Euler's Eyes
2026-10-11
Inequalities in triangles
In every triangle, each side is smaller than the sum of the other two. The ancients already joked that “even a donkey knows this”.
Who discovered
The triangular inequality is the proposition I.20 of the The Elements of Euclid (c. 300 BC); the relationship between the largest side and the largest angle is in propositions I.18 and I.19. The idea that the segment is the shortest path between two points was intuitive long before; Euclides’ merit was to demonstrate it based on previous results.
The story
Proclus (5th century AD) says that the Epicureans mocked Euclid for proving the triangular inequality: even a donkey, they said, knows that the straight path to the hay is shorter than going around it. Proclus responded that perceiving a fact is not the same as knowing why it is true. The demonstration shows that the result depends on properties of the plane — and therefore may no longer be valid in other geometries, such as the Earth’s surface.
In the 20th century, the triangular inequality became one of the three rules that define any “distance” in mathematics. It appears in analysis, statistics and computing, far from triangles.
In everyday life
- Shortcuts in the grass. The trails that form through squares and gardens show that people know, in practice, that the diagonal is shorter than both sides.
- Do three sticks form a triangle? Only if the largest is smaller than the sum of the other two. 2 cm, 3 cm and 6 cm rods do not close.
- Delivery routes. Logistics algorithms use triangle inequality to rule out deviations that can never be shorter.
- Roofs. In a pair of scissors, the side opposite the greatest angle is the longest beam — exactly what propositions I.18 and I.19 say.
Sources: Elements, proposition I.20 · Elements, proposition I.18 · Proclus (MacTutor).
Theory and constructions
TeoremaThe greatest angle is opposite the greatest side
If two sides of a triangle are not congruent, then the angles opposite them are not congruent, and the largest angle is opposite the largest side: $BC > AC \Rightarrow \hat{A} > \hat{B}$.
Reciprocal: the largest angle is opposed by the largest side: $\hat{A} > \hat{B} \Rightarrow a > b$.
View the demos
Direct. Take $D$ into $\overline{BC}$ with $\overline{CD} \equiv \overline{CA}$. As $BC > AC$, $D$ is internal to the angle $C\hat{A}B$, therefore $C\hat{A}B > C\hat{A}D$. $\triangle CAD$ is isosceles with base $\overline{AD}$: $C\hat{A}D \equiv C\hat{D}A$. And $C\hat{D}A$ is the external angle of $\triangle ABD$, therefore $C\hat{D}A > A\hat{B}D$. Joining: $C\hat{A}B > C\hat{D}A > A\hat{B}C$, that is, $\hat{A} > \hat{B}$.
Reciprocal. If it were $BC < AC$, by the direct theorem we would have $\hat{A} < \hat{B}$; if it were $BC \equiv AC$, by isosceles triangle, $\hat{A} \equiv \hat{B}$. Both contradict the hypothesis, so $BC > AC$.
TeoremaTriangular inequality
In every triangle, each side is smaller than the sum of the other two: $a < b + c$.
Equivalently, each side is greater than the difference of the other two. The three conditions $a < b + c$, $b < a + c$ and $c < a + b$ are summarized in
$$|b - c| < a < b + c.$$View the demo
Take $D$ on the ray opposite $\overrightarrow{AC}$ with $\overline{AD} \equiv \overline{AB}$. Then $DC = AC + AB$ (2). $\triangle ABD$ is isosceles with base $\overline{BD}$: $A\hat{D}B \equiv A\hat{B}D$. As $A$ is internal to the angle $C\hat{B}D$, we have $C\hat{B}D > A\hat{B}D \equiv C\hat{D}B$. In $\triangle BCD$, the largest angle is opposed by the largest side: $BC < DC$. With (2), $BC < AC + AB$.