Sum of Rational Area and Medial Area gives rise to four Irrational Straight Lines

== Theorem ==
== Proof ==
:
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 :
:$EH$ 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 :
:$HK$ 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 :
:$EG : HI = EH : HK$
Therefore from :
:$EH$ 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 :
: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.
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 :
: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.
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 :
: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.
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 :
: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.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory