hcju/proofwikiqa huggingface dataset:
:400px From {{EuclidPropLink|book = X|prop = 35|title = Construction of Components of Side of Sum of Medial Areas}}: :Let $AB$ and $BC$ be two straight lines incommensurable in square such that: :: $AB^2 + BC^2$ is medial :: $AB \cdot BC$ is medial :: $AB \cdot BC$ is incommensurable with the $AB^2 + BC^2$. Let $DE$ be a rational straight line. Using {{EuclidPropLink|book = I|prop = 45|title = Construction of Parallelogram in Given Angle equal to Given Polygon}}: :Let $DF$ be a rectangle set out on $DE$ equal to $AB^2 + BC^2$ :Let $GH$ be a rectangle set out on $DE$ equal to $2 \cdot AB \cdot BC$. From {{EuclidPropLink|book = II|prop = 4|title = Square of Sum}}: :The rectangle $DH$ equals $AC^2$. Since $AB^2 + BC^2$ is medial, $DF$ is medial. We have that $DF$ is applied to the rational straight line $DE$. Therefore from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$DG$ is rational and incommensurable in length with $DE$. For the same reason $GK$ is rational and incommensurable in length with $GF$, that is, $DE$. Since $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$: :$DF$ is incommensurable with $GH$. 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}}: :$DG$ is also incommensurable in length with $GK$. Also, $DG$ and $GK$ are rational. Therefore $DG$ and $GK$ are rational straight lines which are commensurable in square only. Therefore $DK$ is irrational. From {{EuclidPropLink|book = X|prop = 36|title = Binomial is Irrational}}: :$DK$ is binomial. But $DE$ is rational. Therefore from {{EuclidDefLink|X|4|Rational Area}}: :$DH$ is irrational. Also, the side of the square equal to $DH$ is irrational. But $AC$ is the side of the square equal to $DH$. Therefore $AC$ is irrational.   Such a straight line is called the '''side of the sum of two medial areas'''. {{qed}} {{Euclid Note|41|X}}
Let $AB$ be a binomial straight line. Let $AB$ be divided into its terms at $C$. That is, let the terms of $AB$ be $AC$ and $CB$. By definition of binomial, $AC$ and $CB$ are rational straight lines commensurable in square only. Suppose there exists a point $D$ on $AB$ different from $C$, such that $AD$ and $DB$ are also rational straight lines which are commensurable in square only. Let $AC \ne DB$, otherwise $AD = BC$ and $AB$ will be divided in the same way by $D$ as it is $C$. Similarly, the points $C$ and $D$ are not equidistant from the point of bisection of $AB$. 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: :$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$ and so: :$\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right) = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ As: : $AC$ and $CB$ are rational straight lines and : $AD$ and $DB$ are rational straight lines it follows by definition that $AC^2$, $CB^2$, $AD^2$ and $DB^2$ are all rational areas. Thus: :$\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right)$ is a rational area. Thus $2 AC \cdot CB - 2 AD \cdot DB$ is also a rational area. But by definition, $2 AC \cdot CB$ and $2 AD \cdot DB$ are both medial. But from {{EuclidPropLink|book = X|prop = 26|title = Medial Area not greater than Medial Area by Rational Area}} this cannot be the case. From this contradiction it follows that $D$ cannot divide $AB$ into rational straight lines which are commensurable in square only. {{qed}} {{Euclid Note|42|X}}
Let $AB$ be a first 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 rational rectangle. Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$. From {{EuclidPropLink|book = II|prop = 4|title = Square of Sum}}: :$AB^2 = \paren {AC + CB}^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$ and: :$AB^2 = \paren {AD + DB}^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$ and so: :$\paren {AC^2 + CB^2} - \paren {AD^2 + DB^2} = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ By definition, $AC \cdot CB$ and $AD \cdot DB$ are both rational areas. So by {{EuclidPropLink|book = II|prop = 4|title = Product of Rationally Expressible Numbers is Rational}}: :$2 AC \cdot CB$ and $2 AD \cdot DB$ are both rational areas. But $AB$ Thus $2 AC \cdot CB - 2 AD \cdot DB$ is also a rational area. As: :$AC$ and $CB$ are medial straight lines and :$AD$ and $DB$ are medial straight lines it follows by definition that $AC^2$, $CB^2$, $AD^2$ and $DB^2$ are all medial areas. But from {{EuclidPropLink|book = X|prop = 26|title = Medial Area not greater than Medial Area by Rational Area}} this cannot be the case. From this contradiction it follows that $D$ cannot divide $AB$ into medial straight lines commensurable in square only and containing a rational rectangle. {{qed}} {{Euclid Note|43|X}}
Let $AB$ be a major straight line. Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that: :$AC$ and $CB$ are incommensurable in square :$AC^2 + CB^2$ is rational :$AC$ and $CB$ contain a medial rectangle. Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$. 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: :$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$ and so: :$\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right) = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ But $AB^2 + CB^2$ and $AD^2 + DB^2$ are rational. So $\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right)$ is rational. Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational. But $2 \cdot AD \cdot DB$ and $2 \cdot AC \cdot CB$ are medial. From {{EuclidPropLink|book = X|prop = 26|title = Medial Area not greater than Medial Area by Rational Area}} this cannot be the case. So there can be no such $D$. {{qed}} {{Euclid Note|45|X}}
