Euclid's WorkshopBook I
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Proposition 5 of 48 • Theorem

In isosceles triangles the angles at the base equal one another.

If a triangle has two equal sides (isosceles), then the two angles opposite those equal sides are also equal.

Before You Read

Draw an isosceles triangle—one where two sides are the same length. Now look at the two angles at the base (the angles opposite the equal sides). Do they look equal? They should. But can you prove it? That's the challenge. This was historically where students either crossed the bridge into real mathematical thinking or turned back.

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A B C D E

What Euclid Is Doing

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

Approach: Euclid's construction: Extend AB beyond B to point D, and extend AC beyond C to point E. Make BD = CE (using Prop 3). Now we have two large triangles (ADC and AEB) and two smaller triangles (BDC and CEB) to work with.

Conclusion: Here's the proof chain: First, AD = AE because AD = AB + BD, AE = AC + CE, and AB = AC (given), BD = CE (constructed). In triangles ADC and AEB: AD = AE (just shown), AC = AB (given), and ∠DAC = ∠EAB (same angle at A). By SAS (Prop 4), △ADC ≅ △AEB. Therefore DC = EB and ∠ACD = ∠ABE. Now examine triangles BDC and CEB: BD = CE (constructed), DC = EB (just proved), ∠BDC = ∠CEB (from the congruence above). By SAS (Prop 4), △BDC ≅ △CEB. Therefore ∠DBC = ∠ECB (the exterior base angles are equal). Finally, ∠ABC = ∠ACB because they are supplements of equal angles to the straight lines ABD and ACE. ✓

Key Moves

  1. Given: Triangle ABC with AB = AC (isosceles)
  2. Extend AB to D and AC to E; make BD = CE (Proposition 3)
  3. AD = AE because AB + BD = AC + CE (given AB = AC, constructed BD = CE)
  4. Triangles ADC and AEB: AD = AE, AC = AB, ∠DAC = ∠EAB (same angle)
  5. By SAS (Prop 4): △ADC ≅ △AEB, so DC = EB and ∠ACD = ∠ABE
  6. Triangles BDC and CEB: BD = CE, DC = EB, ∠BDC = ∠CEB
  7. By SAS (Prop 4): △BDC ≅ △CEB, so ∠DBC = ∠ECB
  8. Therefore ∠ABC = ∠ACB (supplements of equal angles) ✓

Try It Yourself

Draw several isosceles triangles with different proportions—a tall skinny one, a short wide one, one that's almost equilateral. In each case, measure the base angles. Are they always equal? Now try the converse: can you draw a triangle where two sides are equal but the base angles are NOT equal?

Proof Challenge

Available Justifications

1.

Given: Triangle ABC with AB = AC (isosceles)

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2.

Extend AB to D and AC to E.

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3.

Make BD = CE.

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4.

Since AB = AC and BD = CE, then AB + BD = AC + CE.

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5.

Therefore AD = AE.

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6.

In triangles ADC and AEB: AD = AE, AC = AB, and ∠DAC = ∠EAB (the angle between the same two lines through A). Therefore DC = EB and ∠ACD = ∠ABE.

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7.

In triangles BDC and CEB: BD = CE, DC = EB, and ∠BDC = ∠CEB. Therefore ∠DBC = ∠ECB.

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8.

Subtracting the equal exterior angles from the equal whole angles gives ∠ABC = ∠ACB.

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 5.

  • ✓Lesson Plan (prop-05-lesson-plan.pdf)
  • ✓Student Worksheet (prop-05-worksheet.pdf)
  • ✓Answer Key (prop-05-answer-key.pdf)
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Why It Matters

This is called the 'Pons Asinorum' (Bridge of Asses) because historically it was where students who couldn't handle geometric reasoning would get stuck and drop out. It's the first theorem that requires real deductive thinking.

Modern connection: Isosceles triangles appear everywhere—in architecture (A-frames), in design (symmetrical logos), in nature (butterflies). The equal base angles create visual and structural balance.

Historical note: The name 'Pons Asinorum' may also refer to the diagram looking like a bridge, or to it being a bridge to the harder proofs that follow. Either way, it's been a milestone in geometry education for centuries.

Discussion Questions

  • Why is this called the 'Bridge of Asses'? What makes it a turning point?
  • Can you think of a visual or physical way to see that the base angles must be equal?
  • What would it mean if an isosceles triangle had UNEQUAL base angles?
Euclid's Original Proof
Let ABC be an isosceles triangle having the side AB equal to the side AC; and let AB, AC be produced further to D and E. [Post. 2]

I say that the angle ABC is equal to the angle ACB, and the angle DBC to the angle ECB.

Make BD equal to CE; [I. 3] and let the straight lines DC, EB be joined. [Post. 1]

Then, since AD is equal to AE and AB to AC, the two sides DA, AC are equal to the two sides EA, AB, respectively; and they contain a common angle, the angle DAE. Therefore the base DC is equal to the base EB, and the triangle ADC is equal to the triangle AEB, and the remaining angles will be equal. [I. 4]

Since the whole AD is equal to the whole AE, and AB is equal to AC, the remainder BD is equal to the remainder CE. [C.N. 3]

Now DC was proved equal to EB; therefore the two sides BD, DC are equal to the two sides CE, EB; and the angle BDC is equal to the angle CEB; and the base BC is common. Therefore the triangle BDC is equal to the triangle CEB. [I. 4]

Therefore the angle DBC is equal to the angle ECB, and the angle BCD to the angle CBE.

Since the whole angle ABE was proved equal to the angle ACD, and the angle CBE equals angle BCD, the remaining angle ABC equals the remaining angle ACB. [C.N. 3]

Therefore etc. (Being) what it was required to prove.

What's Next

Proposition 5 proves that equal sides force equal angles. The natural question: does it work the other way? If two angles are equal, must the opposite sides be equal? That's Proposition 6—the converse—and Euclid proves it using a clever contradiction argument.