Euclid's WorkshopBook I
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Proposition 16 of 48 • Theorem

In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles.

When you extend one side of a triangle, the exterior angle formed is larger than either of the two non-adjacent interior angles.

Before You Read

Extend one side of any triangle beyond a vertex. The angle formed outside the triangle at that vertex is called the exterior angle. Do you think it must be larger than either of the two interior angles on the far side of the triangle? Try a few extreme cases—a very flat triangle, a very tall one.

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B C A D E F M N G

What Euclid Is Doing

Setup: We have triangle ABC, and we extend side BC past C to a point D. This creates an exterior angle ACD. We want to prove that this exterior angle is greater than either of the two remote interior angles: ∠BAC and ∠ABC.

Approach: Euclid's strategy is to construct a point F such that ∠ECF (a part of the exterior angle) equals ∠BAC exactly. He does this by bisecting AC at E, drawing BE, and extending it to F so that EF = BE. This creates two congruent triangles (by SAS), which transfers the interior angle to a piece of the exterior angle.

Conclusion: Bisect AC at E (Prop 10), so AE = CE. Extend BE to F with EF = BE (Prop 3). Now triangles AEB and CEF have: AE = CE, BE = EF, and ∠AEB = ∠CEF (vertical angles, Prop 15). By SAS (Prop 4), the triangles are congruent, so ∠BAE = ∠ECF. But ∠ACD > ∠ECF because ∠ECF is only a part of ∠ACD (Common Notion 5). Therefore ∠ACD > ∠BAC. ✓ For the second inequality: bisect BC at M (Prop 10), so BM = CM. Draw AM and extend it to N with MN = AM (Prop 3). Draw CN (Postulate 1). Now triangles AMB and CMN have: BM = CM, AM = MN, and ∠AMB = ∠CMN (vertical angles, Prop 15). By SAS (Prop 4), the triangles are congruent, so ∠ABC = ∠MCN. But ∠ACD > ∠MCN because ∠MCN is only a part of ∠ACD (Common Notion 5). Therefore ∠ACD > ∠ABC. ✓

Key Moves

  1. Extend BC to D, forming exterior angle ACD
  2. Bisect AC at E so that AE = CE (Proposition 10)
  3. Draw BE and extend it to F so that EF = BE (Postulate 2, Proposition 3)
  4. Draw CF (Postulate 1)
  5. In triangles AEB and CEF: AE = CE, BE = EF, and ∠AEB = ∠CEF (vertical angles, Proposition 15)
  6. By SAS (Proposition 4), triangles AEB and CEF are congruent, so ∠BAE = ∠ECF
  7. ∠ACD > ∠ECF since ∠ECF is a part of ∠ACD (Common Notion 5), so ∠ACD > ∠BAC ✓
  8. For the second inequality: bisect BC at M so that BM = CM (Proposition 10)
  9. Draw AM and extend it to N so that MN = AM (Postulate 2, Proposition 3)
  10. Draw CN (Postulate 1)
  11. In triangles AMB and CMN: BM = CM, AM = MN, and ∠AMB = ∠CMN (vertical angles, Proposition 15)
  12. By SAS (Proposition 4), triangles AMB and CMN are congruent, so ∠ABC = ∠MCN
  13. ∠ACD > ∠MCN since ∠MCN is a part of ∠ACD (Common Notion 5), so ∠ACD > ∠ABC ✓

Try It Yourself

Draw a triangle and extend one side to create an exterior angle. Measure the exterior angle and the two non-adjacent interior angles. Is the exterior angle always strictly greater than each? Now try the same experiment on a very obtuse triangle and on a nearly equilateral one. Does the inequality always hold?

Proof Challenge

Available Justifications

1.

Bisect AC at point E

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2.

Draw BE and extend it to F where EF = BE

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3.

Draw line CF

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4.

∠AEB = ∠CEF (vertical angles)

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5.

AE = CE, BE = EF, ∠AEB = ∠CEF, so △AEB ≅ △CEF by SAS

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6.

∠BAC = ∠ECF, and ∠ACD > ∠ECF, so ∠ACD > ∠BAC

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7.

Now bisect BC at point M

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8.

Draw AM and extend it to N where MN = AM

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9.

Draw line CN

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10.

BM = CM, AM = MN, ∠AMB = ∠CMN (vertical angles), so △AMB ≅ △CMN by SAS

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11.

∠ABC = ∠MCN, and ∠ACD > ∠MCN (whole > part), so ∠ACD > ∠ABC

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 16.

  • ✓Lesson Plan (prop-16-lesson-plan.pdf)
  • ✓Student Worksheet (prop-16-worksheet.pdf)
  • ✓Answer Key (prop-16-answer-key.pdf)
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Why It Matters

This is the first inequality result in the Elements. Everything before this was about equality—equal sides, equal angles, congruent triangles. Now Euclid proves something is strictly greater than something else, opening the door to comparison arguments that dominate the rest of Book I.

Modern connection: The exterior angle theorem is foundational in navigation and surveying. When you sight two landmarks from a position and then move to a new position, the exterior angle relationship helps determine relative distances. It is also key in proving that parallel lines exist.

Historical note: This proposition only holds in Euclidean (flat) geometry. On a sphere, you can construct triangles where the exterior angle equals or is even less than a remote interior angle. This makes Prop 16 one of the results that breaks down in non-Euclidean geometry.

Discussion Questions

  • Why does Euclid bisect AC specifically, rather than some other segment? Could the proof work by bisecting BC instead?
  • The proof constructs point F as a mirror image of B through E. What geometric transformation is happening here?
  • This is the first inequality in the Elements. Why do you think Euclid waited until Proposition 16 to introduce inequalities?
Euclid's Original Proof
In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.

Let ABC be a triangle, and let one side of it BC be produced to D;

I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC.

Let AC be bisected at E [I.10], and let BE be joined and produced in a straight line to F;

let EF be made equal to BE [I.3], let FC be joined [Post. 1], and let AC be drawn through to G [Post. 2].

Then, since AE is equal to EC, and BE to EF, the two sides AE, EB are equal to the two sides CE, EF respectively;

and the angle AEB is equal to the angle FEC, for they are vertical angles. [I.15]

Therefore the base AB is equal to the base FC, and the triangle AEB is equal to the triangle FEC, and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I.4]

therefore the angle BAE is equal to the angle ECF.

But the angle ECD is greater than the angle ECF; [C.N. 5]

therefore the angle ACD is greater than the angle BAE.

Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I.15], can be proved greater than the angle ABC as well.

Therefore etc.

Q.E.D.

What's Next

Proposition 16 tells us the exterior angle beats either remote interior angle. Proposition 17 is a quick but important corollary: since the exterior angle is less than two right angles, and it's bigger than each of the two remote interior angles, any two angles of a triangle must sum to less than two right angles. Watch how efficiently Euclid derives it.