Construction of Third Proportional Straight Line

Theorem

Given any two straight lines of given length $a$ and $b$, it is possible to construct a third straight line of length $c$ such that $a : b = b : c$.

In the words of Euclid:

To two given straight lines to find a third proportional.

(The Elements: Book $\text{VI}$: Proposition $11$)

Construction

Let $AB, AC$ be the two given straight lines.

Let them be placed to contain any angle.

Let $AB$ be produced to $D$, and $AC$ be produced to $E$.

Let $BD$ be constructed equal to $AC$.

Join $BC$ and construct $DE$ parallel to $BC$.

Then $CE$ is the required third proportional line.

Proof

We have that:

$BC \parallel DE$

So from Parallel Transversal Theorem:

$AB : BD = AC : CE$

But $BD = AC$ and so:

$AB : AC = AC : CE$

as required.

$\blacksquare$

Historical Note

This proof is Proposition $11$ of Book $\text{VI}$ of Euclid's The Elements.
It is a special case of Proposition $12$ of Book $\text{VI} $: Construction of Fourth Proportional Straight Line.

Sources

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