:300px Let $AB$ be an apotome. Let $CD$ be commensurable in length with $AB$. It is to be demonstrated that: :$CD$ is an apotome and: :the order of $CD$ is the same as the order of $AB$. Let $BE$ be the annex of $CD$. This can be constructed uniquely by Construction of Apotome is Unique. By definition, $AE$ and $EB$ are rational straight lines which are commensurable in square only. From {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$BE : DF = AB : CD$ From {{EuclidPropLink|book = V|prop = 12|title = Sum of Components of Equal Ratios}}: :$AE : CF = AB : CD$ and so from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = BE : DF$ But $AB$ is commensurable in length with $CD$. Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AE$ is commensurable in length with $CF$ and: :$BE$ is commensurable in length with $DF$. We have that $AE$ and $EB$ are rational straight lines which are commensurable in square only. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: : $CF$ and $FD$ are rational straight lines which are commensurable in square only. Thus $CD$ is an apotome. {{qed|lemma}} It remains to be shown that the order of $CD$ is the same as the order of $AB$. We have that: :$AE : CF = BE : DF$ So from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ We have that: : $AE^2 = EB^2 + \lambda^2$ where either: :$\lambda$ is commensurable in length with $AE$ or: :$\lambda$ is incommensurable in length with $AE$. Suppose $\lambda$ is commensurable in length with $AE$. Then by {{EuclidPropLink|book = X|prop = 14|title = Commensurability of Squares on Proportional Straight Lines}}: :$CF^2 > FD^2 + \mu^2$ where $\mu$ is commensurable in length with $CF$. Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$CD$ is commensurable in length with $\alpha$. Similarly, let $BE$ be commensurable in length with $\alpha$. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$DF$ is commensurable in length with $\alpha$. Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$. Then by {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :neither $CF$ nor $FD$ will be commensurable in length with $\alpha$. Let $\lambda$ be incommensurable in length with $AE$. Then by {{EuclidPropLink|book = X|prop = 14|title = Commensurability of Squares on Proportional Straight Lines}}: :$CF^2 > FD^2 + \mu^2$ where $\mu$ is incommensurable in length with $CF$. Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$CD$ is commensurable in length with $\alpha$. Similarly, let $BE$ be commensurable in length with $\alpha$. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$DF$ is commensurable in length with $\alpha$. Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$. Then by {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :neither $CF$ nor $FD$ will be commensurable in length with $\alpha$. It follows that $CD$ is an apotome of the same as the order as $AB$. {{qed}} {{Euclid Note|103|X}}