Let $AB$ be a side of a rational plus a medial area. Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that: :$AC$ and $CB$ are incommensurable in square :$AC^2 + CB^2$ is medial :$AC$ and $CB$ contain a rational rectangle. From {{EuclidPropLink|book = X|prop = 19|title = Product of Rationally Expressible Numbers is Rational}}: :$2 \cdot AC \cdot CB$ is a rational rectangle. Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$. From {{EuclidPropLink|book = II|prop = 4|title = Square of Sum}}: :$AB^2 = \paren {AC + CB}^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$ and: :$AB^2 = \paren {AD + DB}^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$ and so: :$\paren {AC^2 + CB^2} - \paren {AD^2 + DB^2} = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ Since $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational: :$\paren {AC^2 + CB^2} - \paren {AD^2 + DB^2}$ is rational. But $\paren {AC^2 + CB^2}$ and $\paren {AD^2 + DB^2}$ are medial. From {{EuclidPropLink|book = X|prop = 26|title = Medial Area not greater than Medial Area by Rational Area}} this cannot be the case. So there can be no such $D$. {{qed}} {{Euclid Note|46|X}}
:400px Let $AB$ be a side of a sum of two medial areas. Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that: :$AC$ and $CB$ are incommensurable in square :$AC^2 + CB^2$ is medial :$AC$ and $CB$ contain a medial rectangle :$AC \cdot CB$ is incommensurable with $AC^2 + CB^2$. Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$. Let $AC > BD$. 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 $EG$ be a rectangle set out on $EF$ equal to $AC^2 + CB^2$ and :Let $HK$ be a rectangle set out on $EF$ equal to $2 \cdot AC \cdot CB$. Therefore $EK$ is a rectangle equal to $AB^2$. Using {{EuclidPropLink|book = I|prop = 45|title = Construction of Parallelogram in Given Angle equal to Given Polygon}}: :Let $EL$ be a rectangle set out on $EF$ equal to $AD^2 + DB^2$ The remainder $MK$ is a rectangle set out on $EF$ equal to $2 \cdot AD \cdot DB$. By hypothesis, $AC^2 + CB^2$ is medial. We have that $AC^2 + CB^2$ is applied to a rational straight line $EF$. So from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$HE$ is rational and incommensurable in length with $EF$. For the same reason: :$HN$ is rational and incommensurable in length with $EF$. Since $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$: :$EG$ is incommensurable with $GN$. 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 with $HN$. Also, $EH$ and $HN$ are rational. Therefore $EH$ and $HN$ are rational straight lines commensurable in square only. Therefore from {{EuclidPropLink|book = X|prop = 36|title = Binomial is Irrational}}: :$EN$ is a binomial straight line divided at $H$. Using the same argument we can deduce: :$EN$ is a binomial straight line divided at $M$. But this contradicts {{EuclidPropLink|book = X|prop = 42|title = Binomial Straight Line is Divisible into Terms Uniquely}}. So there can be no such $D$. {{qed}} {{Euclid Note|47|X}}
:300px Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line. Using {{EuclidPropLink|book = X|prop = 28|qual = Lemma 1|title = Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater}}, let: : $AB : BC = m^2 : n^2$ where $m$ and $n$ are numbers such that $m^2 - n^2$ is not square. Let $D$ be a rational straight line. Let $EF$ be constructed commensurable in length with $D$. Then $EF$ is also a rational straight line. Using {{EuclidPropLink|book = X|prop = 6|qual = Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let: :$BA : AC = EF^2 : FG^2$ where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line. But: :$AB : AC = m^2 : \left({m^2 - n^2}\right)$ where $\left({m^2 - n^2}\right)$ is not square. So from {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: : $EF^2 : FG^2 = m^2 : \left({m^2 - n^2}\right)$ $EF$ and $FG$ are commensurable in square. But from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: $EF$ and $FG$  are incommensurable in length. Therefore $EF$ and $FG$ are rational straight lines which are commensurable in square only. Therefore by definition $EG$ is a binomial. We have that: :$BA : AC = EF^2 : FG^2$ while: :$BA > AC$ Therefore: :$EF^2 > FG^2$ Let: :$FG^2 + H^2 = EF^2$ for some $H$. From {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$AB : BC = EF^2 : H$ But $AB : BC$ has the ratio that a square number has to a square number. Therefore by {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: : $EF$ is commensurable in length with $H$. Therefore $EF^2$ is greater than $FG^2$ by the square on a straight line commensurable with $EF$. We have that: :$EF$ and $FG$ are rational straight lines and: :$EF$ is commensurable in length with $D$. Therefore $EF$ is a first binomial straight line. {{qed}} {{Euclid Note|48|X}}
