Proposition 21 of 48 • Theorem

If from the ends of one of the sides of a triangle two straight lines are constructed meeting within the triangle, then the sum of the straight lines so constructed is less than the sum of the remaining two sides of the triangle, but the constructed straight lines contain a greater angle than the angle contained by the remaining two sides.

If you pick a point inside a triangle and draw lines from the two ends of the base to that point, those two inner lines add up to less than the other two sides of the triangle, but the angle they form at the interior point is larger than the angle at the apex.

Before You Read

Pick any point inside a triangle and connect it to the two ends of the base. Those two inner lines form a bigger angle than the apex of the triangle—but they're shorter in total than the other two sides. Both of those claims are true at once. Does that seem plausible? Can you see intuitively why a shortcut through the interior would also spread open to a wider angle?

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What Euclid Is Doing

Setup: We have triangle ABC, and D is a point inside the triangle. Lines BD and CD are drawn from the endpoints of BC to the interior point D. We need to prove two things: (1) BD + DC < BA + AC, and (2) ∠BDC > ∠BAC.

Approach: For part (1), Euclid extends BD to meet AC at E, creating two auxiliary triangles. The triangle inequality (Prop 20) is applied twice: first to triangle ABE, then to triangle CDE. For part (2), the exterior angle theorem (Prop 16) is applied twice to show the angle at D is greater than the angle at A.

Conclusion: Part (1): Extend BD to meet AC at E. In triangle ABE, BA + AE > BE (Prop 20). Add EC to both sides: BA + AE + EC > BE + EC, so BA + AC > BE + EC. In triangle CDE, DE + EC > DC (Prop 20). So BE + EC = BD + DE + EC > BD + DC. Combining: BA + AC > BE + EC > BD + DC. Part (2): In triangle CDE, ∠BDC is an exterior angle (since BDE is a straight line), so ∠BDC > ∠DCE (Prop 16). In triangle ABE, ∠DEC (= ∠BEA, an exterior angle of ABE) is greater than ∠BAE = ∠BAC (Prop 16). But ∠BDC > ∠DEC (exterior angle of CDE, since ∠BDC > ∠DCE, and ∠DEC is in the same triangle). Actually: ∠BDC is exterior to triangle CDE at D, so ∠BDC > ∠DEC. And ∠DEC is exterior to triangle ABE at E (since BED is a line), so ∠DEC > ∠BAE. Therefore ∠BDC > ∠BAC. ✓

Key Moves

Extend BD to meet AC at point E (Postulate 1, Postulate 2)
In triangle ABE, BA + AE > BE (Proposition 20, triangle inequality)
Add EC to both sides: BA + AC > BE + EC (Common Notion 5, since AE + EC = AC)
In triangle CDE, DE + EC > DC (Proposition 20)
So BE + EC = BD + DE + EC > BD + DC. Combined with the above: BA + AC > BD + DC ✓
For the angle part: ∠BDC is an exterior angle of triangle CDE, so ∠BDC > ∠DEC (Proposition 16)
∠DEC is an exterior angle of triangle ABE, so ∠DEC > ∠BAE = ∠BAC (Proposition 16)
Therefore ∠BDC > ∠BAC ✓

Try It Yourself

Draw a triangle ABC and mark a point D clearly inside it. Measure BD + DC and compare it to BA + AC—which pair is longer? Now measure the angle BDC at your interior point and compare it to angle BAC at the apex. Try moving D closer to the apex and closer to the base: does the relationship always hold?

Proof Challenge

Available Justifications

Extend BD to meet AC at point E

Drag justification

In triangle ABE, AB + AE > BE (triangle inequality)

Drag justification

Add EC to both sides: AB + AE + EC > BE + EC, so AB + AC > BE + EC

Drag justification

In triangle CDE, DE + EC > DC

Drag justification

BD + DC < BE + EC (since BD < BE and we add DC < EC from triangle CDE). Combined with step 3: BD + DC < BA + AC.

Drag justification

∠BDC > ∠BEC (exterior angle of triangle CDE) and ∠BEC > ∠BAC (exterior angle of triangle ABE). Therefore ∠BDC > ∠BAC.

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 21.

✓Lesson Plan (prop-21-lesson-plan.pdf)
✓Student Worksheet (prop-21-worksheet.pdf)
✓Answer Key (prop-21-answer-key.pdf)

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Why It Matters

This proposition reveals a deep property of triangles: interior lines are always shorter in total than the sides they replace, but they spread apart at a wider angle. It captures an inverse relationship between total length and angular spread that recurs throughout geometry.

Modern connection: This result formalizes the intuition behind the 'shortcut': a path through the interior of a triangle is shorter than going around the outside. In optimization and network design, the principle that interior paths are shorter than boundary paths is fundamental to routing algorithms and minimum spanning trees.

Historical note: This is the first proposition in the Elements that proves two distinct conclusions (a length inequality and an angle inequality) about the same configuration. The two-part structure is unusual for Euclid and shows how a single geometric setup can yield multiple insights.

Discussion Questions

The proposition requires D to be strictly inside the triangle. What happens if D is on one of the sides? Does the result still hold?
Part (1) says the inner lines sum to less, and part (2) says they contain a greater angle. Is there an intuitive way to see why these two facts go together?
Could you extend this result to points inside a quadrilateral? What would the analogous statement be?

Euclid's Original Proof

If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.

On BC, one of the sides of the triangle ABC, from its extremities B, C, let the two straight lines BD, DC be constructed meeting within the triangle;

I say that BD, DC are less than the remaining two sides of the triangle BA, AC, but contain an angle BDC greater than the angle BAC.

For let BD be drawn through to E. [Post. 2]

Then, since in any triangle two sides are greater than the remaining one, [I.20]

therefore, in the triangle ABE, the two sides AB, AE are greater than BE.

Let EC be added to each;

therefore BA, AC are greater than BE, EC. [C.N. 5]

Again, since, in the triangle CED, the two sides CE, ED are greater than CD,

let DB be added to each;

therefore CE, EB are greater than CD, DB. [C.N. 5]

But BA, AC were proved greater than BE, EC;

therefore BA, AC are much greater than BD, DC.

Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [I.16]

therefore, in the triangle CDE, the exterior angle BDC is greater than the angle CED.

For the same reason, moreover, in the triangle ABE also, the exterior angle CEB is greater than the angle BAC.

But the angle BDC was proved greater than the angle CEB;

therefore the angle BDC is much greater than the angle BAC.

Therefore etc.

Q.E.D.

What's Next

Proposition 21 tells us that interior lines through a triangle are always shorter in total than the outer sides. The natural next question is: given three lengths satisfying the triangle inequality, can you actually build a triangle with exactly those side lengths? That constructive challenge is exactly what Proposition 22 takes on.