# Root of Area contained by Rational Straight Line and Fifth Binomial

## Theorem

In the words of Euclid:

If an area be contained by a rational straight line and the fifth binomial, the "side" of the area is the irrational straight line called the side of a rational plus a medial area.

## Proof

Let the rectangular area $ABCD$ be contained by the rational straight line $AB$ and the fifth binomial $AD$.

Let $E$ divide $AD$ into its terms such that $AE > ED$.

From Proposition $42$ of Book $\text{X}$: 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 $F$.

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 $AE$.

$AG$ is commensurable in length with $EG$.

Let $GH$, $EK$ and $FL$ be drawn from $G$, $E$ and $F$ parallel to $AB$ and $CD$.

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.

$SQ$ 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 $EF^2$.

$AG : EF = EF : EG$
$AH : EL = EL : KG$

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 $NQ$.

$MR$ 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 $MO$.
$AE$ is incommensurable in length with $AD$.

By:

Proposition $1$ of Book $\text{VI}$: Areas of Triangles and Parallelograms Proportional to Base

and:

Proposition $11$ of Book $\text{X}$: Commensurability of Elements of Proportional Magnitudes

it follows that:

$AH$ is commensurable with $HE$.

That is:

$MN^2$ is incommensurable with $NO^2$.

Therefore $MN$ and $NO$ are incommensurable in square.

We have that $AD$ is a fifth binomial with $ED$ the lesser segment.

Therefore by Book $\text{X (II)}$ Definition $5$: Fifth Binomial:

$ED$ is commensurable in length with $AB$.

But $AE$ is incommensurable in length with $ED$.

$AB$ is incommensurable in length with $AE$.

Therefore, by definition, $AK = MN^2 + NO^2$ is medial.

We have that:

$DE$ is commensurable in length with $AB = EK$

and:

$DE$ is commensurable in length with $EF$.
$EF$ is commensurable in length with $EK$.

We have that $EK$ is a rational straight line.

$EL = MR = MN \cdot NO$ is a rational area.

Therefore $MN$ and $NO$ are straight lines which are incommensurable in square such that:

$MN^2 + NO^2$ is medial
$MN \cdot NO$ is rational.

Therefore by definition, $MO$ is the side of a rational plus a medial area.

$\blacksquare$