Euclid's WorkshopBook I
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Proposition 18 of 48 • Theorem

In any triangle the angle opposite the greater side is greater.

In a triangle, the longer side is always opposite the larger angle.

Before You Read

In a triangle with two sides of different lengths, which interior angle do you expect to be larger—the one opposite the long side or the one opposite the short side? It seems intuitive, but intuition is not proof. Think about how you might prove it using the tools Euclid has assembled.

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

What Euclid Is Doing

Setup: In triangle ABC, side AB is greater than side AC. We want to prove that the angle opposite the longer side (∠ACB) is greater than the angle opposite the shorter side (∠ABC).

Approach: Euclid marks a point D on AB such that AD = AC, creating an isosceles triangle ACD. The base angles of that isosceles triangle are equal (Prop 5). Then he uses the exterior angle theorem (Prop 16) to show that one of those equal angles is already bigger than ∠ABC.

Conclusion: Since AB > AC, mark D on AB so that AD = AC (Prop 3). Draw CD (Postulate 1). In triangle ACD, since AD = AC, we have ∠ACD = ∠ADC (Prop 5). Now ∠ADC is an exterior angle of triangle BCD (since D is between A and B), so ∠ADC > ∠ABC (Prop 16). Also ∠ACB > ∠ACD since ∠ACD is only part of ∠ACB (Common Notion 5). Chaining: ∠ACB > ∠ACD = ∠ADC > ∠ABC. Therefore ∠ACB > ∠ABC. ✓

Key Moves

  1. Since AB > AC, mark point D on AB so that AD = AC (Proposition 3)
  2. Draw CD (Postulate 1), forming isosceles triangle ACD
  3. In triangle ACD: AD = AC, so ∠ACD = ∠ADC (Proposition 5, isosceles triangle theorem)
  4. ∠ADC is an exterior angle of triangle BCD, so ∠ADC > ∠ABC (Proposition 16)
  5. ∠ACB > ∠ACD because ∠ACD is only part of ∠ACB (Common Notion 5)
  6. Chaining the inequalities: ∠ACB > ∠ACD = ∠ADC > ∠ABC
  7. Therefore ∠ACB > ∠ABC — the angle opposite the greater side is greater ✓

Try It Yourself

Draw a triangle where one side is noticeably longer than another. Measure all three sides and all three angles, then pair each side with the angle across from it. Is the longest side always opposite the largest angle? Try several triangles including a right triangle and an obtuse one.

Proof Challenge

Available Justifications

1.

Since AB > AC, mark point D on AB so that AD = AC

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2.

Draw line CD

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3.

Triangle ACD is isosceles (AD = AC), so ∠ACD = ∠ADC

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4.

∠ADC > ∠ABC (exterior angle of triangle BCD)

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5.

∠ACB > ∠ACD > ∠ADC > ∠ABC, so ∠ACB > ∠ABC

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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 18.

  • ✓Lesson Plan (prop-18-lesson-plan.pdf)
  • ✓Student Worksheet (prop-18-worksheet.pdf)
  • ✓Answer Key (prop-18-answer-key.pdf)
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Why It Matters

This proposition establishes the fundamental relationship between sides and angles in a triangle: bigger sides face bigger angles. This is the forward direction; the converse (Prop 19) goes the other way. Together they form a complete correspondence between side lengths and angle sizes.

Modern connection: This principle is used constantly in structural engineering and physics. If one side of a triangular support is longer, the stress concentration at the opposite vertex is greater. It also underpins the law of sines in trigonometry, which quantifies this relationship precisely.

Historical note: Euclid's clever construction of point D to create an isosceles sub-triangle is a technique that reappears throughout the Elements. By reducing the problem to isosceles triangles (where he already has results), he avoids needing any new machinery.

Discussion Questions

  • Why does Euclid place point D on AB rather than extending AC? What would go wrong with a different construction?
  • This proof chains two inequalities together. Can you trace exactly where each link comes from?
  • If two sides of a triangle are equal, what does this proposition tell you about the opposite angles? Does that match Proposition 5?
Euclid's Original Proof
In any triangle the greater side subtends the greater angle.

For let ABC be a triangle having the side AC greater than AB;

I say that the angle ABC is also greater than the angle BCA.

For, since AC is greater than AB, let AD be made equal to AB [I.3], and let BD be joined.

Then, since the angle ADB is an exterior angle of the triangle BCD, it is greater than the interior and opposite angle DCB. [I.16]

But the angle ADB is equal to the angle ABD, since the side AB is equal to AD; [I.5]

therefore the angle ABD is also greater than the angle ACB;

therefore the angle ABC is much greater than the angle ACB. [C.N. 5]

Therefore etc.

Q.E.D.

What's Next

Proposition 18 proves that a greater side forces a greater opposite angle. Proposition 19 proves the converse: a greater angle forces a greater opposite side. Euclid proves it indirectly—by assuming the opposite and showing a contradiction using Proposition 18 itself. Watch for the elegant reversal.