# Side of Area Contained by Rational Straight Line and Fourth Apotome

## Theorem

In the words of Euclid:

If an area be contained by a rational straight line and a fourth apotome, the "side" of the area is minor.

## Proof Let the area $AB$ be contained by the rational straight line $AC$ and the fourth apotome $AD$.

It is to be proved that the "side" of $AB$ is minor.

Let $DG$ be the annex of the fourth apotome $AD$.

Then, by definition:

$AG$ and $GD$ are rational straight lines which are commensurable in square only
$AG$ is commensurable with the rational straight line $AC$
the square on the whole $AG$ is greater than the square on the annex $GD$ by the square on a straight line which is incommensurable in length with $AG$.

Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $GD$ and deficient by a square figure.

that parallelogram divides $AG$ into incommensurable parts.

Let $DG$ be bisected at $E$.

Let the rectangle contained by $AF$ and $FG$ be applied to $AG$ which is equal to the square on $EG$ and deficient by a square figure.

Therefore $AF$ is commensurable with $FG$.

Through $E$, $F$ and $G$ let $EH$, $FI$ and $GK$ be drawn parallel to $AC$.

We have that $AG$ is rational and commensurable in length with $AC$.

the rectangle $AK$ is rational.

We have that $DG$ isincommensurable in length with $AC$, while both are rational.

Therefore from Proposition $21$ of Book $\text{X}$: Medial is Irrational:

the rectangles $DK$ is medial.

We have that $AF$ is incommensurable in length with $FG$.

Therefore from:

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:
$AI$ is incommensurable with $FK$.

Let the square $LM$ be constructed equal to $AI$.

Let the square $NO$ be subtracted from $LM$ having the common angle $\angle LPM$ equal to $FK$.

the squares $LM$ and $NO$ are about the same diameter.

Let $PR$ be the diameter of $LM$ and $NO$.

We have that the rectangle contained by $AF$ and $FG$ equals the square on $EG$.

$AF : EG = EG : FG$

But we also have:

$AF : EG = AI : EK$
$EG : FG = EK : KF$
$AI : EK = EK : FK$

Therefore $EK$ is a mean proportional between $AI$ and $FK$.

$MN$ is a mean proportional between $LM$ and $NO$.

We have that:

$AI$ equals the square $LM$

and:

$KF$ equals the square $NO$.

Therefore:

$MN = EK$

But:

$EK = DH$

and:

$MN = LO$

Therefore $DK$ equals the gnomon $UVW$ and $NO$.

But:

$AK$ equals the squares $LM$ and $NO$.

Therefore the remainder $AB$ equals $ST$.

But $ST$ is the square on $LN$.

Therefore the square on $LN$ equals $AB$.

Therefore $LN$ is the "side" of $AB$.

Now it is to be shown that $LN$ is the irrational straight line called minor.

We have that $AK$ is rational and equal to $LP^2 + PN^2$.

Therefore $LP^2 + PN^2$ is rational.

We have that the rectangle $DK$ is medial.

But $DK$ equals twice the rectangle contained by $LP$ and $PN$.

Therefore twice the rectangle contained by $LP$ and $PN$ is medial.

We have that $AI$ is incommensurable with $FK$.

Therefore $LP^2$ is incommensurable with $PN^2$.

Therefore $LP$ and $PN$ are straight lines which are incommensurable in square such that $LP^2 + PN^2$ is rational and such that $2 \cdot LP \cdot PN$ is medial.

Therefore by definition $LN$ is minor.

But $LN$ is the "side" of the area $AB$.

Hence the result.

$\blacksquare$