hcju/proofwikiqa huggingface dataset:
:300px Let $AB$ be bimedial. Let $CD$ be commensurable in length with $AB$. It is to be shown that $CD$ is bimedial, and that the order of $CD$ is the same as the order of $AB$. Let $AB$ be divided into its medials by $E$. Let $AE$ be the greater medial. By definition, $AE$ and $EB$ are medial straight lines which are commensurable in square only. Using {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$AB : CD = AE : CF$ Therefore by {{EuclidPropLink|book = V|prop = 19|title = Proportional Magnitudes have Proportional Remainders}}: :$EB : FD = AB : CD$ 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: :$EB$ is commensurable in length with $FD$. But by hypothesis $AE$ and $EB$ are medial. Therefore by {{EuclidPropLink|book = X|prop = 23|title = Straight Line Commensurable with Medial Straight Line is Medial}}: : $CF$ and $FD$ are medial. From {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = EB : FD$ Therefore by {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ But by hypothesis $AE$ and $EB$ are commensurable in square only. Therefore by {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$CF$ and $FD$ are commensurable in square only. But $CF$ and $FD$ are medial. Therefore, by definition, $CD$ is bimedial. It remains to be demonstrated that $CD$ is of the same order as $AB$. We have that: :$AE : EB = CF : FD$ Therefore: :$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$ Therefore by {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE^2 : CF^2 = AE \cdot EB : CF \cdot FD$ But: :$AE^2$ is commensurable with $CF^2$. Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$. Suppose $AB$ is a first bimedial. Then $AE \cdot EB$ is rational. It follows that $CF \cdot FD$ is rational. Thus by definition $CD$ is a first bimedial. Suppose otherwise that $AB$ is a second bimedial. Then $AE \cdot EB$ is medial. It follows that $CF \cdot FD$ is medial. Thus by definition $CD$ is a second bimedial. Thus $CD$ is of the same order as $AB$. {{qed}} {{Euclid Note|67|X}}
:300px Let $AB$ be a major. Let $CD$ be commensurable in length with $AB$. It is to be shown that $CD$ is also a major. Let $AB$ be divided into its terms by $E$. Let $AE$ be the greater term. By definition, $AE$ and $EB$ are straight lines such that: : $AE$ and $EB$ are incommensurable in square : $AE^2 + EB^2$ is rational : $AE \cdot EB$ is medial. Using {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$AB : CD = AE : CF$ Therefore by {{EuclidPropLink|book = V|prop = 19|title = Proportional Magnitudes have Proportional Remainders}}: :$EB : FD = AB : CD$ Therefore by {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = EB : FD$ 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: :$EB$ is commensurable in length with $FD$. We have that: :$AE : CF = EB : FD$ Therefore by {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ and by by {{EuclidPropLink|book = V|prop = 18|title = Magnitudes Proportional Separated are Proportional Compounded}}: :$AB : BE = CD : DF$ Therefore by {{EuclidPropLink|book = VI|prop = 20|title = Similar Polygons are composed of Similar Triangles}}: :$AB^2 : BE^2 = CD^2 : DF^2$ Using a similar line of reasoning to the above: :$AB^2 : AE^2 = CD^2 : CF^2$ and: :$AB^2 : AE^2 + EB^2 = CD^2 : CF^2 + FD^2$ Therefore by {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AB^2 : CD^2 = AE^2 + EB^2 : CF^2 + FD^2$ But $AB^2$ is commensurable with $CD^2$. Therefore $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$. We have that $AE^2 + EB^2$ is rational. Therefore $CF^2 + FD^2$ is rational. Similarly, $2 \cdot AE \cdot EB$ is commensurable with $2 \cdot CF \cdot FD$. By {{EuclidPropLink|book = X|prop = 23|qual=Porism|title = Straight Line Commensurable with Medial Straight Line is Medial}}: :$2 \cdot AE \cdot EB$ is medial. Therefore $2 \cdot CF \cdot FD$ is medial. Thus it has been demonstrated that $CF$ and $FD$ are straight lines such that: : $CF$ and $FD$ are incommensurable in square : $CF^2 + FD2$ is rational : $CF \cdot FD$ is medial. Thus by definition $CD$ is a major. {{qed}} {{Euclid Note|68|X}}
