Greater Angle of Triangle Subtended by Greater Side

Theorem

In the words of Euclid:

In any triangle, the greater angle is subtended by the greater side.

(The Elements: Book $\text{I}$: Proposition $19$)

Proof

Let $\triangle ABC$ be a triangle such that $\angle ABC$ is greater than $\angle BCA$.

Suppose $AC$ is not greater than $AB$.

If $AC$ were equal to $AB$, then by Isosceles Triangle has Two Equal Angles, $\angle ABC = \angle BCA$, but they're not so it isn't.

If $AC$ were less than $AB$, then by Greater Side of Triangle Subtends Greater Angle it would follow that $\angle ABC$ is less than $\angle BCA$, but it's not so it isn't.

So $AC$ must be greater than $AB$

Hence the result.

$\blacksquare$

Historical Note

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

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 $1.17$