# Greater Angle of Triangle Subtended by Greater Side

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## Theorem

In the words of Euclid:

(*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 $1.17$