:300px Let $AB$ be the side of a rational plus a medial area. Let $CD$ be commensurable in length with $AB$. It is to be shown that $CD$ is also the side of a rational plus a medial area. Let $AB$ be divided into its terms by $E$. Let $AE > EB$. By definition, $AE$ and $EB$ are straight lines such that: : $AE$ and $EB$ are incommensurable in square : $AE^2 + EB^2$ is medial : $AE \cdot EB$ is rational. Using {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$AB : CD = AE : CF$ Therefore by {{EuclidPropLink|book = V|prop = 19|title = Proportional Magnitudes have Proportional Remainders}}: :$EB : FD = AB : CD$ Therefore by {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = EB : FD$ 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: :$EB$ is commensurable in length with $FD$. We have that: :$AE : CF = EB : FD$ Therefore by {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ and by by {{EuclidPropLink|book = V|prop = 18|title = Magnitudes Proportional Separated are Proportional Compounded}}: :$AB : BE = CD : DF$ Using a similar line of reasoning to {{EuclidPropLink|book = X|prop = 68|title = Straight Line Commensurable with Major Straight Line is Major}}: : $CF$ and $FD$ are incommensurable in square : $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$ : $AE \cdot EB$ is commensurable with $CF \cdot FD$. {{explain|The above needs to be worked through in detail.}} We have that $AE^2 + EB^2$ is medial. Therefore $CF^2 + FD^2$ is medial. Similarly we have that $AE \cdot EB$ is rational. Therefore $CF \cdot FD$ is rational. Hence: : $CF$ and $FD$ are incommensurable in square : $CF^2 + FD^2$ is medial : $CF \cdot FD$ is rational. Thus by definition $CD$ is the side of a rational plus a medial area. {{qed}} {{Euclid Note|69|X}}
:300px Let $AB$ be the side of the sum of two medial areas. Let $CD$ be commensurable in length with $AB$. It is to be shown that $CD$ is also the side of the sum of two medial areas. Let $AB$ be divided into its terms by $E$. Let $AE > EB$. By definition, $AE$ and $EB$ are straight lines such that: : $AE$ and $EB$ are incommensurable in square : $AE^2 + EB^2$ is medial : $AE \cdot EB$ is medial. Using {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$AB : CD = AE : CF$ Therefore by {{EuclidPropLink|book = V|prop = 19|title = Proportional Magnitudes have Proportional Remainders}}: :$EB : FD = AB : CD$ Therefore by {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = EB : FD$ 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: :$EB$ is commensurable in length with $FD$. We have that: :$AE : CF = EB : FD$ Therefore by {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ and by by {{EuclidPropLink|book = V|prop = 18|title = Magnitudes Proportional Separated are Proportional Compounded}}: :$AB : BE = CD : DF$ Using a similar line of reasoning to {{EuclidPropLink|book = X|prop = 68|title = Straight Line Commensurable with Major Straight Line is Major}}: : $CF$ and $FD$ are incommensurable in square : $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$ : $AE \cdot EB$ is commensurable with $CF \cdot FD$. {{explain|The above needs to be worked through in detail.}} We have that $AE^2 + EB^2$ is medial. Therefore $CF^2 + FD^2$ is medial. Similarly we have that $AE \cdot EB$ is medial. Therefore $CF \cdot FD$ is medial. Hence: : $CF$ and $FD$ are incommensurable in square : $CF^2 + FD^2$ is medial : $CF \cdot FD$ is medial. Thus by definition $CD$ is the side of the sum of two medial areas. {{qed}} {{Euclid Note|70|X}}
