:400px Let the area $AB$ be contained by the rational straight line $AC$ and the first apotome $AD$. It is to be proved that the "side" of $AB$ is an apotome. Let $DG$ be the annex of the first apotome $AD$. Then, by definition: :$AG$ and $GD$ are rational straight lines which are commensurable in square only :the whole $AG$ is commensurable with the rational straight line $AC$ :the square on $AG$ is greater than the square on $GD$ by the square on a straight line which is commensurable in length with $AG$. Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $DG$ and deficient by a square figure. From {{EuclidPropLink|book = X|prop = 17|title = Condition for Commensurability of Roots of Quadratic Equation}}: :that parallelogram divides $AG$ into commensurable parts. Let $DG$ be bisected at $E$. Let the rectangle contained by $AF$ and $FG$ be applied to $AG$ which is equal to the square on $EG$ and deficient by a square figure. Therefore $AF$ is commensurable with $FG$. Through $E$, $F$ and $G$ let $EH$, $FI$ and $GK$ be drawn parallel to $AC$. We have that $AF$ is commensurable with $FG$. Therefore from {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$AG$ is commensurable with each of $AF$ and $FG$. But $AG$ is commensurable with $AC$. Therefore from {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :each of the straight lines $AF$ and $FG$ is commensurable in length with $AC$. We have that $AC$ is rational. Therefore each of the straight lines $AF$ and $FG$ is rational. Therefore from {{EuclidPropLink|book = X|prop = 19|title = Product of Rationally Expressible Numbers is Rational}}: :each of the rectangles $AI$ and $FK$ is rational. We have that $DE$ is commensurable in length with $EG$. Therefore from {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$DG$ is also commensurable in length with each of the straight lines $DE$ and $EG$. But $DG$ is rational and incommensurable in length with $AC$. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}:  :each of the straight lines $DE$ and $ED$ is rational and incommensurable in length with $AC$. Therefore from {{EuclidPropLink|book = X|prop = 21|title = Medial is Irrational}}:  :each of the rectangles $DH$ and $EK$ is medial. Let the square $LM$ be constructed equal to $AI$. Let the square $NO$ be subtracted from $LM$ having the common angle $\angle LPM$ equal to $FK$. Therefore from {{EuclidPropLink|book = VI|prop = 26|title = Parallelogram Similar and in Same Angle has Same Diameter}}: :the squares $LM$ and $NO$ are about the same diameter. Let $PR$ be the diameter of $LM$ and $NO$. We have that the rectangle contained by $AF$ and $FG$ equals the square on $EG$. Therefore from {{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}}: :$AF : EG = EG : FG$ But we also have: :$AF : EG = AI : EK$ And from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$EG : FG = EK : KF$ Therefore from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AI : EK = EK : FK$ Therefore $EK$ is a mean proportional between $AI$ and $FK$. But from {{EuclidPropLink|book = X|prop = 54|qual=Lemma|title = Root of Area contained by Rational Straight Line and First Binomial}}: :$MN$ is a mean proportional between $LM$ and $NO$. We have that: :$AI$ equals the square $LM$ and: :$KF$ equals the square $NO$. Therefore: :$MN = EK$ But: :$EK = DH$ and: :$MN = LO$ Therefore $DK$ equals the gnomon $UVW$ and $NO$. But: :$AK$ equals the squares $LM$ and $NO$. Therefore the remainder $AB$ equals $ST$. But $ST$ is the square on $LN$. Therefore the square on $LN$ equals $AB$. Therefore $LN$ is the "side" of $AB$. Now it is to be shown that $LN$ is an apotome. We have that each of the rectangles $AI$ and $FK$ is rational. Also: :$AI = LM$ and: :$FK = NO$ Therefore each of the squares $LM$ and $NO$, that is the squares on $LP$ and $PN$, is rational. Therefore $LP$ and $PN$ are both rational. We have that $DH$ is medial and equals $LO$. Therefore $LO$ is medial. But $NO$ is rational. Therefore $LO$ is incommensurable with $NO$. But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$LO : NO = LP : PN$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$LP$ is incommensurable in length with $PN$. We have that both $LP$ and $PN$ are rational. Thus $LP$ and $PN$ are rational straight lines which are commensurable in square only. Therefore by definition $LN$ is an apotome. But $LN$ is the "side" of the area $AB$. Hence the result. {{qed}} {{Euclid Note|91|X|{{EuclidNoteConverse|prop = 97|title = Square on Apotome applied to Rational Straight Line}}}}