Construction of Golden Section

Construction

 * Euclid-VI-30.png

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 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.