:450px Let $AB$ be a rational area. Let $CD$ be a medial area. It is to be demonstrated that the "side" of their combined area $AD$ is either: :$(1): \quad$ a binomial :$(2): \quad$ a first bimedial :$(3): \quad$ a major or: :$(4): \quad$ a side of a rational plus a medial area. We have that $AB$ is either greater than or less than $CD$. Let $AB > CD$. Let a rational straight line $EF$ be set out. Let the rectangle $EG = AB$ be applied to $EF$, producing $EH$ as breadth. Let the rectangle $HI = DC$ be applied to $EF$, producing $HK$ as breadth. We have that $AB$ is rational and equals $EG$. Therefore $EG$ is also rational. By {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :$EH$ is a rational straight line which is commensurable in length with $EF$. We have that $CD$ is medial and equals $HI$. Therefore $HI$ is also medial. By {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$HK$ is a rational straight line which is incommensurable in length with $EF$. We have that: :$CD$ is medial and: $AB$ is rational. Therefore $AB$ is incommensurable with $CD$. Thus $EG$ is incommensurable with $HI$. From {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$EG : HI = EH : HK$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$EH$ is incommensurable in length with $HK$. Both $EH$ and $HK$ are rational straight lines. Therefore $EH$ and $HK$ are rational straight lines which are commensurable in square only. Therefore, by definition, $EK$ is a binomial straight line which is divided at $H$. We have that: : $AB > CD$ and: : $AB = EG, CD = HI$ Therefore: :$EH > HK$ Thus $EH^2$ is greater than $HK^2$ by either: :the square on a straight line which is commensurable in length with $EH$ or: :the square on a straight line which is incommensurable in length with $EH$. Let $EH^2 = HK^2 + \lambda^2$. First, let $\lambda$ be commensurable in length with $EH$. We have that the greater straight line $HE$ is commensurable in length with the rational straight line $EF$. Then by definition $EK$ is a first binomial. But $EF$ is rational. From {{EuclidPropLink|book = X|prop = 54|title = Root of Area contained by Rational Straight Line and First Binomial}}: :the "side" of the square equal to the rectangle contained by a rational straight line and a first binomial is binomial. Therefore the "side" of $EI$ is binomial. That is, the "side" of $AD$ is binomial. {{qed|lemma}} Next, let $\lambda$ be incommensurable in length with $EH$. As before, the greater straight line $HE$ is commensurable in length with the rational straight line $EF$. Then by definition $EK$ is a fourth binomial. But $EF$ is rational. From {{EuclidPropLink|book = X|prop = 57|title = Root of Area contained by Rational Straight Line and Fourth Binomial}}: :the "side" of the square equal to the rectangle contained by a rational straight line and a fourth binomial is major. Therefore the "side" of $EI$ is major. That is, the "side" of $AD$ is major. {{qed|lemma}} Let $AB < CD$. Then $EG < HI$ and so $EH < HK$. Thus $HK^2$ is greater than $EH^2$ by either: :the square on a straight line which is commensurable in length with $HK$ or: :the square on a straight line which is incommensurable in length with $HK$. Let $HK^2 = EK^2 + \lambda^2$. First, let $\lambda$ be commensurable in length with $HK$. We have that $EH$ is commensurable in length with $EF$. Therefore by definition $EK$ is a second binomial. But $EF$ is rational. From {{EuclidPropLink|book = X|prop = 55|title = Root of Area contained by Rational Straight Line and Second Binomial}}: :the "side" of the square equal to the rectangle contained by a rational straight line and a second binomial is first bimedial. Therefore the "side" of $EI$ is first bimedial. That is, the "side" of $AD$ is first bimedial. {{qed|lemma}} Next, let $\lambda$ be incommensurable in length with $HK$. We have that $EH$ is commensurable in length with $EF$. Therefore by definition $EK$ is a fifth binomial. But $EF$ is rational. From {{EuclidPropLink|book = X|prop = 58|title = Root of Area contained by Rational Straight Line and Fifth Binomial}}: :the "side" of the square equal to the rectangle contained by a rational straight line and a fifth binomial is first bimedial. Therefore the "side" of $EI$ is the side of a rational plus a medial area. That is, the "side" of $AD$ is the side of a rational plus a medial area. {{qed}} {{Euclid Note|71|X}}