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Proposition 19 of 48 • Theorem

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

In a triangle, the larger angle is always opposite the longer side (the converse of Proposition 18).

Before You Read

Proposition 18 said: longer side → larger opposite angle. Now flip it: if one angle of a triangle is larger than another, must the side opposite the bigger angle be longer? It feels like it must be true, but the proof can't just say 'it's the same argument backwards'—Euclid needs a genuinely new approach.

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B C A

What Euclid Is Doing

Setup: In triangle ABC, angle ABC is greater than angle BCA. We want to prove that the side opposite the larger angle (AC) is greater than the side opposite the smaller angle (AB).

Approach: Euclid uses proof by exhaustion: there are only three possibilities for how AC compares to AB. He eliminates two of them using earlier propositions, leaving only the desired conclusion.

Conclusion: There are exactly three cases: AC = AB, AC < AB, or AC > AB. Case 1: If AC = AB, then ∠ABC = ∠ACB by Proposition 5 (isosceles triangle theorem). But we are given ∠ABC > ∠ACB, so this is a contradiction. Case 2: If AC < AB, then AB > AC, so ∠ACB > ∠ABC by Proposition 18. But we are given ∠ABC > ∠ACB, so this is also a contradiction. Case 3: AC > AB. This is the only remaining possibility, so it must be true. Therefore AC > AB. ✓

Key Moves

  1. Assume for contradiction that AC = AB. Then ∠ABC = ∠ACB (Proposition 5). This contradicts ∠ABC > ∠ACB.
  2. Assume for contradiction that AC < AB. Then ∠ACB > ∠ABC (Proposition 18). This also contradicts ∠ABC > ∠ACB.
  3. The only remaining possibility is AC > AB.
  4. Therefore the side opposite the greater angle is greater ✓

Try It Yourself

Draw a triangle with one angle clearly larger than the others—try one close to 90 degrees and one close to 30 degrees. Measure the side opposite each of those angles. Is the side opposite the 90-degree angle always longer than the side opposite the 30-degree angle? Test this with five different triangles.

Proof Challenge

Available Justifications

1.

Assume for contradiction that AC = AB

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2.

Then ∠ABC = ∠ACB — but we assumed ∠ABC > ∠ACB. Contradiction.

Drag justification
3.

Assume for contradiction that AC < AB

Drag justification
4.

Then ∠ACB > ∠ABC — contradicts ∠ABC > ∠ACB. So AC > AB.

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0 of 4 steps completed

Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 19.

  • ✓Lesson Plan (prop-19-lesson-plan.pdf)
  • ✓Student Worksheet (prop-19-worksheet.pdf)
  • ✓Answer Key (prop-19-answer-key.pdf)
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Why It Matters

This is the converse of Proposition 18, and together they establish a complete two-way correspondence: in any triangle, the larger side faces the larger angle, and the larger angle faces the larger side. This pair is essential for the triangle inequality (Prop 20) and many later results.

Modern connection: This side-angle relationship is formalized in trigonometry by the law of sines: a/sin(A) = b/sin(B) = c/sin(C). Since sine is increasing on (0, 180 degrees), a larger angle means a larger sine, and therefore a larger opposite side.

Historical note: The proof by exhaustion (or proof by elimination of cases) used here is one of Euclid's standard techniques. He lists all logical possibilities, disproves all but one, and concludes that the remaining case must hold. This is closely related to proof by contradiction.

Discussion Questions

  • This proof eliminates two cases to arrive at the answer. Is this the same as a proof by contradiction? How is it different?
  • Props 18 and 19 are converses. Does every theorem have a true converse? Can you think of a counterexample?
  • Why can't Euclid prove this directly the way he proved Prop 18? What makes the converse direction harder?
Euclid's Original Proof
In any triangle the greater angle is subtended by the greater side.

Let ABC be a triangle having the angle ABC greater than the angle BCA;

I say that the side AC is also greater than the side AB.

For, if not, AC is either equal to AB or less.

Now AC is not equal to AB; for then the angle ABC would also have been equal to the angle ACB; [I.5]
but it is not;
therefore AC is not equal to AB.

Neither is AC less than AB, for then the angle ABC would also have been less than the angle ACB; [I.18]
but it is not;
therefore AC is not less than AB.

And it was proved that it is not equal either.

Therefore AC is greater than AB.

Therefore etc.

Q.E.D.

What's Next

With both the greater-side and greater-angle results secured, Euclid is ready for one of geometry's most famous theorems: the triangle inequality. Proposition 20 proves that in any triangle, the sum of any two sides is always greater than the third—a fact that bounds how 'collapsed' a triangle can get.