Euclid's WorkshopBook I
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Proposition 20 of 48 • Theorem

In any triangle the sum of any two sides is greater than the remaining one.

In any triangle, the sum of any two sides is always greater than the third side (the triangle inequality).

Before You Read

Can you draw a triangle with sides of length 1, 2, and 10? Try it—you'll find the two short sides can't reach each other to close the triangle. Is it always true that two sides of a triangle together must be longer than the third side, no matter what triangle you draw?

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Basic Constructions (5)
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Perpendiculars & Angles (5)
Exterior Angles & Inequalities (5)
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B C A D

What Euclid Is Doing

Setup: We have triangle ABC and want to prove that BA + AC > BC. (By symmetry, the same argument applies to any pair of sides versus the third.)

Approach: Euclid extends BA to a point D so that AD = AC, making BD = BA + AC. Then he shows that in triangle BCD, the angle at C is larger than the angle at D, which by Prop 19 means the side opposite C (which is BD) is greater than the side opposite D (which is BC).

Conclusion: Extend BA to D where AD = AC (Postulate 2, Proposition 3). Draw CD (Postulate 1). Since AD = AC, triangle ACD is isosceles, so ∠ACD = ∠ADC (Prop 5). Now ∠BCD > ∠ACD (since ∠ACD is part of ∠BCD, Common Notion 5). Therefore ∠BCD > ∠ADC = ∠BDC. In triangle BCD, since ∠BCD > ∠BDC, the side opposite the larger angle is greater: BD > BC (Prop 19). But BD = BA + AD = BA + AC. Therefore BA + AC > BC. ✓

Key Moves

  1. Extend BA past A to point D, with AD = AC (Postulate 2, Proposition 3). Now BD = BA + AC.
  2. Draw CD (Postulate 1), forming triangle BCD
  3. Since AD = AC, triangle ACD is isosceles: ∠ACD = ∠ADC (Proposition 5)
  4. ∠BCD > ∠ACD because ∠ACD is only part of ∠BCD (Common Notion 5)
  5. Therefore ∠BCD > ∠ADC = ∠BDC
  6. In triangle BCD, the side opposite the larger angle is larger: BD > BC (Proposition 19)
  7. But BD = BA + AD = BA + AC. Therefore BA + AC > BC ✓

Try It Yourself

Try to construct triangles with these side triples using a compass and straightedge: (3, 4, 5), (1, 1, 1), (2, 3, 6), (5, 5, 9). Which ones are impossible to close? For each valid triangle, verify with a ruler that the sum of every pair of sides exceeds the third. Can you find a 'near miss' where the sum just barely exceeds?

Proof Challenge

Available Justifications

1.

Extend BA to point D where AD = AC

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2.

Draw line CD

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3.

In isosceles triangle ACD, ∠ACD = ∠ADC

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4.

∠BCD > ∠ACD = ∠BDC, so in triangle BCD, BD > BC

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5.

BD = BA + AD = BA + AC, so BA + AC > BC

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0 of 5 steps completed

Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 20.

  • ✓Lesson Plan (prop-20-lesson-plan.pdf)
  • ✓Student Worksheet (prop-20-worksheet.pdf)
  • ✓Answer Key (prop-20-answer-key.pdf)
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Why It Matters

The triangle inequality is one of the most important results in all of geometry. It tells you when three lengths can form a triangle and when they cannot. It is the foundation for the concept of distance in metric spaces, the most general setting for geometry in modern mathematics.

Modern connection: The triangle inequality is an axiom of every metric space in modern mathematics. GPS systems, network routing, and machine learning distance functions all depend on it. In everyday life, it is the reason the shortest path between two points is a straight line — any detour through a third point makes the journey longer.

Historical note: The Epicureans reportedly mocked this proposition, saying it was 'evident even to an ass' — a donkey walking toward food goes in a straight line, not via two sides of a triangle. Proclus defended Euclid, noting that what is obvious to perception still requires logical proof. The distinction between intuition and proof remains central to mathematics.

Discussion Questions

  • The Epicureans said this was obvious even to a donkey. Do you agree it is obvious? Is it still worth proving?
  • When does equality hold — that is, when does BA + AC = BC exactly? What does that 'triangle' look like?
  • The triangle inequality defines what a 'metric space' is. Can you think of a way to measure distance where the triangle inequality fails?
Euclid's Original Proof
In any triangle two sides taken together in any manner are greater than the remaining one.

For let ABC be a triangle;

I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one, namely BA, AC greater than BC; AB, BC greater than AC; BC, CA greater than AB.

For let BA be drawn through to the point D, let DA be made equal to CA [I.3], and let DC be joined. [Post. 1]

Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD. [I.5]

Therefore the angle BCD is greater than the angle ADC. [C.N. 5]

And, since DCB is a triangle having the angle BCD greater than the angle BDC, and the greater angle is subtended by the greater side, [I.19]

therefore DB is greater than BC.

But DA is equal to AC;

therefore BA, AC are greater than BC.

Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.

Therefore etc.

Q.E.D.

What's Next

The triangle inequality feels obvious but required careful proof. Proposition 21 pushes deeper: if you draw two lines from the endpoints of a triangle's base to an interior point, their total length is less than the sum of the other two sides—yet the angle they form is larger. A subtle and surprising result awaits.