Euclid’s Elements — Book XIII, Proposition 2

From a 5-to-1 Square Ratio to the Golden Division

In Book XIII of Euclid’s Elements, Proposition 2 explores a fascinating connection between the golden ratio and a special property of squares.

It essentially shows how, if a certain relationship between the squares of two segments of a straight line holds, the line can be divided in extreme and mean ratio — the essence of the golden ratio.

---

Proposition Statement

If the square on a straight line is five times the square on one of its segments, then when twice that segment is divided in extreme and mean ratio, the greater segment equals the remaining part of the original straight line.

---

Geometric Setup

• Let AB be a straight line.

• On AB, take a segment AC such that:

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

• Let $CD = 2 \cdot AC$.

• The goal is to prove:
  When CD is divided in extreme and mean ratio, the greater segment equals CB (the remaining part of AB).

---

Key Result

Euclid proves that:

$$\frac{CD}{CB}=\frac{CB}{B\mathrm{D}}$$

Thus, CB is the greater segment of CD when CD is cut in extreme and mean ratio — what we now call the golden ratio.

---

Sketch of the Proof

Euclid’s proof uses geometric algebra, combining relationships between squares and rectangles:

1. Square Proportions
   Given $AB^2 = 5 \cdot AC^2$, the relationship between AB and AC links naturally to the golden ratio.

2. Double Segment
   Since $CD = 2 \cdot AC$, we analyze the relationship between the square on CD and the square on AC:

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

   This becomes essential later.

3. Equality of Areas
   Through constructing squares on AB and CD, and using properties of gnomons, Euclid shows:

   $$CD\cdot DB=C{B}^2$$

   Which implies:

   $$\frac{CD}{CB}=\frac{CB}{B\mathrm{D}}$$

4. Comparison of Segments
   Since $CD > CB$, the conclusion is that CB is the greater segment when CD is divided in extreme and mean ratio.

---

Golden Ratio Connection

This proposition ties directly to the golden ratio $\phi$.
If a line is divided so that:

$$\frac{CD}{CB}=\frac{CB}{BD},$$

then:

$$\varphi =\frac{CD}{CB}=\frac{CB}{BD}=\frac{1+\sqrt{5}}{2}\mathrm{.}$$

Thus, the condition $AB^2 = 5 \cdot AC^2$ leads naturally to a division involving the golden ratio.