Root of Area contained by Rational Straight Line and First Binomial/Lemma

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Lemma to Root of Area contained by Rational Straight Line and First Binomial

In the words of Euclid:

Let there be two squares $AB, BC$, and let them be placed so that $DB$ is in a straight line with $BE$; therefore $FB$ is also in a straight line with $BG$.
Let the parallelogram $AC$ be completed; I say that $AC$ is a square, that $DG$ is a mean proportional between $AB, BC$, and further, that $DC$ is a mean proportional between $AC, CB$.

(The Elements: Book $\text{X}$: Proposition $54$ : Lemma)


Proof

Euclid-X-54-Lemma.png

We have that $DB = BF$ and $BE = BG$.

Thus $DE = FG$.

But from Proposition $34$ of Book $\text{I} $: Opposite Sides and Angles of Parallelogram are Equal:

$DE = AH = KC$

and:

$FG = AK = HC$.

Therefore:

$AH = AK = HC = KC$

Therefore the parallelogram $AKCH$ is equilateral.

The parallelogram $AKCH$ is also rectangular.

Therefore by definition $AKCH$ is a square.

We have that:

$FB : BG = DB : BE$

and:

$FB : BG = AB : DG$

and from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:

$DB : BE = DG : BC$

it follows from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:

$AB : DG = DG : BC$

Therefore $DG$ is a mean proportional between $AB$ and $BC$.


We also have that:

$AD : DK = KG : GC$

and so from Proposition $1$ of Book $\text{V} $: Magnitudes Proportional Separated are Proportional Compounded:

$AK : KD = KC : CG$

while:

$AK : KD = AC : CD$

and from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:

$KC : CG = DC : CB$

it follows from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:

$AC : DC = DC : BC$

Therefore $DC$ is a mean proportional between $AC$ and $BC$.

$\blacksquare$


Historical Note

This proof is Proposition $54$ of Book $\text{X}$ of Euclid's The Elements.
It can be argued that this lemma is not original to Euclid.


Sources