Greater Side of Triangle Subtends Greater Angle

Theorem

In the words of Euclid:

In any triangle, the greater side subtends the greater angle.

(The Elements: Book $\text{I}$: Proposition $18$)


Proof

Let $\triangle ABC$ be a triangle such that $AC$ is greater than $AB$.

Let $AD$ be made equal to $AB$.

Let $BD$ be joined.

Then $\angle ADB$ is an exterior angle of the triangle $\triangle BCD$.

Therefore $\angle ADB$ is greater than $\angle ACB$.

As $AD = AB$, the triangle $\triangle ABD$ is isosceles.

From Isosceles Triangle has Two Equal Angles, $\angle ADB = \angle ABD$.

Therefore $\angle ABD$ is greater than $\angle ACB$.

Therefore, as $\angle ABC = \angle ABD + \angle DBC$, it follows that $\angle ABC$ is greater than $\angle ACB$.


Hence the result.

$\blacksquare$


Historical Note

This proof is Proposition $18$ of Book $\text{I}$ of Euclid's The Elements.
It is the converse of Proposition $19$: Greater Angle of Triangle Subtended by Greater Side.


Sources

  • 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions
  • 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.16$
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