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