:500px Let the rectangular area $ABCD$ be contained by the rational straight line $AB$ and the sixth binomial $AD$. Let $E$ divide $AD$ into its terms such that $AE > ED$. Let $E$ divide $AD$ into its terms such that $AE > ED$. From {{EuclidPropLink|book = X|prop = 42|title = Binomial Straight Line is Divisible into Terms Uniquely}} this is possible at only one place. 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 $AE$. 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 $AD$ is a third binomial straight line. Therefore $AE$ and $ED$ are rational straight lines which are commensurable in square only. 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 $MO$. We have that $EA$ is incommensurable in length with $AB$. Therefore $EA$ and $AB$ are rational straight lines which are commensurable in square only. Therefore, by definition, $AK = MN^2 + NO^2$ is medial. We have that: :$ED$ is incommensurable in length with $AB = EK$ So by {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$FE$ is incommensurable in length with $EK$. Therefore $FE$ and $EK$ are rational straight lines which are commensurable in square only. Therefore by definition: :$EL = MR = MN \cdot NO$ is a medial area. Since: :$AE$ is incommensurable in length with $EF$ by: : {{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}}: :$AK$ is incommensurable with $EL$. But: :$AK = MN^2 + NO^2$ and: :$EL = MN \cdot NO$ Therefore: :$MN^2 + NO^2$ is incommensurable with $MN \cdot NO$. So by definition, $MO$ is the side of the sum of two medial areas. {{qed}} {{Euclid Note|59|X|{{EuclidNoteConverse|prop=65|title=Square on Side of Sum of two Medial Area applied to Rational Straight Line}}}}