Construction of Part of Line

Theorem

In the words of Euclid:

From a given straight line to cut off a prescribed part.

(The Elements: Book $\text{VI}$: Proposition $9$)

Construction

Let $AB$ be the given straight line.

From $A$ draw $AC$ making any angle with $AB$.

On $AC$, take any point $D$ and make $AC$ the same multiple of $AD$ that $AB$ is to be of the required part which is to be cut from it.

Join $BC$ and draw $DE$ parallel to it.

Then $AE$ is the required part of $AB$.

Proof

We have that $ED$ is parallel to one of the sides of $\triangle ABC$.

So from Parallel Transversal Theorem:

$CD : DA = BE : EA$

From Magnitudes Proportional Separated are Proportional Compounded:

$CA : AD = BA : AE$

From Proportional Magnitudes are Proportional Alternately:

$CA : BA = AD : AE$

From Ratio Equals its Multiples:

$AD : AE = n \cdot AD : n \cdot AE$

where $n$ is the number of times $AD$ is contained in $AC$.

From Equality of Ratios is Transitive:

$AC : AB = n \cdot AD : n \cdot AE$

So from Relative Sizes of Components of Ratios it follows that:

$AB = n \cdot AE$

$\blacksquare$

Historical Note

This proof is Proposition $9$ of Book $\text{VI}$ of Euclid's The Elements.
It is in fact only a particular case of Proposition $10$ of Book $\text{VI} $: Construction of Similarly Cut Straight Line.
The proof given here is that given by Robert Simson, with a refinement by Camerer, as Euclid's original was unnecessarily specific.

Sources

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