Construction of Golden Section

Theorem

In the words of Euclid:

To cut a given finite straight line in extreme and mean ratio.

(The Elements: Book $\text{VI}$: Proposition $30$)

Construction

Let $AB$ be the given finite straight line.

Construct the square $\Box BC$ on $AB$.

Using Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram, construct $\Box CD$ equal to $\Box BC$ and exceeding by the figure $\Box AD$ similar to $\Box BC$.

Then $E$ is the point at which $AB$ is cut so that $AB : AE = AE : EB$.

Proof

As $\Box BC$ is a square, then $\Box AD$ is also a square.

Since $\Box CD = \Box BC$, we subtract $\Box CE$ from each.

Therefore $\Box BF = \Box AD$.

From Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional‎, in $\Box BF$ and $\Box AD$ the sides about the equal angles are reciprocally proportional.

Therefore $FE : ED = AE : EB$.

But $FE = AB$ and $ED = AE$.

So $BA : AE = AE : EB$.

Also $AB > AE$ and so $AE > EB$.

Hence the result.

$\blacksquare$

Historical Note

This proof is Proposition $30$ of Book $\text{VI}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VI}$. Propositions
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): golden section