:300px Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line. Using {{EuclidPropLink|book = X|prop = 28|qual = Lemma 1|title = Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater}}, let: : $AB : BC = m^2 : n^2$ where $m$ and $n$ are numbers such that $m^2 - n^2$ is not square. Let $D$ be a rational straight line. Let $EF$ be constructed commensurable in length with $D$. Then $EF$ is also a rational straight line. Using {{EuclidPropLink|book = X|prop = 6|qual = Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let: :$CA : AB = EF^2 : FG^2$ where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line. From {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: : $EF^2 : FG^2 = m^2 : \left({m^2 - n^2}\right)$ $EF$ and $FG$ are commensurable in square. But from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: $EF$ and $FG$  are incommensurable in length. Therefore $EF$ and $FG$ are rational straight lines which are commensurable in square only. Therefore by definition $EG$ is a binomial. We have that: :$BA : AC = GF^2 : FE^2$ while: :$BA > AC$ Therefore: :$GF^2 > FE^2$ Let: :$FE^2 + H^2 = GF^2$ for some $H$. From {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$AB : BC = FG^2 : H$ But $AB : BC$ has the ratio that a square number has to a square number. Therefore by {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$FG$ is commensurable in length with $H$. Therefore $FG^2$ is greater than $FE^2$ by the square on a straight line commensurable with $EF$. We have that: :$EF$ and $FG$ are rational straight lines which are commensurable in square only and: :$EF$ is commensurable in length with $D$. Therefore $EG$ is a second binomial straight line. {{qed}} {{Euclid Note|49|X}}
:300px Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line. Let neither $AB : AC$ nor $AB : BC$ be the ratio which a square number has to a square number. Let $D$ be a rational straight line. Let $EF$ be constructed commensurable in length with $D$. Then $EF$ is also a rational straight line. Using {{EuclidPropLink|book = X|prop = 6|qual = Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let: :$AB : AC = EF^2 : FG^2$ where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line. From {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$EF$ and $FG$ are commensurable in square. Therefore $FG$ is also a rational straight line. But from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: $EF$ and $FG$  are incommensurable in length. Therefore $EF$ and $FG$ are rational straight lines which are commensurable in square only. Therefore by definition $EG$ is a binomial. Since: :$AB : AC = EF^2 : FG^2$ while: :$BA > AC$ Therefore: :$EF^2 > FG^2$ Let: :$FG^2 + H^2 = EF^2$ for some $H$. From {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$AB : BC = EF^2 : H^2$ But $AB : BC$ is not the ratio that a square number has to a square number. Therefore $EF^2 : H^2$ is not the ratio that a square number has to a square number. Therefore by {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$EF$ is incommensurable in length with $H$. Therefore $EF^2 > GF^2$ by the square on a straight line which is incommensurable in length with $EF$. We have that: :$EF$ and $FG$ are rational straight lines which are commensurable in square only and: :$EF$ is commensurable in length with $D$. Therefore $EG$ is a fourth binomial straight line. {{qed}} {{Euclid Note|51|X}}
:300px Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line. Let neither $AB : AC$ nor $AB : BC$ be the ratio which a square number has to a square number. Let $D$ be a rational straight line. Let $EF$ be constructed commensurable in length with $D$. Then $EF$ is also a rational straight line. Using {{EuclidPropLink|book = X|prop = 6|qual = Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let: :$CA : AB = EF^2 : FG^2$ where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line. From {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$EF$ and $FG$ are commensurable in square. But from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: $EF$ and $FG$  are incommensurable in length. Therefore $EF$ and $FG$ are rational straight lines which are commensurable in square only. Therefore by definition $EG$ is a binomial. Since: :$CA : AB = EF^2 : FG^2$ while: :$AB > CA$ Therefore: :$GF^2 > FE^2$ Let: :$EF^2 + H^2 = GF^2$ for some $H$. From {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$AB : BC = GF^2 : H^2$ But $AB : BC$ is not the ratio that a square number has to a square number. Therefore $FG^2 : H^2$ is not the ratio that a square number has to a square number. Therefore by {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$FG$ is incommensurable in length with $H$. Therefore $FG^2 > EF^2$ by the square on a straight line which is incommensurable in length with $FG$. We have that: :$EF$ and $FG$ are rational straight lines which are commensurable in square only and: :the lesser term $EF$ is commensurable in length with $D$. Therefore $EG$ is a fifth binomial straight line. {{qed}} {{Euclid Note|52|X}}
