Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio

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Theorem

In the words of Euclid:

If the square on a straight line be five times the square on a segment of it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line.

(The Elements: Book $\text{XIII}$: Proposition $2$)


Lemma

In the words of Euclid:

That the double of $AC$ is greater than $BC$ is to be proved thus.

(The Elements: Book $\text{XIII}$: Proposition $2$ : Lemma)


Proof

Euclid-XIII-2.png

Let the square on $AB$ be five times the square on $AC$.

Let $CD = 2 \cdot AC$.

It is to be demonstrated that when $CD$ is cut in extreme and mean ratio, the greater segment is $CB$.


Let the squares $AF$ and $CG$ be described on $AB$ and $CF$ respectively.

Let the figure $AF$ be drawn.

Let $BE$ be produced from $FB$.

We have that:

$BA^2 = 5 \cdot AC^2$

That is:

$AF = 5 \cdot AH$

Therefore the gnomon $MNO$ is $4$ times $AH$.

We have that:

$DC = 2 \cdot CA$

Therefore:

$DC^2 = 4 \cdot CA^2$

that is:

$CG = 4 \cdot AH$

But:

$MNO = 4 \cdot AH$

Therefore:

$MNO = CG$

We have that:

$DC = CK$

and:

$AC = CH$

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

$KB = 2 \cdot BH$

But we also have:

$LH + HB = 2 \cdot BH$

Therefore:

$KB = LH + HB$

But:

$MNO = CG$

Therefore:

$HF = BG$

and as $CD = DG$:

$BG = CD \cdot DB$

Also:

$HF = CB^2$

Therefore:

$CD \cdot DB = CB^2$

Therefore:

$DC : CB = CB : BD$

But:

$DC > CB$

therefore:

$CB > BD$

Therefore, when $CD$ is cut in extreme and mean ratio, the greater segment is $CB$.

$\blacksquare$


Historical Note

This proof is Proposition $2$ of Book $\text{XIII}$ of Euclid's The Elements.
It is the converse of Proposition $1$: Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio.


Sources