# Square on Straight Line which produces Medial Whole with Rational Area applied to Rational Straight Line

## Theorem

In the words of Euclid:

The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.

## Proof

Let $CD$ be a rational straight line.

Let the rectangle $CE$ be applied to $CD$ equal to $AB^2$ producing $CF$ as breadth.

It is to be demonstrated that $CF$ is a fifth apotome.

Let $BG$ be the annex to $AB$.

Therefore, by definition, $AG$ and $GB$ are straight lines which are incommensurable in square which make $AG^2 + GB^2$ medial but $2 \cdot AG \cdot GB$ rational.

Let the rectangle $CH$ be applied to $CD$ equal to the square on $AG$, producing $CK$ as breadth.

Let the rectangle $KL$ be applied to $CD$ equal to the square on $BG$, producing $KM$ as breadth.

Then the whole $CL$ is equal to the squares on $AG$ and $GB$.

But $AG^2 + GB^2$ medial.

Therefore $CL$ is medial.

We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth.

$CM$ is rational and incommensurable in length with $CD$.

We have that:

$CL = AG^2 + GB^2$

and:

$AB^2 = CE$
$2 \cdot AG \cdot GB = FL$

Let $FM$ be bisected at the point $N$.

Let $NO$ be drawn through $N$ parallel to $CD$.

Therefore each of the rectangles $FO$ and $LN$ is equal to the rectangle contained by $AB$ and $GB$.

We have that $2 \cdot AG \cdot GB$ is rational.

Therefore $FL$ is rational.

Also $FL$ is applied to the rational straight line $FE$, producing $FM$ as breadth.

$FM$ is rational and commensurable in length with $CD$.

We have that $CL$ is medial and $FL$ is rational.

Therefore $CL$ is incommensurable with $FL$.

$CL : FL = CM : FM$
$CM$ is incommensurable in length with $FM$.

But both $CM$ and $FM$ are rational.

Therefore $CM$ and $FM$ are rational straight lines which are commensurable in square only.

Therefore, by definition, $CF$ is an apotome.

It remains to be shown that $CF$ is a fifth apotome.

It can be proved similarly that:

$CK \cdot KM = NM^2 = \dfrac {FM^2} 4$

We have that $AG^2$ is incommensurable with $GB^2$.

We also have:

$CH = AG^2$

and:

$KL = BG^2$

Therefore $CH$ is incommensurable with $KL$.

$CH : KL = CK : KM$
$CK$ is incommensurable with $KM$.

We have that:

$CM$ and $MF$ are unequal straight lines

and:

the rectangle $CK \cdot KM$ has been applied to $CM$ equal to $\dfrac {FM^2} 4$ and deficient by a square figure

while:

$CK$ is incommensurable with $KM$.
$CM^2$ is greater than $MF^2$ by the square on a straight line which is incommensurable in length with $CM$.

Also the annex $FM$ is commensurable in length with the rational straight line $CD$.

Therefore, by definition, $CF$ is a fifth apotome.

$\blacksquare$
The text is a detailed geometric proof from Euclid's work, discussing the properties of a specific type of straight line and its relationship to other geometric elements. It does not contain any explicit mathematical questions or answers in the form of word problems or calculations.
no_question
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Square on Straight Line which produces Medial Whole with Rational Area applied to Rational Straight Line

## Theorem

In the words of Euclid:

The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.

## Proof

Let $CD$ be a rational straight line.

Let the rectangle $CE$ be applied to $CD$ equal to $AB^2$ producing $CF$ as breadth.

It is to be demonstrated that $CF$ is a fifth apotome.

Let $BG$ be the annex to $AB$.

Therefore, by definition, $AG$ and $GB$ are straight lines which are incommensurable in square which make $AG^2 + GB^2$ medial but $2 \cdot AG \cdot GB$ rational.

Let the rectangle $CH$ be applied to $CD$ equal to the square on $AG$, producing $CK$ as breadth.

Let the rectangle $KL$ be applied to $CD$ equal to the square on $BG$, producing $KM$ as breadth.

Then the whole $CL$ is equal to the squares on $AG$ and $GB$.

But $AG^2 + GB^2$ medial.

Therefore $CL$ is medial.

We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth.

$CM$ is rational and incommensurable in length with $CD$.

We have that:

$CL = AG^2 + GB^2$

and:

$AB^2 = CE$
$2 \cdot AG \cdot GB = FL$

Let $FM$ be bisected at the point $N$.

Let $NO$ be drawn through $N$ parallel to $CD$.

Therefore each of the rectangles $FO$ and $LN$ is equal to the rectangle contained by $AB$ and $GB$.

We have that $2 \cdot AG \cdot GB$ is rational.

Therefore $FL$ is rational.

Also $FL$ is applied to the rational straight line $FE$, producing $FM$ as breadth.

$FM$ is rational and commensurable in length with $CD$.

We have that $CL$ is medial and $FL$ is rational.

Therefore $CL$ is incommensurable with $FL$.

$CL : FL = CM : FM$
$CM$ is incommensurable in length with $FM$.

But both $CM$ and $FM$ are rational.

Therefore $CM$ and $FM$ are rational straight lines which are commensurable in square only.

Therefore, by definition, $CF$ is an apotome.

It remains to be shown that $CF$ is a fifth apotome.

It can be proved similarly that:

$CK \cdot KM = NM^2 = \dfrac {FM^2} 4$

We have that $AG^2$ is incommensurable with $GB^2$.

We also have:

$CH = AG^2$

and:

$KL = BG^2$

Therefore $CH$ is incommensurable with $KL$.

$CH : KL = CK : KM$
$CK$ is incommensurable with $KM$.

We have that:

$CM$ and $MF$ are unequal straight lines

and:

the rectangle $CK \cdot KM$ has been applied to $CM$ equal to $\dfrac {FM^2} 4$ and deficient by a square figure

while:

$CK$ is incommensurable with $KM$.
$CM^2$ is greater than $MF^2$ by the square on a straight line which is incommensurable in length with $CM$.

Also the annex $FM$ is commensurable in length with the rational straight line $CD$.

Therefore, by definition, $CF$ is a fifth apotome.

$\blacksquare$
user
The text is a detailed geometric proof from Euclid's work, discussing the properties of a specific type of straight line and its relationship to other geometric elements. It does not contain any explicit mathematical questions or answers in the form of word problems or calculations.

TYPE: NO_QUESTION