:400px Let $A$ be a rational straight line. Let $BD$ be an apotome. Let $BD \cdot KH$ be the rectangle on $BD$ equal to $A^2$. It is to be demonstrated that: :$KH$ is a binomial straight line whose terms are commensurable with the terms of $BD$ and in the same ratio and: :the order of $KH$ is the same as the order of $BD$. Let $DC$ be the annex of $BD$. By the definition of apotome, $BC$ and $CD$ are rational straight lines which are commensurable in square only. Let $BC \cdot G$ be the rectangle on $BC$ equal to $A^2$. But $A^2$ is rational. Therefore $BC \cdot G$ is also rational. We have that $BC \cdot G$ has been applied to a rational straight line $BC$. Thus from {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :$G$ is rational and commensurable in length with $BC$. We have that: :$BC \cdot G = BD \cdot KH$ Therefore by {{EuclidPropLink|book = VI|prop = 16|title = Rectangles Contained by Proportional Straight Lines}}: :$CD \cdot BD = KH \cdot G$ But: :$BC > BD$ 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: :$KH > G$ Let $KE = G$. Then $KE$ is commensurable in length with $BC$. We have that: :$CB : BD = HK : KE$ So from {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$CD : BD = HF : FE$ Let it be contrived that $KH : HE = HF : FE$. Therefore from {{EuclidPropLink|book = V|prop = 19|title = Proportional Magnitudes have Proportional Remainders}}: :$KF : FH = KH : HE$ That is: :$KF : FH = BC : CD$ But $BC$ and $CD$ are commensurable in square only. Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$KF$ and $FH$ are commensurable in square only. We have that: :$KH : HE = KF : FH$ and: :$KH : HE = HF : FE$ Therefore from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$KF : HE = HF : FE$ So from {{EuclidDefLink|V|9|Duplicate Ratio}}: :$KF : FE = KF^2 : FH^2$ But $KF$ and $FH$ are commensurable in square. So $KF^2$ is commensurable with $FH^2$. Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$KF$ is commensurable in length with $FE$. So from {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$KF$ is commensurable in length with $KE$. But $KE$ is rational and commensurable in length with $BC$. Therefore from {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$KF$ is rational and commensurable in length with $BC$. We have that: :$BC : CD = KF : FH$ Thus from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$BC : KF = DC : FH$ But $BC$ is commensurable in length with $KF$. Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$FH$ is commensurable in length with $CD$. But $BC$ and $CD$ are rational straight lines which are commensurable in square only. Therefore $KF$ and $FH$ are rational straight lines which are commensurable in square only. Therefore, by definition, $KH$ is binomial. {{qed|lemma}} We have that: :$BC^2 = CD^2 + \lambda^2$ where either: :$\lambda$ is commensurable in length with $BC$ or: :$\lambda$ is incommensurable in length with $BC$. First suppose $\lambda$ is commensurable in length with $BC$. Then by {{EuclidPropLink|book = X|prop = 14|title = Commensurability of Squares on Proportional Straight Lines}}: : $KF^2 = FH^2 + \mu^2$ where $\mu$ is commensurable in length with $KF$. Let $BC$ 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: :$KF$ is commensurable in length with $\alpha$. Let $CD$ be commensurable in length with $\alpha$. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$FH$ is commensurable in length with $\alpha$. Let neither $BC$ nor $CD$ be commensurable in length with $\alpha$. Then neither $KF$ nor $FH$ is commensurable in length with $\alpha$. Next suppose $\lambda$ is incommensurable in length with $BC$. Then by {{EuclidPropLink|book = X|prop = 14|title = Commensurability of Squares on Proportional Straight Lines}}: :$KF^2 = FH^2 + \mu^2$ where $\mu$ is incommensurable in length with $KF$. Let $BC$ 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: :$KF$ is commensurable in length with $\alpha$. Let $CD$ be commensurable in length with $\alpha$. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$FH$ is commensurable in length with $\alpha$. Let neither $BC$ nor $CD$ be commensurable in length with $\alpha$. Then neither $KF$ nor $FH$ is commensurable in length with $\alpha$. It follows that: :the terms of $KH$ are commensurable with the terms of $BD$ and in the same ratio and: :the order of $KH$ is the same as the order of $BD$. {{qed}} {{Euclid Note|113|X}}