Greater Side of Triangle Subtends Greater Angle
Theorem
In the words of Euclid:
(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 $1.16$