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

Proof

 * Euclid-XIII-5.png

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 :
 * $BA > AD$

Hence the result as stated.