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

## 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.

### Lemma

In the words of Euclid:

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

## Proof

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:

$AC = CH$
$KB = 2 \cdot BH$

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 $CB$.

$\blacksquare$