:400px Let $AB$ be a second apotome of a medial straight line. Let $CD$ be a rational straight line. Let the rectangle $CE$ be applied to $CD$ equal to $AB^2$ producing $CF$ as breadth. It is to be demonstrated that $CF$ is a third apotome. Let $BG$ be the annex to $AB$. Therefore, by definition, $AG$ and $GB$ are medial straight lines which are commensurable in square only which contain a medial rectangle. Let the rectangle $CH$ be applied to $CD$ equal to the square on $AG$, producing $CK$ as breadth. Let the rectangle $KL$ be applied to $CD$ equal to the square on $BG$, producing $KM$ as breadth. Then the whole $CL$ is equal to the squares on $AG$ and $GB$. 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: :$CL$ is medial. We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth. Therefore by {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$CM$ is rational and incommensurable in length with $CD$. We have that: :$CL = AG^2 + GB^2$ and: :$AB^2 = CE$ Therefore by {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :$2 \cdot AG \cdot GB = FL$ Let $FM$ be bisected at the point $N$. Let $NO$ be drawn through $N$ parallel to $CD$. Therefore each of the rectangles $FO$ and $LN$ is equal to the rectangle contained by $AB$ and $GB$. We have that the to the rectangle contained by $AB$ and $GB$ is medial. Therefore $FL$ is medial. Also $FL$ is applied to the rational straight line $FE$, producing $FM$ as breadth. Therefore by {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$FM$ is rational and incommensurable in length with $CD$. We have that $AG$ and $GB$ are commensurable in square only. Therefore $AG$ is incommensurable in length with $GB$. 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}}: we have that: :$AG^2$ is incommensurable with the rectangle contained by $AB$ and $GB$. But the squares on $AG$ and $GB$ are commensurable with the square on $AG$. Also $2 \cdot AG \cdot GB$ is commensurable with $AG \cdot GB$. So from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :the squares on $AG$ and $GB$ are incommensurable with $2 \cdot AG \cdot GB$. We have that: :$CL = AG^2 + GB^2$ and: :$FL = 2 \cdot AG \cdot GB$ Therefore $CL$ is incommensurable with $FL$. But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$CL : FL = CM : FM$ Therefore by {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$CM$ is incommensurable in length with $FM$. But both $CM$ and $FM$ are rational. Therefore $CM$ and $FM$ are rational straight lines which are commensurable in square only. Therefore, by definition, $CF$ is an apotome. It remains to be shown that $CF$ is a third apotome. We have that the squares on $AG$ and $GB$ are commensurable with the square on $AG$. Therefore $CH$ is commensurable 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}}: we have that: :$CK$ is commensurable with $KM$. We have that: :the rectangle contained by $AG$ and $GB$ is a mean proportional between the squares on $AG$ and $GB$. and: :$CH = AG^2$ and: :$KL = BG^2$ and: :$NL = AG \cdot GB$ Therefore $NL$ is a mean proportional between $CH$ and $KL$. Therefore: :$CH : NL = NL : KL$ But we also have: :$CH : NL = CK : NM$ and from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$NL : KL = NM : KM$ Therefore by {{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}}: :$CK \cdot KM = NM^2 = \dfrac {FM^2} 4$ We have that: :$CM$ and $MF$ are unequal straight lines and: :the rectangle $CK \cdot KM$ has been applied to $CM$ equal to $\dfrac {FM^2} 4$ and deficient by a square figure while: :$CK$ is commensurable with $KM$. Therefore from {{EuclidPropLink|book = X|prop = 17|title = Condition for Commensurability of Roots of Quadratic Equation}}: :$CM^2$ is greater than $MF^2$ by the square on a straight line which is commensurable in length with $CM$. Also neither $CM$ nor $FM$ is commensurable in length with the rational straight line $CD$. Therefore, by definition, $CF$ is a third apotome. {{qed}} {{Euclid Note|99|X|{{EuclidNoteConverse|prop=93|title=Side of Area Contained by Rational Straight Line and Third Apotome}}}}