:450px Let $AB$ and $CD$ be medial areas. It is to be demonstrated that the "side" of their combined area $AD$ is either: :$(1): \quad$ a second bimedial or: :$(2): \quad$ a side of the sum of two medial areas. WLOG let $AB > CD$. Let a rational straight line $EF$ be set out. Let the rectangle $EG = AB$ be applied to $EF$, producing $EH$ as breadth. Let the rectangle $HI = DC$ be applied to $EF$, producing $HK$ as breadth. We have that $AB$ and $CD$ are medial areas and equals $EG$ and $HI$ respectively. Therefore $EG$ and $HI$ are medial areas. By {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$EH$ and $HK$ are rational straight lines which are both incommensurable in length with $EF$. We have that: :$AB$ is incommensurable with $CD$ and: :$AB = EG, CD = HI$ Therefore: :$EG$ is incommensurable with $HI$. But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$EG : HI = EH : HK$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$EH$ is incommensurable in Length with $HK$. Therefore $EH$ and $HK$ are rational straight lines which are commensurable in square only. Therefore, by definition, $EK$ is a binomial straight line which is divided at $H$. We have that: :$AB > CD$ and: :$AB = EG, CD = HI$ Therefore: :$EH > HK$ Thus $EH^2$ is greater than $HK^2$ by either: :the square on a straight line which is commensurable in length with $EH$ or: :the square on a straight line which is incommensurable in length with $EH$. Let $EH^2 = HK^2 + \lambda^2$. First, let $\lambda$ be commensurable in length with $EH$. Neither of $EH$ or $HK$ is commensurable in length with the rational straight line $EF$. Therefore $EK$ is a third binomial. But $EF$ is rational. From {{EuclidPropLink|book = X|prop = 56|title = Root of Area contained by Rational Straight Line and Third Binomial}}: :the "side" of the square equal to the rectangle contained by a rational straight line and a first binomial is second bimedial. Therefore the "side" of $EI$ is second bimedial. That is, the "side" of $AD$ is second bimedial. {{qed|lemma}} Next, let $\lambda$ be incommensurable in length with $EH$. Neither of $EH$ or $HK$ is commensurable in length with the rational straight line $EF$. Therefore $EK$ is a sixth binomial. But $EF$ is rational. From {{EuclidPropLink|book = X|prop = 59|title = Root of Area contained by Rational Straight Line and Sixth Binomial}}: :the "side" of the square equal to the rectangle contained by a rational straight line and a first binomial is the side of the sum of two medial areas. Therefore the "side" of $EI$ is the side of the sum of two medial areas. That is, the "side" of $AD$ is the side of the sum of two medial areas. {{qed}} {{Euclid Note|72|X}}
:300px Let $AB$ be a rational straight line. Let a rational straight line $BC$ which is commensurable in square only with $AB$ be cut off from $AB$. We have that $AB$ is incommensurable in length with $BC$. We also have: :$AB : BC = AB^2 : AB \cdot AC$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AB^2$ is incommensurable with $AB \cdot AC$. But from {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$AB^2 + BC^2$ is commensurable with $AB^2$. and from {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$2 \cdot AB \cdot AC$ is commensurable with $AB \cdot AC$. From {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :$AB^2 + BC^2 = 2 \cdot AB \cdot AC + CA^2$ and so from: :{{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}} and: :{{EuclidPropLink|book = X|prop = 16|title = Incommensurability of Sum of Incommensurable Magnitudes}} it follows that: :$AB^2 + BC^2$ is incommensurable with $AC^2$. But $AB$ and $BC$ are rational. Therefore $AC$ is irrational. Such a straight line is known as an apotome. {{qed}} {{Euclid Note|73|X}}
