:450px Let $AB$ be the first apotome of a medial straight line. Let $BC$ be added to $AB$ such that: : $AC$ and $CB$ are medial straight lines : $AC$ and $CB$ are commensurable in square only : $AC \cdot CB$ is a medial rectangle. It is to be proved that no other medial straight line can be added to $AB$ which is commensurable in square only with the whole and which contains with the whole a medial rectangle. Suppose $BD$ can be added to $AB$ so as to fulfil the conditions stated. Then by definition of the first apotome of a medial straight line, $AD$ and $DB$ are such that: : $AD$ and $DB$ are medial straight lines : $AD$ and $DB$ are commensurable in square only : $AD \cdot DB$ is a medial rectangle. Let $EF$ be a rational straight line. Let $EG = AC^2 + CB^2$ be applied to $EF$, producing $EM$ as breadth. Let $HG = 2 \cdot AC \cdot CB$ be subtracted from $EG$ producing $HM$ as breadth. Therefore from {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :the remainder $EL$ equals $AB^2$. Thus $AB$ equals the "side" of $EL$. Let $EI = AD^2 + DB^2$ be applied to $EF$, producing $EN$ as breadth. But $EL = AB^2$. Therefore the remainder $HI$ equals $2 \cdot AD \cdot DB$. We have that $AC$ and $CB$ are medial straight lines. Therefore $AC^2$ and $CB^2$ are also medial. Also $AC^2 + CB^2 = EG$. So by: :{{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: :$EG$ is medial. But $EG$ is applied to the rational straight line $EF$, producing $EM$ as breadth. Therefore from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$EM$ is  rational and incommensurable in length with $EF$. We have that $AC \cdot CB$ is a medial rectangle. So from {{EuclidPropLink|book = X|prop = 23|qual=Porism|title = Straight Line Commensurable with Medial Straight Line is Medial}}: :$2 \cdot AC \cdot CB$ is a medial rectangle. But $2 \cdot AC \cdot CB = HG$. Therefore $HG$ is medial. Also, $HG$ has been applied to the rational straight line $EF$, producing $EM$ as breadth. Therefore by {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$HM$ is  rational and incommensurable in length with $EF$. Also, $AC$ and $CB$ are commensurable in square only. Therefore $AC$ and $CB$ are incommensurable in length. But: :$AC : CB = AC^2 : AC \cdot CB$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AC^2$ is incommensurable with $AC \cdot CB$. But $AC^2 + CB^2$ is commensurable with $AC^2$. Also from {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$2 \cdot AC \cdot CB$ is commensurable with $AC \cdot CB$. Therefore by {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$AC^2 + CB^2$ incommensurable with $AC \cdot CB$. We have that: :$EG = AC^2 + CB^2$ and: :$GH = 2 \cdot AC \cdot CB$ Therefore: :$EG$ is incommensurable with $HG$. But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$EG : HG = EM : HM$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$EM$ is incommensurable in length with $MH$. But both $EM$ and $MH$ are rational straight lines. Therefore $EM$ and $MH$ are rational straight lines which are commensurable in square only. Therefore $EH$ is an apotome, and $HM$ an annex to it. Similarly it can be shown that $HN$ is also an annex to it. Therefore we have different straight lines which are annexes to an apotome which are commensurable in square only to the whole. From {{EuclidPropLink|book = X|prop = 79|title = Construction of Apotome is Unique}}, this is impossible. The result follows. {{qed}} {{Euclid Note|81|X}}