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