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

## Theorem

In the words of Euclid:

*If an area be contained by a rational straight line and the second binomial, the "side" of the area is the irrational straight line which is called a first bimedial.*

(*The Elements*: Book $\text{X}$: Proposition $55$)

## Proof

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 Proposition $42$ of Book $\text{X} $: 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 Book $\text{X (II)}$ Definition $2$: Second Binomial, the lesser term $ED$ is commensurable in length with $AB$.

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 $AE$.

By Proposition $17$ of Book $\text{X} $: 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 Proposition $14$ of Book $\text{II} $: 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.

- $SQ$ is a square.

We have that the rectangle contained by $AG$ and $GE$ is equal to $EF^2$.

Then by Proposition $17$ of Book $\text{VI} $: Rectangles Contained by Three Proportional Straight Lines:

- $AG : EF = EF : EG$

and so by Proposition $1$ of Book $\text{VI} $: 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 $NQ$.

- $MR$ 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 Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:

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

As:

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

from Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:

- $AE$ is commensurable in length with each of $AG$ and $GE$.

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

- $AG$ 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:

and:

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 Proposition $13$ of Book $\text{X} $: 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:

and:

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 Proposition $11$ of Book $\text{X} $: 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:

By definition, $MO$ is a first bimedial straight line.

$\blacksquare$

## Historical Note

This proof is Proposition $55$ of Book $\text{X}$ of Euclid's *The Elements*.

It is the converse of Proposition $61$: Square on First Bimedial Straight Line applied to Rational Straight Line.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions