Greater Angle of Triangle Subtended by Greater Side

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In the words of Euclid:

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

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



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.


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.