# Straight Line cut in Extreme and Mean Ratio plus its Greater Segment

## Theorem

In the words of Euclid:

If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment.

## Proof

Let the line $AB$ be cut in extreme and mean ratio at the point $C$.

Let $AC$ be the greater segment.

Let $AD = AC$.

It is to be demonstrated that $DB$ is cut in extreme and mean ratio at the point $A$ where $BA > AD$.

Let the square $AE$ be described on $AB$.

Let the figure be drawn as above.

We have that $AB$ is cut in extreme and mean ratio at $C$.

Therefore from:

Proposition $17$ of Book $\text{VI}$: Rectangles Contained by Three Proportional Straight Lines

and:

Book $\text{VI}$ Definition $3$: Extreme and Mean Ratio

it follows that:

$AB \cdot BC = AC^2$

We have that:

$CE = AB \cdot BC$

and:

$CH = AC^2$

THerefore:

$CH = HC$

But:

$HE = CE$

and:

$DH = HC$

Therefore:

$DH = HE$

Therefore:

$DK = AE$

Thus:

$DK = BD \cdot DA$

That is:

$BD \cdot DA = AB^2$

Therefore:

$DB : BA = BA : AD$

and:

$DB > BA$
$BA > AD$

Hence the result as stated.

$\blacksquare$

## Historical Note

This proof is Proposition $5$ of Book $\text{XIII}$ of Euclid's The Elements.