# 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.*

(*The Elements*: Book $\text{XIII}$: Proposition $5$)

## 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:

and:

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:

- $DB > BA$

Therefore by Proposition $14$ of Book $\text{V} $: Relative Sizes of Components of Ratios:

- $BA > AD$

Hence the result as stated.

$\blacksquare$

## Historical Note

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

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XIII}$. Propositions