:500px Let the rectangular area $ABCD$ be contained by the rational straight line $AB$ and the second binomial $AD$. Let $E$ divide $AD$ into its terms. From {{EuclidPropLink|book = X|prop = 42|title = Binomial Straight Line is Divisible into Terms Uniquely}} this is possible at only one place. Let $AE$ be its greater term. By definition of binomial, $AE$ and $ED$ are rational straight lines which are commensurable in square only. Thus $AE^2$ is greater than $ED^2$ by the square on a straight line which is commensurable with $AE$. By {{EuclidDefLink|X (II)|2|Second Binomial}}, the lesser term $ED$ is commensurable in length with $AB$. Let $ED$ be bisected at $F$. Using {{EuclidPropLink|book = VI|prop = 29|title = Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram}}: :Let a parallelogram be applied to $AE$ equal to $EF^2$ and deficient by a square. Let the rectangle contained by $AG$ and $GE$ equal to $EF^2$ be applied to $AE$. By {{EuclidPropLink|book = X|prop = 17|title = Condition for Commensurability of Roots of Quadratic Equation}}: :$AG$ is commensurable in length with $EG$. Let $GH$, $EK$ and $FL$ be drawn from $G$, $E$ and $F$ parallel to $AB$ and $CD$. Using {{EuclidPropLink|book = II|prop = 14|title = Construction of Square equal to Given Polygon}}: :let the square $SN$ be constructed equal to the parallelogram $AH$ and: :let the square $NQ$ be constructed equal to the parallelogram $GK$. Let $SN$ and $NQ$ be placed so that $MN$ is in a straight line with $NO$. Therefore $RN$ is also in a straight line with $NP$. Let the parallelogram $SQ$ be completed. From {{EuclidPropLink|book = X|prop = 54|qual=Lemma|title = Root of Area contained by Rational Straight Line and First Binomial}}: :$SQ$ is a square. We have that the rectangle contained by $AG$ and $GE$ is equal to $EF^2$. Then by {{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}}: :$AG : EF = EF : EG$ and so by {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$AH : EL = EL : KG$ Therefore $EL$ is a mean proportional between $AH$ and $GK$. But: :$AH = SN$ and: :$GK = NQ$ therefore $EL$ is a mean proportional between $SN$ and $NQ$. But from {{EuclidPropLink|book = X|prop = 54|qual=Lemma|title = Root of Area contained by Rational Straight Line and First Binomial}}: :$MR$ is a mean proportional between $SN$ and $NQ$. Therefore: :$EL = MR$ and so: :$EL = PO$ But: :$AH + GK = SN + NQ$ Therefore: :$AC = SQ$ That is: :$AC$ is an area contained by a rational straight line $AB$ and the first binomial $AD$, while the side of the area $SQ$ is $MO$. It remains to be demonstrated that $MO$ is first bimedial. We have that: :$AE$ is incommensurable in length with $ED$ and: :$ED$ is commensurable in length with $AB$ it follows from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$AE$ is incommensurable in length with $AB$. As: :$AG$ is commensurable in length with $EG$ from {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$AE$ is commensurable in length with each of $AG$ and $GE$. But $AE$ is incommensurable in length with $AB$. So from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$AG$ and $GE$ are incommensurable in length with $AB$. Therefore: :$BA$ and $AG$ are rational straight lines which are commensurable in square only and: :$BA$ and $GE$ are rational straight lines which are commensurable in square only. Therefore by definition rectangles $AH$ and $GK$ are both medial. Therefore squares $SN$ and $NQ$ are both medial. Therefore $MN$ and $NO$ are both medial straight lines. We have that $AG$ is commensurable in length with $GE$. 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: :$AH$ is commensurable with $GK$. That is: :$SN$ is commensurable with $NQ$. That is: :$MN^2$ is commensurable with $NO^2$. We have that: :$AE$ is incommensurable in length with $ED$ and: :$AE$ is commensurable in length with $AG$ and: :$ED$ is commensurable in length with $EF$. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$AG$ is incommensurable in length with $EF$ so that: :$AH$ is incommensurable in length with $EL$. That is: :$SN$ is incommensurable with $MR$. 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: :$PN$ is incommensurable in length with $NR$. That is: :$MN$ is incommensurable in length with $NO$. But $MN$ and $NO$ have been shown to be medial and commensurable in square. Therefore, by definition, $MN$ and $NO$ are medial and commensurable in square only. Finally it remains to be shown that $MN$ and $NO$ contain a rational rectangle. We have by hypothesis: :$DE$ is commensurable in length with each of $AB$ and $EF$. Therefore by {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$EF$ is commensurable in length with each of $EK$. Both $EF$ and $EK$ are rational straight lines. Therefore $EL = MR$ is rational. But $MR$ is the rectangle contained by $MN$ and $MO$. Therefore $MN$ and $NO$ are: : medial straight lines : commensurable in square only : containing a rational rectangle. By definition, $MO$ is a first bimedial straight line. {{qed}} {{Euclid Note|55|X|{{EuclidNoteConverse|prop=61|title=Square on First Bimedial Straight Line applied to Rational Straight Line}}}}