:400px Let the area $AB$ be contained by the rational straight line $AC$ and the fourth apotome $AD$. It is to be proved that the "side" of $AB$ is minor. Let $DG$ be the annex of the fourth apotome $AD$. Then, by definition: :$AG$ and $GD$ are rational straight lines which are commensurable in square only :$AG$ is commensurable with the rational straight line $AC$ :the square on the whole $AG$ is greater than the square on the annex $GD$ by the square on a straight line which is incommensurable in length with $AG$. Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $GD$ and deficient by a square figure. From {{EuclidPropLink|book = X|prop = 18|title = Condition for Incommensurability of Roots of Quadratic Equation}}: :that parallelogram divides $AG$ into incommensurable 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 $AG$ is rational and commensurable in length with $AC$. Therefore from {{EuclidPropLink|book = X|prop = 19|title = Product of Rationally Expressible Numbers is Rational}}:  :the rectangle $AK$ is rational. We have that $DG$ isincommensurable in length with $AC$, while both are  rational. Therefore from {{EuclidPropLink|book = X|prop = 21|title = Medial is Irrational}}:  :the rectangles $DK$ is medial. We have that $AF$ is incommensurable in length with $FG$. Therefore 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}}: :$AI$ is incommensurable with $FK$. 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 the irrational straight line called minor. We have that $AK$ is rational and equal to $LP^2 + PN^2$. Therefore $LP^2 + PN^2$ is rational. We have that the rectangle $DK$ is medial. But $DK$ equals twice the rectangle contained by $LP$ and $PN$. Therefore twice the rectangle contained by $LP$ and $PN$ is medial. We have that $AI$ is incommensurable with $FK$. Therefore $LP^2$ is incommensurable with $PN^2$. Therefore $LP$ and $PN$ are straight lines which are incommensurable in square such that $LP^2 + PN^2$ is rational and such that $2 \cdot LP \cdot PN$ is medial. Therefore by definition $LN$ is minor. But $LN$ is the "side" of the area $AB$. Hence the result. {{qed}} {{Euclid Note|94|X|{{EuclidNoteConverse|prop = 100|title = Square on Minor Straight Line applied to Rational Straight Line}}}}