:400px Let $AB$ be a medial straight line. Let a medial straight line $CB$ such that: :$CB$ is commensurable in square only with $AB$ :the rectangle contained by $AB$ and $BC$ is medial be cut off from $AB$. Let $DI$ be a rational straight line. Using {{EuclidPropLink|book = I|prop = 45|title = Construction of Parallelogram in Given Angle equal to Given Polygon}}: :Let $DE$ be a parallelogram set out on $DI$ equal to $AB^2 + BC^2$. Let its breadth be $DG$. Similarly: :Let $DH$ be a parallelogram set out on $DI$ equal to $2 \cdot AB \cdot BC$. From {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :$FE = AC^2$ We have that $AB^2$ and $BC^2$ are medial areas which are commensurable. Therefore from: :{{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: :$DE$ is medial. We have that $DE$ has been applied to the rational straight line $DI$ producing $DG$ as breadth. So from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$DG$ is rational and incommensurable in length with $DI$. We have that $AB \cdot BC$ is medial. So from{{EuclidPropLink|book = X|prop = 23|qual=Porism|title = Straight Line Commensurable with Medial Straight Line is Medial}}: :$2 \cdot AB \cdot BC$ is medial. But $2 \cdot AB \cdot BC = DH$. Therefore $DH$ is medial. We have that $DH$ has been applied to the rational straight line $DI$ producing $DF$ as breadth. So from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$DF$ is rational and incommensurable in length with $DI$. We have that $AB$ and $BC$ are commensurable in square only. Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AB$ is incommensurable in length with $BC$. But from {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$AB^2 + BC^2$ are commensurable with $AB^2$ and from {{EuclidPropLink|book = X|prop = 6|title = Commensurability of Sum of Commensurable Magnitudes}}: :$2 \cdot AB \cdot BC$ is commensurable with $AB \cdot BC$ therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$2 \cdot AB \cdot BC$ is incommensurable with $AB^2 + BC^2$. But: :$DE = AB^2 + BC^2$ and: :$DH = 2 \cdot AB \cdot BC$ and so $DE$ is incommensurable with $DH$. But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$DE : DH = GD : DF$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$GD$ is incommensurable with $DF$. But both $GD$ and $DF$ are rational. Therefore $GD$ and $DF$ are rational straight lines which are commensurable in square only. Therefore, by definition, $FG$ is an apotome. But $DI$ is rational. From {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :a rectangle contained by a rational and an irrational straight line is irrational. Hence its "side" is irrational. But $AC$ is the "side" of $FE$. Therefore $AC$ is irrational. Such a straight line is known as a second apotome of a medial. {{qed}} {{Euclid Note|75|X}}