Square on Second Apotome of Medial Straight Line applied to Rational Straight Line
Theorem
In the words of Euclid:
- The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome.
(The Elements: Book $\text{X}$: Proposition $99$)
Proof
Let $AB$ be a second apotome of a medial straight line.
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 third apotome.
Let $BG$ be the annex to $AB$.
Therefore, by definition, $AG$ and $GB$ are medial straight lines which are commensurable in square only which contain a medial rectangle.
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 $GB$.
From:
and:
it follows that:
- $CL$ is medial.
We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth.
Therefore by Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:
- $CM$ is rational and incommensurable in length with $CD$.
We have that:
- $CL = AG^2 + GB^2$
and:
- $AB^2 = CE$
Therefore by Proposition $7$ of Book $\text{II} $: Square of Difference:
- $2 \cdot AG \cdot GB = FL$
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 the to the rectangle contained by $AB$ and $GB$ is medial.
Therefore $FL$ is medial.
Also $FL$ is applied to the rational straight line $FE$, producing $FM$ as breadth.
Therefore by Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:
- $FM$ is rational and incommensurable in length with $CD$.
We have that $AG$ and $GB$ are commensurable in square only.
Therefore $AG$ is incommensurable in length with $GB$.
So from:
and:
we have that:
- $AG^2$ is incommensurable with the rectangle contained by $AB$ and $GB$.
But the squares on $AG$ and $GB$ are commensurable with the square on $AG$.
Also $2 \cdot AG \cdot GB$ is commensurable with $AG \cdot GB$.
- the squares on $AG$ and $GB$ are incommensurable with $2 \cdot AG \cdot GB$.
We have that:
- $CL = AG^2 + GB^2$
and:
- $FL = 2 \cdot AG \cdot GB$
Therefore $CL$ is incommensurable with $FL$.
But from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:
- $CL : FL = CM : FM$
Therefore by Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $CM$ 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 third apotome.
We have that the squares on $AG$ and $GB$ are commensurable with the square on $AG$.
Therefore $CH$ is commensurable with $KL$.
So from:
and:
we have that:
- $CK$ is commensurable with $KM$.
We have that:
- the rectangle contained by $AG$ and $GB$ is a mean proportional between the squares on $AG$ and $GB$.
and:
- $CH = AG^2$
and:
- $KL = BG^2$
and:
- $NL = AG \cdot GB$
Therefore $NL$ is a mean proportional between $CH$ and $KL$.
Therefore:
- $CH : NL = NL : KL$
But we also have:
- $CH : NL = CK : NM$
and from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:
- $NL : KL = NM : KM$
Therefore by Proposition $17$ of Book $\text{VI} $: Rectangles Contained by Three Proportional Straight Lines:
- $CK \cdot KM = NM^2 = \dfrac {FM^2} 4$
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 commensurable with $KM$.
Therefore from Proposition $17$ of Book $\text{X} $: Condition for Commensurability of Roots of Quadratic Equation:
- $CM^2$ is greater than $MF^2$ by the square on a straight line which is commensurable in length with $CM$.
Also neither $CM$ nor $FM$ is commensurable in length with the rational straight line $CD$.
Therefore, by definition, $CF$ is a third apotome.
$\blacksquare$
Historical Note
This proof is Proposition $99$ of Book $\text{X}$ of Euclid's The Elements.
It is the converse of Proposition $93$: Side of Area Contained by Rational Straight Line and Third Apotome.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions