Let $AB$ be a second bimedial straight line. Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that: :$AC$ and $CB$ are medial straight lines :$AC$ and $CB$ are commensurable in square only :$AC$ and $CB$ contain a medial rectangle. :400px Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$. WLOG let $AC > DB$. From {{EuclidPropLink|book = X|prop = 42|qual = Lemma|title = Binomial Straight Line is Divisible into Terms Uniquely}}: :$AD^2 + DB^2 < AC^2 + CB^2$ From {{EuclidPropLink|book = II|prop = 4|title = Square of Sum}}: :$AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$ and: :$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$ Let $EF$ be a rational straight line. Using {{EuclidPropLink|book = I|prop = 45|title = Construction of Parallelogram in Given Angle equal to Given Polygon}}: :Let $EK$ be a rectangle set out on $EF$ equal to the square on $AB$. Let $EG = EK - AC^2 - CB^2$. Then $HK = 2 \cdot AC \cdot CB$. Similarly, let $EL = AD^2 + DB^2$. Then $HL = 2 \cdot AD \cdot DB$. Since the squares on $AC$ and $CB$ are by definition medial, it follows that $EG$ is medial. But $EG$ is applied to the rational straight line $EF$. Therefore from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$EH$ is a rational straight line which is incommensurable in length with $EF$. Using the same reasoning: :$HN$ is a rational straight line which is incommensurable in length with $EF$. We have that $AC$ and $CB$ are medial straight lines which are commensurable in square only. Therefore $AC$ and $CB$ are incommensurable in length. But: :$AC : CB = AC^2 : AC \cdot CB$ So from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AC^2$ is incommensurable with $AC \cdot CB$. But we have that $AC$ and $CB$ are commensurable in square. So from {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$AC^2 + CB^2$ is commensurable in square with $AC^2$. From {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$2 \cdot AC \cdot CB$ is commensurable with $AC \cdot CB$. So from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$. We have: : $EG = AC^2 + CB^2$ and: : $HK = 2 \cdot AC \cdot CB$ Therefore $EG$ is incommensurable with $HK$. So by: : {{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: : $EH$ is incommensurable in length with $HN$. We also have that $EH$ and $HN$ are rational. Therefore $EH$ and $HN$ are rational straight lines which are commensurable in square only. But from {{EuclidPropLink|book = X|prop = 36|title = Binomial is Irrational}}: :Two rational straight lines commensurable in square only, when added together, form an irrational straight line which is a binomial. Therefore $EN$ is a binomial straight line which is divided at $H$. In the same way $EM$ and $MN$ can be shown to be rational straight lines commensurable in square only. Thus $EN$ is a binomial straight line which is divided into terms in two places: $H$ and $M$. We have that: :$AC^2 + CB^2 > AD^2 + DB^2$ Also: :$AD^2 + DB^2 > 2 \cdot AD \cdot DB$ Thus $EG = AC^2 + CB^2 > 2 \cdot AD \cdot DB = MK$ Thus $EH > MN$. But from {{EuclidPropLink|book = X|prop = 42|title = Binomial Straight Line is Divisible into Terms Uniquely}}: :$EN$ can be divided into its terms in only one way. This contradicts the conclusion that has been drawn: that $H$ and $M$ divide $EH$ into terms in two different ways. So the supposition that $AB$ can be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$ must be false. {{qed}} {{Euclid Note|44|X}}