:300px Let $AB$ be a major straight line divided at $C$. Let $AC > CB$. Let $DE$ be a rational straight line. Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth. It is to be demonstrated that $DG$ is a fourth binomial straight line. From {{EuclidPropLink|book = II|prop = 4|title = Square of Sum}}: :$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$ Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$. Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$. Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$. Let $MG$ be bisected at $N$. Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing). Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$. We have that $AB$ is a major which has been divided at $C$. Therefore, by definition, $AC$ and $CB$ are straight lines which are incommensurable in square such that: :$AC^2 + CB^2$ is rational :$AC \cdot CB$ is a medial rectangle. Since $AC^2 + CB^2$ is rational, it follows that $DL$ is rational. Therefore from {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :$DM$ is rational and commensurable in length with $DE$. We have that: :$2 \cdot AC \cdot CB$, which equals $MF$, is medial. We also have that $MF$ is applied to the rational straight line $ML$. Therefore from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$MG$ is rational and incommensurable in length with $DE$. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$DM$ is incommensurable in length with $MG$. That is, $DM$ and $MG$ are rational straight lines which are incommensurable in length with each other. Therefore by definition $DG$ is binomial. It remains to be proved that $DG$ is a fourth binomial straight line. From {{EuclidPropLink|book = X|prop = 60|qual = Lemma|title = Square on Binomial Straight Line applied to Rational Straight Line}}: :$AC^2 + CB^2 > 2 \cdot AC \cdot CB$ Therefore $DL > MF$. From {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$DM > MG$ Also $DK \cdot KM = MN^2$. Because: :$AC^2$ is incommensurable with $CB^2$ it follows that: :$DH$ is incommensurable with $KL$. So from: :{{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: :$DK$ is incommensurable in length with $KM$. So from {{EuclidPropLink|book = X|prop = 18|title = Condition for Incommensurability of Roots of Quadratic Equation}}: :$DM^2$ is greater than $MG^2$ by the square on a straight line incommensurable in length with $DM$. Also: :$DM$ and $MG$ are rational straight lines which are commensurable in square only with each other and: :$DM$ is commensurable in length with the rational straight line $DE$. Therefore $DG$ is a fourth binomial straight line. {{qed}} {{Euclid Note|63|X|{{EuclidNoteConverse|prop = 57|title = Root of Area contained by Rational Straight Line and Fourth Binomial}}}}
:300px Let $AB$ be the side of a rational plus a medial area divided at $C$. Let $AC > CB$. Let $DE$ be a rational straight line. Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth. It is to be demonstrated that $DG$ is a fifth binomial straight line. From {{EuclidPropLink|book = II|prop = 4|title = Square of Sum}}: :$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$ Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$. Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$. Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$. Let $MG$ be bisected at $N$. Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing). Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$. We have that $AB$ is the side of a rational plus a medial area which has been divided at $C$. Therefore, by definition, $AC$ and $CB$ are straight lines which are incommensurable in square such that: :$AC^2 + CB^2$ is medial :$AC \cdot CB$ is a rational rectangle. Since $AC^2 + CB^2$ is medial, $DL$ is medial. Therefore from {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$DM$ is rational and incommensurable in length with $DE$. We have that $2 \cdot AC \cdot CB$, that is $MF$ is rational. Therefore from {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :$MG$ is rational and commensurable in length with $DE$. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: $DM$ is incommensurable in length with $MG$ That is, $DM$ and $MG$ are rational straight lines which are  commensurable in square only. Therefore by definition $DG$ is binomial. It remains to be proved that $DG$ is a fifth binomial straight line. From {{EuclidPropLink|book = X|prop = 60|qual = Lemma|title = Square on Binomial Straight Line applied to Rational Straight Line}}: :$AC^2 + CB^2 > 2 \cdot AC \cdot CB$ Therefore $DL > MF$. From {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$DM > MG$ Also $DK \cdot KM = MN^2$. Because: :$AC^2$ is incommensurable with $CB^2$ it follows that: :$DH$ is incommensurable with $KL$. So from: :{{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: :$DK$ is incommensurable in length with $KM$. So from {{EuclidPropLink|book = X|prop = 18|title = Condition for Incommensurability of Roots of Quadratic Equation}}: :$DM^2$ is greater than $MG^2$ by the square on a straight line incommensurable in length with $DM$. Also: :$DM$ and $MG$ are rational straight lines which are commensurable in square only with each other and: :$DM$ is commensurable in length with the rational straight line $DE$. Therefore $DG$ is a fifth binomial straight line. {{qed}} {{Euclid Note|64|X|{{EuclidNoteConverse|prop = 58|title = Root of Area contained by Rational Straight Line and Fifth Binomial}}}}