# 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 $1.16$