Construction of Part of Line

Construction

 * Euclid-VI-9.png

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:
 * $AB = n \cdot AE$