Euclid’s Elements — Book XIII, Proposition 12

An Equilateral Triangle in a Circle: A Triple-Square Relationship

In Book XIII of Euclid’s Elements, Proposition 12 establishes a beautiful relationship between an equilateral triangle inscribed in a circle and the radius of that circle.

It shows that the square on the side of the triangle is three times the square on the radius.

This result connects circle geometry with the properties of regular polygons and leads directly to deeper results involving the tetrahedron and icosahedron later in the book.

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Proposition Statement

If an equilateral triangle is inscribed in a circle, then the square on the side of the triangle equals three times the square on the radius of the circle.

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Geometric Setup

• Let ABC be a circle.

• Inscribe within it an equilateral triangle ABC.

• Let D be the center of the circle.

• Join AD (a radius) and extend it to E on the circumference.

• Join BE as well.

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Key Result

Euclid proves that:

$$A{B}^2=3\cdot D{E}^2$$

where:

• AB = side of the equilateral triangle,

• DE = radius of the circle.

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Sketch of the Proof

The proof cleverly combines properties of circles, polygons, and the Pythagorean theorem:

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1. Relationship Between the Triangle and the Hexagon

• Since the triangle ABC is equilateral, the entire circle is divided into three equal arcs.

• Therefore, the arc BE is exactly one-sixth of the circumference, meaning BE is the side of a regular hexagon.

• A fundamental property of hexagons inscribed in a circle is:

  $$BE=DE,$$

  i.e., the side of a regular hexagon equals the circle’s radius.

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2. Doubling the Radius

• Since AE passes through the circle’s center, it equals 2 · DE.

• Therefore, the square on AE is:

  $$A{E}^2=4\cdot D{E}^2\mathrm{.}$$

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3. Applying the Pythagorean Theorem

• In triangle ABE, by the Pythagorean theorem:

  $$A{E}^2=A{B}^2+B{E}^2\mathrm{.}$$

• Substituting $AE^2 = 4 \cdot DE^2$ and $BE = DE$:

  $$4\cdot D{E}^2=A{B}^2+D{E}^2\mathrm{.}$$

• Simplifying:

  $$A{B}^2=3\cdot D{E}^2\mathrm{.}$$

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Final Identity

$$\boxed{{\displaystyle A{B}^2=3\cdot D{E}^{\mathrm{2}}}}$$

This shows that the side of the equilateral triangle is related to the radius by:

$$AB = \sqrt{3} \cdot DE.$$

Summary

When an equilateral triangle is inscribed in a circle,
the square on its side equals three times the square on the radius.

$$\boxed{{\displaystyle A{B}^2=3\cdot D{E}^{\mathrm{2}}}}$$

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Why This Matters

This elegant relationship is a stepping stone to deeper results in Book XIII concerning:

• The construction of regular solids,

• The geometry of equilateral triangles in spheres,

• And the role of radical proportions in classical geometry.

It also explains why the circumradius of an equilateral triangle is always:

$$R=\frac{AB}{\sqrt{3}},$$

a formula still widely used in modern geometry.