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 $a : b = b : c$.

As Euclid defined it:

To two given straight lines to find a third proportional.

(The Elements: Book 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 Line in Triangle Cuts Sides Proportionally $AB : BD = AC : CE$.

But $BD = AC$ and so $AB : AC = AC : CE$ as required.

$\blacksquare$

Historical Note

This is Proposition 11 of Book VI of Euclid's The Elements.

This theorem is a special case of Book VI Proposition 12: Construction of Fourth Proportional Straight Line.