:300px Let $AB$ be a medial straight line. Let a medial straight line $BC$ such that: :$BC$ is commensurable in square only with $AB$ :the rectangle contained by $AB$ and $BC$ is rational be cut off from $AB$. We have that $AB$ and $BC$ are medial. So by definition $AB^2$ and $BC^2$ are both medial. But $2 \cdot AB \cdot BC$ is rational. Therefore: :$AB^2$ and $BC^2$ are incommensurable with $2 \cdot AB \cdot BC$. From {{EuclidPropLink|book = X|prop = 16|title = Incommensurability of Sum of Incommensurable Magnitudes}}: :if $AB$ is incommensurable with either $AC$ or $CB$, $AC$ and $CB$ are incommensurable with each other. Therefore by {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :$2 \cdot AB \cdot BC$ is incommensurable with $AC^2$. Therefore $AC^2$ is irrational. Therefore by definition $AC$ is irrational. Such a straight line is known as a first apotome of a medial. {{qed}} {{Euclid Note|74|X}}
:300px Let $AB$ be a straight line. Let a straight line $BC$ such that: :$BC$ is incommensurable in square with $AB$ :$AB^2 + BC^2$ is rational :the rectangle contained by $AB$ and $BC$ is medial be cut off from $AB$. We have that: :$AB^2 + BC^2$ is rational while: :$2 \cdot AB \cdot BC$ is medial. Therefore $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$. From: :{{EuclidPropLink|book = II|prop = 7|title = Square of Difference}} and: :{{EuclidPropLink|book = X|prop = 16|title = Incommensurability of Sum of Incommensurable Magnitudes}} it follows that: :$AB^2 + BC^2$ is incommensurable with $AC^2$. But $AB^2 + BC^2$ is rational. Therefore $AC^2$ is irrational. Therefore $AC$ is irrational. Such a straight line is known as a minor. {{qed}} {{Euclid Note|76|X}}
:300px Let $AB$ be a straight line. Let a straight line $BC$ such that: :$BC$ is incommensurable in square with $AB$ :$AB^2 + BC^2$ is medial :the rectangle contained by $AB$ and $BC$ is rational be cut off from $AB$. We have that: :$AB^2 + BC^2$ is medial while: :$2 \cdot AB \cdot BC$ is rational. Therefore $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$. From: :{{EuclidPropLink|book = II|prop = 7|title = Square of Difference}} and: :{{EuclidPropLink|book = X|prop = 16|title = Incommensurability of Sum of Incommensurable Magnitudes}} it follows that: :$2 \cdot AB \cdot BC$ is incommensurable with $AC^2$. But $2 \cdot AB \cdot BC$ is rational. Therefore $AC^2$ is irrational. Therefore $AC$ is irrational. Such a straight line is known as '''that which produces with a rational area a medial whole'''. {{qed}} {{Euclid Note|77|X}}
:400px Let $AB$ be a straight line. Let a straight line $BC$ such that: :$BC$ is incommensurable in square with $AB$ :$AB^2 + BC^2$ is medial :twice the rectangle contained by $AB$ and $BC$ is medial :$AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$ 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 $AB^2 + BC^2$ is medial and equal to $DE$. Therefore $DE$ 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 $2 \cdot AB \cdot BC$ is medial. But $2 \cdot AB \cdot BC = DH$. Therefore $DH$ is medial. We have that $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$. Therefore $DE$ is incommensurable with $DH$. But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$DE : DH = DG : DF$ Therefore by {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$DG$ 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. 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 '''that which produces with a medial area a medial whole'''. {{qed}} {{Euclid Note|78|X}}
:300px Let $AB$ be an apotome. Let $BC$ be added to $AB$ so that $AC$ and $CB$ are rational straight lines which are commensurable in square only. It is to be proved that no other rational straight line can be added to $AB$ which is commensurable in square only with the whole. Suppose $BD$ can be added to $AB$ so as to fulfil the conditions stated. Then $AD$ and $DB$ are rational straight lines which are commensurable in square only. From {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :$AD^2 + DB^2 - 2 \cdot AD \cdot DB = AC^2 + CB^2 - 2 \cdot AC \cdot CB = AB^2$ Therefore: :$AD^2 + DB^2 - AC^2 + CB^2 = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ But $AD^2 + DB^2$ and $AC^2 + CB^2$ are both rational. Therefore $AD^2 + DB^2 - AC^2 + CB^2$ is rational. Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational. But from {{EuclidPropLink|book = X|prop = 21|title = Medial is Irrational}}: :$2 \cdot AD \cdot DB$ and $2 \cdot AC \cdot CB$ are both medial. From {{EuclidPropLink|book = X|prop = 26|title = Medial Area not greater than Medial Area by Rational Area}}: :a medial area does not exceed a medial area by a rational area. Therefore no other rational straight line can be added to $AB$ which is commensurable in square only with the whole. Therefore only one rational straight line can be added to $AB$ which is commensurable in square only with the whole. {{qed}} {{Euclid Note|79|X}}
:300px 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 rational 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 rational 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 rational rectangle. From {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :$AD^2 + DB^2 - 2 \cdot AD \cdot DB = AC^2 + CB^2 - 2 \cdot AC \cdot CB = AB^2$ Therefore: :$AD^2 + DB^2 - AC^2 + CB^2 = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ But $AD \cdot DB$ and $AC \cdot CB$ are both rational. Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational. Therefore $AD^2 + DB^2 - AC^2 + CB^2$ is rational. But 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}}: both $AD^2 + DB^2$ and $AC^2 + CB^2$ is medial. By {{EuclidPropLink|book = X|prop = 26|title = Medial Area not greater than Medial Area by Rational Area}} this cannot happen. Therefore only one medial straight line can be added to $AB$ which is commensurable in square only with the whole and which contains with the whole a rational rectangle. {{qed}} {{Euclid Note|80|X}}
:300px Let $AB$ be a minor. Let $BC$ be added to $AB$ such that: : $AC$ and $CB$ are incommensurable in square : $AC^2 + CB^2$ is rational : $2 \cdot AC \cdot CB$ is a medial rectangle. It is to be proved that no other straight line can be added to $AB$ which fulfils these conditions. Suppose $BD$, different from $BC$, can be added to $AB$ such that: :$AD$ and $DB$ are incommensurable in square :$AD^2 + DB^2$ is rational :$2 \cdot AD \cdot DB$ is a medial rectangle. From {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :$AD^2 + DB^2 - 2 \cdot AD \cdot DB = AC^2 + CB^2 - 2 \cdot AC \cdot CB = AB^2$ Therefore: :$AD^2 + DB^2 - AC^2 + CB^2 = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ But :$AD^2 + DB^2$ and $AC^2 + CB^2$ are both rational. Therefore $AD^2 + DB^2 - AC^2 + CB^2$ is rational. Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational. But by definition of medial area, this is impossible. The result follows. {{qed}} {{Euclid Note|82|X}}
:300px Let $AB$ be a straight line which produces with a rational area a medial whole. Let $BC$ be added to $AB$ such that: : $AC$ and $CB$ are incommensurable in square : $AC^2 + CB^2$ is medial : $2 \cdot AC \cdot CB$ is a rational] rectangle. It is to be proved that no other straight line can be added to $AB$ which fulfils these conditions. Suppose $BD$, different from $BC$, can be added to $AB$ such that: :$AD$ and $DB$ are incommensurable in square :$AD^2 + DB^2$ is [[Definition:Medial Area|medial] :$2 \cdot AD \cdot DB$ is a rational] rectangle. From {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :$AD^2 + DB^2 - 2 \cdot AD \cdot DB = AC^2 + CB^2 - 2 \cdot AC \cdot CB = AB^2$ Therefore: :$AD^2 + DB^2 - AC^2 + CB^2 = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ But $2 \cdot AD \cdot DB$ and $2 \cdot AC \cdot CB$ are both rational. Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational. Therefore $AD^2 + DB^2 - AC^2 + CB^2$ is rational. But 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}}: both $AD^2 + DB^2$ and $AC^2 + CB^2$ is medial. By {{EuclidPropLink|book = X|prop = 26|title = Medial Area not greater than Medial Area by Rational Area}} this cannot happen. Hence the result. {{qed}} {{Euclid Note|83|X}}