Euclid’s Elements — Book XIII, Proposition 1

Extreme and Mean Ratio and a Surprising Relationship Between Squares

In Book XIII of Euclid’s Elements, we enter the fascinating world of the extreme and mean ratio — what we now call the golden ratio.

In Proposition 1, Euclid establishes an elegant relationship involving the greater segment of a line cut in extreme and mean ratio, the whole line, and its half.

---

Proposition Statement

If a straight line is cut in extreme and mean ratio, then the square on the greater segment plus the half of the whole equals five times the square on the half.

---

Geometric Setup

• Let a straight line AB be cut in extreme and mean ratio at point C.

• By definition:

  $$\frac{AB}{AC}=\frac{AC}{C\mathrm{B}}$$

  with AC being the greater segment.

• Extend the line to point D so that:

  $$AD=\frac{1}{2}AB$$

• The goal is to prove:

  $$C{D}^2=5\cdot A{D}^2$$

---

Key Result

Euclid proves the following identity:

$$C{D}^2=5\cdot A{D}^2$$

In words:

The square on the greater segment plus half the whole is exactly five times the square on the half.

---

Sketch of the Proof

Euclid’s proof uses geometric algebra — expressing algebraic identities through areas of rectangles and squares:

1. Extreme and Mean Ratio Property
   Since AB is cut in extreme and mean ratio at C:

   $$AB\cdot BC=A{C}^2$$

   This relationship is the cornerstone of the entire proof.

2. Construction of Squares and Rectangles

   • Squares are constructed on the line segments.

   • Various rectangles and gnomons (L-shaped figures) are introduced to compare areas.

3. Half of the Whole
   Since AD is half of AB, relationships between the constructed squares are derived:

   $$A{B}^2=4\cdot A{D}^2$$

4. Combination of Equal Areas
   By manipulating the constructed squares and rectangles, Euclid arrives at the final equality:

   $$C{D}^2=5\cdot A{D}^2$$

---

Mathematical Significance

This proposition establishes a fundamental connection between:

• Golden ratio geometry and

• Pythagorean relationships between squares.

In modern terms, if a line AB is divided in the golden ratio, then:

$$AC=\varphi \cdot CB$$

and

$$C{D}^2=5\cdot A{D}^2,$$

where $\phi = \dfrac{1+\sqrt{5}}{2}$.

This beautifully links the golden ratio, square relationships, and pentagon geometry — themes that reappear throughout Book XIII.

---

Summary

If a line is cut in extreme and mean ratio, the square on the greater segment plus half the whole is five times the square on the half.

$$\boxed{{\displaystyle C{D}^2=5\cdot A{D}^{\mathrm{2}}}}$$