Euclid's WorkshopBook I
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Proposition 6 of 48 • Theorem

If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.

If a triangle has two equal angles, then the sides opposite those angles are also equal—making it isosceles. (This is the converse of Proposition 5.)

Before You Read

Proposition 5 told us: equal sides force equal base angles. Now flip it around—if you know two angles of a triangle are equal, must the opposite sides be equal too? Try to convince yourself one way or the other before reading on.

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A B C D

What Euclid Is Doing

Setup: Triangle ABC has ∠ABC = ∠ACB (the base angles are equal). We want to prove AB = AC—that the triangle is isosceles.

Approach: Here's the key insight: there's no direct way to go from equal angles to equal sides, so Euclid flips the question. Instead of proving the sides ARE equal, he asks: what if they AREN'T? If AB were longer than AC, you could snip off a piece DB equal to AC. But then triangle DBC would match the whole triangle ACB by SAS—a part equaling the whole. That's absurd, so the assumption was wrong and the sides must have been equal all along. This is our first taste of proof by contradiction.

Conclusion: The assumption that AB ≠ AC leads to contradiction. Therefore AB = AC, and the triangle is isosceles.

Key Moves

  1. Given: ∠ABC = ∠ACB. We want to show AB = AC.
  2. Assume for contradiction: AB ≠ AC. Then one is greater—say AB > AC.
  3. Cut off DB from AB so that DB = AC (Proposition 3)
  4. Join DC (Postulate 1)
  5. In triangles DBC and ACB: DB = AC (construction), BC = CB (common side), ∠DBC = ∠ACB (given)
  6. By SAS (Proposition 4): △DBC ≅ △ACB
  7. But △DBC is a part of △ACB, so a part equals the whole—impossible (Common Notion 5)
  8. Therefore AB = AC ✓

Try It Yourself

Draw a triangle and use a protractor to make two of its angles the same. Now measure the sides opposite those angles. Are they equal? Try it with different angle sizes—30°, 45°, 60°. Can you ever make the angles equal but the sides different?

Proof Challenge

Available Justifications

1.

Given: Triangle ABC with ∠ABC = ∠ACB. Assume AB ≠ AC, and suppose AB > AC.

Given
2.

Cut off DB from BA such that DB = AC.

Drag justification
3.

Join DC.

Drag justification
4.

In triangles DBC and ACB: DB = AC, BC = BC, and ∠DBC = ∠ACB.

Setup
5.

Therefore △DBC ≅ △ACB.

Drag justification
6.

But △DBC is a part of △ACB, so the whole cannot equal the part.

Drag justification
7.

Therefore the assumption AB ≠ AC is false, so AB = AC.

Conclusion
0 of 4 steps completed

Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 6.

  • ✓Lesson Plan (prop-06-lesson-plan.pdf)
  • ✓Student Worksheet (prop-06-worksheet.pdf)
  • ✓Answer Key (prop-06-answer-key.pdf)
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Why It Matters

This is the converse of Proposition 5. Together they say: equal sides ↔ equal angles. This bidirectional relationship is powerful—you can go from sides to angles or angles to sides.

Modern connection: This is the logical foundation for symmetry detection. If you can measure that two angles are equal, you know the opposite sides must be equal too—useful in surveying, construction, and any measurement scenario.

Historical note: Proof by contradiction (reductio ad absurdum) appears here for the first time in the Elements. Euclid assumes the opposite of what he wants to prove, then shows this leads to impossibility.

Discussion Questions

  • How is this proposition related to Proposition 5? Why do we need both?
  • Why does Euclid use proof by contradiction here instead of a direct proof?
  • If a triangle has all three angles equal, what can you conclude about its sides?
Euclid's Original Proof
Given triangle ABC with ∠ABC = ∠ACB.

Assume, for contradiction, that AB ≠ AC. Then one must be greater—say AB > AC.

Cut off DB from AB equal to AC [I.3]. Join DC [Post. 1].

Now in triangles DBC and ACB: DB = AC (by construction), BC is common, and ∠DBC = ∠ACB (given). So △DBC ≅ △ACB [I.4].

But △DBC is part of △ABC, so a part equals the whole—which is absurd [C.N. 5].

Therefore AB is not unequal to AC; hence AB = AC.

Q.E.D.

What's Next

We now know that equal angles force equal sides (and vice versa). Next comes a technical but important question: if you have a base line and two specific lengths, can two DIFFERENT triangles sit on the same side of that base? Proposition 7 says no—and that uniqueness result is the key to unlocking SSS congruence.