2 = EH \cdot BD$. But $A^2$ is rational. Therefore $EH \cdot BD$ is also rational. We have that $EH \cdot BD$ is applied to the rational straight line $BD$. Therefore from {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :$EH$ is rational and commensurable in length with $BD$. We have that $EK$ is rational and commensurable in length with $EH$. So $EK$ is rational and commensurable in length with $BD$. We have that: :$CD : DB = FK : KE$ and: :$CD$ and $DB$ are straight lines which are commensurable in square only. Therefore by {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$FK$ and $KE$ are straight lines which are commensurable in square only. But $KE$ is rational. Therefore $FK$ is also rational. Therefore $FK$ and $KE$ are rational straight lines which are commensurable in square only. Therefore $EF$ is an apotome. {{qed|lemma}} We have that: : $CD^2 = DB^2 + \lambda^2$ where either: :$\lambda$ is commensurable in length with $CD$ or: :$\lambda$ is incommensurable in length with $CD$. First suppose $\lambda$ is commensurable in length with $CD$. Then by {{EuclidPropLink|book = X|prop = 14|title = Commensurability of Squares on Proportional Straight Lines}}: : $FK^2 = KE^2 + \mu^2$ where $\mu$ is commensurable in length with $FK$. Let $CD$ be commensurable in length with a rational straight line $\alpha$ which has been set out. Then by: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}} and: :{{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}} it follows that: :$FK$ is commensurable in length with $\alpha$. Let $BD$ be commensurable in length with $\alpha$. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$KE$ is commensurable in length with $\alpha$. Let neither $CD$ nor $BD$ be commensurable in length with $\alpha$. Then neither $FK$ nor $KE$ is commensurable in length with $\alpha$. Next suppose $\lambda$ is incommensurable in length with $HF$. Then by {{EuclidPropLink|book = X|prop = 14|title = Commensurability of Squares on Proportional Straight Lines}}: :$FK^2 = KE^2 + \mu^2$ where $\mu$ is incommensurable in length with $FK$. Let $CD$ be commensurable in length with a rational straight line $\alpha$ which has been set out. Then by: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}} and: :{{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}} it follows that: :$FK$ is commensurable in length with $\alpha$. Let $BD$ be commensurable in length with $\alpha$. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$KE$ is commensurable in length with $\alpha$. Let neither $CD$ nor $BD$ be commensurable in length with $\alpha$. Then neither $FK$ nor $KE$ is commensurable in length with $\alpha$. It follows that: :the terms of $EF$ are commensurable with the terms of $BD$ and in the same ratio and: :the order of $EF$ is the same as the order of $BD$. {{qed}} {{Euclid Note|112|X}}