text: # 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.
  
  $\blacksquare$
explanation: The provided text is a detailed mathematical proof from Euclid's work, specifically dealing with the properties of a fifth apotome. The text does not contain a question but rather presents a theorem and its proof. The theorem states that the square on the straight line which produces with a rational area a medial whole, when applied to a rational straight line, produces as breadth a fifth apotome. The proof involves several steps and geometric constructions to demonstrate this property.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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.
  
  $\blacksquare$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a detailed mathematical proof from Euclid's work, specifically dealing with the properties of a fifth apotome. The text does not contain a question but rather presents a theorem and its proof. The theorem states that the square on the straight line which produces with a rational area a medial whole, when applied to a rational straight line, produces as breadth a fifth apotome. The proof involves several steps and geometric constructions to demonstrate this property.
  
  Label: no-question